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77,175 |
In calculus classes it is sometimes said that the tangent line to a curve at a point is the line that we get by "zooming in" on that point with an infinitely powerful microscope. This explanation never really translates into a formal definition - we instead approximate the tangent line by secant lines. I seem to have found a way to obtain tangent lines (and more) by taking "zooming in" seriously. Example 1 Take the curve $y = x(x-1)(x+1)$. I want to find an equation for the tangent line to this curve at the origin. So I zoom in on the origin with a microscope of magnification power $c$ (i.e. I stretch both vertically and horizontally by a factor of $c$) to obtain $\frac{y}{c} = \frac{x}{c}(\frac{x}{c} - 1)(\frac{x}{c}+1)$. Multiplying through by $c$ I have $y = x(\frac{x}{c} - 1)(\frac{x}{c}+1) $ Now letting my magnification power go to infinity I have $y = -x$ Which is the correct answer. Example 2 Take the curve $y = x^2$. I want to find an equation for the tangent line to this curve at the point (3,9). I first rewrite the equation as $(y-9) + 9= ((x-3) + 3)^2$ so that I am focusing on the appropriate point. To zoom on this point with magnification $c$ I have $\frac{y-9}{c} + 9 = (\frac{x-3}{c} + 3)^2$. $\frac{y-9}{c} + 9 = \frac{(x-3)^2}{c^2} + 6\frac{x-3}{c} + 9 $
Multiplying through by $c$ I have $y - 9 = \frac{(x-3)^2}{c} + 6(x-3) $ Now letting my magnification power $c$ go to infinity I have $y - 9 = 6(x-3)$ Which is the correct answer. Example 3 Here is the example which actually motivated me to consider this at all: Take the curve $y^2 = x^2(1 - x)$. This is a cubic curve with a singularity at the origin, and so it doesn't really have a well defined tangent line. It sort of looks like it should have two tangent lines (y = x, and y = -x), but it is a little bit tricky to formalize this. Let's see what "zooming in" does: $\frac{y^2}{c^2} = \frac{x^2}{c^2}(1 - \frac{x}{c})$ $y^2 = x^2(1 - \frac{x}{c})$ Letting $c$ go to infinity I have $y^2 = x^2$, or $(y-x)(y+x) = 0$, which is the pair of lines I desired. My Questions Do any books take this approach when developing the derivative? I would imagine that algebraic geometers do this kind of thing formally. Is there a more rigorous analogue of the prestidigitation I engage in above? Where would I look to read up on such things? p.s. It would be nice to illustrate each of these examples with a little movie of the "zooming in" process, but I am not sure how to put such things on MO. Any hints?
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In algebraic geometry, this construction is known as the tangent cone to the graph. More generally, suppose we have the zero set of any polynomial $f(x,y) = 0$, and assume $f(0,0)=0$. Then we can write $f(x,y) = a_m (x,y) + a_{m+1}(x,y) +a_{m+2}(x,y) +\cdots$ where $a_i(x,y)$ is a homogeneous polynomial of degree $i$ and $a_m$ is nonzero. The zero set of $a_m$ is called the tangent cone to the curve at the origin. It is a product of $m$ linear forms (over $\mathbb{C}$), and $m=1$ exactly when the zero set is smooth at the origin. In this case, the tangent cone coincides with the tangent space. From your point of view, when we substitute $x\mapsto x/c$ and $y\mapsto y/c$ it is clear that the term left in the limit is $a_m$. We can of course find tangent cones at other points of the zero set by changing coordinates. In general, for a smooth function $f$ you should be able to take a multivariate Taylor expansion and read off the tangent cone from the lowest degree part. This is where the difficulty comes in for actually defining the tangent line in terms of the tangent cone in a calculus class, as computing the Taylor expansion demands we already have a notion of derivative. This difficulty is obviously not seen in the case of polynomials, although recentering the Taylor expansion of a polynomial at a different point is perhaps easiest done with the aid of derivatives. Higher dimensional analogues are also available without any real work, although in the singular case the tangent cone is much more interesting than just a union of hyperplanes: it will be a cone over some variety. The homogeneous polynomial $a_m(x_1,\ldots,x_n)$ typically doesn't factor into a product of linear forms when $n>2$. Tangent cones are treated in any reasonable introduction to algebraic geometry, such as Harris' "First course" book or Shafarevich.
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77,177 |
Let A be a quasinilpotent operator on a Hilbert space and let $A^{*}A$ have finite spectrum. Does then follow, that A is nilpotent ?
|
In algebraic geometry, this construction is known as the tangent cone to the graph. More generally, suppose we have the zero set of any polynomial $f(x,y) = 0$, and assume $f(0,0)=0$. Then we can write $f(x,y) = a_m (x,y) + a_{m+1}(x,y) +a_{m+2}(x,y) +\cdots$ where $a_i(x,y)$ is a homogeneous polynomial of degree $i$ and $a_m$ is nonzero. The zero set of $a_m$ is called the tangent cone to the curve at the origin. It is a product of $m$ linear forms (over $\mathbb{C}$), and $m=1$ exactly when the zero set is smooth at the origin. In this case, the tangent cone coincides with the tangent space. From your point of view, when we substitute $x\mapsto x/c$ and $y\mapsto y/c$ it is clear that the term left in the limit is $a_m$. We can of course find tangent cones at other points of the zero set by changing coordinates. In general, for a smooth function $f$ you should be able to take a multivariate Taylor expansion and read off the tangent cone from the lowest degree part. This is where the difficulty comes in for actually defining the tangent line in terms of the tangent cone in a calculus class, as computing the Taylor expansion demands we already have a notion of derivative. This difficulty is obviously not seen in the case of polynomials, although recentering the Taylor expansion of a polynomial at a different point is perhaps easiest done with the aid of derivatives. Higher dimensional analogues are also available without any real work, although in the singular case the tangent cone is much more interesting than just a union of hyperplanes: it will be a cone over some variety. The homogeneous polynomial $a_m(x_1,\ldots,x_n)$ typically doesn't factor into a product of linear forms when $n>2$. Tangent cones are treated in any reasonable introduction to algebraic geometry, such as Harris' "First course" book or Shafarevich.
|
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77,195 |
I have a friend who is very biased against algebraic geometry altogether. He says it's because it's about polynomials and he hates polynomials. I try to tell him about modern algebraic geometry, scheme theory, and especially the relative approach, things like algebraic spaces and stacks, etc, but he still thinks it sounds stupid. This stuff is very appealing for me and I think it's one of the most beautiful theories of math and that's enough for me to love it, but in our last talk about this he asked me well how has the modern view of algebraic geometry been useful or given cool results in math outside of algebraic geometry itself. I guess since I couldn't convince him that just studying itself was interesting, he wanted to know why else he'd want to study it if he isn't going to be an algebraic geometer. But I found myself unable to give him a good answer that involved anything outside of algebraic geometry or number theory (which he dislikes even more than polynomials). He really likes algebraic topology and homotopy theory and says he wants to learn more about the categorical approaches to algebraic topology and is also interested in differential and noncommutative geometry because of their applications to mathematical physics. I know that recently there's been a lot of overlap between algebraic topology/homotopy theory and algebraic geometry (A1 homotopy theory and such), and applications of algebraic geometry to string theory/mirror symmetry and the Konstevich school of noncommutative geometry. However, I am far from qualified to explain any of these things and have only picked up enough to know they will be extremely interesting to me when I get to the point that I can understand them, but that's not a satisfactory answer for him. I don't know enough to really explain how modern algebraic geometry has affected math outside of itself and number theory enough to spark interest in someone who doesn't just find it intrinsically interesting. So my question are specifically as follows: How would one explain how the modern view of algebraic geometry has affected or inspired or in any way advanced math outside of algebraic geometry and number theory? How would one explain why modern algebraic geometry is useful and interesting for someone who's not at all interested in classical algebraic geometry or number theory? Specifically why should someone who wants to learn modern algebraic topology/homotopy theory care or appreciate modern algebraic geometry? I'm not sure if this should be CW or not so tell me if it should.
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As others have suggested, your friend is getting it backwards. He's like a hammer asking what a carpenter is useful for. Given a field (of mathematics, say), there are typically some fields that are more structured than it and others that are less structured. In mathematics, people often say the more structured ones are 'harder', and the less structured are 'softer'. For instance, in increasing order of hardness, we have sets, topological spaces, topological manifolds, differential manifolds, complex manifolds, complex algebraic varieties, algebraic varieties over the rational numbers, integral algebraic varieties. These are in a linear order, but if you throw in other subjects, you'll get a non-linear one. (p-adic algebraic geometry and Riemannian geometry immediately come to mind.) (I think Gromov has some remarks at the end of an ICM address where he talks about this and gives other examples. Also, don't confuse 'harder' and 'softer' in this sense with what they mean in the sciences, which is essentially 'more precise' and 'less precise'. For instance, in science people say that biology is softer than chemistry. In fact, the two meanings are opposites because in science, more structured objects are less amenable to a precise analysis. But this typically isn't the case in mathematics.) Now given a subject S and a harder subject H, it's usually true that most objects in S don't admit the structure of an object in H. For instance, most topological manifolds don't admit a complex structure. On the other hand, for the objects of S that do admit such a structure, their theory from the point of view of H is typically much richer than that from the point of view of S. For instance, the study of Riemann surfaces as topological spaces is less rich than their study as complex manifolds. You might say that softer subjects are broad and flexible and harder ones are rich and rigid. Mathematicians tend to view subjects that are softer than their specialty as general nonsense, and harder ones as excessively particular. This is not to say that a soft field is easier or less interesting than a harder one. Even if it is true that the directly analogous question in the soft subject is easier (e.g. classify Riemann surfaces topologically rather than holomorphically), it just means that the people in the soft subject can move on and study more sophisticated objects. So they just get stuck later rather than sooner. For instance, over the past 50 years, a big fraction of the best number theorists have been studying elliptic curves over number fields. Now elliptic curves over the complex numbers are much easier (I think there hasn't been much new since the 19th century), so the algebraic geometers just moved on to higher genus or higher dimension and are grappling with the issues there, issues that are way out of reach in the presence of arithmetic structure. Now my main point here is that soft subjects were typically invented to break up the study of harder ones into smaller pieces. (This is surely something of a creation myth, but one with a fair amount of truth.) For instance, the real numbers were invented to break up the study of polynomial equations into two steps: when a polynomial has a real solution and when that real solution is rational. I know very little about modern analysis, but I think that much of it was invented to do the same with differential equations. You first find solutions in some soft sense and then see whether it comes from a solution in the harder sense of original interest. So the role of soft subjects is to aid in the study of harder ones---people usually don't ask for applications of partial differential equations to the study of topological vector spaces, but it's considered a mark of respectability to ask for the opposite. Similarly, no one talks about applications of engineering to mathematics. Since algebraic geometry is at the hard end of the spectrum above, there aren't many fields in which it is natural to ask for applications. Number theory, or arithmetic algebraic geometry, is harder and of course there are zillions of applications there, but that's not what your friend wants. Just about all mathematicians work in a subject that is softer than some and harder than others (and if you include non-mathematical subjects, then all mathematicians do). That's all good---it takes a whole food chain to make an ecosystem. But it's backwards to ask about the nutritional value of something that typically eats you . [This picture of mathematics is of course simplistic. There are examples of hard subjects with applications to softer ones. See Donu Arapura's answer, for example. There are also applications of arithmetic algebraic geometry to complex algebraic geometry. For instance, Grothendieck's proof of the Ax-Grothendieck theorem, or the proof of the decomposition theorem for perverse sheaves using the theory of weights and the Weil conjectures. But I think it's fair to say that such applications are the exception---and are prized because of it---rather than the rule.]
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77,278 |
Could someone please recommend a good introductory text on Galois representations? In particular, something that might help with reading Serre's "Abelian l-Adic Representations and Elliptic Curves" and "Propriétés galoisiennes des points d'ordre fini des courbes elliptiques".
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Kevin Ventullo's suggestion of Silverman's book is a very good one. The first examples
of Galois representations in nature are Tate modules of elliptic curves, and if you haven't
read about them in Silverman's book, you should. If you have read Silverman's book, a nice paper to read is Serre and Tate's "On the good reduction of abelian varieties". It is a research paper, not a text-book, and is at a higher level than Silverman (especially in its use of algebraic geometry), but it has the merit of being short and beautifully written, and uses Galois representation techniques throughout. One fantastic paper is Swinnerton-Dyer's article in Lecture Notes 350. Here he
explains various things about the Galois representations attached to modular forms.
The existence of the Galois representations is taken as a black box, but he explains
the Galois theoretic significance of various congruences on the coefficients of the
modular forms. Reading it is a good way to get a concrete feeling of what Galois
representations are and how you can think about and argue with them. Another source is Ken Ribet's article "Galois representations attached to modular forms
with nebentypus" (or something like that) in one of the later Antwerp volumes. It presupposes some understanding of modular forms, but this would be wise to obtain anyway if you want to learn about elliptic curves, and again demonstrates lots of Galois representation techniques. It would be a good sequel to Swinnerton-Dyer's article. Yet another good article to read is Ribet's "Converse to Herbrand's criterion" article, which is a real classic. It is reasonably accessible if you know class field theory, know a little bit about Jacobians (or are willing to take some results on faith, using your knowledge of elliptic curves as an intuitive guide), and something about modular forms. Mazur recently wrote a very nice article surveying Ribet's, available here on his web-site. One problem with reading Serre is that he uses $p$-adic Hodge theory in a strong way,
but his language is a bit old-fashioned and out-dated (he was writing at a time when this
theory was in its infancy); what he calls "locally algebraic" representations would
now be called Hodge--Tate representations. To learn the modern formulation of and perspective on $p$-adic Hodge theory you can look at Laurent Berger's various exposes, available on his web-site. (This will tell you much more than you need to know for
Serre's book, though.) For a two page introduction to Galois representation theory,
you could read Mark Kisin's What is ... a Galois representation? for a two-page
introduction. Yet another source is the Fermat's Last Theorem book (Cornell--Silverman--Stevens), which
has many articles related to Galois representations, some more accessible than others. The article of Taylor that Chandan mentioned in a comment is also very nice, although it
moves at a fairly rapid clip if you haven't seen any of it before. Serre's article in Duke 54, in which he explains his conjecture about the modularity
of 2-dimensional mod p Galois representations, is also very beautiful, and involves
various concrete computations which could be helpful One last remark: if you do want to understand Galois representations, you will need to have
a good understanding of the structure of the Galois groups of local fields (as described e.g. in Serre's book "Local fields"), in particular the role of the Frobenius element,
of the inertia subgroup, and of the significance of tame and wild inertia.
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77,383 |
I asked myself the following question when I was student just for curiosity. I asked a bit around (my professor, some researchers that I know), but nobody was able to give me an answer. So maybe it is just that nobody thought enough about that, or maybe it is not a stupid question. Question: Do there exist two Banach spaces, one separable and one non-separable, having isomorphic dual spaces? Note: isomorphic in the sense that there exists a linear homeomorphism between the two.
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The duals of $C[0,1]$ and of $C[0,1]\oplus_\infty c_0(\Bbb{R})$ are isometrically isomorphic.
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77,635 |
Maybe it would be helpful for me to summarize the little bit I
think know. A 2D CFT assigns a Hilbert space ${\cal H}$ to a circle and
an operator $$A(X): {\cal H}^{\otimes n}\rightarrow {\cal H}^{\otimes m}$$ to a Riemann surface $X$ with $n$ incoming boundaries and $m$ outgoing boundaries. This data is subject to natural conditions
arising from the sewing of surfaces. Here is how I understand the relation to string theory. The
Hilbert space ${\cal H}$ might be the space of functions on the
configuration space of a string sitting in a manifold $M$ . So ${\cal H}=L^2(Maps(S^1,M))$ with some suitable restrictions on the maps.
It is natural then that the functions on the configuration space
of $n$ circles is ${\cal H}^{\otimes n}$ . Now we consider $n$ strings
evolving into $m$ strings. There are many ways to do this, one for
each Riemann surface $X$ as above. When $X$ is fixed, $A(X)$ is the evolution operator, usually described in terms of some path
integral over maps from $X$ into $M$ involving a conformally
invariant functional. All this makes a modicum of sense. So ${\cal H}$ is
the Hilbert space arising from quantization of the cotangent
bundle of $Map(S^1,M)$ , while $A$ describes time evolution. So in
this sense, such a conformal field theory appears to be the
quantization of the classical string. I guess what is missing in
my description up to here is the prescription for turning
functions on $T^*Map(S^1,M)$ into operators on ${\cal H}$ . To my
deficient understanding, maybe this situation corresponds to
having quantized only the Hamiltonian. Now, what I was really wondering was this: When I was in graduate
school, I remember frequently hearing the phrase: CFT theory is the space of classical solutions to string
theory. Does this make some sense? And if so, what does it mean? This phrase has been hindering my understanding of conformal field theory ever since, making me feel like my grasp of physics is all wrong.
According to the paragraphs above, my naive formulation would have
been: Quantization of a string theory gives rise to a CFT. What is wrong with this naive point of view? If you could provide
some enlightenment on this, you'll have resolved a long-standing
cognitive itch in the back of my mind. Thanks in advance. Added: As Jose suggests, I could simply be remembering incorrectly, or misunderstood before what I heard. That, in fact, is what I had hoped to be the case. But read, for example, the first page of Moore and Seiberg's famous paper "Classical and quantum conformal field theory": Link Added again: To quote Moore and Seiberg more precisely, the second sentence of the paper reads 'Conformal field theories are classical solutions of the string equations of motion.' Now, I might attempt to understand this as follows. When the Riemann surface is $$S^1\times [0,t]$$ (with the conformal structure induced by the standard metric)
one interprets $$A(S^1\times [0,t])$$ as $$e^{itH}.$$ Thus, when applied to a vector $\psi_0\in {\cal H}$ , the theory would generate a solution to Schroedinger's equation $$\frac{d}{dt}\psi =iH \psi$$ with initial condition $\psi_0$ as $t$ varies. So one might think of the various $A(X)$ as $X$ varies as being 'generalized solutions' to Schroedinger's equation for a quantized string. I suppose I could get used to such an idea (if correct). But then, the question remains: why do they (and others) say classical solutions? Is there some kind of second quantization in mind with this usage? Added, 11 October: Even at the risk of boring the experts, I will have one more go. Jeff Harvey seems to indicate the following. We can think of $Map(X, M)$ as the fields in a non-linear sigma model on $X$ , provisionally thought of as 1+1 dimensional spacetime. However, it seems that one can also associate to the situation a space of fields on $M$ (the string fields?). If we denote by ${\cal F}$ this space of fields, it seems that there is a functional $S$ on ${\cal F}$ with the property that the extrema of $S$ (the 'string equations of motion') can be interpreted as the $A(X)$ . From this perspective, my main question might then be 'what is ${\cal F}$ ?' Since I think of fields on $M$ as being sections of some bundle on $M$ , I can't see how to get such a thing out of maps from $X$ to $M$ . Thank you very much for your patience with these ignorant questions. Final addition, 11 October: Thanks to the kind guidance of Jose, Aaron, and especially Jeff, I think I have some kind of an understanding of the situation.
I will attempt to summarize it now, superficial as my knowledge obviously is. I don't wish to waste more of the experts' time on this question. However, I am hoping that truly egregious errors will offend their sensibilities enough to elicit at least a cry of outrage, enabling me to improve my poor understanding. I apologize in advance for putting down even more statements that are either trivial or wrong. As far as I can tell, the sense of Moore and Seiberg's sentence is as in my second addition: it is referring to second quantization. Recall that in this process, the single particle wave functions become the classical fields, and Schroedinger's equation is the classical equation of motion. Now the truly elementary point that I was missing (as I feared), is that quantization of a 'single particle' string theory cannot give you a conformal field theory. At most, a single string will propagate though space, giving us exactly the operators $A(S^1\times [0,t])$ . If we want operators $$A(X):{\cal H}^{\otimes n}\rightarrow {\cal H}^{\otimes m}$$ corresponding to a Riemann surface with many boundaries, then we are already requiring a theory where particle numbers can change, that is, a quantum field theory coming from second quantization. WIth such a theory in place, of course, the $A(S^1\times [0,t])$ are exactly the solutions to the classical equations of motion, while the general $A(X)$ can be viewed either as 'generalized classical solutions' (I hope this expression is reasonable) or contributions to a perturbation series, as in the field theory of a point particle. So this, I think. already answers my original question. To repeat, because of the changing 'particle number' the operators of conformal field theory cannot be the quantization of a 'single particle' theory. They must be construed as classical evolution operators of some kind of quantum (string) field theory. The part I'm still far, far from understanding even superficially is this: The classical fields in the case of strings would be something like functions on $Map(S^1, M)$ . I haven't the vaguest idea of how to get from this to fields on spacetime. The difficulty surrounding this issue seems to be discussed in the beginning pages of Zwiebach's paper referred to by Jeff, which is quite heavy reading for a pure mathematician like me. Some mention is made of infinitely many fields arising from the situation (alluded to also by Jeff), which perhaps is some way of turning the data of a function on loop space to fields on space(-time).
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One must distinguish between quantum/classical on the string world-sheet and in spacetime.
Both of your statements are basically correct, but should read something like "CFT theory is the space of classical solutions to the spacetime equations of string theory" and "Quantization of the
the world-sheet sigma model of a string theory gives rise to a CFT." In a little more detail,
the sigma-model describing string theory propagation on some manifold M is a 2-dimensional
quantum field theory which in order to describe a consistent string theory must be a conformal
field theory. The "classical limit" of this 2-dimensional field theory is a limit in which some
measure of the curvature of M is small in units of the string tension. To construct a CFT one
must solve the sigma-model exactly, including world-sheet quantum effects. The coupling constants of the sigma-model are fields in spacetime such as the metric $g_{\mu \nu}(X(\sigma))$ on $M$ where $X: \Sigma \rightarrow M$ define the embedding of the string world-sheet $\Sigma$ into $M$. Now there
is also a spacetime theory of these fields. You can think of it as a ``string field theory". At low-energies it can sometimes be usefully approximated by a theory of gravity coupled to some finite number
of quantum fields, but in full generality it is a theory of an infinite number of quantum fields. Roughly speaking, each operator in the CFT gives rise to a field in spacetime. The spacetime string field theory lives
in 10 dimensions for the superstring or 26 dimensions for the bosonic string and it also has a classical limit. The classical limit is $g_s \rightarrow 0$ where $g_s$ is
a dimensionless coupling constant. It appears in perturbative string theory as a factor which
weights the contribution of a Riemann surface by the Euler number of the surface. It can also be
thought of as the constant (in spacetime) mode of a scalar spacetime field known as the dilaton. The main point is that there are two notions of classical/quantum in string theory, one involving
the world-sheet theory, the other the spacetime theory. In order to avoid confusion one must be clear which is being discussed. Unfortunately string theorists often assume it is clear from the context. In response to the further question about the space of string fields, I would suggest that you have a look at the introductory material in http://arXiv.org/pdf/hep-th/9305026 . You may also find http://arXiv.org/pdf/hep-th/0509129 useful. I should add that while string field theory has had some success recently in the description of D-brane states, it is not widely thought to be a completely satisfactory definition of non-perturbative string theory.
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|
77,730 |
Suppose that there are $n$ vertices, we want to construct a regular graph with degree $p$, which, of course, is less than $n$. My question is how many possible such graphs can we get?
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McKay and Wormald conjectured that the number of simple $d$-regular graphs of order $n$ is asymptotically
$$\sqrt 2 e^{1/4} (\lambda^\lambda(1-\lambda)^{1-\lambda})^{\binom n2}\binom{n-1}{d}^n,$$
where $\lambda=d/(n-1)$ and $d=d(n)$ is any integer function of $n$ with $1\le d\le n-2$ and $dn$ even. Bender and Canfield, and independently Wormald, proved this for bounded $d$ in 1978, and Bollobás extended this to $d=O(\sqrt{\log n})$ in 1980. McKay and Wormald proved the conjecture in 1990-1991 for $\min\{d,n-d\}=o(n^{1/2})$ [1], and $\min\{d,n-d\}>cn/\log n$ for constant $c>2/3$ [2]. These remain the best results. The gap between these ranges remains unproved, though the computer says the conjecture is surely true there too. The formula apart from the $\sqrt2e^{1/4}$ has a simple combinatorial interpretation, and the universality of the constant $\sqrt2e^{1/4}$ is an enigma crying out for an explanation. Incidentally this conjecture is for labelled regular graphs. For isomorphism classes, divide by $n!$ for $3\le d\le n-4$, since in that range almost all regular graphs have trivial automorphism groups (references on request). For $d=0,1,2,n-3,n-2,n-1$, this isn't true. [1] Combinatorica, 11 (1991) 369-382. http://cs.anu.edu.au/~bdm/papers/nickcount.pdf [2] European J. Combin., 11 (1990) 565-580. http://cs.anu.edu.au/~bdm/papers/highdeg.pdf ADDED in 2018: The "gap between those ranges" mentioned above was filled by Anita Liebenau and Nick Wormald [3]. Another proof, by Mikhail Isaev and myself, is not ready for distribution yet. [3] https://arxiv.org/abs/1702.08373
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77,934 |
There is a trend, for some people, to study representations of quivers. The setting of the problem is undoubtedly natural, but representations of quivers are present in the literature for already >40 years. Are there any connections of this trend with other Maths? For, it seems like it is a self-contained topic and basically I wonder why people study quivers so much -- in a sense everything becomes clear after the initial results of Gabriel and the old result of Yuri Drozd about wild/tame dichotomy, and these things ought to become boring. BTW, about the dichotomy theorem -- is it really necessary to study so hard whether a given problem is of tame or wild (representation) type? In particular, why some people try to lift tame/wild things to curves and surfaces -- would that really yield something interesting in geometry? (I am currently attending lectures about these things and unfortunately we were not told a single word about motivation, and when I tried to learn from the lecturer if this is really "top" Maths as he claims, he basically replied "this is important because I am doing this")
|
In addition to being a nice example for abelian, $A_{\infty}$ and Calabi–Yau categories, and being a prototypical example for Generalized Donaldson–Thomas Invariants and the Wall Crossing Phenomenon , quivers have a lot of applications in various different fields. Since the question is asking for applications in addition to representation theory, I'm listing a few cases. Most prominent is Algebraic Geometry, particularly Moduli problems and GIT (read motivations in Reineke's article ) and Video lectures on quivers by Reineke at Newton Institute, Cambridge. Recently, a correspondence has been proposed between Gromov–Witten invariants and Quivers. ( Pandharipande–Gross ). Also in physics applications in String Theory , Supersymmetry , Black Holes and Particle physics . Relation with quantum dilogarithm, number theory, and cluster algebras, read e.g. this review by Keller. Also through the work of Nakajima, there is a relation to Instantons of Yang–Mills theory, Hitchin Moduli spaces and the theory of Hyperkähler manifolds.
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|
77,986 |
Pell equations can be solved using continued fractions. I have heard that some elliptic curves can be "solved" using continued fractions. Is this true? Which Diophantine equations other than Pell equations can be solved for rational or integer points using continued fractions? If there are others, what are some good references? Edit: Professor Elkies has given an excellent response as to the role of continued fractions in solving general Diophantine equations including elliptic curves. What are some other methods to solve the Diophantine equations $$X^2 - \Delta Y^2 = 4 Z^3$$ and $$18 x y + x^2 y^2 - 4 x^3 - 4 y^3 - 27 = D z^2 ?$$
|
[edited to insert paragraph on Cornacchia and point-counting] Continued fractions, or (more-or-less) equivalently the Euclidean algorithm, can be used to find small integer solutions of linear Diophantine equations $ax+by=c$, and integer solutions of quadratic equations such as $x^2-Dy^2=\pm1$ ("Pell"). Continued fractions in themselves won't find rational points on elliptic curves, but there's a technique using Heegner points that calculates a close real approximation to a rational point, which is then recovered from a continued fraction — this is possible because the recovery problem amounts to finding a small integer solution of a linear Diophantine equation. My paper Noam D. Elkies: Heegner point computations, Lecture Notes in Computer Science 877 (proceedings of ANTS-1, 5/94; L.M. Adleman and M.-D. Huang, eds.), 122-133. might have been the first to describe this approach. Another application of continued fractions is Cornacchia's algorithm to solve $x^2+Dy^2=m$ for large $m>0$ coprime to $D$, given $x/y \bmod m$ which is a square root of $-D \bmod m$. This has an application to counting points on elliptic curves $E\bmod p$ for $E$ such as $y^2 = x^3 + b$ or $y^2 = x^3 + ax$ for which the CM field ${\bf Q}(\sqrt{-D})$ is known: the count (including the point at infinity) is $p+1-t$ where $t^2+Du^2=4p$ for some integers $t$ and $u$, and this determines $t$ up to an ambiguity of at most $6$ possibilities that in practice is readily resolved. The necessary square root mod $p$ is readily found in random polynomial time, though it is a persistent embarrassment that we cannot extract square roots modulo a large prime in deterministic polynomial time without assuming something like the extended Riemann hypothesis. Indeed the application that Schoof gave to motivate his polynomial-time algorithm to compute $t$ for any elliptic curve mod $p$ was to recover a square root of $-D \bmod p$ for small $D$ ! (Though this would never be done in practice because the exponent in Schoof's algorithm is much larger than for the randomized algorithm.) The reference for Schoof's paper is René Schoof: Elliptic Curves over Finite Fields and the Computation of Square Roots $\bmod p$, Math. Comp. 44 (#170, April 1985), 483-494. A natural generalization of the Euclidean algorithm to higher dimensions is the LLL algorithm and other techniques for lattice basis reduction (LBR), which have found various other Diophantine uses, including some other techniques for finding rational points on elliptic curves; another of my papers describes some of these Diophantine applications of LBR.
|
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|
78,247 |
In the one-dimensional case the Langlands program is equivalent to the class field theory and the two-dimensional case implies the Taniyama-Shimura conjecture. I would like to know: are there any other important consequences of the Langlands program?
|
There are many, many consequences of the general Langlands program (which I'll interpret to mean both functoriality for automorphic forms and reciprocity between Galois representations and automorphic forms). Some of these are: The Selberg $1/4$ conjecture. The Ramanujan conjecture for cuspforms on $GL_n$ over arbitrary number fields. Modularity of elliptic curves over arbitrary number fields. (Indeed, Langlands reciprocity
is essentially the statement that all Galois representations coming from geometry are attached to automorphic forms.) Analogues of Sato--Tate for Frobenius eigenvalues on the $\ell$-adic cohomology of arbitrary varieties over number fields.
|
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|
78,400 |
One modern POV on model categories is that they are presentations of $(\infty, 1)$-categories (namely, given a model category, you obtain an $\infty$-category by localizing at the category of weak equivalences; better, if you have a simplicial model category, is to take the homotopy coherent nerve of the fibrant-cofibrant objects). What other functions, then, do model categories serve today? I understand that getting the theory of $\infty$-categories off the ground (as in HTT, for instance) requires a significant use of a plethora of model structures. However, if we assume that there exists a good model of $(\infty, 1)$-categories that satisfies the properties we want (e.g. that mapping sets are replaced with mapping spaces, limits and colimits are determined by homotopy limits of spaces), how are model categories useful? I suppose one example would be computing hom-spaces: a simplicial model category gives you a nice way of finding the mapping space between two objects in the associated localization. However, in practice one only considers cofibrant or fibrant objects in the $\infty$-category in the first place, as in Lurie's construction of the derived $\infty$-category (basically, one considers the category of projective complexes -- for the bounded-above case, anyway -- and makes that into a simplicial category, and then takes the homotopy coherent nerve). One example where having a model structure seems to buy something is the theorem that $E_\infty$ ring spectra can be modeled by 1-categorical commutative algebras in an appropriate monoidal model category of spectra (in DAG 2 there is a general result to this effect), and that you can straighten things out to avoid coherence homotopies. I don't really know anything about $E_\infty$-ring spectra, but I'm not sure how helpful this is when one has a good theory of monoidal objects in $\infty$-categories.
|
I find some of this exchange truly depressing. There is a subject of ``brave
new algebra''and there are myriads of past and present constructions and calculations that
depend on having concrete and specific constructions. People who actually compute
anything do not use $(\infty,1)$ categories when doing so. To lay down a
challenge, they would be of little or no use there. One can sometimes use
$(\infty,1)$ categories to construct things not easily constructed otherwise,
and then one can compute things about them (e.g. work of Behrens and Lawson). But the tools of computation are not $(\infty,1)$ categorical, and often not even
model categorical. People should learn some serious computations, do some
themselves, before totally immersing themselves in the formal theory. Note that
$(\infty,1)$ categories are in principle intermediate between the point-set level
and the homotopy category level. It is easy to translate into $(\infty,1)$
categories from the point-set level, whether from model categories or from
something weaker. Then one can work in $(\infty,1)$ categories. But the
translation back out to the "old-fashioned'' world that some writers seem to
imagine expendable lands in homotopy categories. That is fine if that is all
that one needs, but one often needs a good deal more. One must be eclectic. Just
one old man's view.
|
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|
78,404 |
Let $R$ be a commutative ring with unit and let $q$ be an ideal of $R$. There is thus a natural map $SL(n,R) \rightarrow SL(n,R/q)$ for all $n$. This map is surjective if $SL(n,R/q)$ is generated by elementary matrices, but I very much doubt that it is surjective in general (though I don't know any examples). My questions are as follows. Can someone give me an example of a ring $R$ and an ideal $q$ of $R$ such that the map $SL(n,R) \rightarrow SL(n,R/q)$ is not surjective for any $n$? I'd like the examples to be as nice as possible. For instance, it would be great to have an example where $R$ is Noetherian and has finite Krull dimension. What conditions can I put on $R$ and $q$ to assure that this map is surjective, at least for large $n$?
|
A sort of universal example: Let $R$ be the polynomial ring $\mathbb Z[x_{11},x_{12},x_{21},x_{22}]$ and let $q$ be the ideal generated by $x_{11}x_{22}-x_{12}x_{21}-1$. The obvious element of $SL_2(R/q)$ does not come from $SL_2(R)$. You can see this by comparing with the example of the ring $\mathbb R[u,v]$ and the ideal generated by $u^2+v^2-1$, using the ring map taking $(x_{11},x_{12},x_{21},x_{22})$ to $(u,v,-v,u)$. If the resulting matrix came from an element of $SL_2(\mathbb R[u,v]$), then topologically the corresponding map from the circle in $\mathbb R^2$ defined by $u^2+v^2=1$ to $SL_2(\mathbb R)$ would extend to a continuous map $\mathbb R^2\to SL_2(\mathbb R)$, which it doesn't. This example persists to $SL_n$ for $n>2$.
|
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|
78,423 |
How far can one go in proving facts about projective space using just its universal property? Can one prove Serre's theorem on generation by global sections, calculate cohomology, classify all invertible line bundles on projective space? I don't like many proofs of some basic technical facts very aesthetic because one has to consider homogeneous prime ideals, homogeneous localizations, etc. Do there exist nice clean conceptual proofs which avoid the above unpleasantries? If you include references in your answer it would be very helpful, thanks.
|
I think the answer is "probably not." The reason is that projective space has two universal properties which are used to prove different kinds of things about it. One of these is the slick universal property you like, and the other is the clunky one which results in unpleasantries. Though each universal property implies the other (since it uniquely identifies projective space), it seems unlikely to me that you can effectively do anything if you try to avoid one of them altogether. One universal property makes it easy to understand maps to projective space: $$
Hom(T,\mathbb P^n) = \{\mathcal O_T^{n+1}\twoheadrightarrow \mathcal L| \mathcal L\text{ a line bundle}\}
$$ Without bending over backwards (i.e.~reproducing the usual theory), I'd be surprised if you could use this universal property to even prove that there are no non-constant regular functions on $\mathbb P^n$. I expect constructions that naturally pull back along morphisms (e.g. line bundles, regular functions) to behave like morphisms from projective space, so it would be strange if you could attack such constructions with this universal property. Another universal property makes it easy to understand maps from projective space: $Hom(\mathbb P^n,T)$ is the equalizer of the two restriction maps $Hom(\coprod_{i=0}^n \mathbb A^n,T)\rightrightarrows Hom(\coprod_{i,j}\mathbb A^{n-1}\times (\mathbb A-0),T)$. I guess this is the one that you don't like, but we're lucky to have it since it actually makes it possible to make sense of projective space having Zariski local properties (e.g. being smooth, $n$-dimensional, etc.), and thereby makes it possible to do geometry on it.
|
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|
78,471 |
Let $D$ be a co-complete category and $C$ be a small category. For a functor $F:C^{op}\times C \to D$ one defines the co-end $$
\int^{c\in C} F(c,c)
$$
as the co-equalizer of
$$
\coprod_{c\to c'}F(c,c'){\longrightarrow\atop\longrightarrow}\coprod_{c\in C}F(c,c).
$$
It is the indexed co-limit $\mbox{colim}_W F$ where the weight is the functor $W:C^{op}\times C \to Set$ given by $Hom(-,-)$. I have two strongly related questions regarding this definition. First, what's the intuition behind this construction? Can I think of it as a kind of ''fattened'' colimit? Second, why is the integral sign used for this? Can ordinary integration be related to this construction?
|
Martin's comment is right on the money; in particular, the best way to get a feeling for coends is through the many examples where they appear, such as (generalized) tensor products. But from an abstract point of view, coends can be considered as universal " extranatural transformations ", and the ubiquity of coends (and ends) is explained by the ubiquity of such extranatural transformations. Let me take a specific example before getting into the abstract aspects. Let's consider the example of geometric realization of simplicial sets, as a left adjoint to the singularization functor $S: Top \to Set^{\Delta^{op}}$. Recall that if $Y$ is a space, then $S(Y)$ is the simplicial set $\Delta^{op} \to Set$ whose value at the ordinal $[n]$ (with $n+1$ points) is defined by $$S(Y)([n]) = \hom_{Top}(\sigma_n, Y)$$ where $\sigma_n$ is the $n$-dimensional affine simplex seen as a topological space. We are interested in constructing a left adjoint $R$ to $S$, so that for any simplicial set $X$, the set of natural transformations $$X \to S(Y)$$ is in natural bijective correspondence with continuous maps $R(X) \to Y$. The way to do this is to "bend" a natural transformation $$X([n]) \to \hom_{Top}(\sigma_n, Y)$$ (a family of maps natural in $[n] \in \Delta$) into another family $$\phi_n: X([n]) \times \sigma_n \to Y$$ of continuous maps indexed by $n$. This family has a property intimately related to the naturality of the first family; it is called "extranaturality". It means that given any morphism $f: [m] \to [n]$, the composite $$X([n]) \times \sigma_m \stackrel{X[f] \times id}{\to} X([m]) \times \sigma_m \stackrel{\phi_m}{\to} Y$$ equals the morphism $$X([n]) \times \sigma_m \stackrel{id \times \sigma_f}{\to} X([n]) \times \sigma_n \stackrel{\phi_n}{\to} Y;$$ this precisely mirrors the naturality of the first family in $n$. Thus, what we are after is an extranatural transformation $$X([n]) \times \sigma_n \to R(X)$$ with the universal property that given any extranatural transformation $\phi_n$ as above, there exists a unique map $R(X) \to Y$ making the evident triangle commute (for each $n$). This is of course the coend $$R(X) = \int^n X([n]) \times \sigma_n$$ and the appropriate construction in terms of coproducts and coequalizers that you indicated in your question is exactly what is required to construct the universal extranatural transformation. This is easily abstracted. Given any functor $F: C^{op} \times C \to D$, one can define what it means for a family of maps $F(c, c) \to d$ (for fixed $d$) to be extranatural in $c$, and the coend is again described as a universal extranatural transformation. In nature, such transformations almost invariably arise by "bending" a natural transformation into an extranatural (also called "dinatural") one. The tensor product mentioned by Martin fits into this pattern: thinking of a left $R$-module map of the form $$M \to \hom_{Ab}(N, A)$$ ($M$ a left $R$-module, $N$ a right $R$-module, $A$ an abelian group; the hom acquires a left $R$-module structure) as a $Ab$-enriched natural transformation between functors of the form $R \to Ab$ (where a ring $R$ is viewed as a one-object $Ab$-enriched category), we can "bend" this map into a map $$M \otimes N \to A$$ which is extranatural with respect to scalar actions, and the quotient $M \otimes N \to M \otimes_R N$ is the universal such extranatural map. But this only scratches the surface of possibilities for this type of situation. Finally, I second Martin's remark on the traditional integral notation -- not too much should be made of this, except that weighted colimits are primary examples of coends, and there are interchange isomorphisms which are reminiscent of Fubini theorems; this is touched upon in Categories for the Working Mathematician.
|
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|
78,501 |
Suppose that $G$ is a finite group, acting via homeomorphisms on $B^n$, the closed $n$-dimensional ball. Does $G$ have a fixed point? A fixed point for $G$ is a point $p \in B^n$ where for all $g \in G$ we have $g\cdot p = p$.
Notice that the answer is "yes" if $G$ is cyclic, by the Brouwer fixed point theorem. Notice that the answer is "not necessarily" if $G$ is infinite. If it helps, in my application I have that the action is piecewise linear. First I thought this was obvious, then I googled around, then I read about Smith theory, and now I'm posting here.
|
The answer is no . A fixed point free action of the finite group $A_5$ on a $n$ -cell was constructed by Floyd and Richardson in their paper An action of a finite group on an n-cell without stationary points , Bull. Amer. Math. Soc. Volume 65 , Number 2 (1959), 73-76. For some non-existence results, you can see the paper by Parris Finite groups without fixed-point-free actions on a disk , Michigan Math. J. Volume 20 , Issue 4 (1974), 349-351.
|
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|
78,505 |
Let $\mathcal{D} \approx \mathbb{P}^{\delta_d}$ be the space of homogeneous
degree $d$ polynomials in three vriables, where
$\delta_d = \frac{d(d+3)}{2}$. Let
$$ X \subset \mathcal{D} \times \mathbb{P}^2$$ be a smooth embedded complex submanifold, not necessarily closed.
Given a point $p\in \mathbb{P}^2$, we get a hyperplane
$$\tilde{H}_p \in \mathcal{D} \times \mathbb{P}^2.$$
Note that a point $p$ first of all gives a hyperplane $H_p$
in $\mathcal{D}$ (which is the space of degree $d$
polynomials passing through the point $p$). This gives us a hyperplane
$$ \tilde{H}_p := H_p\times \mathbb{P}^2 \in \mathcal{D} \times \mathbb{P}^2.$$
Let us further define $H_p^* \subset H_p$ to be the
space of degree $d$ curves such that $p$ is a smooth point of the
curve. Similarly define
$$ \tilde{H}_p^* := H_p^*\times \mathbb{P}^2 \in \mathcal{D} \times \mathbb{P}^2.$$ Is it true that for almost all choices of $p \in \mathbb{P}^2$,
$\tilde{H}^*_p$ is transverse to $X$?
|
The answer is no . A fixed point free action of the finite group $A_5$ on a $n$ -cell was constructed by Floyd and Richardson in their paper An action of a finite group on an n-cell without stationary points , Bull. Amer. Math. Soc. Volume 65 , Number 2 (1959), 73-76. For some non-existence results, you can see the paper by Parris Finite groups without fixed-point-free actions on a disk , Michigan Math. J. Volume 20 , Issue 4 (1974), 349-351.
|
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|
78,813 |
What could be a reference about binomial expansions for non-commutative elements? Specifically, where can I find a closed formula for the expansion of $(A+B)^n$ where $[A,B]=C$ and $[C,A]=[C,B]=0$? I've found some ideas about that and also a proof using PDE's in the following website: link . But I haven't found a such formula in a published scientific paper or book.
|
I don't know if you prefer a particular presentation of the formula, but this is essentially covered by the Baker-Campbell-Hausdorff formula, or actually it's dual, Zassenhaus formula , which in your case reduces to
$$e^{(A+B)t}=e^{At}e^{Bt}e^{-[A,B]t^2/2},$$
where one side is the generating function for $(A+B)^n$ while the other has terms of the form $f(n,m,p)A^nB^mC^p$. The binomial theorem here is given by equating the coefficients of $t^n$ on both sides.
$$(A+B)^n=\sum_{n\equiv k\pmod{2}} \left(\sum_{r=0}^k \binom{k}{r}A^rB^{k-r}\right)\left(-\frac{C}{2}\right)^{\frac{n-k}{2}}\frac{n!}{k!(\frac{n-k}{2})!}$$
|
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|
78,827 |
Suppose we play a chess-variant, where any finite number of pieces are allowed, and the board is as large as we wish, but only two kings in total. And there is no 50 move-rule, no castling and no captures and no pawn-moves. Then is it possible to have a a pair of positions A and B, such that we can go from configuration A to B by legal moves, but not from B to A? No stalemates or checkmates allowed, the game must be extendable atleast two moves in both direction from both A and B. Two configurations are different also if the pieces are in the same places but it is a different player to move.
|
Here's an "irreversible chess" construction that's fundamentally different from the ones so far based on Ed Dean's scheme. The essential pieces and pawns are in boldface : Position A: White Kh1, Ra1, Nd1, pawns b2,b3,c3,d2,e3,f2 ,g2,h7, Bg8; Black Kh8, Bb1 , Bg1, pawns f7,g7,h2. (source: janko.at ) Position D = Position A after 1...Ba2 2 Rc1 Bb1, i.e. with the Rook on c1 and White to move: (source: janko.at ) I call it D rather than B because the two intermediate positions can be called B and C, and then each arrow in A $\rightarrow$ B $\rightarrow$ C $\rightarrow$ D is irreversible. [This was also possible in some of the previous examples, including Ed's original one; I don't think we've seen "A $\rightarrow$ B $\rightarrow$ C $\rightarrow$ D $\rightarrow$ E" yet.] In Position A, the rook and a1 and bishop on b1 can reversibly roam the board. But after 1...Ba2 (Position B) the only locally reversible continuation is 2 Rc1 (Position C; if 2 Rb1 Black has no reversible reply) 2...Bb1 (Position D) 3 Rc2 Ba2 4 Rc1 Bb1 and we're back to D; the White rook and Black bishop can no longer escape the corner because they keep getting in each other's way. The previous constructions all exploit the special behavior of kings, which must not be in check on the opponents' move. This new approach does not need kings at all — it would still work if we removed the kings and their un-boldfaced retinues, except that the problem as posed required each side to have a king. The key ingredient here instead of the check rule is move alternation: if either side were allowed to skip a turn it would be easy to get back to Position A — whereas in several of the check-based constructions, skipping turns would not help as long as neither side is allowed to make a move (even as part of an unanswered series) that leaves its own king in check.
|
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|
78,830 |
Yesterday I attended a seminar talk titled "Cluster presentation of reflection groups", and before it I tried to ask Google what is a "cluster presentation" --- but all one can find on this request is about databases or social studies. Another example: try to find something on "geometrical groups" and not "geometric group theory". I think this is a very common problem, when you are interested in some area you don't know anything about and trying to find some texts on it --- but it's very hard if you don't know what exactly to look for. Of course if you have someone to ask, it's not a problem anymore, but usually you don't. So the question is: how to start learning an advanced topic if you don't have any texts or guidelines about it (and have nobody to ask except for MO community)?
|
Here's an "irreversible chess" construction that's fundamentally different from the ones so far based on Ed Dean's scheme. The essential pieces and pawns are in boldface : Position A: White Kh1, Ra1, Nd1, pawns b2,b3,c3,d2,e3,f2 ,g2,h7, Bg8; Black Kh8, Bb1 , Bg1, pawns f7,g7,h2. (source: janko.at ) Position D = Position A after 1...Ba2 2 Rc1 Bb1, i.e. with the Rook on c1 and White to move: (source: janko.at ) I call it D rather than B because the two intermediate positions can be called B and C, and then each arrow in A $\rightarrow$ B $\rightarrow$ C $\rightarrow$ D is irreversible. [This was also possible in some of the previous examples, including Ed's original one; I don't think we've seen "A $\rightarrow$ B $\rightarrow$ C $\rightarrow$ D $\rightarrow$ E" yet.] In Position A, the rook and a1 and bishop on b1 can reversibly roam the board. But after 1...Ba2 (Position B) the only locally reversible continuation is 2 Rc1 (Position C; if 2 Rb1 Black has no reversible reply) 2...Bb1 (Position D) 3 Rc2 Ba2 4 Rc1 Bb1 and we're back to D; the White rook and Black bishop can no longer escape the corner because they keep getting in each other's way. The previous constructions all exploit the special behavior of kings, which must not be in check on the opponents' move. This new approach does not need kings at all — it would still work if we removed the kings and their un-boldfaced retinues, except that the problem as posed required each side to have a king. The key ingredient here instead of the check rule is move alternation: if either side were allowed to skip a turn it would be easy to get back to Position A — whereas in several of the check-based constructions, skipping turns would not help as long as neither side is allowed to make a move (even as part of an unanswered series) that leaves its own king in check.
|
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|
79,004 |
Let $H$ be an $(\infty,1)$-topos (seen as a generalization of the homotopy category of spaces). You can define the suspension of an object $X$ as the (homotopy) pushout of $*\leftarrow X \to *$, hence you can define inductively the spheres $\mathbb{S}^n$ (the sphere of dimension $-1$ is the initial object of $H$ and the sphere of dimension $n+1$ is the suspension of the sphere of dimension $n$). You can also define the loop spaces of a pointed object as the (homotopy) pullback of $*\to X \leftarrow *$. It will be itself pointed (because there is an obvious commutative diagram with a $1$ instead of $\Omega{}X$, so there is (I think) an arrow between this $1$ and $\Omega{}X$). Then, given two integers $n, k$, you can define $\pi_k(\mathbb{S}^n)$ as the set of connected components (global elements up to homotopy) of the $k$-fold loop space of the $n$-sphere (I don’t know if this definition is one of the two described in the nlab ) Is there a natural group structure on $\pi_k(\mathbb{S}^n)$? Is there something known about these groups in general? For example, Are they completely known for some $H$? Is it always true that $\pi_k(\mathbb{S}^n)$ is trivial for $k<n$ and isomorphic to $\mathbb{Z}$ for $k=n$? Are they isomorphic (or related in some way) to the usual homotopy groups of spheres? Addition: What if you assume that $H$ is a cohesive $(\infty,1)$-topos? (see here for the nLab page)
|
If $H$ is the terminal category (=sheaves on the empty space), then $\pi_k^HS^n$ (notation for homotopy groups of "spheres" in $H$) is known! The slice category $H=\mathrm{Spaces}/B$ is an $(\infty,1)$-topos. The homotopy groups of spheres in this setting amount to the homotopy groups of the space $\mathrm{map}(B,S^n)$ of unbased maps (with basepoint at a constant map $B\to S^n$).
This shows that $\pi_k^HS^n$ need not be trivial if $k<n$. This also provides non-trivial examples in which $\pi_k^HS^n$ is isomorphic to the "usual" homotopy groups of spheres (e.g., if $B=BG$ for $G$ a finite group, by Miller's theorem.) If $f: H\to H'$ is a geometric morphism, then the pullback functor $f^*: H'\to H$ induces a homomorphism $\pi_k^{H'}(S^n)\to \pi_k^{H}(S^n)$. In particular, if $H$ has a point (a geometric morphism $\mathrm{Spaces}\to H$), then $\pi_kS^n$ is a summand of $\pi_k^HS^n$. Edit. As I understand it, if $H$ is cohesive, then $p^\*: \mathrm{Spaces} \to H$ is supposed to be fully faithful, where $p:H\to\mathrm{Spaces}$ is the unique geometric morphism. Spheres are in the image of $p^\*$, so it ought to follow that that $\pi_k^H S^n = \pi_kS^n$. The only example of cohesive topos I understand is $H=s\mathrm{Spaces}$ (simplicial spaces), and it is certainly true in this case.
|
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|
79,044 |
Writing a CV makes me paranoid that I'm failing to abide by unwritten rules. Of course CVs are flexible to capture the diversity of accomplishments someone might have. But there must be plenty of things a hiring committee absolutely expects. So I'm interested in anything that must be on a CV — whose omission would raise a red flag — of a mathematician looking for an academic job. "Obvious" answers are welcome. Even things which sound obvious like "your name." What is obvious to someone who has read and evaluated lots of CVs is different from someone preparing one for the first time. In your answer, please also be fairly specific about scope: have you served on hiring committees? for what types of positions? in the US or Europe or? (Note: As suggested in the comments, it is very good to ask people "in the know" directly for such career advice. One reason for asking this question on MO is to have more open, less clubby answers — there is an echo chamber effect when you ask a bunch of people in the same subcommunity.)
|
From my perspective, the critical question isn't what must be included on your CV, but rather what mustn't, since that seems to be the more common problem (judging by the ones I see). What I'm about to describe is based on my experience at a U.S. research lab; I imagine it generalizes quite a bit beyond that, but I can't say how far, and it is certainly country-specific. I'll discuss five rules below, with some overlap between them. Of course these rules are not absolute (except for the last one), but you certainly shouldn't break them without thinking carefully about it and deciding there's a good reason to do so. (1) Your CV should represent you as a professional mathematician. Anything that is not relevant to your professional life should be left out. For example, you should generally not describe non-math-related summer or part-time jobs, hobbies, side interests outside of mathematics and related fields, etc. If there's something unusually interesting or impressive (you published a novel or are a chess champion) or that displays relevant skills (you write free software in your spare time), it's OK to mention it, but just briefly and not in a prominent position. I've seen some hair-raising violations of this rule, in which applicants devoted considerable space to things that have nothing to do with working as a mathematician. Nobody is going to reject your application just because you put something weird in your CV, but it's not good for your image as a professional. (2) Your CV shouldn't include anything unless you think the search committee might need or want to know it. For example, contact information is valuable, as is anything that can legitimately help judge your application. However, in the U.S. you should not list your age or birthdate, your marital status, information about your children, or your religion (unless you are applying to a religious institution). I realize this is common in some countries, and of course people will be understanding about that, but it comes across strangely to give people information they don't want and shouldn't be influenced by. (3) You should try not to seem desperate to impress, particularly with awards and distinctions. Some people provide enormous lists of very minor distinctions, sometimes with no relevance to research/teaching/service (for example, a college scholarship from a local business club). Coming across as insecure can make you seem less attractive: an ambitious department wants to hire people who are marginally too good for them, not people who are trying hard to be good enough. As a rule of thumb, when you get your Ph.D. and apply for your first job, it's OK to list any substantive distinction from grad school. You can list a few undergraduate honors, but only if they are impressive (Putnam fellow or major university-wide prize, yes; random scholarship, no). You shouldn't list high school honors at all (well, just maybe an IMO medal, but be careful not to look like you consider it your proudest achievement). (4) Be sure not to give the impression you are trying to obfuscate anything. I don't just mean you should tell the truth, but also that you should be clear and straightforward. For example, people sometimes feel bad about not having enough items to list in their publication or talk sections, and it can be tempting to reorganize the CV to try to obscure this. For example, you could replace the "publications" section with a "research" section in which you list not just publications but also talks and poster presentations, or even current/future research topics. This is a bad idea, since it can look like you are trying to make the information less accessible, and then everything on your CV will be looked at more skeptically. Instead, you want to make it easy to understand your CV and easy to see that you aren't doing anything tricky. (5) Don't lie. Don't say a paper will appear in a journal until it has been accepted, even if you are sure it will be. Don't say a paper is submitted until it is, even if you plan to submit it by the time the committee meets. Don't call something a preprint until it is written down and ready to distribute (you can say "in preparation" before then, but many people will ignore this since it is unverifiable). Don't say you have received a fellowship or prize if you haven't. You'd think all these things go without saying, but I've seen a couple of people get caught on one of them. You really don't want to be the person who gets asked for a copy of their preprint and can't produce one.
|
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|
79,113 |
I wondered whether there were an infinite number of
palindromic primes written in binary (11, 101, 111, 10001, 11111, 1001001, 1101011, ...)
and quickly discovered that it is unknown
( OEIS A117697 ).
Indeed, even though almost all palindromes in any base are composite,
whether there are an infinite number of palindromic primes in any base is unknown
( Wolfram article ). Earlier
(in the MO question, "Why are this operator’s primes the Sophie Germain primes?" ),
I learned that
it is unknown if there are an infinite number of Sophie Germain Primes .
In addition, is not known if there are an infinite number of
Mersenne primes,
Fibonacci primes ( OEIS A005478 ), Wilson primes , Cullen primes ,
not to mention prime twins, quadruplets, sextuplets, and $k$ -tuples .
No doubt this list of our ignorance could be extended. It appears to the naïve (me) that there is no nontrivial restriction on the
primes for which we know there remain an infinite number in the restricted sequence. Q1 . Is this superficial perception in fact true? Q2 . If so, is there any high-level reason why it is so difficult
to prove these statements? Or is each difficult for its own idiosyncratic
reason? I ask this out of curiosity, without expert knowledge of number theory.
Thanks for enlightening me! Questions Answered . Thanks for the wonderfully rich and informative answers!
Essentially both questions have been answered: My superficial perception ( Q1 )
is not in fact accurate, as detailed in the examples provided by quid, Anthony Quas, and Joël,
augmented by comments by several.
A high-level reason ( Q2 ) explaining the difficulty in the examples I listed was nicely encapsulated
by Frank Thorne, enriched by appended comments. Thanks!
|
To give a vague answer, I think these questions are difficult because they mix multiplicative conditions (being prime) and additive conditions (as in the twin prime case). Basically all results about primes that I can think of come down to unique factorization of the integers. For example, the zeta function is given as $$\zeta(s) = \sum_n n^{-s} = \prod_p (1 - p^{-s})^{-1}.$$ The right hand side is why the zeta function tells you about prime numbers, but the left hand side is what typically helps you prove theorems. For example, Riemann noticed that the left side looked like something similar to what Poisson summation is good for, and hence proved analytic continuation and the functional equation. On a simpler level, one nice proof that there are infinitely many primes is to observe that $\sum_n 1/n$ diverges, by elementary calculus, and therefore the right hand side diverges for $s = 1$ as well. Gerhard Paseman suggested looking at arithmetic progressions, and I think this is an extremely instructive example. Looking at the sum of $n^{-s}$ restricted to an arithmetic progression, you don't have any equation like the above. Conversely, if you take a product over only the primes $p$ in some arithmetic progression, you don't get anything nice like the left side. However, if you let $\chi$ be a Dirichlet character, e.g., a homomorphism from $(\mathbb{Z}/N)^{\times}$ to $\mathbb{C}$, then you get the Dirichlet $L$-function $$L(s, \chi) = \sum_n \chi(n) n^{-s} = \prod_p (1 - \chi(p) p^{-s})^{-1}.$$ In some way this is forcing a round peg into a square hole: the arithmetic progression condition couldn't be handled directly. But it can be written as a linear combination of Dirichlet characters, and once you force everything to be multiplicative, the machinery (Poisson summation, etc.) all works. So in other words, IMHO, the question isn't "why is the twin prime conjecture difficult", but "why can we prove anything about the primes at all?" Our toolbox is, in my experience, still pretty limited.
|
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|
79,371 |
This talk by Jinhyun Park connects a lot of interesting themes, making me curious to read more about that. Do you know where?
|
This circle of topics is certainly one of my favourite surprising connections in mathematics. I will try to outline what little I understand of the big picture. Apologies for the length. Hilbert's 3rd problem and Dehn complexes: As is well-known, Hilbert's 3rd problem asked for examples of tetrahedra of equal volume which are not scissors congruent, and Dehn solved it using the invariant now named after him. However, this only gave rise to further questions: what about higher-dimensional euclidean spaces, and what about the other classical geometries? By now, I guess the proper objects to consider are the Sah algebra and Dehn complexes I will outline. To study scissors congruences, it is best to consider the scissors congruence groups for all dimensions at the same time. Join of simplices induces a product, dimension a grading. In the spherical case, the Dehn invariants provide a coproduct, and the duality of spherical simplices gives an involution, making the total spherical scissors congruences into a graded Hopf algebra. This object is called Sah algebra . The coproduct statement does not work in the other geometries since the dihedral-angle part of the Dehn invariant always introduces a spherical geometry component. However, in the other geometries, we still get comodule structures over the spherical scissors congruences. It is also possible (and interesting) to consider the graded cobar complexes for these modules over the coalgebra of spherical scissors congruences. These complexes were introduced by Goncharov who called them Dehn complexes (in the JAMS paper discussed below). The Hopf algebra and comodule picture is discussed in the following book (not easy to get but very much worth reading): C.-H. Sah. Hilbert's third problem: scissors congruence. Research notes in mathematics 33, Pitman, 1969. Another book concerning scissors congruences (also very much worth reading but does not discuss the Hopf algebra view) is: J.-L. Dupont. Scissors congruences, group homology and characteristic classes. Nankai tracts in mathematics 1, World Scientific, 2001. A very general form of Hilbert's 3rd problem can then be formulated for all the classical geometries: are Dehn invariant and volume sufficient to completely characterize the scissors congruence class of a polytope? This is only known for euclidean space in dimensions $\leq 4$, for hyperbolic and spherical not even in dimension $3$. Even more general, one can wonder about the exact computation of the cohomology of the Dehn complexes. Goncharov's conjectures then relate the above Dehn complexes to K-theoretic stuff. In the spherical and hyperbolic cases, Goncharov conjectures that the Dehn complexes compute the ($+1$ and $-1$-eigenspaces of complex conjugation on the) weight-graded pieces of algebraic K-theory of $\mathbb{C}$. This is formulated and discussed in A. Goncharov. Volumes of hyperbolic manifolds and mixed Tate motives. J. Amer. Math. Soc. 12 (1999), 569-618. Maybe it would also make sense to expect an explicit quasi-isomorphism between Bloch's cycle complexes (computing motivic cohomology) and the Dehn complexes.
I am not sure if a formulation like this exists in the literature, but probably the above conjecture says something like ``the Sah algebra should be isomorphic to the motivic Galois group of mixed Tate motives over $\mathbb{C}$'' (only that we do not have the latter). Anyway, the conjecture is known in weight 2 due to the work of Dupont, Sah, Bloch, Suslin (this goes under the name of motivic weight $2$ complex, dilogarithm complex etc). The trilogarithm case has been worked out by Goncharov, see e.g. A. Goncharov. Geometry of configurations, polylogarithms and motivic cohomology. Adv. Math. 144 (1995), 197-318. Regulators, volumes and number theory Goncharov's conjecture translates the regulator on K-theory to the volume of hyperbolic simplices (with Dehn invariant $0$). From this point of view, Goncharov's conjecture is a far-reaching generalization of the fact that volumes of hyperbolic 3-manifolds can be expressed in terms of dilogarithms. Under Goncharov's conjecture, the generalized version of Hilbert's 3rd problem for hyperbolic spaces and spheres translates into an injectivity conjecture for K-theoretic regulators due to Ramakrishnan. Currently, it is not even known if $K_3(\mathbb{C})$ (which is relevant for scissors congruences in $\mathbb{H}^3$) is bigger than $K_3(\overline{\mathbb{Q}})$ (which we understand in terms of Borel regulators). This is the number theory appearing in the title (and might be related to other conjectures on the description of the period algebra which I know nothing about). Probably the consequences relating regulators and volumes actually motivated the conjecture in the first place. Certainly the relation between K-theory and polylogarithms is part of the motivation for Bloch's definition of mixed Tate motives as modules over a suitable Lie algebra. Euclidean geometry and cyclic homology: So far, we have talked about the relation between the noneuclidean geometries and algebraic K-theory. Now, infinitesimally, hyperbolic space looks like euclidean space. As a consequence, euclidean Dehn complexes are the ``tangent spaces'' to hyperbolic Dehn complexes. Again, Goncharov has a precise conjecture how the euclidean Dehn complex should be related to mixed Tate motives over the dual numbers $\mathbb{C}[\epsilon]/(\epsilon^2)$, cf. A. Goncharov. Euclidean scissors congruence groups and mixed Tate motives over dual numbers. Math. Res. Lett. 11 (2004), 771-784. This is how we get a relation to cyclic homology, because cyclic homology is related to the tangent space of algebraic K-theory. This way, the euclidean part of Hilbert's 3rd problem involves talking about additive Chow groups (about which you can learn by reading e.g. the papers of Park, but also Cathelineau, Bloch, Esnault,...). You see the appearance of groups related to cyclic homology by reformulating the classical Dehn-Sydler theorem as an exact sequence:
$$
0\to \mathbb{R}\to\operatorname{SC}(\mathbb{E}^3)\stackrel{D}{\longrightarrow}\mathbb{R}\otimes S^1\to\Omega^1_{\mathbb{R}/\mathbb{Q}}\to 0,
$$
where the map $D$ is the Dehn invariant from 3-dimensional scissors congruences (mapping a three-simplex to the sum of edge lengths tensor dihedral angles). The kernel of the Dehn invariant is detected by the volume map to $\mathbb{R}$, and the cokernel of the Dehn invariant is a group of Kähler differentials. Further reading: check the works of Sah, Dupont, Cathelineau, Goncharov,... There is too much literature to mention in this already oversized answer.
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|
79,493 |
Is it true that for any finite nontrivial group G , there exist two inequivalent irreducible representations of G over the complex numbers that have the same degree. If so, is there an easy proof? If not, what is the smallest counterexample? Note: Any counterexample group must be perfect, because if the abelianization is nontrivial, we get multiple irreducible representations of degree one. [EDIT: Further, as Colin Reid notes in the comment, a minimal counterexample must be a simple (non-abelian) group]. This whittles down our search considerably. The general expressions for the degrees of irreducible representations for the families of simple groups that I've checked suggests that there is plenty of repetition of degrees in these cases.
|
It seems that the answer is yes. A MathSciNet search brought up the paper Y. Berkovich, D. Chillag, and M. Herzog, Finite groups in which the degrees of the nonlinear irreducible characters are distinct , Proc. Amer. Math. Soc. 115 (1992), 955–959. In it you can find a characterization of groups whose nonlinear irreducible characters have distinct degrees. In particular, such a group can't be perfect (see Lemma 1), and so will always have multiple linear characters as was noted in the OP. The proof, however, relies on the classification of finite simple groups, so is not "easy". Addendum: I took a closer look at the related literature and happened across the following interesting result, which I figured was worth sharing. (It can also be used to give an affirmative answer to the original question.) Theorem. Let $G$ be a nontrivial finite group. If the character table of $G$ has a column or row containing distinct rational entries, then $G$ must be isomorphic to either $S_2$ or $S_3$. The reference is M. Bianchi, D. Chillag, A. Gillio, Finite groups with many values in a column or a row of the character table , Publ. Math. Debrecen 69 (2006), no. 3, 281–290. The result from the classification of finite simple groups used in the Berkovich–Chillag–Herzog paper is also used here (in very much the same spirit).
|
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|
79,542 |
In ordinary category theory, the notion of limit in a category $C$ is usually formulated with a category (of indices) $J$ and a functor $F:J\to C$ (a diagram in $C$), and a limit of this diagram is something satisfying some universal property. In the context of quasi-categories (I looked in the nLab and in Higher Topos Theory ), the category $J$ is replaced by a simplicial set $K$ and the functor $F$ is replaced by a map of simplicial sets $K\to C$ (a quasi-category is a particular simplicial set). But this definition is specific of quasi-categories, if you take another model for $(\infty,1)$-categories, it does not make sense to talk about a map between a simplicial set and an $(\infty,1)$-category. Is there a definition of limits for $(\infty,1)$-categories independent of the model? In particular, can we replace $K$ by an $(\infty,1)$-category and $F$ with an $(\infty,1)$-functor? If we can, why does everybody take a simplicial set instead (for quasi-categories)?
|
Definitions I think there is a definition that should fit into most models of $(\infty,1)$-categories. If you want an "elevator speech" answer, it's: Definition. A limit of a diagram $\mathcal D \to \mathcal C$ is a terminal object in the $(\infty,1)$-category of objects living over $\mathcal D$. This is the (almost naive) generalization of one definition for limits in usual category theory. But let me elaborate on this definition. (Full disclosure, I almost always work with quasicategories, so I anticipate that a better answer can be given by someone who's worked with all models. I hope this answer will be helpful regardless. Also, when you say "limit," I assume you mean homotopy limit. I sometimes distinguish between the two since not everybody is happy when I use classical terminology with an implicit "$\infty$" or "homotopy" before every word.) Morally speaking, a good model for "$(\infty,1)$-categories" should have definitions for the following ideas: Mapping spaces. That is, given an $(\infty,1)$-category
$\mathcal C$, between any two
objects $X,Y$ of $\mathcal C$, a
topological space of morphisms
${\mathcal C}(X,Y)$. This is a
fairly obvious pre-requisite because $(\infty,1)$-categories are supposed to be like categories enriched in spaces. Terminal objects. Morally, these are objects $\ast$ such that
for any other object $Y$ in your
$(\infty,1)$-category $\mathcal C$,
the mapping space ${\mathcal
C}(Y,\ast)$ is contractible. There may be more subtle issues involved in defining terminal objects properly, depending on your model, but at least in the case of quasi-categories, it turns out this moral definition is perfectly fine as an actual one. (See Corollary 1.2.12.5 of HTT.) Under/Over-Categories, aka Cone Categories. Given two $(\infty,1)$-categories $\mathcal C$ and $\mathcal D$, an $(\infty,1)$-category of ( $(\infty,1)$-) functors between them. And in our discussion, we specifically want the following: Given a diagram ${\mathcal D} \to \mathcal C$, a good notion of an ( $(\infty,1)$-)category whose objects are functors from $\ast \star {\mathcal D}$ to $\mathcal C$, where $\ast \star {\mathcal D}$ is the category obtained by affixing an initial object to $\mathcal D$. This is the same thing as the category of objects of $\mathcal C$ equipped with a map to the diagram $\mathcal D \to \mathcal C$. You can see why this third point, about cone categories, is so simple in the quasi-category model. It is as simple as defining the join of simplicial sets, and knowing what the mapping space is between simplicial sets. Anyhow, if you believe that your model (whatever it is) has definitions for the above three things, you can define a limit to be a terminal object in a cone category. You can dually define colimits as initial objects in an undercategory. Actually proving that (co)limits are preserved. I assume you wanted an answer that was more specific about actually computing (homotopy) limits using different models (complete Segal spaces, quasi-categories, Kan simplicial categories, et cetera) but I'm afraid I don't know much about comparing homotopy limit computations in different models. Lurie does, however, prove in HTT (Theorem 4.2.4.1) that the usual homotopy (co)limits you'd compute in a category enriched over Kan complexes will agree with the homotopy (co)limits you'd compute in the quasi-category model. So that's a good start! And if you believe in the equivalences between different models of $(\infty,1)$-categories (see for instance Julie Bergner's "A Survey of $(\infty,1)$-Categories" ) then the equivalences should preserve initial objects of cone categories, so this would be an argument that all models preserve (homotopy) (co)limits. Why "everybody" takes simplicial sets. Actually, a lot of people prefer to use other models like the Segal space model. But you can see that with the combinatorics of quasi-categories, a lot of things can be defined and proved fairly cleanly, as I pointed out in some of my commentary above. So that's one advantage of Joyal's quasi-category model. But there are many situations in which the space of objects is so naturally a space that you might prefer a model which isn't based on weak Kan complexes. For instance, in Galatius-Madsen-Tillman-Weiss , they think of the category of cobordisms as a category with a space of objects and a space of morphisms. This model might make it easier, for instance, to compute the classifying space of an $(\infty,1)$-category. And if you were interested in computing a (co)limit of a functor mapping such an $(\infty,1)$-category into another, you wouldn't want to say that your diagram comes from a simplicial set. Simplicial Sets as the Diagram Also, it seems you're interested in why Lurie takes as the diagram a map $\mathcal{D} \to \mathcal C$ in which $\mathcal D$ is a simplicial set . I don't think I would take "simplicial set" as the important idea here; I think it's more that $\mathcal D$ is an $(\infty,1)$-category, so for Lurie, it's a weak Kan complex, and in particular a simplicial set. I'm not sure (as Moosbrugger alludes to above) why the generality of an arbitrary simplicial set is so important, but in general I think we would take a diagram $\mathcal D \to \mathcal C$ to be a map of $(\infty,1)$-categories . That is, $\mathcal D$ being a simplicial set isn't so important for this definition. It just needs to be an $(\infty,1)$-category in whatever model you're using, and in the Lurie example, you should probably think of it as a quasi-category, rather than just an arbitrary simplicial set. And you can always replace an arbitrary simplicial set with a weak Kan complex (these are the fibrant-cofibrant objects in the Joyal model category.)
|
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|
79,546 |
Let's consider algebraic curves over a fixed algebraically closed field $K$. It's well known, that every smooth elliptic curve (genus $g = 1$) can be embedded in a quadric surface in $\mathbb{P}^3$. This fact follows simply from the Riemann–Roch theorem. More generally, for smooth hyperelliptic curves of higher genus ($g \ge 2$) it's known that such curves can be embedded in a quadric in weighted projective space $\mathbb{P}(1,1,g)$, see, for example, work of D. Eisenbud . (So, in the case $g=2$ we have embedding in $\mathbb{P}^4$). But, can any smooth hyperelliptic curve $H$ be embedded in a quadric surface in $\mathbb{P}^3$? It's natural question, because, by the definition we have mophfism $\phi:H \to \mathbb{P}^1$ of degree 2. I think, this problem is connected with topics "Families of hyperelliptic curves and double covers of quadric surface" and Quotient Surface of A Hyperelliptic Involution , but I don't get it. (I'm interested in the case of any algebraicaly closed field and in the case of finite field).
|
Yes. Let $C$ be a hyperelliptic curve of genus $g$, and let $L$ be a general line bundle of degree $g+1$. By Riemann-Roch, $\dim|L| = 1$ and $|L|$ is base-point free, so the complete series $|L|$ gives a degree $g+1$ map to $\mathbb{P}^1$. Then the product of this map and the degree $2$ map $C\to \mathbb{P}^1$ gives a map $f:C\to \mathbb{P}^1 \times \mathbb{P}^1$, whose image is a curve of type $(2,g+1)$. But $\mathbb{P}^1\times \mathbb{P}^1$ is just a quadric in $\mathbb{P}^3$. To see the map is an embedding, it will suffice to show that it is birational. Indeed, the image has arithmetic genus $g$ by adjunction on a quadric surface. But it also has geometric genus $g$ since $C$ is its normalization. Thus the image is smooth if it is reduced. Finally we must see that the map is birational. The only way the map fails to be injective is if some divisor of $|L|$ contains a pair of points conjugate under the hyperelliptic involution. But in this case the assumption that $\dim |L|=1$ implies that $|L| = g_2^1 + p_1+\cdots+ p_{g-1}$, where $g_2^1$ is the hyperelliptic series and $p_1,\ldots,p_{g-1}$ are base points. Since $L$ was general, this isn't true, and we're done. Here's a cute (although trivial) kind of partial converse: If $g$ is prime, then any smooth curve $C$ of genus $g$ which embeds in a smooth quadric is hyperelliptic.
|
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|
79,583 |
Consider the 3-simplex, or tetrahedron, in 3-space. Regardless of the positions of the vertices, every point in the simplex lies on a chord between two non-adjacent edges of the simplex. Or, equivalently, every interior point lies along a straight line segment which intersects two non-adjacent edges. When is this property true of other convex (or non-convex) polyhedra? How does this property extend to the general $N$-simplex?
|
The question asks whether every point $v$ in the interior of a 3-polytope $P \ $ is on an interval between two edge-points. This is easy. Project the edges of $P$ onto a unit sphere centered at $v$. Call the resulting graph $G$ blue . Take the opposite $-G$ and call this graph red . Clearly red and blue graphs intersect, since otherwise one must lie in the face of another, which is impossible since $v$ is interior. Thus the line through the intersection point and $v$ is as desired. As for higher dimensions, this is clearly not possible already for dim-reasons. We are talking about 2-parametric family of intervals, which cannot possibly cover the interior of a $d$-polytope, for $d\ge 4$.
|
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|
79,685 |
The question is contained in the title; I mean the standard axioms ZFC. The wiki link: Riemann hypothesis . There are finite algorithms allowing one to decide if there are non-trivial zeroes of the $\zeta$ -function in the domains whose union exhausts the whole strip $0<\Re z<1$ , but this does not seem to be the obstacle for undecidability. Are there other arguments?
|
I do not know anything about zero-finding algorithms for $\zeta$, so I will make only one small remark which doesn't require such knowledge: If the Riemann Hypothesis is false, then it is provably false (in ZFC, or any similar system). This is because Robin's theorem tells us that the Riemann hypothesis is equivalent to the assertion that, for every natural $n \geq 5041$, the sum of the divisors of $n$ is less than $e^{\gamma} n \log{\log{n}}$; since there are programs which calculate this latter quantity to arbitrary precision, and thus can verify whether this inequality holds for any given $n$, we find that the Riemann hypothesis is a $\Pi_1$ statement: it is equivalent to the assertion that some computer program never outputs "NO" on any input. (Although not familiar with the proofs of Robin's theorem, etc., I assume they can be carried out in ZFC, and thus establish the relevant equivalence within ZFC.). There may be more direct ways to establish that the Riemann hypothesis is a $\Pi_1$ statement, such as by knowledge of algorithms which enumerate to arbitrary precision the zeros of $\zeta$, but at any rate, there is this one. Accordingly, if the Riemann hypothesis is false, then the relevant computer program does output "NO" on some input, from which it would follow that ZFC proves that that computer program outputs "NO" on that input, and thus ZFC would prove the Riemann hypothesis to be false. The possibility still remains, however, as far as I know, that the Riemann hypothesis may be true but unprovable in ZFC.
|
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|
79,741 |
If $G \neq \lbrace 1 \rbrace$ is a finite group with classifying space $BG$
then there are infinitely many i such that $H^i(BG,\mathbb{Z}) \neq 0$. This
can be found, for example, there: Non-vanishing of group cohomology in sufficiently high degree As a consequence, the CW-complex $BG$ (unique up to homotopy) can not be of finite dimension. Question: Are there alternative proofs for this observation. In particular, I would be interested in knowing if there is a purely topological proof without homological algebra.
|
I believe there's an argument using Euler characteristic. Let $G$ be a finite group, $BG=K(G,1)$ the classifying space, and $EG=\widetilde{BG}$ the universal cover, which is contractible. Then $\chi(EG)=1$. Now, if $H_{\ast}(G)=H_{\ast}(BG)$ were finite, then $\chi(BG)$ would be an integer (use whatever field coefficients you prefer). But by the multiplicativity of Euler characteristic, then $\chi(BG)|G|=\chi(EG)=1$, so $\chi(BG)=1/|G|$, a contradiction. I forget who this argument is attributed to. Also, one may see geometrically that any finite group has a classifying space with finitely many cells in each dimension, so if $H_{\ast}(BG)$ is infinite, it must be non-vanishing in infinitely many dimensions (i.e. not infinite rank in a single dimension).
|
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|
79,959 |
Let $k$ be a field, and $A$ a $k$-domain, so that the fraction field of $A$ has transcendence degree $n$ over $k$. If $A$ is finitely-generated over $k$, then $A$ has Krull dimension $n$ (Theorem A in Eisenbud). However, if $A$ is infinitely-generated, then it is possible for the dimension of $A$ to be less than the transcendence degree of its fraction field. Take, for example, rational functions in one variable $A=k(x)$. Dimension 0, transcendence degree 1. Is it always true that the dimension of $A$ is less than or equal to $n$?
|
It looks to me like the answer is yes . Fix any strictly increasing chain of primes $P_0 \subsetneq P_1 \subsetneq \cdots \subsetneq P_m$ in $A$ of length $m$; we'll prove that $m \leq n$. Choose elements $x_i \in P_i \setminus P_{i-1}$ for $i = 1,\dots,m$. Let $B \subseteq A$ be the $k$-subalgebra generated by $\{x_1,\dots,x_m\}$. The primes $Q_i = P_i \cap B$ form a strictly increasing chain of length $m$ in $B$, because $x_i \in Q_i \setminus Q_{i-1}$. The fraction field $Q(B)$ of $B$ lives inside of $Q(A)$ and therefore has $k$-transcendence degree at most $n$. Because $B$ is finitely generated, this means that it has Krull dimension at most $n$. It now follows that $m \leq n$.
|
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|
80,056 |
I am toying with the idea of using slides (Beamer package) in a third year math course I will teach next semester. As this would be my first attempt at this, I would like to gather ideas about the possible pros and cons of doing this. Obviously, slides make it possible to produce and show clear graphs/pictures (which will be particularly advantageous in the course I will teach) and doing all sorts of things with them; things that are difficult to carry out on a board. On the other hand, there seems to be something about writing slowly with a chalk on a board that makes it easier to follow and understand (I have to add the disclaimer that here I am just relying on only a few reports I have from students and colleagues). It would very much like to hear about your experiences with using slides in the classroom, possible pitfalls that you may have noticed, and ways you have found to optimize this. I am aware that the question may be a bit like the one with Japanese chalks, once discussed here, but I think as using slides in the classroom is becoming more and more common in many other fields, it may be helpful to probe the advantages and disadvantages of using them in math classrooms as well.
|
I think you already touched on the two main points: pretty pictures are so much better than anything done on a chalkboard is the pro, but you cannot decently unwind any argument on slides. I've used them intensively, I do it a lot less now. (Here's a con you did forget about: they take a lot of time to prepare, even when you're only revising them.) If the room lends itself well to it, the hybrid method is best: use the slides only when they beat the board. Rooms that have a screen in the corner, rather than in front of the board, are best for this. Also, it seems that it's easier to fall asleep to slides than to a lecture, so be aware of that. Make sure that the room is never too dark (the quality of the screen material can be critical here too: good screens should be readable in full light). And switching your routine, never showing slides for too long, helps keeping the students awake.
|
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|
80,061 |
Let $X$ be a complex, projective, nonsingular variety. We also understand it as a Kähler Manifold. My question now is, when people say $c_1(X) < 0$, what exactly do they mean? Let me elaborate. In this paper , it is said that Yau's inequality $$ (-1)^n c_1^n \le (-1)^n \frac{2(n+1)}{n} c_2 c_1^{n-2} $$ holds under the condition that $c_1(X) < 0$. I would have thought that this is equivalent to $K_X$ being effective. In the original paper , Yau requires $X$ to have ample canonical class, however. Now, I am wondering: For the above equality to hold, do I need $K_X$ to be ample or does it suffice for $K_X$ to be effective?
|
I think you already touched on the two main points: pretty pictures are so much better than anything done on a chalkboard is the pro, but you cannot decently unwind any argument on slides. I've used them intensively, I do it a lot less now. (Here's a con you did forget about: they take a lot of time to prepare, even when you're only revising them.) If the room lends itself well to it, the hybrid method is best: use the slides only when they beat the board. Rooms that have a screen in the corner, rather than in front of the board, are best for this. Also, it seems that it's easier to fall asleep to slides than to a lecture, so be aware of that. Make sure that the room is never too dark (the quality of the screen material can be critical here too: good screens should be readable in full light). And switching your routine, never showing slides for too long, helps keeping the students awake.
|
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"https://mathoverflow.net/questions/80061",
"https://mathoverflow.net",
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|
80,081 |
I'm trying to get a grasp on what it means for a manifold to be spin. My question is, roughly: What are some "good" (in the sense of illustrating the concept) examples of manifolds which are spin (or not spin) (and why)? For comparison, I'd consider the cylinder and the mobius strip to be "good" examples of orientable (or not) bundles. I've read the answers to Classical geometric interpretation of spinors which are helpful, but I'd like specific examples (non-examples) to think about.
|
There's the traditional obstruction-theoretic perspective. Orientability means the tangent bundle trivializes over a 1-skeleton. Dually you could think of that as saying the complement of a co-dimension $2$ subcomplex has a trivial tangent bundle. So admitting a spin structure is the same, but it will be the tangent bundle trivializes over a 2-skeleton, dually the complement of a co-dimension three subcomplex admits a trivial tangent bundle. A surface is orientable if and only if it contains no Moebius bands -- a regular neighbourhood of any simple closed curve must be a cylinder. In higher dimensions this translates into a manifold being orientable if and only if it contains no twisted bundles $D^{n-1} \rtimes S^1$, i.e. regular neighbourhoods of simple closed curves are diffeomorphic to $D^{n-1} \times S^1$. For spin structures there's something very similar. Of course, a surface admits a spin structure if and only if it is orientable. It's a more interesting notion in higher dimensions. The statement there is the manifold is orientable, and if you take a regular neighbourhood of any surface in the manifold, then it has a trivial tangent bundle. So manifolds like $\mathbb RP^3$ are perfectly valid spin manifolds -- $\mathbb RP^3$ contains $\mathbb RP^2$ but the total space of its normal bundle has a perfectly trivializable tangent bundle. Technically, the condition is a little stronger than that -- you can trivialize the tangent bundle of the complement of a co-dimension $3$ subset. So not only can you trivialize the total spaces of normal bundles of surfaces, but even the regular neighbourhoods of unions of surfaces. So if you want a manifold that isn't spin, the archetype would be a vector bundle over a surface so that the total space does not have a trivializable tangent bundle. Take the $D^2$-bundle over $S^2$ with Euler Class $\chi$. I think this happens if and only if $\chi$ is even. I suppose you have more entertaining examples when dealing with the regular neighbourhood of a 2-complex that isn't itself a manifold. edit: Milnor's "Spin structures on manifolds" in L'Enseignement Mathematique Vol 9 (1963) is an excellent reference for most of the above. I don't believe he goes into all the descriptions above since I think he wants to keep the article simple. The Poincare duality interpretation above is a very standard mode of thinking that's employed throughout much of low-dimensional topology. Kirby's book on 4-manifolds is a nice place to look for this material. Specifically, R. Kirby "The topology of 4-manifolds" Springer-Verlag (1989). A more modern reference would be Gompf and Stipsicz, but again I don't think they use all the above descriptions. Milnor and Stasheff's "Characteristic Classes" describes most of the basic constructions involved above, in the obstruction theory section. In a couple of months I'll be putting up a paper on the arXiv that gives some very combinatorial ways of describing spin and spin^c-structures on manifolds (mostly for computer implementation). I hope that will be a good reference, too! But the paper is still unreadable.
|
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|
80,125 |
Prove/ Disprove: Let $n$ be a positive integer. Let $A$, $B$ be two $n \times n$ square matrices over the complex numbers. If $AB = BA$ and $\ker A = \ker A^2$ and $\ker B = \ker B^2$
then $\ker AB = \ker A + \ker B$. (Recall that $\ker A$ is the set of all vectors $v$ such that $Av = 0$.) Background: I am teaching linear algebra this semester. I did not like the standard proof of the Jordan canonical form I found in the textbooks, and thought I could prove it differently, directly from the axioms for a vector space, without using either the determinant, or the classification theorem for finite abelian groups. If the statement above is true, I believe I have a proof for the Jordan canonical form for $T$ by setting $A = (T-\lambda_1I)^{n_1}$ and $B=(T-\lambda_2I)^{n_2}$ for appropriate $n_1$ and $n_2$. Note 1: If $A = B = \left( \begin{array}{cc}
0 & 1\\\0 & 0 \end{array} \right)$ then $AB = BA$ but $\ker AB \neq \ker A + \ker B$. Note 2: It is easy to find $A, B$ such that $\ker A = \ker A^2$ and $\ker B = \ker B^2$ and $B$ maps a vector outside $\ker A + \ker B$ to $\ker A$, so that $\ker AB \neq \ker A + \ker B$. Hence, both conditions are necessary.
|
Since $\ker A = \ker A^2$, the map $\bar{A} : V/\ker A \to V/\ker A$ is injective. Since $V/\ker A$ is finite dimensional, this map is surjective. So for any $x \in V$ we can find $y \in V$ and $z \in \ker A$ such that $x = Ay + z$. Now suppose $ABx = 0$ and let $x = Ay + z$ as above. Then $0 = ABx = ABAy + ABz = A^2By + BAz = A^2By$ because $AB = BA$ and $Az = 0$. So $By \in \ker A^2 = \ker A$ whence $ABy = 0$. So $B(Ay) = 0$ and $Ay \in \ker B$. Hence $x = Ay + z \in \ker B + \ker A$.
|
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|
80,150 |
I apologize for the somewhat vague question: there may be multiple answers but I think this is phrased in such a way that precise answers are possible. Let $\mathfrak{g}$ be a semisimple Lie algebra (say over $\mathbb{C}$) and $\mathfrak{h} \subset \mathfrak{g}$ a Cartan subalgebra. All the references I have seen which study the representation theory of $\mathfrak{g}$ in detail make use of the half-sum of positive roots, which is an element of $\mathfrak{h}^\ast$: e.g. Gaitsgory's notes on the category O introduce the "dotted action" of the Weyl group on $\mathfrak{h}^\ast$, the definition of which involves this half-sum. Is there a good general explanation of why this element of $\mathfrak{h}^\ast$ is important? The alternative, I suppose, is that it is simply convenient in various situations, but this is rather unsatisfying.
|
I don't think there is a one-line answer to this question, since it depends a lot on the direction from which you approach semi-simple Lie theory. For one thing, it's probably best at first to emphasize just integral weights, among which the dominant ones parametrize irreducible finite dimensional representations. Here the weight usually denoted $\rho$ plays a ubiquitous role in the classical Weyl theory, but that too can be developed in a number of different ways. (There was some early experimentation with the notation; the alternative symbol $\delta$ also had widespread use before the Bourbaki preference for $\rho$ started to take over in 1968.) While it's important in proofs of the Weyl character formula to view $\rho$ as the half-sum of positive roots (given a fixed positive or simple system), it's also essential to identify it with the sum of fundamental dominant weights for many purposes. In this guise it's the smallest regular dominant weight, fixed by no element of the Weyl group except the identity. When passing from integral weights to line bundles on an associated flag variety $G/B$ (with $B$ a Borel subgroup associated to positive roots relative to a fixed maximal torus which it contains), the weight $\rho$ has the distinction of defining an ample line bundle. This property is crucial in geometric approaches to Weyl's formula, as well as in spin-offs in prime characteristic due to Andersen and others. Ultimately the importance of the weight $\rho$ is probably appreciated best in the setting of representation theory, where the finite dimensional theory is enriched by treatment of highest weight modules in more generality and the shift by $\rho$ is again ubiquitous. By the way, the convenient "dot" notation
$w \cdot \lambda := w(\lambda +\rho) - \rho$ is apparently due to Robert Moody. In the earlier literature the more awkward full notation appears, or else is replaced in the Paris notation by a hidden $\rho$-shift. None of what I've said is a complete answer to the question asked, but in any case it's more than a matter of "convenience" to emphasize $\rho$.
|
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|
80,364 |
Leibniz was a noted polymath who was deeply interested in philosophy as well as mathematics, among other things. From my mathematical readings I have the impression that Leibniz's stature as a mathematician has grown in the last fifty years as some of his philosophically oriented mathematical ideas have connected with modern mathematicians and mathematics. That because of Leibniz's philosophical reflections, he foresaw aspects or parts of modern mathematics. Can anyone elaborate on these connections and recommend any references? EDIT, Will Jagy. Editing mostly to bump this to the front of active. It is evident that Jacques and Sergey have good, substantial answers in mind. Please do not answer unless you have read Leibniz at length. I kind of liked philosophy in high school and college, or thought I did. Recently, I read one page of Spinoza and gave up.
|
Abraham Robinson explicitly referred to Leibniz's idea of infinitesimal quantities when developing non-standard analysis in 1960's. Wikipedia article has a quotation from his book
Robinson, Abraham (1996). Non-standard analysis (Revised edition ed.). Princeton University Press. ISBN 0-691-04490-2. Added: the idea of expressing logic in an algebraic way is credited to Leibniz; see e.g. the following article in Stanford Encyclopedia of Philosophy: http://plato.stanford.edu/entries/leibniz-logic-influence/#DisLeiMatLog Added: Saul Kripke introduced a semantics of possible worlds (really, relational semantics) for modal logic. http://en.wikipedia.org/wiki/Modal_logic#Semantics The idea of possible worlds precedes Leibniz, but he devoted a lot of consideration to it. Ironically, his claim that our existing world is the best out of possible ones is perhaps most known from the ridicule it received in Voltaire's "Candide". Oh wait, this is Math Overflow...
|
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|
80,595 |
The standard definition for simplicial homotopy groups only works for Kan complexes (cf. http://ncatlab.org/nlab/show/simplicial+homotopy+group ). I learned that the hard way, when I tried to compute a very simple example, i.e. the homotopy group of the boundary of the standard 2-simplex.
My naive idea to actually compute simplicial homotopy groups for arbitrary simplicial sets was taking the fibrant replacement. But obviously we need a model structure for that. Then again, a weak equivalence in the usual model structure for simplicial sets is precisely a weak equivalence of the geometric realization.(cf. http://ncatlab.org/nlab/show/model+structure+on+simplicial+sets ) As I understand it so far, the only satisfactory way to talk about simplicial homotopy groups requires the notion of "classical" homotopy groups. Hence my question: why does it still make sense, to actually talk about simplicial homotopy groups in the first place?
|
To compute the homotopy groups of a simplicial set $X$ , you need to be able to construct a weak equivalence $X \to Y$ where $Y$ is a Kan complex, and then compute the homotopy groups of $Y$ using the definitions you were discussing. This might seem circular - you need to detect if $X \to Y$ is an equivalence. However, you can construct $Y$ directly using certain more elementary equivalences. Specifically, for a map of a horn $\Lambda \to X$ we can form the pushout of the diagram $\Delta \leftarrow \Lambda \rightarrow X$ , called $X'$ ; on geometric realizations this is homotopy equivalent because you can construct an explicit retraction. The class of maps $X \to Y$ generated by such pushouts is called the family of anodyne extensions of $X$ , and you can always construct an anodyne extension which is a Kan complex by the small object argument (you keep gluing in solutions to every possible horn-filling problem). If you want a more canonical answer, there is also Kan's $\operatorname{Ex}^\infty$ construction . If you want another reference, there are Kan's older papers, and my recollection is that Joyal and Tierney has quite a number of details as well.
|
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|
80,627 |
Koszul duality Given a finite-dimensional $k$-vector space $V$ (I am happy taking $k = \mathbb{C}$ anywhere in the following if it makes a difference) and a subspace $R \subseteq V \otimes V$, we can form the quadratic algebra $$A = A(V,R) = T(V)/ \langle R \rangle,$$
where $\langle R \rangle$ is the 2-sided ideal in the tensor algebra generated by $R$. We can then form the quadratic algebra $A^! = A(V^*, R^\perp)$, where
$$ R^\perp = \{ \phi \in V^* \otimes V^* \mid \phi(R) = 0 \}, $$
and we have identified $V^* \otimes V^*$ with $(V \otimes V)^*$. This algebra $A^!$ is also quadratic by construction, and is known as the Koszul dual of $A$. It's pretty clear that $(A^!)^! \simeq A$. One example of this is given by the symmetric and exterior algebras of a vector space and its dual, i.e. for a finite-dimensional vector space $V$, we have
$$ S(V)^! \simeq \Lambda(V^*), \quad \Lambda(V)^! \simeq S(V^*). $$ Clifford and Weyl algebras Now suppose that $V$ is even-dimensional, say $\mathrm{dim}_\mathbb{C}(V) = 2n$, and let $h: V \otimes V \to k$ be a nondegenerate symmetric bilinear form on $V$. The Clifford algebra is the algebra
$$ \mathrm{Cl}(V,h) = T(V)/\langle x - h (x) \mid x \in S^2(V) \rangle, $$
and this can be viewed as a deformation of the exterior algebra in the sense that the Clifford algebra is naturally filtered and the associated graded is $\Lambda(V)$. If $h$ is nondegenerate, then (over $\mathbb{C}$, at least) we can show that $\mathrm{Cl}(V,h) \simeq M_{2^n}(\mathbb{C})$. If we take instead a nondegenerate alternating (i.e. symplectic) form $g:V \otimes V \to k$, then we can form the Weyl algebra $$ A_n = A_n(V,g) = T(V)/\langle x - g(x) \mid x \in \Lambda^2(V) \rangle. $$
This too has a natural filtration from the tensor algebra, and the associated graded is $S(V)$. These two deformations share some features in common. For instance, the Weyl algebra is isomorphic to the algebra of polynomial differential operators on $\mathbb{C}[x_1, \dots, x_n]$, and one can think of the Clifford algebra as being a $\mathbb{Z}/2$-graded analogue of that via creation and annihilation operators on $\Lambda(V)$. Both algebras are simple. Main question Is there any sort of non-quadratic Koszul duality that relates the Clifford and Weyl algebras?
|
Non-homogeneous Koszul duality is now well-understood. Here are a few references: I guess the original reference is L. E. Positsel′ski˘ı. Nonhomogeneous
quadratic duality and curvature.
Funktsional. Anal. i Prilozhen.,
27:57–66, 96, 1993. for a more systematic study you can have alook at A. Polishchuk and L. Positselski.
Quadratic algebras, volume 37 of
University Lecture Series. American
Mathematical Society, Providence, RI,
2005. As far as I remember the new book of Loday and Vallette discusses this too (see $\S 3.6$). You can find the statement that Weyl and Clifford algebras are Koszul in the inhomogenous sens in this paper of Braverman-Gaistgory ($\S 5.3$). Nevertheless, as it is said in Leonid Positselski's comment, Weyl and Clifford algebras are not Koszul dual to each other. The reason is that inhomogeneous Koszul duality is inhomogeneous! quadratic-linear algebras are dual to DG quadratic algebras (e.g. the universal envelopping algebra of a Lie algebra is Koszul dual its Chevalley-Eilenberg algebra). quadratic--linear-constant algebra (e.g. Weyl or Clifford, for which there is even no linear part) are dual to curved quadratic DG algebras. E.g. for the Weyl algebra $\mathcal W_{(V,\omega)}$, its Kozsul dual is the pair $(\wedge(V^*),\omega)$ where the symplectic form $\omega$ is viewed as a curvature (a degree 2 element) in the exterior algebra.
|
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|
80,667 |
It seems that the normalizer of $H=\mathrm{GL}(n,\mathbf Z)$ in $G=\mathrm{GL}(n,\mathbf Q)$ is "almost" equal to itself, that is,
$$
N_G(\mathrm{GL}(n,\mathbf Z))=Z(G) \cdot \mathrm{GL}(n,\mathbf Z)
$$
where $Z(G)$ is the centre of $G$ (one may guess so by applying the description of automorphisms of groups $\mathrm{GL}(n,\mathbf Z)$ by Hua and Reiner). Is there, however, a simpler and direct proof/disproof of this fact? More generally, for which integral domains $R$ it is known that $\mathrm{GL}(n,R)$ "almost" coincides with its normalizer in the group $\mathrm{GL}(n,Q(R))$ where $Q(R)$ is the quotient field of $R?$ (The question has been earlier posted at mathunderflow ).
|
Let $g \in GL(n,\mathbb Q)$ normalize $GL(n,\mathbb Z)$. Consider the lattice
$g(\mathbb Z^n) \subset \mathbb Q^n$; it is preserved by $GL(n,\mathbb Z)$. Replacing
$g$ by $gz$ for some appropriate scalar matrix $z$, we may assume that
$g(\mathbb Z^n) \subset {\mathbb Z}^n$, but that $g(\mathbb Z^n)\not\subset p \mathbb Z^n$
for any prime $p$. Suppose now that $p$ divides the index $[\mathbb Z^n:g(\mathbb Z^n)]$. Then
the image of $g(\mathbb Z^n)$ is a proper subspace of $\mathbb F_p^n$ (by the assumption
that the $p$ divides the index) which is non-zero (by the assumption that $g(\mathbb Z^n)$
is not contained in $p\mathbb Z^n$). It is preserved by $GL(n,\mathbb F_p)$.
[Added: As tomasz points out in a comment below, $GL(n,\mathbb Z)$ does not surject
onto $GL(n,\mathbb F_p)$. However, its image does contain $SL(n,\mathbb F_p)$,
so the argument below goes through, if we replace $GL(n,\mathbb F_p)$ by $SL(n,\mathbb F_p)$.] But this is a contradiction, since $\mathbb F_p^n$ is an irreducible $GL(n,\mathbb F_p)$-representation. Consequently, no such $p$ exists, and so $g(\mathbb Z^n) = \mathbb Z^n$.
Thus $g \in GL(n,\mathbb Z)$, and so we have shown that the normalizer of $GL(n,\mathbb Z)$ is equal to $Z(G) \cdot GL(n,\mathbb Z),$ as required. This argument (assuming that it's correct!) extends at least to the case when $R$ is a PID.
|
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|
80,724 |
Anytime I see an $n!$ in some formula, my instinct is to look for the symmetric group on $n$ letters coming in somewhere. I have never done this seriously with the $n!$ in Taylor's theorem. Question: Is there some way to see the $n!$ in Taylor's theorem coming naturally from a symmetry group? Possible lead: Here is a definition of $f^{(n)}(a)$ which does not depend on finding earlier derivatives: Let $g: \mathbb{R}^n \rightarrow \mathbb{R}$ be defined by $g(x_1,x_2,x_3, ..., x_n)$ is the lead coefficient of the unique $n^{th}$ degree polynomial passing through $(a, f(a)), (x_1, f(x_1)),(x_2, f(x_2)),...,(x_n, f(x_n))$. Then $f^{(n)}(a)$ is $1/n!$ times the limit of $g(x_1,x_2,x_3, ..., x_n)$ as $(x_1,x_2,x_3, ..., x_n)$ approaches $(a,a,a,...,a)$. Could the $1/n!$ be related to the symmetry of $g$ under exchange of coordinates?
|
There must be many ways to think of this. Here's one: The symmetric group is involved with homogeneous polynomials of degree $n$ because they correspond to symmetric multilinear functions of $n$ variables, and division by $n$ factorial appears when recovering the former from the latter.
For example, a homogeneous polynomial map $f:V\to W$ of degree $3$ determines a map $F:V\times V\times V\to W$ by
$$
F(x,y,z)=f(x+y+z)-f(x+y)-f(x+z)-f(y+z)
+f(x)+f(y)+f(z)-f(0),
$$
and $f(x)$ can be recovered as
$$\frac{F(x,x,x)}{6}.
$$ The same formula for $F(x,y,z)$ can be applied to other functions $f(x)$. Applied to a polynomial of degree $<3$ it gives zero. Applied to a polynomial of degree $\le3$ it gives the multilinear function that corresponds to the purely cubic part of $f$. Applied to any function at all it gives a symmetric function of $(x,y,z)$ that vanishes when $x$ or $y$ or $z$ is zero. The foundation of all differential calculus is the process that takes $f$, subtracts $f(0)$, and then makes the best linear approximation of the result. If you apply this process in all three variables to $f(x+y+z)$, you are first making $F(x,y,z)$ and then (assuming that $f$ was sufficiently smooth) making best linear approximation in all variables. If you then set $x=y=z$ and divide by $3$ factorial, you get the third term in the Taylor series of $f$. Of course, if you perform the linearizations in the three variables one after another then you see this trilinear map as a derivative of a derivative of a derivative. The reason I look at it this way is that a homotopical categorical analogue of this plays a big role in my "functor calculus". The analogue there of division by $n$ factorial is a homotopy orbit spectrum for an action of the symmetric group. In a little more detail: Let $V$ and $W$ be model categories in which filtered homotopy colimits commute with finite homotopy limits. Assume that $W$ is stable, meaning that the final object is equivalent to initial object and that homotopy pushout squares are the same as homotopy pullback squares. For example, $W$ might be the category of spectra and $V$ might be spaces or spectra. Consider functors $f:V\to W$ that are homotopy-invariant (preserve weak equivalences). Say that $f$ has degree $\le 1$ if it preserves homotopy pushout squares. Say that it has degree $\le d$ if it takes those $(d+1)$-dimensional square diagrams all of whose $2$-dimensional faces are homotopy pushouts to homotopy pushout cubes, those in which the last object is the homotopy colimit of the others. Call $f$ reduced if it takes the trivial object $\star$ to itself (up to equivalence). Call it linear if it has degree $\le 1$ and is reduced. There is a linearization process that takes a reduced functor and makes the universal linear functor under it: basically $hocolim \Omega^nf\Sigma^n$. This can be easily generalized to make something called $P_1$ (first polynomial approximation) which takes any functor $f$, reduced or not, and makes the universal degree $\le 1$ functor $P_1f$ under it. (If we first reduce $f$ by taking the homotopy fiber of $f\to f(\star)$ and then linearize, it's the same as first doing $P_1$ and then reducing.) It can be further generalized to make something called $P_d$ ($d$th polynomial approximation) which takes any functor $f$ and makes the universal degree $\le d$ functor $P_df$ under it. The homotopy fiber of the canonical map $P_df\to P_{d-1}f$ is always a degree $\le d$ functor such that $P_{d-1}$ of it is trivial. Call such functors homogeneous of degree $d$. When $d=1$ this is the same as linear. There is the functor of $d$ variables $(X_1,\dots,X_d)\mapsto f(X_1+\dots +X_d)$, where $+$ means (derived) coproduct. This is symmetric in the sense that there are isomorphisms when you permute the inputs, satisfying the obvious identities. Reduce it in all variables simultaneously, by first making a cubical diagram consisting of the objects $f(X_S)$, coproduct of all the $X_i$ for $i\in S$, and then looking at the homotopy fiber of the canonical map from $f(X_1+\dots +X_d)$ to the holim of all the rest. I call this reduced symmetric functor of $d$ variables the $d$th crosseffect of $f$. If we simultaneously linearize it in all variables, we get a symmetric multilinear functor. If $f$ has degree $\le (d-1)$ then its $d$th crosseffect is contractible. If $f$ is homogeneous of degree $d$ then its $d$th crosseffect is already multilinear, and this way of making a symmetric multilinear functor from a homogeneous degree $d$ functor can be inverted: $f(X)$ is the homotopy orbit object for the action of the $d$th symmetric group on $F(X,\dots ,X)$, where the action on $F(X,\dots ,X)$ is the one that you get from the symmetry on $F(X_1,\dots ,X_d)$. For general $f$ the multilinearization of the crosseffect is the same symmetric multilinear functor that corresponds in this way to the $d$th homogeneous part of $f$, i.e. to the homotopy fiber of $P_df\to P_{d-1}f$. Again, by linearizing in one variable at a time you can work out a sense in which the multilinear functor corresponding to the $d$th homogeneous part is in fact an iterated derivative. But in this context it is important to have the option of performing linearization in all variables simultaneously, so to speak, in order that the result should be symmetric in the sense that is required for reconstructing the corresponding homogeneous functor. (Here symmetry is a structure, not a property.)
|
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|
80,797 |
I'm working through Ravi Vakil's notes on algebraic geometry, and I'm at exercise 2.3.R, which states that, if there are maps of objects $X_1\rightarrow Y,X_2\rightarrow Y,Y\rightarrow Z$, then $\begin{matrix}X_1\times_YX_2\rightarrow&X_1\times_ZX_2\\\\\downarrow&\downarrow\\\\Y\rightarrow&Y\times_ZY\end{matrix}$ is a fibered/Cartesian square, assuming all relevant fibered products exist. I believe I have written a full proof, but a thing or two bothered me. In writing down the fibered square for $Y\times_ZY$, I guess I assumed that the maps to both $Y$ terms were the same. But I'm sure there are other ways of building this fibered product. Also, a similar situation in seeing what the map along the bottom of the magic square is; I wrote it down assuming that $Y$ mapped to itself via the identity and then applying the universal property, but again, I'm sure that's not the only way. The map on the right side of the square... Is there some sort of natural map there? I mean, I guess you could look at the maps $X_1\times_ZX_2\rightarrow X_1\rightarrow Y$ (and similarly for the other factor) and again appealing to the universal property... But there are, again, other ways to do this (I think..). Am I wrong? Or could this really be a lot more precise?
|
This cartesian square is important for establishing some basic results about base change in algebraic geometry (although, of course, it holds in every category). The maps are constructed as follows: $X_1 \times_Y X_2 \to X_1 \times_Z X_2$ corresponds to a pair of maps $X_1 \times_Y X_2 \to X_i$ whose composition to $Z$ is the same; well just take the projections from the fiber product and remark that since their composition to $Y$ is the same, the same is true for $Z$. The map $X_1 \times_Z X_2 \to Y \times_Z Y$ is induced by the two maps $X_i \to Z$. The map $Y \to Y \times_Z Y$ is the diagonal map, which is defined to correspond to the identity of $Y$ in both factors. Finally, the morphism $X_1 \times_Y X_2 \to Y$ is just the natural map. So to sum up: Every morphism in the diagram is defined canonically. Of course there are other choices possible, but no other choice does make sense. I would like to make a digression which makes this cartesian square even more clear and deduces it from a more general result, namely that limits commute with limits. Besides, the general result will also yield other canonical isomorphisms which occur often in algebraic geometry. Typically, these isomorphisms are proven separatedly, but as you will see, they are all just corollaries of the following: Lemma . In an arbitrary category, consider the following commutative diagram: $\begin{matrix} X_1 & \longrightarrow & X_0 & \longleftarrow & X_2 \\\\
\downarrow & & \downarrow & & \downarrow \\\\\
S_1 & \longrightarrow & S_0 & \longleftarrow & S_2 \\\\
\uparrow & & \uparrow & & \uparrow \\\\\
Y_1 & \longrightarrow & Y_0 & \longleftarrow & Y_2
\end{matrix}$ Assuming that all the fiber products exist, then we have $(X_1 \times_{S_1} Y_1) \times_{X_0 \times_{S_0} Y_0} (X_2 \times_{S_2} Y_2) = (X_1 \times_{X_0} X_2) \times_{S_1 \times_{S_0} S_2} (Y_1 \times_{Y_0} Y_2)$ In order to make sense of the fiber products, we use, of course, the only possible maps. For example, the two squares on the left yield the map $X_1 \times_{S_1} Y_1 \to X_0 \times_{S_0} Y_0$. The Lemma may be memorized as follows: The horizontal fiber product of the vertical fiber products equals the vertical fiber product of the horizontal fiber products. Now as for the proof of the Lemma, just use the Yoneda Lemma to reduce it to the case of the category of sets, where you can really see this equation immediately. It isn't necessary to draw any arrows and verify the universal property by hand; or rather you encode these arrows as elements. The first corollary of the lemma is the "cancelling law": For morphisms $X \to T \to S$ and $Y \to S$, we have $X \times_T (T \times_S Y) \cong X \times_S Y$. Proof: Apply the lemma to: \begin{matrix} X & \longrightarrow & T & \longleftarrow & T \\\\
\downarrow & & \downarrow & & \downarrow \\\\\
S & \longrightarrow & S & \longleftarrow & S \\\\
\uparrow & & \uparrow & & \uparrow \\\\\
S & \longrightarrow & S & \longleftarrow & Y
\end{matrix} The second one is the "Magic square" of the initial question: For morphisms $X \to S$, $Y \to S$, $S \to T$, there is a cartesian square $\begin{matrix} X \times_S Y & \longrightarrow & X \times_T Y \\\\
\downarrow & & \downarrow \\\\ S & \longrightarrow & S \times_T S \end{matrix}$ Proof: Apply the lemma to: \begin{matrix} S & \longrightarrow & S & \longleftarrow & X \\\\
\downarrow & & \downarrow & & \downarrow \\\\\
S & \longrightarrow & T & \longleftarrow & T \\\\
\uparrow & & \uparrow & & \uparrow \\\\\
S & \longrightarrow & S & \longleftarrow & Y
\end{matrix}
|
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|
80,846 |
A fundamental result in three-dimensional smooth topology, which in computer jargon we might refer to as "a primitive", is the statement that any ( $C^\infty$ ) diffeomorphism of the two-sphere $S^2$ extends to a diffeomorphism of the closed three-ball $D^3$ . Equivalently: $\mathrm{Diff}(S^2)$ is connected. This theorem was first proven by Munkres [Mich. Math. Jour. 7 (1960), 193-197]. Later, Smale proved the stronger result that $\mathrm{Diff}(S^2)$ has the homotopy type of $O(3)$ [Proc. AMS 10 (1959), 621-626]. Another proof of Smale's result is given by Cerf in the appendix to [Sur les difféomorphismes de la sphère de dimension trois ( $Γ_4=0$ ), Lecture Notes in Mathematics, No. 53. Springer-Verlag, Berlin-New York 1968]. Question 1: Are there any other known proofs of the statement that any diffeomorphism of the two-sphere $S^2$ extends to a diffeomorphism of the closed three-ball $D^3$ ? There are two reasons I'm not fully happy with the proofs I cited above. Smale's proof and Cerf's proof show much more and use what looks to me like "too much machinery" for just the " $\mathrm{Diff}(S^2)$ is connected" statement, and, in particular, machinery which seems outside basic differential topology (maybe I'm wrong; I haven't gone into them in much detail). Munkres's proof has a number of back-references to another of his papers [Ann. Math. 72(3) (1960), 521-554], and corners need to be smoothed over and over and over and over again to get an honest smooth isotopy between a given diffeomorphism of $S^2$ and the identity. What is worse, it seems difficult to extract an algorithm from Munkres's proof (Lemma 1.1 looks non-constructive - I wouldn't know how to extract a concrete diffeomorphism out of its proof), which brings me to my second question: Question 2: How could I implement an extension of a smooth diffeomorphism of the two-sphere to the three-ball? To make things really concrete, let's say I had an image of the surface of the earth which I deformed by some strange diffeomorphism $f$ of $S^2$ . How (by computer) could I smoothly deform it back to the usual picture of the earth? One dimension down, maybe one way to do it might be to "relax a diffeomorphism of a circle gradually using the heat equation" (see Greg Kuperberg's comment here ). Does this work one dimension up? I couldn't figure this out, but I don't see an obvious obstruction- not in dimension three. Or maybe there's a slick way of implementing Munkres's proof by lifting an orientation-preserving diffeomorphism of $S^2$ to $\mathrm{Spin}(3)$ or something... I really have no idea. Note, though, that other proofs that diffeomorphisms of $S^1$ extend to $D^2$ clearly seem to fail in dimension three... in particular, trying to use some sort of Alexander trick to comb all the "bad parts" of the diffeomorphism into a small disc and shrink that disc to a point will not give rise to a smooth isotopy. Finally, Morris Hirsch says in a footnote on Page 38 of The Collected Papers of Stephen Smale: "Around this time [1959] an outline of a proof attributed to Kneser was circulating by word of mouth; it was based on an alleged version of the Riemann Mapping Theorem which gives smoothness at the boundary of smooth Jordan domains, and smooth dependence on parameters. I do not know if such a proof was ever published." Question 3: Was such a proof ever published? Is there anything else to be said about this proof outline? Edit: Actually, I'd like to add even a fourth question: Question 4: Are there any "second generation" detailed expositions of any of the above proofs?
|
More a survey of related things than an answer, but here goes. Let's write $D(n)$ for the space of compactly supported diffeomorphisms $\mathbb R^n\to \mathbb R^n$. A reasonable guess might be that this is contractible, but this is true only for very small values of $n$. The space $Diff(S^n)$ of diffeomorphisms $S^n\to S^n$ contains the Lie group $O(n+1)$. A reasonable guess might be that the inclusion of $O(n+1)$ is a homotopy equivalence, but no. In fact, this is equivalent to the first guess: The subgroup of $Diff(S^n)$ consisting of diffeomorphisms supported in the complement of a given point may be identified with $D(n)$, and the multiplication map $D(n)\times O(n+1)\to Diff(S^n)$ (which is not a group homomorphism) is an equivalence. You can see this by comparing $O(n+1)$ with the space of cosets $Diff(S^n)/D(n)$. Thus $D(n)$ is what you might call the exotic part of the homotopy type of $Diff(S^n)$. It is also equivalent to the space of all diffeomorphisms $D^n\to D^n$ fixing the boundary pointwise. Introduce one more player: the space of all compactly supported diffeomorphisms $\mathbb R^n\times \lbrack 0,\infty )\to \mathbb R^n\times \lbrack 0,\infty )$. Call this $P(n)$. It fibers over $D(n)$, and the fiber is equivalent to $D(n+1)$. It is equivalent to the space of "pseudoisotopies" of $D^n$. The statement that every diffeomorphism of $S^n$ extends to $D^{n+1}$ means precisely that $P(n)\to D(n)$ induces a surjection on components. This is in principle weaker than the statement that $D(n)$ is connected, but it turns out that (at least for most values of $n$, maybe all?) $P(n)$ is connected. In low dimensions the story is this: $D(0)$ is a point. $P(0)=D(1)$ is contractible because it is convex: convex linear combinations of order-preserving
diffeomorphisms of the line are again order-preserving diffeomorphisms. $P(1)\sim D(2)$ is contractible. I am aware of two approaches to this: (1) I believe that when Smale (re-)proved this he used the Poincare-Bendixson Theorem. The crux is that, given a compactly supported field of tangent lines in the half-plane $y\ge 0$ transverse to the line $y=0$, if you follow it from $y=0$ you will get all the way up and not get trapped in some spiral. (2) Complex analysis. I always imagine that the following works, but I'm not sure of the details: The space of Riemannian metrics on $S^2$ is contractible because it is convex. The space of conformal structures on $S^2$ is contractible because its product with the space of positive functions on $S^2$ is that space of metrics. The group $Diff(S^2)$ acts on this contractible space. It acts transitively, by the uniformization theorem, and presumably in a strong enough sense that this implies that the subgroup preserving the standard conformal structure is equivalent to the whole. This subgroup (Moebius transformations plus complex conjugation) is equivalent to its maximal compact subgroup $O(3)$. Thus $P(2)\sim D(3)$. Smale conjectured that this is contractible. Hatcher proved it. Thus $P(3)\sim D(4)$. I don't actually know anything about this space. For large values of $n$ (I forget how large): The space $P(n)$ is connected, by a theorem of Cerf. But this does not mean at all that $D(n)$ is connected: we have an exact sequence $\dots \to\pi_1D(n)\to \pi_0D(n+1)\to \pi_0P(n)\to \pi_0D(n)$. $\pi_0D(n)$ is the Kervaire-Milnor group of homotopy $(n+1)$-spheres, known to be finite and frequently nontrivial: You can make a smooth manifold homeomorphic to $S^{n+1}$ from any element of $\pi_0D(n)$ by gluing two hemispheres together. By the $h$-cobordism theorem, you get all homotopy spheres in this way; and it's elementary to see that two elements give the same thing if and only if they differ by something in the image of $\pi_0P(n)$. There is an important map $P(n)\to P(n+1)$. Hatcher showed, and Igusa proved, that it is about $\frac{n}{3}$-connected. In the stable range $P(n)$ is essentially the Waldhausen $K$-theory of a point, which is rationally the same as the algebraic $K$-theory of $\mathbb Z$, and this implies plenty of elements of infinite order in the homotopy groups of $P(n)$ and $D(n)$ in degrees less than about $\frac{n}{3}$.
|
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|
80,861 |
Hi, Let $G$ be an algebraic reductive group over an algebraically closed field $k$, $T$ a maximal torus and $B = TU$ a Borel subgroup containing it. I'm interested in computing $H^*(G/U,\mathcal O_{G/U})$ [corrected typo; I had written $B/U$] (coherent cohomology) (in terms of the representation theory of $G$?). I suppose this is well known, but I can't find it anywhere.... Any suggestions? Thanks!
|
More a survey of related things than an answer, but here goes. Let's write $D(n)$ for the space of compactly supported diffeomorphisms $\mathbb R^n\to \mathbb R^n$. A reasonable guess might be that this is contractible, but this is true only for very small values of $n$. The space $Diff(S^n)$ of diffeomorphisms $S^n\to S^n$ contains the Lie group $O(n+1)$. A reasonable guess might be that the inclusion of $O(n+1)$ is a homotopy equivalence, but no. In fact, this is equivalent to the first guess: The subgroup of $Diff(S^n)$ consisting of diffeomorphisms supported in the complement of a given point may be identified with $D(n)$, and the multiplication map $D(n)\times O(n+1)\to Diff(S^n)$ (which is not a group homomorphism) is an equivalence. You can see this by comparing $O(n+1)$ with the space of cosets $Diff(S^n)/D(n)$. Thus $D(n)$ is what you might call the exotic part of the homotopy type of $Diff(S^n)$. It is also equivalent to the space of all diffeomorphisms $D^n\to D^n$ fixing the boundary pointwise. Introduce one more player: the space of all compactly supported diffeomorphisms $\mathbb R^n\times \lbrack 0,\infty )\to \mathbb R^n\times \lbrack 0,\infty )$. Call this $P(n)$. It fibers over $D(n)$, and the fiber is equivalent to $D(n+1)$. It is equivalent to the space of "pseudoisotopies" of $D^n$. The statement that every diffeomorphism of $S^n$ extends to $D^{n+1}$ means precisely that $P(n)\to D(n)$ induces a surjection on components. This is in principle weaker than the statement that $D(n)$ is connected, but it turns out that (at least for most values of $n$, maybe all?) $P(n)$ is connected. In low dimensions the story is this: $D(0)$ is a point. $P(0)=D(1)$ is contractible because it is convex: convex linear combinations of order-preserving
diffeomorphisms of the line are again order-preserving diffeomorphisms. $P(1)\sim D(2)$ is contractible. I am aware of two approaches to this: (1) I believe that when Smale (re-)proved this he used the Poincare-Bendixson Theorem. The crux is that, given a compactly supported field of tangent lines in the half-plane $y\ge 0$ transverse to the line $y=0$, if you follow it from $y=0$ you will get all the way up and not get trapped in some spiral. (2) Complex analysis. I always imagine that the following works, but I'm not sure of the details: The space of Riemannian metrics on $S^2$ is contractible because it is convex. The space of conformal structures on $S^2$ is contractible because its product with the space of positive functions on $S^2$ is that space of metrics. The group $Diff(S^2)$ acts on this contractible space. It acts transitively, by the uniformization theorem, and presumably in a strong enough sense that this implies that the subgroup preserving the standard conformal structure is equivalent to the whole. This subgroup (Moebius transformations plus complex conjugation) is equivalent to its maximal compact subgroup $O(3)$. Thus $P(2)\sim D(3)$. Smale conjectured that this is contractible. Hatcher proved it. Thus $P(3)\sim D(4)$. I don't actually know anything about this space. For large values of $n$ (I forget how large): The space $P(n)$ is connected, by a theorem of Cerf. But this does not mean at all that $D(n)$ is connected: we have an exact sequence $\dots \to\pi_1D(n)\to \pi_0D(n+1)\to \pi_0P(n)\to \pi_0D(n)$. $\pi_0D(n)$ is the Kervaire-Milnor group of homotopy $(n+1)$-spheres, known to be finite and frequently nontrivial: You can make a smooth manifold homeomorphic to $S^{n+1}$ from any element of $\pi_0D(n)$ by gluing two hemispheres together. By the $h$-cobordism theorem, you get all homotopy spheres in this way; and it's elementary to see that two elements give the same thing if and only if they differ by something in the image of $\pi_0P(n)$. There is an important map $P(n)\to P(n+1)$. Hatcher showed, and Igusa proved, that it is about $\frac{n}{3}$-connected. In the stable range $P(n)$ is essentially the Waldhausen $K$-theory of a point, which is rationally the same as the algebraic $K$-theory of $\mathbb Z$, and this implies plenty of elements of infinite order in the homotopy groups of $P(n)$ and $D(n)$ in degrees less than about $\frac{n}{3}$.
|
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|
80,881 |
This is probably something five-year-old physicists know, but here goes: Is there a standard methodology for computing Fourier transforms of things like $\log |x|$? Wolfram Alpha will happily give an answer (involving a delta function), but actually trying to do this yourself (by parts) gives horribly divergent-looking terms (the question which actually came up had $x$ be a vector in $\mathbb{R}^3,$ where the divergent terms are even more horrible than in the one-dimensional case (I am referring to the technique of just cutting off the function at some large $R;$ there are obviously other techniques, like weighting the integrand by an exponential weight (so you are computing a combination of Fourier and Laplace transforms), then computing the analytic continuation at $0,$ but all these should give the same answer,and there should be a not-totally-ad-hoc way of doing this, one should think...
|
The umbrella legitimization of many such Fourier transforms is as tempered _distributions_ (where the sense of "distribution" is not the probability sense, but in the sense of Laurent Schwartz). The various "regularization" tricks amount to approaching the given distribution in the "weak *-topology" on distributions, by more tractable functions. Fourier transform on tempered distributions is (provably) continuous, so we conclude that all these trick must yield the same outcome. [Edit in response to comment:] The "how to compute" (once we know that any device succeeds) is non-trivial, insofar as it is not clear a-priori how explicit an outcome could be expected. The first volume of Gelfand-Graev-et-alia's "Generalized Functions" does many illuminating examples, mostly computed via meromorphic continuation. The simplest family of examples is probably $|x|^s$. Here, the homogeneity and rotational symmetry, and the fact that Fourier transform respects these (in suitable senses), promise that the Fourier transform of $|x|^s$ on $\mathbb R^n$ is a constant multiple of $|x|^{-n-s}$, for $-n<\Re(s)<0$ to assure local integrability (of both). The constant multiple is determined (for example) by integrating against Gaussians. Then use the fact that the derivative of $|x|^s$ in $s$ multiplies it by $\log|x|$, and set $s=0$. This is the nice way logarithms can arise. The implicit claim that we can do complex analysis with distribution-valued functions was legitimized by Schwartz, and is pervasive in Gelfand-et-alia. Products of $|x|^s$ by harmonic polynomials can be treated almost identically, using the repn theory of the orthogonal group on harmonic polynomials. That is, very often, some sort of _unique_characterization_ of the tempered distribution, and of its image under Fourier Transform, reduce the computation to determination of the relevant constant! Edit: oops, as Bazin notes, the exponent is not $n-s$ but $-n-s$, and adjust the local integrability assertion. (Adjusted above.)
|
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|
80,966 |
I wonder if it there exists a topological compact group $G$ (by compact, I mean Hausdorff and quasi-compact) and a non-zero group morphism
$\phi : G \to \mathbb{Z}$ (without assuming any topological condition on this morphism). For compact Lie groups, using the exponential map, the answers is no, but in general I don't know.
|
The answer is no in general, but this is a rather deep fact. Theorem: (Nikolov, Segal) If $G$ is any compact Hausdorff topological group, then every finitely generated (abstract) quotient of $G$ is finite. N. Nikolov and D. Segal, Generators and commutators in finite groups; abstract quotients of compact groups , arXiv, http://arxiv.org/abs/1102.3037
|
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81,128 |
I rarely find modern research papers (on mathematics) that are less than 5 pages long. However, recently I came across a couple of mathematical research papers from the 1960/1970's that were very short (only 2-4 pages long). The authors of both papers solved very specific problems, and stopped writing (I guess) as soon as they were done. This made me realize that I very much like short papers! When possible, I will strive to do the same with my own future papers. Sometimes one finds short papers that are entitled "A note on...". Since I am first of all not a native English speaker and also only a junior mathematician, I would like to know what you think about my questions/thoughts: 1) Is it a good (or bad?) idea to try to let the title of the paper reflect the fact that it is short? Are there any standard ways of doing this? 2) A paper which is entitled "A note on...", is it expected to be short? How short? Can a 50-pages long paper be "A note on..." or would that not be customary? 3) At least to me "A note on..." could give the impression that the paper is a survey article where no new material is presented, but I guess this need not be the case. When is it appropriate to entitle a paper "A note on..."? 4) When is it appropriate to entitle a paper "On the ..."?
|
I feel the answer is NEVER. You must describe the content of the article, not the length. Some journals publish notes separately from regular papers, and often even encourage their submission by offering speedy refereeing and publication (even the Annals encourages "short", i.e. under 20 pp. papers). Anyhow, if your paper is published as a note it will have "Note" written on it anyway, so no need to be redundant. More generally, you should emphasize not the length but the content. If you prove that all tennis balls are white make the title "All tennis balls are white". If you prove that some tennis balls are white, title your note "On white tennis balls", or "New examples of white tennis balls" or whatever. If your note is a new simple proof, and this is what you want to emphasize, make the title "Short proof that all tennis balls are white". If there was a conjecture that all tennis balls were white and you found a counterexample, use "Not all tennis balls are white". If you study further color properties of white tennis balls, use "A remark on white tennis balls". You see the idea. On the other hand, if you wrote a survey, it important to emphasize that, regardless whether it's long or short. That's because this is a property of the content and style of presentation. For example, "A survey on white tennis balls" or "White tennis balls, a survey in colored pictures", etc. In fact, if your title is "A short survey on tennis balls colors", that would mean that your survey is short in content, as in "brief, incomplete", rather than in length - an important info for the reader to know.
|
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|
81,342 |
I'm giving a talk for the seminar of the PhD students of my math departement. I actually work on Berkovich spaces and arithmetic geometry but, of course, I cannot really talk about that to an audience that includes probabilists, computer scientists and so on. I'd rather like to do an introduction to $p$-adic numbers and $p$-adic analysis. I think these kind of things come up to be really cool when you work on it even just for a short time, but I have the ambitious aim to show them something nice and elementary whose statement will be understood by everyone (of course the proof may also be really hard, but there I could give them just its general idea). In other words the question is: if I had prepared something about Galois theory I would have finished with the application to resolubility of polynomial or compass and straightedge constructions; if it had been something about modular forms, it would have been for sure Fermat's last theorem; with 3-surfaces it would have been Poincaré conjecture and so on. What if it's about p-adic numbers or p-adic analysis? I thought about results on valuations of roots of polynomials, but it seems to me already too complicated ( par ailleurs since I'm introducing valuations at the beginning of the talk, it won't turn out to be an application to something they already knew).
|
Introduce ${\mathbf Z}_p$ as "formal" infinite base $p$ expansions where you add and multiply by carrying (any other description will probably take too long and not be concrete). Show them the series for $-1$ in ${\mathbf Z}_3$ is $2 + 2\cdot 3 + 2 \cdot 3^2 + 2 \cdot 3^3 + \cdots$ by adding 1 to that, carrying, and killing off a new term at each step so the sum is 0. Then emphasize the idea that in ${\mathbf Z}_p$ the number $p$ is small and redo the previous computation with geometric series: $2/(1 - 3) = 2/(-2) = -1$ . Show ${\mathbf Z}_3$ contains a square root of 7: $1 + 3 + 3^2 + 2 \cdot 3^4 + 2 \cdot 3^5 + \cdots$ . (To explain why $p$ being prime is important, say the $p$ -adic integers form an integral domain, and for a contrast you could define $Z_{10}$ in a similar way and say there is a number $x$ in ${Z}_{10}$ besides 0 and 1 satisfying $x^2 = x$ : $x = 5 + 2\cdot 10 + 6\cdot 10^2 + 9\cdot 10^4 + 8\cdot 10^5 + \cdots$ . Compute the first few digits of $x^2$ to check this works. This is related to the elementary school question of finding integers whose square ends in themselves: $5^2$ ends in 5, $25^2$ ends in 25, $625^2$ ends in 625, and so on. The 10-adic solution of $x^2 = x$ which I wrote the initial expansion of above packages all of this information into one number.) You could introduce a topology on ${\mathbf Z}_p$ where numbers are close if a long string of initial digits are the same and make a metric from this too. Then indicate how this makes ${\mathbf Z}_p$ compact by a sequential argument, in the same spirit in which $[0,1]$ is compact by an argument with decimal expansions. A key new feature here, which those with experience only in real and complex analysis haven't seen before, is that ${\mathbf Z}_p$ is a compact ring . In ordinary geometry there are plenty of compact groups, but no compact rings. A nice application of this compactness is the finiteness of integral solutions to certain equations. For example, $x^2 - 7y^2 = 1$ has an infinite number of integral solutions, but $x^3 - 7y^3 = 1$ has just two integral solutions: (1,0) and (2,1). One way to prove this finiteness is to use $3$ -adic power series. By algebraic methods one can show that if integers $x$ and $y$ satisfy $x^3 - 7y^3 = 1$ then $x - y\sqrt[3]{7} = (2-\sqrt[3]{7})^n$ for some integer $n$ ; the solutions $(x,y) = (1,0)$ and $(2,1)$ correspond to $n = 0$ and $n = 1$ , respectively. When you expand $(2-\sqrt[3]{7})^n$ (say, by the binomial theorem when $n > 0$ ) you get a formula $a_n + b_n\sqrt[3]{7} + c_n\sqrt[3]{49}$ with integer coefficients $a_n, b_n, c_n$ and we want $c_n = 0$ . That is a very strong constraint, since normally you don't expect $c_n = 0$ . There is an exponential formula for $c_n$ in terms of $n$ : $$
c_n = \frac{1}{21}\left(\sqrt[3]{7}(2 - \sqrt[3]{7})^n + \omega\sqrt[3]{7}(2 - \omega\sqrt[3]{7})^n + \omega^2\sqrt[3]{7}(2 - \omega^2\sqrt[3]{7})^n\right),
$$ where $\omega$ is a cube root of unity. In the same way $a^x$ can be expanded as a real power series in $x$ when $a > 0$ , the formula for $c_n$ above can be expanded into a $3$ -adically convergent power series in $n$ by interpreting $\omega$ and $\sqrt[3]{7}$ to be a cube root of unity and a cube root of 7 in the 3-adics. (Strictly speaking you need to pass to a finite extension of the 3-adics to pick up a cube root of unity, but let's gloss over that point.) Asking for $c_n$ to be 0 is then asking for $n$ to be an integral root of a 3-adic power series. Just as a nonzero analytic function on a closed (hence compact ) disc in ${\mathbf C}$ has finitely many roots in the disc, a 3-adic power series that converges on ${\mathbf Z}_3$ has finitely many roots in ${\mathbf Z}_3$ . Since ${\mathbf Z}$ is inside of ${\mathbf Z}_3$ this implies in particular that there are finitely many roots in ${\mathbf Z}$ . To make this result effective (i.e., to know $n=0$ and $n=1$ are the only 3-adic roots of that power series), you need techniques to bound the number of $3$ -adic roots of a $3$ -adic power series, and that goes beyond the scope of your talk. :) Details are written, for instance, in section 6.4.7 of Henri Cohen's "Number Theory I: Tools and Diophantine Equations". This technique of proving effective finiteness theorems for integral solutions of Diophantine equations by $p$ -adic methods goes back to work of Strassman and Skolem and later developments in this direction are due to Chabauty and Coleman. However, even without an effective bound it's striking to see that the finiteness of integral solutions to an equation can be explained by an argument inspired by finiteness of zeros of a holomorphic function in a compact set, by sticking the integers into a compact domain on which a (3-adic) power series converges. Here is a cute application of $p$ -adic continuity of polynomials on $\mathbf Q$ (as a contrast to continuity of polynomials on $\mathbf Q$ in the "usual" topology your audience will know). It was the subject of a math.stackexchange question here . If we write out $\sqrt{1+x}$ as a power series it is $$
\sqrt{1+x} = \sum_{n \geq 0} \binom{1/2}{n}x^n = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \frac{7}{256}x^5 - \frac{21}{1024}x^6 + \cdots
$$ and a striking feature is that the coefficients all have denominators that are powers of 2.
It's not a surprise there is some power of 2 in the denominator considering a formula for the coefficient of $x^n$ is $$
\binom{1/2}{n} = \frac{(1/2)(1/2-1)(1/2-2)\cdots(1/2-n+1)}{n!},
$$ but why are there are no other primes in the denominator when you simplify the right side?
One answer, which doesn't use anything $p$ -adic, is to grind out the algebra and check that $$
\binom{1/2}{n} = \frac{(-1)^{n-1}}{2^{2n-1}}\left(\binom{2n-2}{n-1} - \binom{2n-2}{n}\right).
$$ The binomial coefficients on the right are integers so the denominator is a power of 2. (That difference of binomial coefficients has a combinatorial interpretation as the $(n-1)$ -th Catalan number.) Here is a slicker explanation of why the denominator is a power of 2: instead of directly seeing the power of 2 in the denominator of the rational number $\binom{1/2}{n}$ , we will instead show for any prime $p$ other than 2 that $\binom{1/2}{n}$ is a $p$ -adic integer, so it has no $p$ in its denominator as a reduced fraction. For odd $p$ we have the $p$ -adic limit formula $\frac{1}{2} = \lim_{k \rightarrow \infty} \frac{p^k+1}{2}$ , and the terms in that limit sequence are integers.
The polynomial function $\binom{X}{n} \in {\mathbf Q}[X]$ is continuous in the $p$ -adic topology, just as polynomials in ${\mathbf Q}[X]$ are continuous in the "usual" topology (in the reals), so by $p$ -adic continuity $$
\binom{1/2}{n} = \lim_{k \rightarrow \infty} \binom{(p^k+1)/2}{n}.
$$ Every binomial coefficient on the right is a positive integer for large $k$ , so $\binom{1/2}{n}$ is a $p$ -adic limit of integers and therefore is a $p$ -adic integer. This means $\binom{1/2}{n}$ has no $p$ in its denominator. This holds for all odd $p$ , so the only prime in the denominator of $\binom{1/2}{n}$ is 2. The same basic idea shows for any nonzero rational number $r$ that the primes in the denominator of $\binom{r}{n}$ are limited to the primes in the denominator of $r$ . For instance, the only primes in the denominator of $\binom{14/75}{n}$ are 3 and 5 because, with experience, one can see with $p$ -adic limits that this fraction is $p$ -adically integral for any $p$ other than 3 or 5. I doubt you're going to find a proof of that fact by some "explicit formula" method like the first proof I indicated for $\binom{1/2}{n}$ only having 2's in its denominator.
|
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|
81,402 |
More accurately, the question should be: Is it known that $S^6,$ the 6-dimensional sphere, is $not$ a (proper) complex algebraic variety, or algebraic space? And is there a reference? It's easy to see that it cannot be projective (as $H^2=0$), but I don't see how it can violate the usual general properties of algebraic varieties [e.g. by Chow's lemma one can get a projective (and smooth, by resolution of singularities) variety lying over and birational to it, but the cohomology groups can get larger when we go up...]. Maybe one needs some theory of classification of 3-folds (if there is one such theory). Knowing that it cannot be algebraic will somehow make the analogous question of complex structure much sharper (by the way, it does have an almost complex structure). Thank you.
|
Suppose that $X$ is a smooth complete positive-dimensional algebraic space over $\mathbb C$. Then $\mathrm H^2(X, \mathbb Q)$ can not be 0. In fact, every algebraic space contains an open dense subscheme. Let $U \subseteq X$ be an open affine subscheme; by a well known-result, the complement $C$ of $U$ has pure codimension 1; consider $C$ as a divisor, with its reduced scheme structure. I claim that the cohomology class of $C$ can not be 0. Let $Y \to X$ a birational morphism, where $Y$ is a smooth projective variety; this exists, by Chow's lemma for algebraic spaces, and resolution of singularities. The pullback of $C$ is a non-zero effective divisor on $Y$, so its cohomology class is not 0. Since formation of cohomology classes of divisors is compatible with pullbacks, this implies the result. This says very little about the problem of existence of a complex structure. Complete algebraic spaces, or, if you prefer, Moishezon manifolds, are not so distant from projective varieties; it seems clear to me that if $S^6$ has a complex structure, this will be a very exotic animal, very far away from the world of algebraic geometry. [Edit]: why does $C$ have codimension 1? This is Corollaire 21.12.7 of EGAIV, when $X$ is a scheme. In general you can reduce to the case of a scheme by considering an étale morphism $Y \to X$, where $Y$ is an affine scheme. This map is affine, because $X$ is separated, so the inverse image of $U$ in $Y$ is affine.
|
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|
81,443 |
Can anyone help me with references to the current fastest algorithms for counting the exact sum of primes less than some number n? I'm specifically curious about the best case running times, of course. I'm pretty familiar with the various fast algorithms for prime counting, but I'm having a harder time tracking down sum of prime algorithms...
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Deléglise-Dusart-Roblot [1] give an algorithm which determines $\pi(x,k,l)$, the number of primes up to $x$ that are congruent to $l$ modulo $k,$ in time $O(x^{2/3}/\log^2x).$ Using this algorithm to count the number of primes in all residue classes $k<2\log x$ takes
$$1+\sum_{p<2\log x}(p-2)\sim\frac{2\log^2x}{\log\log x}$$
invocations of Deléglise-Dusart-Roblot for a total of $O(x^{2/3}/\log\log x)$ time. This allows one to determine the value of $\sum_{p\le x}p$ mod all primes up to $2\log x$ and hence, by the Prime Number Theorem and Chinese Remainder Theorem, the value of the sum mod $\exp(\vartheta(2\log x))=x^2(1+o(1)).$ Together with bounds on the value of $\sum_{p\le x}p$ [2], this allows the computation of the sum. Note that the primes slightly beyond $2\log x$ may be required depending on the value of $\vartheta(2\log x).$ Practically speaking, except for $x$ tiny, $2\log x+\log x/\log\log x$ suffices. This does not change the asymptotics. I do not know if it is possible to modify the Lagarias-Odlyzko analytic formula [3] to count in residue classes. If so, this would allow an $O(x^{1/2+o(1)})$ algorithm. References [1] Marc Deléglise, Pierre Dusart, and Xavier-François Roblot, Counting primes in residue classes , Mathematics of Computation 73 :247 (2004), pp. 1565-1575. doi 10.1.1.100.779 [2] Nilotpal Kanti Sinha, On the asymptotic expansion of the sum of the first n primes (2010). [3] J. C. Lagarias and A. M. Odlyzko, Computing $\pi(x)$: An analytic method , Journal of Algorithms 8 (1987), pp. 173-191.
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|
81,501 |
So let $R$ be a discrete valuation ring and let $X$ be a scheme which is proper and flat over $R$. Let $X_s$ denote the special fiber of $X$. So intuitively, when somebody says that a curve $X$ is semistable I kind of equate this in my mind with the property that $X_s$ has only ordinary double points as singularities. Q1 : So in general (i.e. in higher dimension) what is the geometrical meaning for a scheme
to be semistable? On the Galois representation side we have a very precise definition of what semistable means using Fontaine's ring $\mathbf{B}_{st}$. Q2 If there is a precise answer to Q1 , is there a good reference (more on the intuitive side than on the technical side) where one shows the equivalence (under suitable assumptions) that the geometrical definition coincides with the galois representation one?
|
In your setting $X$ is semi-stable means that its special fiber $X_s$ is a reduced divisor with normal crossings on $X$ . The link with Galois representations is very deep. In fact in general only one implication is known, namely if $X$ is semi-stable then its associated Galois representation is semi-stable. This was known as (a consequence of) the conjecture $C_{\mathrm{st}}$ of Fontaine and Jannsen. There are now at least three different proofs of this conjecture. One was given by the Japanese school (Hyodo, Kato, Tsuji), see Tsuji's survey in Astérisque 279. Another was given by Faltings using his theory of almost étale extensions. Recently Niziol gave another proof using $K$ -theory. The converse implication seems very difficult in general. For abelian varieties this was proved by Coleman-Iovita (Duke Math. 1999) and Breuil (Annals of Math. 2000). For curves this follows from Deligne-Mumford's theorem that a curve is semi-stable if and only if its Jacobian is semi-stable (Publ. Math. IHÉS 1969) (see Mathieu Romagny's article ). Faltings also has a result that the Galois representations associated to proper schemes over $K = \operatorname{Frac}(R)$ are de Rham (and thus potentially semi-stable). So if we knew the converse implication in general, then we would deduce that every scheme is potentially semi-stable (in the sense that it acquires semi-stable reduction after a finite extension), but this is not known in general.
|
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|
81,626 |
Recently a colleague and I needed to use the fact that the natural map $SL_2(\mathbb{Z}) \rightarrow SL_2(\mathbb{Z}/N\mathbb{Z})$ is surjective for each $N$. I happily chugged my way through an elementary proof, but my colleague pointed out to me that this is a consequence of strong approximation. After browsing through some references (including Ch. 7 of Platonov and Rapinchuk, and one of Kneser's papers) I now cheerfully agree with my colleague. Indeed, as P+R describe, strong approximation is "the algebro-geometric version of the Chinese Remainder Theorem". But what about the proofs? To this particular novice in this area, the proofs seemed rather difficult -- involving a variety of cohomology calculations, nontrivial theorems about Lie groups, and much else besides. I confess I did not attempt to read them closely. But perhaps this is because the theorems are proved in a great deal of generality, or perhaps such arguments are regarded as routine by experts. My question is this: Is the proof of strong approximation genuinely difficult? For example, can a reasonably simple proof be given for $SL_n$ which does not involve too much machinery, but which illustrates the concepts behind the general case?
|
There are two very-different questions here: the best arguments for surjectivity of the natural maps $SL(n,R)\rightarrow SL(n,R/I)$, and about Strong Approximation. While it is true that Strong Approximation more-than implies these surjectivities in situations that are not quite elementary, it is serious overkill, I think. The most direct proofs of these surjectivities may very well strike one as needlessly and unilluminatingly messy, but I think this is a result of underestimating the issue. It's not as serious as (edit: "the strong version of") Strong Approximation, by far, but is not an easy exercise, either. The arguments in Rosenberg cannot be simplified much, apart from removing $K$-theory terminology, if one insists. A possibly irreducibly-minimal argument for $SL(2,\hbox{PID})$ is on-line at http://www.math.umn.edu/~garrett/m/mfms/notes/07b_surjectivity.pdf In any case, once one reconciles oneself to the not-quite-triviality of the surjectivity issue, the proofs one reasonably finds may seem less unreasonable. The question(s) about proofs of (edit: "the strong forms of") Strong Approximation are in a different league. (Indeed, as in one comment or answer, the surjectivity issue by itself does not need adeles or cohomology.) In the late 1930s, Eichler had results amounting to Strong Approximation for simple algebras over number fields. M. Kneser's work starting about 1960 (and with two articles in the Boulder conference) mostly addressed orthogonal groups and classical groups other than SL(n) or unit groups of simple algebras. For the latter, including SL(n), the argument is essentially linear algebra and elementary analysis, with some clevernesses. (The argument in my old Hilbert Modular Forms book was (if I recall correctly) a distillation of remarks in Eichler's old papers, and remarks of Kneser. The SL(n) case was not the main focus of any of that, since, despite its non-triviality, Strong Approximation for SL(n), even over number fields, is much easier than the orthogonal group case. Perhaps I'll put a version of that argument on-line sometime soon, since the book is out of print, and there is no longer any electronic version...) In brief, the reason the Platonov-Rapinchuk iconicized version uses cohomology is to address the orthogonal-group or unitary-group cases, and others with more genuine arithmetic content than has the SL(n) case. That is, again, the SL(n) case of Strong Approximation is "trivial" by comparison to the orthogonal-group case, although it itself is vastly more serious than the surjectivity question. [EDIT 2:] Following up @LYX's note, indeed, for the usual invocations of Strong Approximation for SL(n), the Bourbaki Comm. Alg. citation suffices. Unsurprisingly, it does argue on elementary matrices (thus implicitly a bit of K-theory), and is purely algebraic, in the sense that finiteness of residue fields is irrelevant. One "usual" invocation is to know that $SL(n,F_\infty)SL(n,F)$ is dense in $SL(n,\mathbb A_F)$ for a number field $F$, where $F_\infty$ is the product of archimedean completions ($F_\infty=F \otimes_{\mathbb Q} \mathbb R$). The application to automorphic forms is that quotients of this adele group are "merely" quotients of the Lie group $SL(n,F_\infty)$. The stronger version replaces $F_\infty$ with $F_v$ for any place $v$. That is, the general case does not specially suppress reference to archimedean places. (The argument I gave in my Hilbert modular forms book did suppress archimedean places, exactly because of the specific motivations. Nevertheless, "my" argument there did more resemble the viewpoint that would extend to the stronger assertion. The date on Bourbaki's Comm. Alg. is 1985, which might explain why, from late 1970s to mid-1980's, I saw no discussion of any form of "Approximation" other than Kneser and Eichler. Kneser's paper does also mention Rosenlicht and Steinberg for discussions of more-interesting cases than SL(n).) The Kneser article "Strong Approximation" in the Boulder Conference (Proc Symp Pure Math AMS IX, 1966, pp 187-196 cites Eichler 1938 J. Reine und Angew. Math "Allgemeine..." as treating the simple algebra case, excepting the definite quaternion algebra case (in which SA does not hold). Thus, Eichler's case was already more complicated than SL(n). In summary: any form of the SL(n) case is vastly simpler than orthogonal-group or other cases. There is a bifurcation already for SL(n), namely, whether or not archimedean places are suppressed (thus, giving a "purely algebraic" result). The more delicate result can be relevant for "Shimura curves", made from quaternion division algebras over totally real fields, split at only a single real place, to know that the adelic construction is a single classical quotient.
|
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|
81,732 |
The question Let $f$ be a nonconstant polynomial over $\mathbb{C}$. Let's say that a point $c \in \mathbb{C}$ is unusual for $f$ if every root $x$ of $f(x) - c$ is repeated. Can $f$ have more than one unusual point? Short remarks There can be exactly one unusual point, e.g. if $f(x) = x^2$. There can be none, e.g. if $f(x) = x^3 + 3x$. There are at most $\deg(f) - 1$ unusual points, since every unusual point is the image under $f$ of a critical point. The hypotheses "nonconstant" and "over $\mathbb{C}$" could be varied. I added them to rule out cases like the following: over a field $k$ of characteristic $p$, every point of $k$ is unusual for $x^p$ (since the derivative of $x^p$ is $0$). I'd be happy to change the hypotheses to "nonconstant polynomial over an algebraically closed field of characteristic 0". Maybe something like "polynomial whose derivative is nonzero, over an algebraically closed field" would also be sensible. Weak reason to expect the answer to be "no" Perhaps there's a very short answer to my question: I could be overlooking something elementary. But in case it's not so easy, I'll give a flimsy argument for why we might expect the answer to be "no" - that is, for why we might expect every polynomial to have at most one unusual point. My question is equivalent to: is there a nonconstant polynomial $f$ over $\mathbb{C}$ for which $1$ and $-1$ are both unusual? For $1$ to be unusual means that every root of $f(x) - 1$ is also a root of $f'(x)$. Writing $d = \deg(f)$, this is equivalent to
$$
(f(x) - 1) \mid f'(x)^d.
$$
Similarly, for $-1$ to be unusual means that $(f(x) + 1) \mid f'(x)^d$. Both together are equivalent to
$$
(f(x)^2 - 1) \mid f'(x)^d,
$$
that is,
$$
(f(x)^2 - 1)\cdot g(x) = f'(x)^d
$$
for some $g(x) \in \mathbb{C}[x]$. This forces $\deg(g) = d(d - 3)$ (and so $d \geq 3$). So, can we find $f$ and $g$ satisfying the last displayed equation? Comparing coefficients, what we have here is a system of $d(d - 1) = d^2 - d$ equations in $(\deg(f) + 1) + (\deg(g) + 1) = d^2 - 2d + 2$ unknowns. There are $d - 2$ more equations than unknowns, and $d \geq 3$, so a first guess is that it can't be done.
|
This is impossible by the Mason-Stothers theorem (which holds over any algebraically closed field of characteristic zero). We want to find $f, g, h$ such that $f + g = h$ where $g$ is a constant and $f, h$ have all of their roots repeated. If $g$ is nonzero, $f, h$ must be relatively prime. Letting $d = \deg f$, it follows that $fgh$ has at most $d$ roots, but by Mason-Stothers $fgh$ must have at least $d+1$ roots; contradiction.
|
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|
81,740 |
I am looking for a modern introduction to spectra that improves on the treatment by Adams in his "Stable Homotopy and Generalized Homology" notes (by improves I mean taking into account what has been learned since the notes were written). In particular I'm interested in a source that covers some of the variations on Spectra (CW spectra, symmetric spectra, other types/categories of spectra etc.) and ring spectra. Question: What is a good introduction to the modern point of view on spectra? I am particularly interested in the stable/unstable Adams' spectral sequence but the source need not take that as a goal. As an aside I'll point out that notes from Hatcher in his unfinished book on spectral sequences has a short but nice, clear and concrete introduction to spectra. It does not go into the detail and depth I need.
|
[I'm a novice, and this got posted out of order: it answers Bak's question below.] Sure, I can provide that. The cited reference was published in 1995, which
was well before details of symmetric or orthogonal spectra were available,
so it gives a fair amount of background but only refers to EKMM spectra for
a modern category. There is a paper (Mandell, May, Schwede, Shipley) that
compares all choices except EKMM, and there are various papers that compare
those choices with EKMM, starting with a paper by Schwede. Those papers are
maybe more technical than you want. A recent survey paper compares the various
approaches philosophically: see Sections 11 and 12 of my paper What precisely are $E_{\infty}$ ring spaces and $E_{\infty}$ ring spectra?
Geometry \& Topology Monographs 16(2009), 215--282. That gives references and is fairly independent of Sections 1-10. It starts
with a theorem (11.1) of Gaunce Lewis explaining that there is no ideal choice
of category: if you assume your category has all the good properties
you want, you reach a contradiction. The incompatibility comes when you ask
for a homotopically meaningful symmetric monoidal structure on your category
of spectra that also has a homotopically meaningful monoidal adjunction
$(\Sigma^{\infty},\Omega^{\infty})$ relating spaces and spectra. I'm old-fashioned
maybe, but I think spaces are still kind of important. EKMM comes as close as possible to having such an adjunction, with the related
advantage that all objects are fibrant and the related disadvantage that the sphere
spectrum is not cofibrant. Symmetric and orthogonal spectra have the advantage that
they are significantly easier to define and the sphere spectrum is cofibrant. The simplicial version of symmetric spectra has the advantage that it is especially
well-suited to adaptation to the motivic world. Orthogonal spectra have the
advantage that they are much better suited for equivariant and parametrized
generalizations than symmetric spectra. Common features are captured by the
web of Quillen equivalences relating not just all known constructions but
all possible ``good'' model categories of spectra: there is an axiomatization,
due to Shipley; symmetric spectra play a privileged role in the proof.
|
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|
81,910 |
Write $X_N$ for this blow up. Place the N points in 'general position' as needed. Then $X_6$ embeds in $CP^2$ as a smooth cubic surface. (See, eg, Griffiths and Harris.) But there is no other $N$ (except $N=0$)
for which $X_N$ embeds in $CP^3$.
(Proof: The topology of the blow-up
disagrees with that of a smooth surface of degree $d$ in $CP^3$. (Gompf-Stipsisz p. 21.) On the other hand, $X_N$ embeds in $CP^5$ simply because any
smooth algebraic surface $X$ so embeds. (Harris, `Algebraic Geometry, a first course', p. 193.) Embarrassingly, I don't even know the answer for $N=1$ where $X_1$ is the 1st Hirzebruch surface! (I'm betting it does embed.) Motivation: This question began in an attempt to better understand the 27 lines on the cubic
and my initial surprise at how the construction described in GH of $X_6$ yielded
a smooth surface in $CP^3$, and how all such surfaces arise through that construction by varying the 6 points. I am hoping answers might help me understand the moduli of blow-ups as I move the N points about the plane, and orient me as a novice to algebraic surfaces.
|
For $N=1$ the answer is yes: the embedding into ${\mathbb P}^4$ is given by the linear system of conics through the blown up point (the image has degree $d=3$).
For $N=5$, the system of cubics through the 5 points gives an embedding ($d=4$). ADDED: here are 2 slightly less obvious examples:
For $N=8$ one can take quartics with an assigned double point and 7 simple base points ($d=5$).
For $N=10$ take the quintics with 3 assigned double points and 7 simple base points ($d=6$; I did not check all the details here, because it's very boring, but I'm sure that it works). In general, giving a satisfactory answer to your question seems very hard. There is a numerical equality,
the so-called "double point formula" (Hartshorne, "Algebraic geometry", p.434), which is satisfied by all smooth surfaces of ${\mathbb P}^4$:
$$d^2-10d-5HK+12\chi-2K^2=0,$$
where $H$ is the hyperplane section, $d=H^2$ is the degree, $K$ is the canonical divisor and $\chi$ the Euler characteristic of ${\mathcal O}_{X_N}$.
In our case the formula becomes:
$$d^2-10d-5HK+2N-6=0.$$
In addition there is result by G. Ellingsrud and C. Peskine [Invent. Math. 95 (1989), no. 1, 1--11] saying that only finitely many components of the Hilbert scheme of smooth surfaces in ${\mathbb P}^4$ contain smooth rational surfaces.
So in principle it should be possible to classify all the smooth rational surfaces in ${\mathbb P}^4$. In practice, it is known that the degree is $\le 76$,
[Cook, An improved bound for the degree of smooth surfaces in P4 not of general type.
Compositio Math. 102 (1996)] and it is conjectured that $d\le 15$ (examples with $d=15$ do exist). There are also papers by several authors (Ranestad, Schreyer, Popescu, and others) that classify the smooth rational surfaces of ${\mathbb P}^4$ of degree $\le 11$. In these papers you can find examples of the kind you are looking for. For instance there are examples with $d=10$ and $N=18$.
|
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|
81,939 |
I am curious about how much descriptive set theory is involved in inner model theory. For instance Shoenfield's absoluteness result is based on the construction of the Shoenfield tree which projection is $\aleph_1$-Suslin. Also the Schoenfield tree is homogeneous, meaning the direct limit $M_x$ of the ultrapowers by the measures $\mu_{x\upharpoonright n}$ is wellfounded. The measures $\mu_{x\upharpoonright n}$ are defined on the sections of the tree. We also have that $L(\mathbb{R}) \vDash AD$ is equiconsistent with $ZFC+$ there are infinitely many Woodin cardinals. Descriptive set theory talks a lot about homogeneously Suslin sets and homogeneous trees (they pave the way for determinacy results) but these concepts seem themselves to be very important for inner model theory (just a simple fact: a set $X$ is homogeneously Suslin iff $X$ is continuously reducible to the wellfoundedness of towers of measures). The Martin Solovay tree is what gives $\Sigma^1_3$ absoluteness between $V$ and a generic extension $V[G]$ assuming measurability. Also, the Kechris-Martin Theorem has a purely descriptive theoretic proof and a purely inner model theoretic proof. A theorem of Woodin states that $(\Sigma^2_1)^{Hom_{\infty}}$ sentences are absolute for set forcing if there are arbitrarily large Woodin cardinals. My question is why are there so many links between descriptive set theory and inner model theory? I would love to hear from an expert about the intuition as to what is really going on. The relationship between both field does not seems "ad hoc", it appears as though there is very deep beautiful and natural structure. I apologize in advance for any vagueness in my question. Thx.
|
I think that in some ways you have answered your question yourself: we see that to prove properties about sets, say within the projective hierarchy, we need representations of those sets of reals as trees , but moreover nice trees with certain properties (and the homogeneous trees you mention in particular with measures attachable to them). To get the latter involves measurable cardinals at least; and to get the determinacy of PD or AD we need to be able to shift those trees around in more subtle ways to prove that complements of nice trees, and projections of those complements etc, also have nice properties, and this requires Woodin cardinals etc... Conversely we find descriptive set theoretic arguments involved in analysing the mouse components that go into the making of the higher inner models, in particular this is necessary for the so-called core model induction. To construct such models one has to have something of an inductive process to do so, and to start off, this involves an analysis of the lower levels of a putative model, and the description of those levels is, or can be seen as, descriptive set-theoretic on the sets of reals of the model. It just begins to look inevitable that DST and inner model theory will thus be inextricably linked.
Thus the fact that the Kechris-Martin theorem has two styles of proof starts to look like two sides of the same coin. Whilst I am not really pretending that this is a comprehensive or full answer to your question, (I hesitated to answer this as I am not setting myself up as "an expert" who you ask for!)
one could add that
one aspect at least, is the following reply to the (simpler, side) question that is often asked "Why large cardinals" or "Why are such large trees, and concomitantly, large sets, measures etc needed for these analyses?" It is possible to see this, at least in the determinacy arena as simply an extension of what is needed to analyse Borel: Friedman showed that we would need iterations of the power set operation (and Collection to collect together the resulting sets)
with roughly speaking the number of iterations proceeding stepwise through the complexity of the Borel sets to show their determinacy (thus for $\alpha \geq \omega$ one would need $\alpha$ many iterates of power set etc to get $Det(\Sigma^0_\alpha)$). Martin's proof of Borel determinacy showed that Friedman had it exactly right: we already need larger and larger trees, or spaces if you like, in which to unravel Borel sets at a certain level and show their clopen representation (and hence determined status) in a corresponding larger space.
So: we exhaust the power of ZF to show there are enough good trees to establish Borel determinacy. To get higher levels of determinacy we are going to need to go beyond ZF and use strong axioms of infinity, i.e. large cardinals. Thus the whole enterprise spans a spectrum through building models of fragments of second order number theory second order number (up to $Det(\Sigma^0_3$) say, through models of fragments ZF of certain uncountable ordinal height (Borel Determinacy, so we use "weak" inner models of fragments of ZF of set height here), and then full height of the ordinals models of ZF, (for $Det(\Pi^1_1)$) and now comes the same with inner models for larger cardinals, albeit more sophisticated arguments.
|
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|
81,942 |
In C. McMullen's Uniformly Diophantine numbers in a fixed real quadratic field generalized Fibonacci sequence are defined as follows: $f_0=0,f_1=1,f_m=tf_{m-1}-nf_{m-2}$ where some fixed $t\in \mathbb Z$ and $n$ is $+1$ or $-1$ and $t^2-4n>0$.
For example, for $t=1,n=-1$ we get the usual Fibonacci sequence. My question: Does there exist $t,n$ such that the resulting Fibonacci sequence has infinitely primes in it?
I think that it is conjectured to hold for the usual Fibonacci sequence.
A weaker assertion: Does there exist $t,n$ such that the resulting Fibonacci has infinitely many elements with a large prime divisor, e.g., infinitely many $m$'s such that $p_m|f_m$, $p_m$ prime and $\frac{p_m}{f_m}>C$ for some C>0? A related paper (which does not contain any answer to the above questions) is
(By Y. Bugeaud F. Luca, M. Mignotte et S. Siksek) On Fibonacci numbers with few prime divisors . I'll be happy to know about any reference that deal with these generalized sequences.
|
I think that in some ways you have answered your question yourself: we see that to prove properties about sets, say within the projective hierarchy, we need representations of those sets of reals as trees , but moreover nice trees with certain properties (and the homogeneous trees you mention in particular with measures attachable to them). To get the latter involves measurable cardinals at least; and to get the determinacy of PD or AD we need to be able to shift those trees around in more subtle ways to prove that complements of nice trees, and projections of those complements etc, also have nice properties, and this requires Woodin cardinals etc... Conversely we find descriptive set theoretic arguments involved in analysing the mouse components that go into the making of the higher inner models, in particular this is necessary for the so-called core model induction. To construct such models one has to have something of an inductive process to do so, and to start off, this involves an analysis of the lower levels of a putative model, and the description of those levels is, or can be seen as, descriptive set-theoretic on the sets of reals of the model. It just begins to look inevitable that DST and inner model theory will thus be inextricably linked.
Thus the fact that the Kechris-Martin theorem has two styles of proof starts to look like two sides of the same coin. Whilst I am not really pretending that this is a comprehensive or full answer to your question, (I hesitated to answer this as I am not setting myself up as "an expert" who you ask for!)
one could add that
one aspect at least, is the following reply to the (simpler, side) question that is often asked "Why large cardinals" or "Why are such large trees, and concomitantly, large sets, measures etc needed for these analyses?" It is possible to see this, at least in the determinacy arena as simply an extension of what is needed to analyse Borel: Friedman showed that we would need iterations of the power set operation (and Collection to collect together the resulting sets)
with roughly speaking the number of iterations proceeding stepwise through the complexity of the Borel sets to show their determinacy (thus for $\alpha \geq \omega$ one would need $\alpha$ many iterates of power set etc to get $Det(\Sigma^0_\alpha)$). Martin's proof of Borel determinacy showed that Friedman had it exactly right: we already need larger and larger trees, or spaces if you like, in which to unravel Borel sets at a certain level and show their clopen representation (and hence determined status) in a corresponding larger space.
So: we exhaust the power of ZF to show there are enough good trees to establish Borel determinacy. To get higher levels of determinacy we are going to need to go beyond ZF and use strong axioms of infinity, i.e. large cardinals. Thus the whole enterprise spans a spectrum through building models of fragments of second order number theory second order number (up to $Det(\Sigma^0_3$) say, through models of fragments ZF of certain uncountable ordinal height (Borel Determinacy, so we use "weak" inner models of fragments of ZF of set height here), and then full height of the ordinals models of ZF, (for $Det(\Pi^1_1)$) and now comes the same with inner models for larger cardinals, albeit more sophisticated arguments.
|
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"https://mathoverflow.net/questions/81942",
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|
81,960 |
The question briefly: Can one explain the "Dzhanibekov effect" (see youtube videos from space station or comments below) on the basis of the standard rigid body dynamics using Euler's equations? (Or explain that this is impossible and that would be yet another focus ... ) Here are more details. See these curious videos from a space station: https://www.youtube.com/watch?v=L2o9eBl_Gzw Or in the original form which is more striking (~30 seconds after the start): https://www.youtube.com/watch?v=dL6Pt1O_gSE A mathematical description of the model is the standard rigid body motion in empty space (since we are in outer space), which (as is well-known) can be decomposed into a center of mass motion and a rotation part described by Euler's equations. (Euler's equations can be seen as a geodesic flow for some left-invariant metric on SO(3) - as V.I. Arnold taught.)
Let us forget about the center of mass motion. The phenomena shown in the videos is the following - a rigid body is rotating around the axis and then SUDDENLY the rotation axis CHANGES ITS POSITION by 180 degrees. And this happens periodically with some time period. I am a little puzzled how to explain this. If we admit that the motion of this rigid body is EXACTLY the rotation around the axis this is for sure IMPOSSIBLE - Euler's equations predict that such a motion will continue forever. However in the real world there is nothing exact so it might be related to some instability effect - that we are around an unstable equilibrium - and go away from it after some time... Can it be so? If it so, it is however not clear why it happens so quickly and we get exactly the rotation axis changing its position by 180 degrees... Vladimir Dzhanibekov who observed this in 1985 is a famous Russian cosmonaut https://en.wikipedia.org/wiki/Vladimir_Dzhanibekov He was in space 5 times (as far as I know he is the champion in this).
In 1985 his "mission impossible" was saving the Soviet space station "Salut-7" which due to some problems had run out of control... If this can be explained I think it can be a beautiful illustration for students in lectures
on rigid body mechanics... Also if one would want to add some humor in such a lecture one may add that our Earth is this kind of finger nut so it might also do such things ...
(Maybe it really can?)
|
One can see this effect qualitatively from Newtonian first principles such as $F=ma$ (as opposed to Hamiltonian or Lagrangian principles, such as conservation of energy and angular momentum) by looking at a degenerate case, when one moment of inertia is very small and the other two are very close to each other. More specifically, consider a thin rigid unit disk, initially oriented in the $xy$ plane and centred at the origin $(0,0,0)$ . We make the "spherical cow" hypotheses that this disk has infinitesimal thickness and mass, but infinite rigidity. On this disk, we place heavy point masses of equal mass $M$ at the points $(1,0,0)$ and $(-1,0,0)$ on the $x$ axis, and light point masses of equal mass $m$ at the points $(0,1,0)$ and $(0,-1,0)$ on the $y$ axis. Here $0 < m \ll M$ , i.e. $m$ should be viewed as negligible with respect to $M$ . (The moments of inertia are then $2m, 2M, 2(m+M)$ , though we will not explicitly use these moments in the analysis below.) We now set up the unstable equilibrium by rotating the disk around the $y$ axis. Thus, the light $m$ -masses stay fixed on the $y$ -axis, while the heavy $M$ -masses rotate in the $xz$ -plane. This is in equilibrium: there are no net forces on the $m$ -masses, while the rigid disk exerts a centripetal force on the $M$ -masses that keeps them moving in a circular motion on the $xz$ -plane. We can view this equilibrium in rotating coordinates, matching the motion of the $M$ -masses. (Imagine a camera viewing the disk, rotating around the $y$ -axis at exactly the same rate as the disk is rotating.) In this rotating frame, the disk is now stationary (so the $m$ -masses are stuck on the $y$ -axis at $(0,\pm 1,0)$ and the $M$ -masses are stuck on the $x$ -axis at $(\pm 1,0,0)$ ), but there is a centrifugal force exerted on all bodies proportional to the distance to the $y$ -axis. The $m$ -masses are on the $y$ -axis and thus experience no centrifugal force; but the $M$ -masses are away from the $y$ -axis and thus experience a centrifugal force, which is then balanced out by the centripetal forces of the rigid disk. Now let us perturb the disk a bit, so that the $m$ -masses and $M$ -masses are knocked a little bit out of position (but keeping the centre of mass fixed at $(0,0,0)$ ). In particular, the $m$ -masses are knocked away from the $y$ -axis and now experience a little bit of centrifugal force. On the other hand, the rigid disk forces the light $m$ -masses to remain orthogonal to the heavy $M$ -masses, by exerting tension forces between the masses. In the regime where $m$ is negligible compared to $M$ , these tension forces will barely budge the heavy $M$ masses (which therefore remain essentially fixed at $(\pm 1,0,0)$ in the rotating frame), so the effect of these tension forces is to constrain the $m$ -masses to lie in the $yz$ -plane (up to negligible errors which we now ignore). Rigidity also keeps the $m$ -masses at a unit distance from the origin, and antipodal to each other, so the $m$ -masses are now constrained to be antipodal points on the unit circle in the $yz$ -plane. However, other than this, rigidity imposes no further constraints on the location of the $m$ -masses, which can then move freely as antipodal points in this unit circle. The effect of centrifugal force in the rotating frame is now clear: if an $m$ -mass (and its antipode) is perturbed to be a little bit off the $y$ -axis in this unit circle with no initial velocity, then centrifugal force will nudge it a little further off the $y$ -axis, slowly at first but with inexorable acceleration. Eventually it will shoot across the unit circle and then approach the antipode of its previous position. At this point the centrifugal forces act to slow the $m$ -masses down, reversing all the previous acceleration, until one ends up with no velocity at a small distance from the antipode. The process then repeats itself (imagine a marble rolling frictionlessly between two equally tall hills, starting from a position very close to the peak of one of the hills). UPDATE, September 2019 : Due to renewed interest in this question, I will expand upon my 2014 comment regarding why the above analysis appears at first glance to also lead to the incorrect conclusion that the disk rotation is also unstable if it one instead rotates around the $x$ -axis (so that it is now the $M$ -masses that are stationary and the $m$ -masses that are rotating), or equivalently if one swaps the location of the $m$ -masses and $M$ -masses (which we will not do here to try to reduce confusion). The reason for this is that centrifugal force $F_{\mathrm{Cent}} = -m \Omega \times \Omega \times r$ is only one of two inertial forces that are introduced when one is in a steadily rotating frame. The other inertial force introduced is the Coriolis force $F_{\mathrm{Cor}} = -2 m \Omega \times v$ , which acts on moving bodies in the rotating reference frame in a direction orthogonal to the motion (and to the axis of rotation). Strictly speaking, one has to take into account the effect of both inertial forces when performing Newtonian mechanics in a steadily rotating frame. As it turns out, the Coriolis force has a negligible impact on the dynamics when rotating around the $y$ -axis, but dominates the dynamics when rotating around the $x$ -axis, which is why the preceding discussion is accurate in the former case but not the latter. In more detail: suppose we are rotating around the $y$ -axis as in the above discussion. In the rotating frame of reference, and starting with a configuration slightly out of equilibrium, we have as before that the $m$ -masses experience a little bit of centrifugal force and begin to slide away from the $y$ -axis and into the rest of the $yz$ -plane. When doing so, they will then experience some Coriolis force in a direction parallel to the $x$ -axis (which direction it is depends on the orientation of the rotation, as per the right hand rule formula for the cross product). However, due to the rigidity of the disk, as mediated by tension forces within the disk, it is not possible for the $m$ -masses to actually move in the $x$ -direction without also moving the much heavier $M$ -masses. But the magnitude of the Coriolis force is proportional to the small mass $m$ rather than the large mass $M$ , so by Newton's law $F=Ma$ for the $M$ -masses, the Coriolis force (or more precisely, the tension force produced in response to the Coriolis force) actually barely affects the motion of the $M$ -masses, which basically stay put on the $x$ -axis, and the $m$ -masses therefore remain essentially constrained to the $yz$ -plane and cannot actually experience any significant motion in the direction of the Coriolis force. (This is what the sentence in the original explanation regarding how tension forces "barely budge" the $M$ -masses was referring too, albeit somewhat obliquely.) The analysis of the original explanation now proceeds as before. In contrast, suppose that the disk is close to the stable equilibrium state when it rotates around the $x$ -axis. Working in a rotating frame around this axis, the $M$ -masses are near the $x$ -axis of rotation, while the $m$ -masses lie near the $y$ -axis. As before, the $M$ -masses experience centrifugal force and thus begin drifting slightly away from the $x$ -axis into the $xz$ -plane. But then the Coriolis force on these masses kicks in, which is now proportional to the heavy mass $M$ rather than the light mass $m$ . As such, the rigid tension forces connecting the $M$ -masses to the much lighter $m$ -masses offer very little resistance to the Coriolis force, and the motion of the $M$ -mass begins to rotate out of the $xz$ -plane, constantly experiencing Coriolis acceleration in a direction orthogonal to its motion. This effect tends to make the $M$ -mass rotate in a tight circle, and basically neutralises the net effect of the centrifugal force by rotating any outward motion away from the $x$ -axis back into inward motion. The end result is that in the rotating frame, the disk wobbles a bit around its equilibrium state, but does not dramatically depart from it, which is what one would expect from a stable equilbrium (imagine now a marble rolling frictionlessly around the trough of a valley).
|
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|
81,967 |
Consider the $n$-dimensional sphere $S^n$. I'm especially interested in the $n=4$ case. The Hilbert space $L^2(S^n)$ can be decomposed into a direct sum of eigenspaces of the Laplacian, which are finite dimensional. I'm looking for non-isometric conformal transformations $$f: S^n \to S^n$$ s.t. for some $\lambda, \mu > 0$ if $\psi$ is an eigenvector of the Laplacian with eigenvalue $\alpha < \lambda$ then $f(\psi)$ is a sum of eigenvectors with eigenvalues $< \mu$. Do such $f$ exist? If so, is it possibly to classify them?
|
One can see this effect qualitatively from Newtonian first principles such as $F=ma$ (as opposed to Hamiltonian or Lagrangian principles, such as conservation of energy and angular momentum) by looking at a degenerate case, when one moment of inertia is very small and the other two are very close to each other. More specifically, consider a thin rigid unit disk, initially oriented in the $xy$ plane and centred at the origin $(0,0,0)$ . We make the "spherical cow" hypotheses that this disk has infinitesimal thickness and mass, but infinite rigidity. On this disk, we place heavy point masses of equal mass $M$ at the points $(1,0,0)$ and $(-1,0,0)$ on the $x$ axis, and light point masses of equal mass $m$ at the points $(0,1,0)$ and $(0,-1,0)$ on the $y$ axis. Here $0 < m \ll M$ , i.e. $m$ should be viewed as negligible with respect to $M$ . (The moments of inertia are then $2m, 2M, 2(m+M)$ , though we will not explicitly use these moments in the analysis below.) We now set up the unstable equilibrium by rotating the disk around the $y$ axis. Thus, the light $m$ -masses stay fixed on the $y$ -axis, while the heavy $M$ -masses rotate in the $xz$ -plane. This is in equilibrium: there are no net forces on the $m$ -masses, while the rigid disk exerts a centripetal force on the $M$ -masses that keeps them moving in a circular motion on the $xz$ -plane. We can view this equilibrium in rotating coordinates, matching the motion of the $M$ -masses. (Imagine a camera viewing the disk, rotating around the $y$ -axis at exactly the same rate as the disk is rotating.) In this rotating frame, the disk is now stationary (so the $m$ -masses are stuck on the $y$ -axis at $(0,\pm 1,0)$ and the $M$ -masses are stuck on the $x$ -axis at $(\pm 1,0,0)$ ), but there is a centrifugal force exerted on all bodies proportional to the distance to the $y$ -axis. The $m$ -masses are on the $y$ -axis and thus experience no centrifugal force; but the $M$ -masses are away from the $y$ -axis and thus experience a centrifugal force, which is then balanced out by the centripetal forces of the rigid disk. Now let us perturb the disk a bit, so that the $m$ -masses and $M$ -masses are knocked a little bit out of position (but keeping the centre of mass fixed at $(0,0,0)$ ). In particular, the $m$ -masses are knocked away from the $y$ -axis and now experience a little bit of centrifugal force. On the other hand, the rigid disk forces the light $m$ -masses to remain orthogonal to the heavy $M$ -masses, by exerting tension forces between the masses. In the regime where $m$ is negligible compared to $M$ , these tension forces will barely budge the heavy $M$ masses (which therefore remain essentially fixed at $(\pm 1,0,0)$ in the rotating frame), so the effect of these tension forces is to constrain the $m$ -masses to lie in the $yz$ -plane (up to negligible errors which we now ignore). Rigidity also keeps the $m$ -masses at a unit distance from the origin, and antipodal to each other, so the $m$ -masses are now constrained to be antipodal points on the unit circle in the $yz$ -plane. However, other than this, rigidity imposes no further constraints on the location of the $m$ -masses, which can then move freely as antipodal points in this unit circle. The effect of centrifugal force in the rotating frame is now clear: if an $m$ -mass (and its antipode) is perturbed to be a little bit off the $y$ -axis in this unit circle with no initial velocity, then centrifugal force will nudge it a little further off the $y$ -axis, slowly at first but with inexorable acceleration. Eventually it will shoot across the unit circle and then approach the antipode of its previous position. At this point the centrifugal forces act to slow the $m$ -masses down, reversing all the previous acceleration, until one ends up with no velocity at a small distance from the antipode. The process then repeats itself (imagine a marble rolling frictionlessly between two equally tall hills, starting from a position very close to the peak of one of the hills). UPDATE, September 2019 : Due to renewed interest in this question, I will expand upon my 2014 comment regarding why the above analysis appears at first glance to also lead to the incorrect conclusion that the disk rotation is also unstable if it one instead rotates around the $x$ -axis (so that it is now the $M$ -masses that are stationary and the $m$ -masses that are rotating), or equivalently if one swaps the location of the $m$ -masses and $M$ -masses (which we will not do here to try to reduce confusion). The reason for this is that centrifugal force $F_{\mathrm{Cent}} = -m \Omega \times \Omega \times r$ is only one of two inertial forces that are introduced when one is in a steadily rotating frame. The other inertial force introduced is the Coriolis force $F_{\mathrm{Cor}} = -2 m \Omega \times v$ , which acts on moving bodies in the rotating reference frame in a direction orthogonal to the motion (and to the axis of rotation). Strictly speaking, one has to take into account the effect of both inertial forces when performing Newtonian mechanics in a steadily rotating frame. As it turns out, the Coriolis force has a negligible impact on the dynamics when rotating around the $y$ -axis, but dominates the dynamics when rotating around the $x$ -axis, which is why the preceding discussion is accurate in the former case but not the latter. In more detail: suppose we are rotating around the $y$ -axis as in the above discussion. In the rotating frame of reference, and starting with a configuration slightly out of equilibrium, we have as before that the $m$ -masses experience a little bit of centrifugal force and begin to slide away from the $y$ -axis and into the rest of the $yz$ -plane. When doing so, they will then experience some Coriolis force in a direction parallel to the $x$ -axis (which direction it is depends on the orientation of the rotation, as per the right hand rule formula for the cross product). However, due to the rigidity of the disk, as mediated by tension forces within the disk, it is not possible for the $m$ -masses to actually move in the $x$ -direction without also moving the much heavier $M$ -masses. But the magnitude of the Coriolis force is proportional to the small mass $m$ rather than the large mass $M$ , so by Newton's law $F=Ma$ for the $M$ -masses, the Coriolis force (or more precisely, the tension force produced in response to the Coriolis force) actually barely affects the motion of the $M$ -masses, which basically stay put on the $x$ -axis, and the $m$ -masses therefore remain essentially constrained to the $yz$ -plane and cannot actually experience any significant motion in the direction of the Coriolis force. (This is what the sentence in the original explanation regarding how tension forces "barely budge" the $M$ -masses was referring too, albeit somewhat obliquely.) The analysis of the original explanation now proceeds as before. In contrast, suppose that the disk is close to the stable equilibrium state when it rotates around the $x$ -axis. Working in a rotating frame around this axis, the $M$ -masses are near the $x$ -axis of rotation, while the $m$ -masses lie near the $y$ -axis. As before, the $M$ -masses experience centrifugal force and thus begin drifting slightly away from the $x$ -axis into the $xz$ -plane. But then the Coriolis force on these masses kicks in, which is now proportional to the heavy mass $M$ rather than the light mass $m$ . As such, the rigid tension forces connecting the $M$ -masses to the much lighter $m$ -masses offer very little resistance to the Coriolis force, and the motion of the $M$ -mass begins to rotate out of the $xz$ -plane, constantly experiencing Coriolis acceleration in a direction orthogonal to its motion. This effect tends to make the $M$ -mass rotate in a tight circle, and basically neutralises the net effect of the centrifugal force by rotating any outward motion away from the $x$ -axis back into inward motion. The end result is that in the rotating frame, the disk wobbles a bit around its equilibrium state, but does not dramatically depart from it, which is what one would expect from a stable equilbrium (imagine now a marble rolling frictionlessly around the trough of a valley).
|
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"https://mathoverflow.net/questions/81967",
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|
82,001 |
Imagine that you are in the following situation: You write up a proof which eventually gets published. There you need a result which is not so well-known but it is contained in another paper P; therefore you just cite it. You read P and come to the conclusion: It's awful. You need plenty of time to insert the details or even correct it. It may also happen that the proof is somewhat too complicated because in your situation it is much easier. Maybe you have found a shorter proof, but based on the ideas in P. Now what do you do? Several options come into my mind: Just cite the paper without any further explanation. Cite the paper but give a short hint how to simplify the arguments. Cite the paper but give a more elaborate explanation of the arguments. Write up the details of the proof of the desired result in your situation and remark somewhere that it was inspired by the paper P. For each option there are pros and cons. For example, you don't want to blow up your proof with material which does not seem be so important. Also, you don't want to bore your readers. This favors the first options. On the other hand, you might want to be sure that the readers understand the argument and don't have to read P. This favors the last options. What do you think, which option is your favorite and why? Also, are there other appropriate options?
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Improving existing proofs is an important and undervalued part of mathematics. We don't just want to know whether something is true; we want to know why it's true. So I think that if you have a better proof of something, you should find a way to share it with the world. Here are a couple of thoughts about the practicalities, to add to Andrew's suggestion about the nLab. First, you could put the simplified proof into an appendix to your paper. I quite like appendices, as both a reader and a writer. Used well, they help to keep the main part of the paper flowing, while providing crucial details to those who want them. Second, it's entirely possible that the author of [Awful 2009] will referee your paper. So whatever you write, you need to keep them sweet. I think this also favours the appendix option.
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82,046 |
I have the following question for which I haven't been able to find any reference or proof. Suppose we know that a univariate polynomial $P(X)$ with integer coefficients is the sum of squares of two polynomials with rational coefficients. Is it true that $P(X)$ must also be the sum of squares of two polynomials with
integer coefficients? For example, take $P(X)=50X^2+14X+1$, then we see that $P(X)=(5X+3/5)^2+(5X+4/5)^2$, but it is also $X^2+(7X+1)^2$. I would greatly appreciate any help pointing me into the right direction. Thanks in advance, and regards,
Guillermo
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Yes. Suppose $n\in \mathbb N$ is minimal so that $P(x)=f_1^2+f_2^2$ , where $nf_1$ and $nf_2$ are in $\mathbb Z[x]$ . Let $p$ be a prime with $p^\alpha||n$ . Since $P\in \mathbb Z[x]$ we have $p^{2\alpha}| (p^\alpha f_1)^2+(p^\alpha f_2)^2$ . Denoting $p^\alpha f_i$ by $g_i$ , and letting $\beta$ be square root of $-1\pmod{p^{2\alpha}}$ (it is not hard to show that this must exist by looking at the coefficients of $g_i$ with lowest $p$ -valuation). We have $g_1^2+g_2^2\equiv 0\pmod{p^{2\alpha}}$ so $g_2^2\equiv (\beta g_1)^2\pmod{p^{2\alpha}}$ so that $p^{2\alpha}| ag_1+bg_2$ for some integers $a,b$ with $a^2+b^2=p^{2\alpha}$ and $(ab,p)=1$ . Now we can take $P(x)=\left(\frac{af_1+bf_2}{p^{\alpha}}\right)^2+\left(\frac{af_2-bf_1}{p^\alpha}\right)^2$ and both polynomials have coefficients with $\nu_p\geq 0$ . Now repeat the procedure with other prime divisors of $n$ until you have polynomials with integer coefficients.
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|
82,177 |
Is there a profinite group $G$ which is not its own profinite completion? Surely not, I thought. But upon looking into it, I found that there is a special name given to a $G$ which is its own profinite completion, namely "strongly complete". And a recent (2003) hard theorem (which according to Wikipedia uses the classification of finite simple groups) due to Nikolov and Segal asserts that, if $G$ is finitely generated (as a topological group), then it is "strongly complete". So the $G$ I'm looking for cannot be topologically finitely generated. An equivalent question to the above is: Is there a profinite group $G$ which admits a non-open subgroup of finite index? Now here's my problem; the only exposure to profinite groups I've had has been in the context of number theory, absolute Galois groups, local fields, etc. In particular, the only non-topologically-finitely-generated profinite group I'm aware of is the absolute Galois group of a number field, say $\mathbb{Q}$. But I reckon the Krull topology demands that the finite index subgroups of $Gal(\bar{\mathbb{Q}}/\mathbb{Q})$ be open. Maybe there is a more 'exotic' example of such a $G$...
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Example taken from Ribes and Zalesskii's book "Profinite groups". Take an infinite set $I$ and a finite group $T$. You can let $G$ be the profinite group $\prod_I T$. Denote its elements by $(g_i)_{i\in I}$. Let $\mathcal F$ be an ultrafilter which contains the filter of all cofinite subsets of $I$. If you denote $H$ to be the subgroup of elements with $\lbrace i\in I \mid g_i=1\rbrace\in \mathcal F$, it is clear that $H$ is proper normal and that it is not open because it is dense and has finite index $|T|$. To show that $H$ has index $|T|$ in $G$ consider all elements $a_t=(t,t,\dots)$ for $t\in T$. For any $g\in G$, consider $I_t=\lbrace i\in I \mid g_i=t\rbrace$. Since we have $\bigcup_{t\in T} I_t=I$
then $I_t\in \mathcal F$ for some $t$, and therefore $ga_t^{-1}\in H$.
|
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|
82,323 |
I am refereeing my first paper and I'm quite excited! But inexperienced and I would like to ask an advice to the Maths Community of MO. Let me tell you that I have already read Refereeing a Paper , but it seems that my question is quite different. Roughly: What is the point after which you get nervous while refereeing a paper? Specifically: I've found many English mistakes. (Well, I'm not a native English and so I can understand. So I am not nervous yet) I've found some maths inaccuracies like "let A be any set"... and then I have discovered that the proof of the first result works only for finite sets. (OK, those are only inaccuracies - I am not nervous yet) There are many references like "we use the notation of [X]", "this result is proved in [X]", where [X] is a BOOK, without specifying a precise section, or the number of the result... should I get this book and read all to find out the correct references? - just thinking of it, makes me a bit nervous.. (most importantly). There is a mathematical more serious mistake. Something that might be fixed, but not obviously (my definition of obvious is three evenings, in this case). I am not saying that the paper is completely wrong but that now... now I'm getting nervous! Now, taking into account that the person who asked me to referee this paper told me: be selective, we accept only 20% of submitted papers, what should you do in these cases? Reject? Ask for a revision? Not getting nervous and try to see if the rest of the piece is good (I'm quite a good guy and I'm doing that at the moment)? Of course I will talk with the editor, but I also would like to know more opinions that might be helpful in future. Update: Thank you very much to everybody for the numerous and helpful comments. Valerio
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My suggestion would be to a. See if the claimed result is interesting. If not, reject. b. If the claimed result is clearly interesting, be honest and say that the paper is clearly not publishable as is (and give your points 1-4, though really already 1 is sufficient), and should be revised before refereeing (this is often a box you can check if use one of the semi-automated review submission systems), but you would be willing to look at it once it is so revised. c. Only get nervous over your own papers.
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82,875 |
Does there exist a set $M \subset \mathbb{R}^2$ which has the following two properties: Forall $x \in \mathbb{R}$ the set $\{y \in \mathbb{R} \mid (x,y) \in M\}$ is countable. Forall $y \in \mathbb{R}$ the set $\{x \in \mathbb{R} \mid (x,y) \notin M\}$ is countable.
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It is a theorem of Sierpinski (Sur un theoreme equivalent a l'hypothese du continu) that the existence of such a set is equivalent to the continuum hypothesis.
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83,026 |
What is the largest possible volume of the convex hull of an open/closed curve of unit length in $\mathbb{R}^3$?
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I believe this problem has been mentioned a few times in the literature, and has been solved for certain restrictions on the curve. For example if the curve has no four coplanar points then the maximal volume is achieved by one turn of a circular helix of height $\frac{1}{\sqrt{3}}$ and base radius $\frac{1}{\pi\sqrt{6}}$, this is due to Egervary. For more references see section A28 of "Unsolved problems in geometry" by H.T. Croft, K.J. Falconer, R.K. Guy. Melzak and Schoenberg have treated the corresponding problem for closed loops (Schoenberg has treated even dimensions), and have given answers under similar restrictions. In no case is a complete answer known.
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|
83,027 |
The number theory community here at University of Michigan is abuzz with talk of this paper recently posted to the arxiv. If you haven't seen it already, the punch line is that the global differences of imaginary parts of zeta zeros show an unusual tendency not to be close to imaginary parts of zeta zeros. "Riemann zeros repel their deltas." The people I have talked to have said that this simple observation appears to be completely new and potentially groundbreaking. Here are my questions: 1) Has anyone independently verified these statistics? (See my histograms below, and also the excellent graphs in Jonathan Bober's answer.) 2) The author discusses the "eñe product" as a source of a mathematical explanation for his observation, and he refers to his own unpublished manuscripts in which he introduces this product. Are these manuscripts available? Does anyone have any reference which defines the eñe product? Finally, any insights into Marco's results would be much appreciated. The histograms (produced in Mathematica) are of the deltas of the first 10000 zeros of $\zeta(s)$ falling in $[100, 110]$ and $[190, 200]$, respectively. The purple bars show the locations of the zeros.
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I have not looked at the paper terribly closely, but my impression is that 1.) The (experimental/statistical) observation is the article is correct, but not really so new. 2.) The author's explanation might not be quite correct. (There doesn't seem to be a great amount of explanation, but what little there is points to the zeros of the zeta function, when the actual explanation seems to be in "shadows" of the zeros on the line Real(s) = 1.) More details: The tendency for gaps between zeros of the zeta function to stay away from the zeros of the zeta function has been observed elsewhere. One place that I know of where this is nicely described is in "Riemann zeros and random matrix theory" by Nina Snaith, Milan Journal of Math, Volume 78, Number 1 . There is a nice graph in that paper, which I feel I have seen elsewhere, but can't seem to find anywhere else right now: (There ought to me some nice lectures about this stuff available from MSRI, but after more than 10 months, the videos still seem to be in post-production.) This sort of behavior seems to be explained by the "L-function ratios conjecture" of Conrey, Farmer, and Zirnbauer ( http://arxiv.org/abs/0711.0718 ), which might be better called the "L-function ratios recipe for making conjectures ." This explanation is not really satisfactory, in that it is a conjecture, but it is not really only a conjecture; it is more like a somewhat plausible sounding "hand-waving" argument that happens to make astoundingly accurate predictions in many cases. Anyway, when you use the ratios conjecture/recipe to compute the "two-point correlation" of the zeros of the zeta function, terms will come up that involve the zeta function on the 1-line. And from the prediction you will expect there to be less zero-gaps of size comparable to the minima of the absolute value of $\zeta(1 + it)$ . Since the minima of $\zeta(1 + it)$ tend to be close to the zeros of $\zeta(1/2 + it)$ , you can "see" the zeros (or shadows of the zeros, as I called them above) in these statistics. (This is why I say that the author's explanation might not be correct.) The ratios prediction (now I am using yet another word) and the phenomenon of zeros influencing gaps, has been seen/tested in other cases. For a somewhat random sample of the unreasonable influence of the zeros of $\zeta(s)$ in other places, see: The beautiful picture (this may be an understatement) on page 54 of Mike Rubinstein's "methods and experiments" paper: http://arxiv.org/abs/math/0412181 Page 16 of Duc Kheim Huynh, Jon Keating, and Nina Snaith's "one level density of elliptic curve L-functions" paper: http://arxiv.org/abs/0811.2304 Page 20 of Ghaith Hiary and Andrew Odlyzko's "Numerical evidence for moments and RMT models" paper: http://arxiv.org/abs/1008.2173 (and, of course, the explanatory text around each of those pictures). As to whether there is anything to this "eñe product," I don't know. It would be nice if there were, but there are rather few details in Marco's paper. For the specific case of the zeta function, it should not be too hard to get longer range histograms of the zero-spacings. Going out to one thousand or so should be enough to distinguish between minima of $\zeta(1 + it)$ and zeros of $\zeta(1/2 + it)$ . -- Update number 2: I happen to have access to lots of zeros of the zeta function computed to high precision by Dave Platt . (These will be more publicly available sometime soon, hopefully, when certain computers have their disk storage upgraded.) I've been wanting to look at them for a while, and I wanted to see a better picture, so I made one. The following picture is a histogram of the differences of the first two billion zeros of the zeta function, restricted to zero spacings between 1000000 and 1000010, along with the Bogomolny-Keating/Conrey-Farmer-Zirnbauer/Conrey-Snaith prediction ("right click", or whatever, to view the image by itself, for higher resolution): zeros-histogram-one-million-two-billion-and-prediction http://sage.math.washington.edu/home/bober/hist_delta_one_million_two_billion_zeros_and_prediction.png Some notes: The histogram has 1024 steps, so a box size of 10/1024. It takes a lot of zeros to make a smooth picture with such a small box size. (There are 57,465,000,000 pairs of zeros "contained" in this histogram.) I haven't been too careful in all of my computations. For example, I don't know exactly which zero was the last one in the histogram (which affects the prediction a little bit), and the formula for the prediction is a little complicated, and I haven't haven't checked it carefully. So I don't know if the prediction is a little larger than the actual values, or if this is my error. The histogram is not normalized! I like it like this, but it hides the fact that the error in the prediction here is generally less than 0.1%, even if I computed it wrong. (The histogram varies by around 4.5%, for comparison.) It can be seen that the zeros (red dots) still have some influence in this range, but it is nowhere near as clear-cut as it is for small spacing size. Specifically, the green line is (or is supposed to be) $$
\frac{10}{1024} \cdot \frac{1}{(2\pi)^2}\Re\Bigg[ 2T\left(\frac{\zeta'}{\zeta}\right)'(1 + iX) - 2T \cdot B(iX) + T\left(\log \frac{T}{2\pi}\right)^2 - 2T\log\frac{T}{2\pi} + 2T
$$ $$
+ \frac{2 \zeta(1 - iX)\zeta(1 + iX)A(iX)}{(2\pi)^{iX}}\left(\frac{T^{1 - iX} - 1}{1 - iX}\right)\Bigg],
$$ where $$
A(s) = \prod_p\left(1 - \frac{1}{p^{1+s}}\right)\left(1 - \frac{2}{p} + \frac{1}{p^{1 + s}}\right)\left(1 - \frac{1}{p}\right)^{-2},
$$ $$
B(s) = \sum_p \left(\frac{\log p}{p^{1 + s} - 1}\right)^2,
$$ $T = 732565723.921443$ (approximately the end of the range of zeros considered) and $X$ runs from 1000000 to 1000010.
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83,092 |
My question, motivated by idle curiosity while sitting in LaGuardia airport, is the following. You've just proved nice result A and it is time to write the paper. What is the real goal of the paper? (a) to get the reader understand why result A is true. (b) to convince the reader result A is true. In some cases it may be feasible to do both. But often there is some tension between (a) and (b). To understand a result one needs global understanding. One may need to do some ugly computations in coordinates or deal with pictures that LaTex doesn't handle. On the other hand to check the truth of a paper it is often easier to have local understanding because humans can only keep so much info in their brain at once. So if (b) is your goal you will work hard to break things up into smaller certifiable statements. That way once the reader has been convinced that Lemma X is true they can work with the statement and forget why it is true. To make it easier to check a proof one is often led to invent new formalisms or language and find coordinate free arguments. This can affect negatively (a) because you may not see the forest because of the trees. I'd like to get the community's opinion. Usual Community Wiki rules are applicable. Edit in hope of reopening. Perhaps if I make things more specific I can get the kind of answer I wanted and things would be less subjective. Suppose I have 2 proofs that finite sets $X$ and $Y$ have the same cardinality. One is proof is a relatively easy computation of the sizes of each set using known identities with binomial coefficients, Stirling numbers, etc. Any decent referee would follow it. The other proof is an involved bijection between $X$ and $Y$ whose details would be involved to check. Space considerations in the journal do not allow for both proofs. Which one should I submit?
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Papers are written so that their author(s) can forget their content and move on to other things. Therefore when you write you should be very careful to put in enough of the big picture and enough of the details so you'd be able to reconstruct your thoughts 10 years later if you'll need to, assuming you'll forget everything but retain some familiarity with some basic principles of mathematics.
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83,097 |
I just finished teaching a freshman calculus course (at an American state university), and one standard topic in the curriculum is related rates. I taught my students to answer questions such as the following (taken, more or less, from the textbook): A man starts walking due north at 5 ft/sec from a Point A. Ten seconds later, a woman starts walking south at 4 ft/sec from a point 20 ft due east of Point A. How fast are they moving apart when the woman has been walking for ten seconds? A 6' man walks away from a 20' lamppost at a speed of 5 ft/sec. How fast is the distance between the tip of his shadow and the top of the post changing when he is 40' away? A baseball player runs from first base to second at 20 ft/sec, and simultaneously another baseball player runs from third base to home at the same speed. How fast are they approaching each other after one second? To put my question bluntly: Who cares ? My students do, but only because they know these questions will appear on their exams. The baseball question (or something very similar) is actually an exercise in Stewart, and I struggled in vain to imagine a situation in which the manager of a baseball team would need to know the answer. This is in stark contrast to many other topics addressed in first-year calculus -- optimization, basic differential equations, etc. -- which are realistic models of questions of natural interest in business, biology, etc. Basically, all the related rates questions seemed to be cooked up in response to the fact that calculus students now knew a method to solve them. My question is in the title. Can anyone share any related rates questions which don't seem quite as contrived, and which might naturally seem interesting and motivated to a typical class of college freshmen? Thank you!
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The skills that students are practicing in related rates problems are: Differentiating a known equation implicitly with respect to time. Interpreting the time derivative of a quantity as a rate of change. The main reason that related rates problems feel so contrived is that calculus books do not want to assume that the students are familiar with any of the equations of science or economics. Every related rates problem inherently involves differentiating a known equation, and the only equations that the calculus book assumes are the equations of geometry. Thus, you can find related rates problems involving various area and volume formulas, related rates problems involving the Pythagorean Theorem or similar triangles, related rates problems involving triangle trigonometry, and so forth. A few of these problems are compelling -- for example, computing the speed of an airplane based on ground observations of its altitude and apparent angular velocity -- but most of them do feel a bit contrived. The reality, of course, is that students are familiar with many of the basic equations and concepts of science and economics, and there's no rule against using these in problems. For example, you can make up all sorts of compelling related rates problems by starting with any physics or chemistry equation and imagining a situation where you might want to take its derivative: The kinetic energy of an object is $K = \frac{1}{2}mv^2$ . If the object is accelerating at a rate of $9.8 \text{m}/\text{s}^2$ , how fast is the kinetic energy increasing when the speed is $30 \;\text{m}/\text{s}$ ? An ideal gas satisfies $PV = nRT$ , where $n$ is the number of moles and $R \approx 8.314\;\; \text{J}\; \text{mol}^{-1} \text{K}^{-1}$ . Give the rate at which the temperature and volume of the gas are increasing, and then ask about the rate of change in pressure when the volume and temperature reach certain amounts. The total energy stored in a capacitor is $\frac{1}{2} Q^2 / C$ , where $Q$ is the amount of charge stored in the capacitor and $C$ is the capacitance. Give the value of $C$ and the rate at which $Q$ is decreasing, and ask about the rate at which the capacitor is losing energy when the energy is a certain amount. In astronomy, the absolute magnitude $M$ of a star is related to its luminosity $L$ by the formula $$
M \;=\; M_{\text{sun}} -\; 2.5\; \log_{10}(L/L_{\text{sun}}).
$$ where $M_{\text{sun}} = 4.75$ and $L_{\text{sun}} = 3.839 \times 10^{26} \text{watts}$ . (Note that, by convention, brighter stars have lower magnitude.) If the absolute magnitude of a variable star is decreasing at a rate of $0.09 / \text{week}$ , how quickly is the luminosity of the star increasing when the magnitude is $3.8$ ? It's easy to make these up: just think of any equation in science or economics whose derivative might be interesting. Wikipedia and/or science textbooks can be helpful for finding equations from a wide variety of fields. Edit: I have compiled a list of these problems in the form that I use them in my classes, and posted them on my professional web page .
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83,363 |
[Title changed, and wording of question tweaked, by YC, because the original title asked a question which seems different from the one people want to answer.] I've read looked at the examples in most category theory books and it normally has little Analysis. Which, is strange as I've even seen lattice theory be used to motivate a whole book on category theory. I was wondering is there a nice application of category theory to functional analysis? It's weird as read that higher category theory is used in Quantum mechanics as it foundation, yet QM has heavy use of Hilbert spaces. Sorry for cross posting to MSE . However, somebody there suggested I posted it here to get better response (well, more than two replies). I'm surprised that most books on category theory have very little mention of Analysis. Especially since Grothendieck originally studied functional analysis.
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In relatively mundane, but intensely useful and practical, ways, the naive-category-theory attitude to characterize things by their interactions with other things, rather than to construct (without letting on what the goal is until after a sequence of mysterious lemmas), is enormously useful to me. E.g., it was a revelation, by now many years ago, to see that the topology on the space of test functions was a colimit (of Frechet spaces). Of course, L. Schwartz already worked in those terms, but, even nowadays, few "introductory functional analysis" books mention such a thing. I was baffled for some time by Rudin's "definition" of the topology on test functions, until it gradually dawned on me that he was constructing a thing which he would gradually prove was the colimit, but, sadly, without every quite admitting this. It is easy to imagine that it was his, and many others', opinion that "categorical notions" were the special purview of algebraic topologists or algebraic geometers, rather than being broadly helpful . Similarly, in situations where a topological vector space is, in truth, a colimit of finite-dimensional ones, it is distressingly-often said that this colimit "has no topology", or "has the discrete topology", ... and thus that we'll ignore the topology. What is true is that it has a unique topology (since finite-dimensional vector spaces over complete non-discrete division rings such as $\mathbb R$ or $\mathbb C$ do, and the colimit is unique, at least if we stay in a category of locally convex tvs's). Also, every linear functional on it is continuous (!). But it certainly is not discrete, because then scalar multiplication wouldn't be continuous, for one thing. But, despite the prevalence of needlessly inaccurate comments on the topology, the fact that all linear maps from it to any other tvs are continuous mostly lets people "get by" regardless. Spaces with topologies given by collections of semi-norms are (projective/filtered) limits of Banach spaces. Doctrinaire functional analysts seem not to say this, but it very nicely organizes several aspects of that situation. An important tangible example is smooth functions on an interval $[a,b]$, which is the limit of the Banach spaces $C^k[a,b]$. Sobolev imbedding shows that the (positively-indexed) $L^2$ Sobolev spaces $H^s[a,b]$ are {\it cofinal} with the $C^k$'s, so have the same limit: $H^\infty[a,b]\approx C^\infty[a,b]$, and such. All very mundane, but clarifying. [Edit:] Partly in response to @Yemon Choi's comments... perhaps nowadays "functional analysts" no longer neglect practical categorical notions, but certainly Rudin and Dunford-Schwartz's "classics" did so. I realize in hindsight that this might have been some "anti-Bourbachiste" reaction. Peter Lax's otherwise very useful relatively recent book does not use any categorical notions. Certainly Riesz-Nagy did not. Eli Stein and co-authors's various books on harmonic analysis didn't speak in any such terms. All this despite L. Schwartz and Grothendieck's publications using such language in the early 1950s. Yosida? Hormander? I do have a copy of Helemskii's book, and it is striking, by comparison, in its use of categorical notions. Perhaps a little too formally-categorical for my taste, but this isn't a book review. :) I've tried to incorporate a characterize-rather-than-construct attitude in my functional analysis notes, and modular forms notes, Lie theory notes, and in my algebra notes, too. Oddly, though, even in the latter case (with "category theory" somehow traditionally pigeon-holed as "algebra") describing an "indeterminate" $x$ in a polynomial ring $k[x]$ as being just a part of the description of a "free algebra in one generator" is typically viewed (by students) as a needless extravagance. This despite my attempt to debunk fuzzier notions of "indeterminate" or "variable". The purported partitioning-up of mathematics into "algebra" and "analysis" and "geometry" and "foundations" seems to have an unfortunate appeal to beginners, perhaps as balm to feelings of inadequacy, by offering an excuse for ignorance or limitations? To be fair (!?!), we might suppose that some tastes genuinely prefer what "we" would perceive as clunky, irrelevant-detail-laden descriptions, and, reciprocally, might describe "our" viewpoint as having lost contact with concrete details (even though I'd disagree). Maybe it's not all completely rational. :)
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83,379 |
For any lie algebra $\mathfrak g$, there is a natural filtration on $U(\mathfrak g)$ by "degree": the filtered piece $U^{\leq n}(\mathfrak g)$ is just the image in $U(\mathfrak g)$ of $\bigoplus_{k=0}^n\mathfrak g^{\otimes k}$. Is there a natural filtration of the quantum group $U_q(\mathfrak g)$ which reduces to this filtration on $U(\mathfrak g)$ when we set $q=1$? (of course, we restrict to the lie algebras $\mathfrak g$ for which one has a definition of $U_q(\mathfrak g)$, e.g. $\mathfrak g=\mathfrak s\mathfrak l_n$). To increase the likelihood of a positive answer to this question, let's use the model for $U_q(\mathfrak g)$ which is often denoted $U_h(\mathfrak g)$, that is, the model over the complete power series ring $\mathbb C[[h]]$. One possible motivation for this question is the following. One definition of the universal enveloping algebra $U(\mathfrak g)$ is that it is a dual of the complete local ring of $G$ at the identity $e\in G$ (where $\operatorname{Lie}(G)=\mathfrak g$). The complete local ring $\hat{\mathcal O}_{G,e}$ comes with a canonical set of quotients $\mathcal O_{G,e}/\mathfrak m_{G,e}^{n+1}$, and corresponding to this is the filtration $U^{\leq n}(\mathfrak g)$ by "degree". Thus we can interpret $U^{\leq n}(\mathfrak g)$ as the part of $U(\mathfrak g)$ consisting of differential operators of order $n$ at $e\in G$. I want to know whether there is a similar filtration/interpretation for quantum groups.
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In relatively mundane, but intensely useful and practical, ways, the naive-category-theory attitude to characterize things by their interactions with other things, rather than to construct (without letting on what the goal is until after a sequence of mysterious lemmas), is enormously useful to me. E.g., it was a revelation, by now many years ago, to see that the topology on the space of test functions was a colimit (of Frechet spaces). Of course, L. Schwartz already worked in those terms, but, even nowadays, few "introductory functional analysis" books mention such a thing. I was baffled for some time by Rudin's "definition" of the topology on test functions, until it gradually dawned on me that he was constructing a thing which he would gradually prove was the colimit, but, sadly, without every quite admitting this. It is easy to imagine that it was his, and many others', opinion that "categorical notions" were the special purview of algebraic topologists or algebraic geometers, rather than being broadly helpful . Similarly, in situations where a topological vector space is, in truth, a colimit of finite-dimensional ones, it is distressingly-often said that this colimit "has no topology", or "has the discrete topology", ... and thus that we'll ignore the topology. What is true is that it has a unique topology (since finite-dimensional vector spaces over complete non-discrete division rings such as $\mathbb R$ or $\mathbb C$ do, and the colimit is unique, at least if we stay in a category of locally convex tvs's). Also, every linear functional on it is continuous (!). But it certainly is not discrete, because then scalar multiplication wouldn't be continuous, for one thing. But, despite the prevalence of needlessly inaccurate comments on the topology, the fact that all linear maps from it to any other tvs are continuous mostly lets people "get by" regardless. Spaces with topologies given by collections of semi-norms are (projective/filtered) limits of Banach spaces. Doctrinaire functional analysts seem not to say this, but it very nicely organizes several aspects of that situation. An important tangible example is smooth functions on an interval $[a,b]$, which is the limit of the Banach spaces $C^k[a,b]$. Sobolev imbedding shows that the (positively-indexed) $L^2$ Sobolev spaces $H^s[a,b]$ are {\it cofinal} with the $C^k$'s, so have the same limit: $H^\infty[a,b]\approx C^\infty[a,b]$, and such. All very mundane, but clarifying. [Edit:] Partly in response to @Yemon Choi's comments... perhaps nowadays "functional analysts" no longer neglect practical categorical notions, but certainly Rudin and Dunford-Schwartz's "classics" did so. I realize in hindsight that this might have been some "anti-Bourbachiste" reaction. Peter Lax's otherwise very useful relatively recent book does not use any categorical notions. Certainly Riesz-Nagy did not. Eli Stein and co-authors's various books on harmonic analysis didn't speak in any such terms. All this despite L. Schwartz and Grothendieck's publications using such language in the early 1950s. Yosida? Hormander? I do have a copy of Helemskii's book, and it is striking, by comparison, in its use of categorical notions. Perhaps a little too formally-categorical for my taste, but this isn't a book review. :) I've tried to incorporate a characterize-rather-than-construct attitude in my functional analysis notes, and modular forms notes, Lie theory notes, and in my algebra notes, too. Oddly, though, even in the latter case (with "category theory" somehow traditionally pigeon-holed as "algebra") describing an "indeterminate" $x$ in a polynomial ring $k[x]$ as being just a part of the description of a "free algebra in one generator" is typically viewed (by students) as a needless extravagance. This despite my attempt to debunk fuzzier notions of "indeterminate" or "variable". The purported partitioning-up of mathematics into "algebra" and "analysis" and "geometry" and "foundations" seems to have an unfortunate appeal to beginners, perhaps as balm to feelings of inadequacy, by offering an excuse for ignorance or limitations? To be fair (!?!), we might suppose that some tastes genuinely prefer what "we" would perceive as clunky, irrelevant-detail-laden descriptions, and, reciprocally, might describe "our" viewpoint as having lost contact with concrete details (even though I'd disagree). Maybe it's not all completely rational. :)
|
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|
83,552 |
Let G be a finite undirected connected graph. A divisor on G is an element of the free abelian group Div(G) on the vertices of G (or an integer-valued function on the vertices.) Summing over all vertices gives a homomorphism from Div(G) to Z which we call degree. For each vertex v, let D(v) be the divisor $d_v v - \sum_{w \sim v} w$ where $d_v$ is the valence of v and $v \sim w$ means " $v$ and $w$ are adjacent." Note that D(v) has degree 0. The subgroup of Div(G) generated by the D(v) is called the group of principal divisors. We denote by Pic(G) the quotient of Div(G) by the group generated by the principal divisors, and by Pic^0(G) the kernel of the degree map from Pic(G) to Z. The notation here suggests that I am really thinking about algebraic curves, not graphs; and that's in some part true! In the work of Matt Baker and his collaborators you can find a really beautiful translation of much of the foundational theory of algebraic curves (Riemann-Roch, Brill-Noether, etc.) into this language. But that's not really what this question is about. Lots of people study this abelian group, maybe most notably statistical physicists and probabilists who study dynamical processes on graphs. In those communities, Pic^0(G) is called the sandpile group, because of its relation with the abelian sandpile model . But that's also not really what this question is about. What this question is about is the following fact: by the matrix-tree theorem, the number of spanning trees of G is equal to |Pic^0(G)|. When one encounters a finite set that has the same cardinality as a finite group, but the set does not have any visible natural group structure, one's fancy lightly turns to thoughts of torsors. So: QUESTION: Is the set S of spanning trees of G naturally a torsor for the sandpile group Pic^0(G)? If so, how can we describe this "sandpile torsor?" (By "naturally" we mean "functorially" -- in particular, this torsor should be equivariant for the automorphism group of G.) That question is rather vague, so let me make it more precise, and at the same time try to argue that in at least some cases the question is not ridiculously speculative. The paper "Chip-Firing and Rotor-Routing on Directed Graphs," by (deep breath) Alexander E. Holroyd, Lionel Levine, Karola Meszaros, Yuval Peres, James Propp and David B. Wilson , contains a very interesting construction of a "local" torsor structure for the sandpile group. Suppose G is a planar graph -- or more generally any graph endowed with a cyclic ordering of the edges incident to each vertex. Then the "rotor-router process" described in Holroyd et al gives S the structure of a Pic^0(G)-torsor! (See Def 3.11 - Cor 3.18) This would seem to answer my question; except that the torsor structure they define depends, a priori, on the choice of a vertex of G. A better way to describe their result is as follows: for each v, let S_v be the set of oriented spanning trees of G with v as root. Then the rotor-router model realizes S_v as a torsor for the group Pic(G) / $\mathbf{Z}$ v. But S_v is naturally identified with S (just forget the orientation) and the natural map Pic^0(G) -> Pic(G) / $\mathbf{Z}$ v is an isomorphism. So for each choice of v, the rotor-router construction endows S with the structure of Pic^0(G)-torsor. Now one can ask: QUESTION (more precise): Are the torsor structures provided by the rotor-router model in fact independent of v? Do they in fact provide a Pic^0(G)-torsor structure on S which is functorial for maps compatible with the cyclic edge-orderings, and in particular for automorphisms of G as a planar graph? If this is false in general, is there some nice class of graphs G for which it's true? REMARK: If you are used to thinking about algebraic curves, like me, your first instinct might be "well, surely if the set of spanning trees is a torsor for Pic^0, it must be Pic^d for some d." But I don't think this can be right. Here's an example: let G be a 4-cycle, which we think of as embedded in the plane. Now the stabilizer of a vertex v in the planar automorphism group of the graph is a group of order 2, generated by a reflection of the square across the diagonal containing v. In particular, you can see instantly that no spanning tree in S is fixed by this group; the involution acts as a double flip on the four spanning trees in S. On the other hand, Pic^d(G) is always going to have a fixed point for this action: namely, the divisor d*v. REMARK 2: Obviously the correct thing to do is to compute a bunch of examples, which might instantly give negative answers to these questions! But it gets a bit tiring to do this by hand; I checked that everything is OK for the complete graph on 3 vertices (in which case the torsor actualy is Pic^1(G)) and then I ran out of steam. But sage has built-in sandpile routines.....
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Answer: The Pic 0 ( G )-torsor structure is independent of the vertex v if and only if G is a planar ribbon graph. This is the main theorem of " Rotor-routing and spanning trees on planar graphs ", by Melody Chan , Thomas Church, and Joshua Grochow , which we just posted to the arXiv [later published in IMRN ]. Quoting from the introduction: The proof is based on three key ideas. First, the rotor-routing action of the sandpile group
on spanning trees can be partially modeled via rotor-routing on unicycles ([HLMPPW, §3]). This is a related dynamical system with the property that rotor-routing becomes periodic, rather than terminating after finitely many steps. The second main idea is that the independence of the sandpile action on spanning trees
can be described in terms of reversibility of cycles. We introduce the notion of reversibility
(previously considered in [HLMPPW] only for planar graphs), and prove that reversibility is a well-defined property of cycles in a ribbon graph. We also establish a relation between reversibility and basepoint-independence. Third, reversibility is closely related to whether a cycle separates the surface corresponding to the ribbon graph into two components. We prove that these conditions are almost equivalent. Moreover, although they are not equivalent for individual cycles, we
prove that all cycles are reversible if and only if all cycles are separating,
in which case the ribbon graph is planar. We're grateful to Jordan for the question, which turned out to have a much more interesting answer than we expected! We're also grateful to Math Overflow for providing a venue for this question.
|
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|
83,585 |
Let me preface this question by saying that I am not an algebraic topologist. Motivation. I was looking with a colleague at the homotopy type of a family of posets and we were able to show using discrete Morse theory that the order complex of this poset was homotopy equivalent to a space with exactly one-cell in dimensions $0$ to $n$ and no cells of higher dimension. Moreover, we can show that the fundamental group is cyclic. Extensive computer computation shows that it is most likely true that the homology in each dimension from $0$ to $n$ is $\mathbb Z$, Thus it smells like our complex is homotopy equivalent to a wedge of spheres, one from each dimension from $1$ to $n$. Before killing ourselves to prove that the homology is as the computer suggests, we wanted to figure out if this data implies the homotopy type of a wedge. Question. Is is true that if one has an $n$-dimensional CW complex $X$ with exactly one cell in each dimension (from $0$ to $n$) such that $\pi_1(X)=\mathbb Z$ and $H^q(X)=\mathbb Z$ for $0\leq q\leq n$, then $X$ has the homotopy type of a wedge $\bigvee_{q=1}^n S^q$?
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No, this is false. For example take $\mathbb{CP}^2\vee S^1\vee S^3$. It admits a cell decomposition and cohomology groups as you describe but clearly has a different homotopy type then a wedge of spheres because the cohomology ring structure is different.
|
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|
83,665 |
Is $\mathbb{R}^n$ homeomorphic to a product $X \times Y$ with $X$ compact and not a point? Bing's Dogbone space is a quotient of $\mathbb{R}^3$ with fibers points and arcs, and whose product with $\mathbb{R}$ is $\mathbb{R}^4$, so it doesn't seem to me to big a stretch to think that it may be possible. Or, is there a notion of dimension which takes care of it swiftly?
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No it is not possible. Suppose that $X\times Y\cong\mathbb{R}^n$. Then, as the product is contractible, both $X$ and $Y$ must be contractible spaces. For any $x\in X$, I'll show that $\lbrace x\rbrace\times Y$ must be an open subset of $\mathbb{R}^n$, which will imply that $\lbrace x\rbrace$ is an open subset of $X$ and, hence, that $X$ is discrete. Discrete contractible spaces consist of a single point. Choose any $p=(x,y)\in X\times Y$. We just need to show that this is contained in the interior of $\lbrace x\rbrace\times Y$.
As the spaces are contractible, there are deformation retractions $H_X\colon X\times[0,1]\to X$ and $H_Y\colon Y\times[0,1]\to Y$ respectively to the points $x,y$. So, $H_X(u,0)=u$, $H_X(u,1)=x$, $H_Y(v,0)=v$, $H_Y(v,1)=y$, for any $u\in X$ and $v\in Y$. Define the deformation retraction $J\colon(X\times Y)\times[0,1]\to X\times Y$ from $X\times Y$ to the point $p=(x,y)$ by
$$
J\left((u,v),t\right)=\begin{cases}
\left(H_X(u,2t),v\right),&\textrm{if }t\le1/2,\cr
\left(x,H_Y(v,2t-1)\right),&\textrm{if }t\ge1/2.
\end{cases}
$$ Identifying $X\times Y$ with $\mathbb{R}^n$, consider the (n-1)-sphere $S_R=\lbrace a\in\mathbb{R}^n\colon\Vert a-p\Vert=R\rbrace$, for any fixed $R > 0$. As $K=X\times\lbrace y\rbrace$ is compact, it will have empty intersection with $S_R$ so long as $R$ is chosen large enough. However, retricted to $S_R\times[0,1]$, $J$ continuously deforms $S_R$ down to the single point $\lbrace p\rbrace$. This implies that $J(S_R\times[0,1])$ contains the open ball of radius $R$ centered at $p$. As $S_R\cap K=\emptyset$, $J(S_R\times[0,1/2])$ is a compact set not containing $p$. So, $J(S_R\times[1/2,1])\subset\lbrace x\rbrace\times Y$ contains a neighborhood of $p$, showing that $\lbrace x\rbrace\times Y$ is open in $\mathbb{R}^n$.
|
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|
83,680 |
Is it possible to prove without Continuum Hypothesis that for every uncountable subset $S$ of $\mathbb{R}$ there is a real number $x$ that splits it into two parts of the same cardinality, i.e. $\left|S \cap (-\infty,x)\right|=\left|S \cap (x,\infty)\right|$? (if the answer to the first question is no) Is this statement equivalent to Continuum Hypothesis?
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No, the statement cannot be proven in ZFC without assuming continuum hypothesis or something similar. In fact, it is equivalent to the statement that there are finitely many cardinalities between $\aleph_0$ and $2^{\aleph_0}$, so it is strictly weaker than the continuum hypothesis. Suppose that there were infinitely many such cardinalities, then you can let $S=\bigcup_{n=1}^\infty S_n$ where $S_n\subseteq(0,1/n)$ has cardinality $\aleph_n$ to obtain a contradition. On the other hand, if there are only finitely many such cardinalities, then $f(x)=\vert S\cap(-\infty,x)\vert$ must achieve its maximum, say $\aleph_n$ ($n > 0$). If $x_0$ is the infimum of the $x\in\mathbb{R}$ such that $f(x)=\aleph_n$ then $S\cap(x_0,\infty)$ has cardinality $\aleph_n$. Choosing $y_k\in\mathbb{R}$ decreasing to $x_0$, the cardinality of $S\cap(y_k,\infty)$ must be $\aleph_n$ for large enough $k$, otherwise $S\cap(x_0,\infty)=\bigcup_k(S\cap(y_k,\infty))$ is a countable union of sets of cardinality less than $\aleph_n$, so is of cardinality less than $\aleph_n$, giving a contradiction. So, $S\cap(-\infty,y_k)$ and $S\cap(y_k,\infty)$ are both of cardinality $\aleph_n$ for $k$ large enough.
|
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|
84,025 |
Recently, some classmates and I were lamenting the fact that our classmates in other disciplines had almost no conception of what we did, despite the large mathematics population at Waterloo. Instead of giving up in the face of a Very Hard Problem, one of us brought up a column popularizing physics that had a brief run in the school paper, and suggested that we author something similar for mathematics. The column will have some particular constraints that seem challenging to satisfy (self-contained week to week, 500-700 words, try to cover at least some of the current research at UW) but this question is a more general one. In looking for resources and guidance to help with the writing we have come across several good discussions of topic . We have also found examples of good popular writing and a general discussion of presenting mathematics to a non-mathematical audience . What we have not found, on MathOverflow or elsewhere, is a popular analogue of the well-answered question "How to write mathematics well?" . A lot of the tactical advice of Knuth, Halmos, and others goes out the window when you answer their first question, "Who is your audience?" with "a general university educated public". What is your advice for writing good mathematics for a popular audience? What holds for all styles of writing and what is article or book specific?
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I get asked this question a lot, so I'll give my pat (but serious) answer: Find yourself an editor. That is, find someone who'll take what you think is crystal clear, engaging prose, and turn it into something that actually is crystal clear and engaging. I couldn't do my job without my editors. (For those who don't know me, I work as a freelance mathematics writer. I report on developments in mathematics for publications such as Science magazine and SIAM News .) It's very easy, especially when writing about a subject you know inside out, to fool yourself into thinking you've explained things so that any fool can understand it; it's equally easy, when reading someone else's writing, to notice they haven't explained things very well. A good editor will, at the least, point out your failings. A great editor will fix them. Now if I can only find someone to edit this posting....
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|
84,074 |
Godel's undecidable sentences in first-order arithmetic were guaranteed to be true, by construction. But are there examples of specific sentences known to be undecidable in first-order arithmetic whose truth values aren't known? I'm thinking, by contrast, of the situation in set theory: CH is undecidable in ZFC, but its truth value is, in some sense, unknown. Paris and Harrington showed the strengthened finite Ramsey theorem is true (in the sense of provable in second-order arithmetic) but undecidable in first-order arithmetic. I'm asking for "natural" examples in this general vein -- but whose truth values haven't yet been settled. EDIT. Let me clarify my interest in the question, which is more philosophical than mathematical. I asked it on the basis of the following passage in Peter Koellner's paper "On the Question of Absolute Undecidability": The above statements of analysis [i.e. all projective sets of reals are Lebesgue measurable] and set theory [i.e. CH] differ from the early arithmetical instances of incompleteness in that their independence does not imply their truth. Moreover, it is not immediately clear whether they are settled at any level of the hierarchy. They are much more serious cases of independence. What I'm asking is whether there are "much more serious cases" of independence even in first-order arithmetic -- and not in the trivial case of full-on ZFC, like V=L, etc. By a sentence with "unknown truth value," I just mean a sentence that hasn't been proved in a theory stronger than first-order arithmetic. (For example, Paris and Harrington proved the strengthened finite Ramsey theorem in second-order arithmetic.)
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Update. I've improved the argument to use only the consistency of $T$. (2/7/12): I corrected some over-statements previously made about Robinson's Q. I claim that for every statement $\varphi$, there is a variant way
to express it, $\psi$, which is equivalent to the original
statement $\varphi$, but which is formally independent of any
particular desired consistent theory $T$. In particular, if $\varphi$ is your favorite natural open question,
whose truth value is unknown, then there is an equivalent
formulation of that question which exhibits formal independence in
the way you had requested. In this sense, every open question is
equivalent to an assertion with the property you have requested. I
take this to reveal certain difficult subtleties with your project. Theorem. Suppose that $\varphi$ is any sentence and $T$ is any consistent theory containing weak arithmetic. Then there is another sentence $\psi$ such that $\text{PA}+\text{Con}(T)$ proves that $\varphi$ and $\psi$ are equivalent. $T$ does not prove $\psi$. $T$ does not prove $\neg\psi$. Proof. Let $R$ be the Rosser sentence for $T$, the self-referential assertion that for any proof of $R$ in $T$, there is a smaller proof of $\neg R$. The Gödel-Rosser theorem establishes that if $T$ is consistent, then $T$ proves neither $R$ nor $\neg R$. Formalizing the first part of this argument shows that $\text{PA}+\text{Con}(T)$ proves that $R$ is not provable in $T$ and hence that $R$ is vacuously true. Formalizing the second part of this argument shows that $\text{Con}(T)$ implies $\text{Con}(T+R)$, and hence by the incompleteness theorem applied to $T+R$, we deduce that $T+R$ does not prove $\text{Con}(T)$. Thus, $T+R$ is a strictly intermediate theory between $T$ and $T+\text{Con}(T)$. Now, let $\psi$ be the assertion $R\to (\text{Con}(T)\wedge \varphi)$. Since $\text{PA}+\text{Con}(T)$ proves $R$, it is easy to see by elementary logic that $\text{PA}+\text{Con}(T)$ proves that $\varphi$ and $\psi$ are equivalent. The statement $\psi$, however, is not provable in $T$, since if it were, then $T+R$ would prove $\text{Con}(T)$, which it does not by our observations above. Conversely, $\psi$ is not refutable in $T$, since
any such refutation would mean that $T$ proves that the hypothesis
of $\psi$ is true and the conclusion false; in particular, it
would require $T$ to prove the Rosser sentence $R$, which it does not by the Gödel-Rosser theorem. QED Note that any instance of non-provability from $T$ will require the consistency of $T$, and so one cannot provide a solution to the problem without assuming the theory is consistent. The observation of the theorem has arisen in some of the philosophical literature you may
have in mind, based on what you said in the question. For example, the claim of the theorem is mentioned in Haim Gaifman's new
paper " On ontology and realism in mathematics ," which we read in my course last semester
on the philosophy of set theory ; see the discussion on page 24 of Gaifman's paper and specifically footnote 35, where he credits a fixed-point argument to Torkel Franzen, and an independent construction to Harvey Friedman. My original argument (see edit history) used the sentence $\text{Con}(T)\to(\text{Con}^2(T)\wedge\varphi)$, where $\text{Con}^2(T)$ is the assertion $\text{Con}(T+\text{Con}(T))$, and worked under the assumption that $\text{Con}^2(T)$ is true, relying on the fact that $T+\text{Con}(T)$ is strictly between $T$ and this stronger theory. The current argument uses the essentially similarly idea that $T+R$ is strictly between $T$ and $T+\text{Con}(T)$, thereby reducing the consistency assumption.
|
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|
84,303 |
When trying to see if a number of the form $n^8-n^4+1$ can be divisible by the square of a prime, I found that it can indeed. The first few values for $n$ are 412, 786, 1417, 1818, 2430, 2640, 2809, 2822, 2899 ... and the first few such primes $p$ (in increasing order) are 73, 97, 193, 241, 313, 337, 409, 433, ... Interestingly enough, the latter is precisely the beginning of this sequence which lists the primes of the form $x^2+24y^2$. I am quite sure that this cannot be a pure coincidence and that some deep number theory must be involved. The number $24$ is not accidental either, as $n^8-n^4+1=\Phi_{24}(n)$, with $\Phi_k(x)$ being the $k$th cyclotomic polynomial. Maybe there is some relation to the field $\mathbb{Q}(\zeta_{24})$... So, the question is if the following is true: Conjecture . A prime $p$ has the form $x^2+24y^2$ if and only if $p^2$ divides $n^8-n^4+1$ for some $n$.
|
Your conjecture is true in the light of the following statements. Proposition 1. A prime $p$ has the form $x^2+24y^2$ if and only if $p\equiv 1\pmod{24}$. Proposition 2. A prime square $p^2$ divides $\Phi_{24}(n)$ for some $n$ if and only if $p\equiv 1\pmod{24}$. Proof of Proposition 1. The four equivalence classes of binary quadratic forms of discriminant $-96$ are represented by $x^2+24y^2$, $3x^2+8y^2$, $4x^2+4xy+7y^2$, $5x^2+2xy+5y^2$. Looking at the values in $(\mathbb{Z}/96\mathbb{Z})^\times$ assumed by these four quadratic forms, we see that they are in four different genera. This means that if $Q(x,y)$ is any of these forms and $p\geq 5$ is any prime, then $Q(x,y)$ represents $p$ if and only if it does so modulo $96$. In particular, $x^2+24y^2$ represents $p$ if and only if $p\equiv 1,25,49,73\pmod{96}$, i.e. when $p\equiv 1\pmod{24}$. Proof of Proposition 2. By Hensel's Lemma, the square of a prime $p\geq 5$ divides $\Phi_{24}(n)$ for some $n$ if and only if $p$ divides $\Phi_{24}(m)$ for some $m$. The latter property holds if and only if $\mathbb{F}_p$ contains a primitive $24$-th root of unity, i.e. when $p\equiv 1\pmod{24}$. References: Rose - A course in number theory; Cox - Primes of the form $x^2+ny^2$
|
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|
84,305 |
I encountered a problem which roughly translates to the following in mathematical terms: Given a directed graph, find an optimal vertex partitioning to maximize the edges inside partitions (strong cohesion) and minimize them between partitions (loose coupling). I've yet to find the exact metric to optimize for but is there an algorithm that can do something like this?
|
Your conjecture is true in the light of the following statements. Proposition 1. A prime $p$ has the form $x^2+24y^2$ if and only if $p\equiv 1\pmod{24}$. Proposition 2. A prime square $p^2$ divides $\Phi_{24}(n)$ for some $n$ if and only if $p\equiv 1\pmod{24}$. Proof of Proposition 1. The four equivalence classes of binary quadratic forms of discriminant $-96$ are represented by $x^2+24y^2$, $3x^2+8y^2$, $4x^2+4xy+7y^2$, $5x^2+2xy+5y^2$. Looking at the values in $(\mathbb{Z}/96\mathbb{Z})^\times$ assumed by these four quadratic forms, we see that they are in four different genera. This means that if $Q(x,y)$ is any of these forms and $p\geq 5$ is any prime, then $Q(x,y)$ represents $p$ if and only if it does so modulo $96$. In particular, $x^2+24y^2$ represents $p$ if and only if $p\equiv 1,25,49,73\pmod{96}$, i.e. when $p\equiv 1\pmod{24}$. Proof of Proposition 2. By Hensel's Lemma, the square of a prime $p\geq 5$ divides $\Phi_{24}(n)$ for some $n$ if and only if $p$ divides $\Phi_{24}(m)$ for some $m$. The latter property holds if and only if $\mathbb{F}_p$ contains a primitive $24$-th root of unity, i.e. when $p\equiv 1\pmod{24}$. References: Rose - A course in number theory; Cox - Primes of the form $x^2+ny^2$
|
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"https://mathoverflow.net/questions/84305",
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|
84,320 |
Surely, $\mathbb{Z}_p$ and $\mathbb{Q}_p$ (and their extensions) are very important for algebra and number theory. Do they have any important applications outside of algebra (that I could easily explain to a student)? Here I do not demand the applications to be (purely) 'mathematical'; for example, I wonder whether p-adic numbers have applications to physics (outside of string theory?). Moreover, I am also interested in those applications that are partially 'algebraic', and yet important for some other parts of mathematics.
|
This won't qualify as something you can explain to undergraduate students, but non-archimedean dynamics has recently seen a number of applications to classical complex dynamics. (Non-archimedean is dynamics over a field with a non-archimedean absolute value, but not specifically an extension of $\mathbb{Q}_p$.) I'll mention one beautiful example, which is a recent theorem of Matt Baker and Laura DeMarco. Let
$$f_c(x) = x^2+c$$
be the usual quadratic polynomial, and for any starting value $a$, let $O_c(a)$ be the forward orbit of $a$ for the map $f_c$. That is,
$$O_c(a) = \{a,f_c(a),f_c^2(a),f_c^3(a),...\}$$
where $f_c^n$ denotes the $n$'th iterate of $f_c$. Theorem : Let $a$ and $b$ be complex numbers with $a^2\ne b^2$. Then
$$\{c\in\mathbb{C} : O_c(a) \text{ and } O_c(b) \text{ are both finite}\}$$
is a finite set. The proof is partly complex dynamics, partly equidistribution theorems (in both the complex and $p$-adic settings), and partly a reduction step in which one works in Berkovich space over a non-archimedean field. Note that the statement of the theorem is purely a statement about complex numbers, but the proof requires non-trivial methods from non-archimedean analysis.
|
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|
84,374 |
This is a part of my answer to this question I think it deserves to be treated separately. Conjecture Let $A$ be the set of all primes from $2$ to $p>19$. Let $q$ be the next prime after $p$. Then $q$ divides $rs+1$ for some $r,s\in A$. I wonder if this conjecture is already known. I checked it for all $p<692,000$. A reformulation of the conjecture is this (motivated by Gjergji Zaimi's comment). Let $p > 19$ be prime. Let $A$ be the set of primes $< p$ considered as a subset of the cyclic multiplicative group $\mathbb{Z}/p\mathbb{Z}^*$. Then the product $A\cdot A$ contains $-1$. It is interesting to know how large $A\cdot A$ is. This seems to be related to Freiman-type results of Green, Tao and others. Also as Timothy Foo pointed out, perhaps Vinogradov's method of trigonometric sums can apply.
|
This answer is a heuristic along the lines of Joro's. We use $p,q,r$ to denote primes. Let $S(p,a)$ denote the number of pairs of primes $(q,r)$ with $q,r\leq p$ and $p|(qr+a)$. We are interested in the case $a=1$, but in general by the orthogonality relations of the characters we have $$ S(p,a)=\frac{1}{\phi(p)}\sum_{r\leq p}\sum_{q\leq p}\sum_{\chi\ \text{mod}\ p}\overline{\chi}(-a)\chi\left(qr\right).\ \ \ \ \ \ \ \ \ \ (1)$$ Rearranging this is $$S(p,a)=\frac{1}{p-1}\sum_{\chi\ \text{mod}\ p}\overline{\chi}(-a)\left(\sum_{q\leq p}\chi\left(q\right)\right)^{2}.$$ We might hope, as is the often the case, that the sums are all very small except when $\chi$ is principal, and that only the principal character contributes. With this in mind we expect $$S(p,a)\approx \frac{1}{p} \text{li}(p)^2.$$ This is the same as conjecturing that $S(p,a)$ does not vary largely between $a$. In particular, if we average over all $a$ modulo $p$, then using (1) along with the orthogonality relation $\sum_{a\text{ mod } p}\sum_{\chi\text{ mod } p}\chi (a)=\phi(p)$, we see that $$\frac{1}{\phi(p)}\sum_{a\text{ mod } p} S(p,a)=\frac{1}{p-1}\sum_{r\leq p}\sum_{q\leq p}1=\frac{1}{p}\pi(p)^2\sim \frac{1}{p}\text{li}(p)^2.$$ Numerically this is remarkably close for $a=1$. Using the calculation done in Joro'sanswer, letting $a=1$ and $p=1000003$ we have $$S(p,1)=6184$$ whereas $$\frac{1}{p} \text{li}(p)^2=6182.307\dots $$ Now all that remains is to understand the sum $$\sum_{q\leq p}\chi\left(q\right)$$ for a character modulo $p$. However, I believe this is very difficult.
|
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|
84,381 |
Direct to the point.
Since now I've looked a lot of presentations of $\infty$-categories, but it seems that the only way to do explicit computations on these objects is via model categories. Is that so, or maybe there are other way of doing computations in $\infty$-categories, and if there such different ways what are they? Edit: I really want to thank all people that answered this question, unfortunately I can choose just one answer.
Btw because here is yet midnight I wish everybody a happy new year.
|
(This is an answer to a question below from Akhil Mathew; he wanted examples of
``explicit computations'' since all he knew were classical 1950s calculations and
abstract theory. My answer is too long for a comment and too short to do justice
to the question.) That is terrible!!! I don't know where to begin, since there are huge masses of explicit calculations in the past half century. You mention infinite loop spaces. There is a theorem that says that the ordinary mod $p$ homology of $\Omega^n\Sigma^n X$ is an explicit functor of the ordinary mod $p$ homology of $X$, with all information (product, Dyer-Lashof operations, coproduct, Steenrod operations) determined. That surely sounds conceptual, but try to find a conceptual proof! We have tons of explicit calculations in the classical and Novikov Adams spectral sequences, without which the solution of the Kervaire invariant problem would be unthinkable. Chromatic theory as a whole is informed by explicit calculations. In unstable homotopy theory, exponent theorems for the homotopy groups of spheres use a remarkable blend of theoretic and calculational techniques. There are tons of explicit calculations in ordinary and generalized cohomology in the past half century. Spin cobordism and its applications to curvature questions is an example of a blend of algebraic topology and differential geometry. I could go on for 100s of pages without pausing for breath, and none of these results use any model category theory, let alone $\infty$ categories. There is a subject out there, with real content.
|
{
"source": [
"https://mathoverflow.net/questions/84381",
"https://mathoverflow.net",
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|
84,414 |
Here is a basic question. When does $H^1_{et}(X,\mathbb{Z})$ vanish? Using the exact sequence of constant etale sheaves $0\rightarrow\mathbb{Z}\rightarrow\mathbb{Q}\rightarrow\mathbb{Q}/\mathbb{Z}\rightarrow 0$, it is enough to show that $H^1_{et}(X,\mathbb{Q})$ vanishes. It is known, for instance by 2.1 of Deninger's 1988 JPAA paper, that $H^1_{et}(X,\mathbb{Q})$ vanishes when $X$ is normal. Note: there are two arguments I think are incorrect that claim to show $H^1_{et}(X,\mathbb{Z})$ always vanishes. The first is that $\mathbb{Z}$ is flasque in the etale topology. This is false. For instance, over the function field $\mathbb{C}(x,y)$, the long exact sequence in cohomology for $0\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}/n\rightarrow 0$ shows that $H^2(\mathbb{C}(x,y),\mathbb{Z})$ is non-zero. So, $\mathbb{Z}$ cannot be flasque. The second argument is that $H^1_{et}(X,\mathbb{Z})=Hom_{cont}(\pi_1^{et}(X),\mathbb{Z})$, where $\pi_1^{et}(X)$ is the etale fundamental group of $X$, which is a profinite group. Since it is profinite, the $Hom$ group above vanishes. But, the claimed equality between $H^1_{et}(X,-)$ and $Hom_{cont}(\pi_1^{et}(X),-)$ only holds for torsion sheaves, as far as I have been able to determine. I am in fact interested in several things. First, either an example of $X$ such that $H^1_{et}(X,\mathbb{Z})$ is non-zero, or a proof that this always vanishes. Second, the same thing but where we only look at affine $X$. In particular, if it exists, I would love to see an example of a commutative ring $R$ where $H^1_{et}(Spec R,\mathbb{Z})$ is non-zero, if this is possible.
|
The standard example is a copy of $\mathbb A^1_k$, where $k$ is an algebraically closed field, with two points glued. In algebraic terms, $X = \mathop{\rm Spec}R$, where $R := k[x,y]/(y^2 - x^3 + x^2)$. Consider the finite morphism $\pi\colon \mathbb A^1 \to X$, which yields an exact sequence
$$
0 \to \mathbb Z_X \to \pi_*\mathbb Z_{\mathbb A^1} \to i_*\mathbb Z_p \to 0,
$$
where $p$ is the singular point of $X$. Since both $\pi_*\mathbb Z_{\mathbb A^1}$ and $i_*\mathbb Z_p$ have trivial étale cohomology , by taking global sections we see that $\mathrm H^1(X, \mathbb Z) = \mathbb Z$.
|
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"https://mathoverflow.net/questions/84414",
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|
84,521 |
I am trying to learn about basic characteristic classes and Generalized Gauss-Bonnet Theorem, and my main reference at the moment is From calculus to cohomology by Madsen & Tornehave. I know the statement of the theorem is as follows: Let $M$ be an even-dimensional compact, oriented smooth manifold, $F^{∇}$ be the curvature of the connection ∇ on a smooth vector bundle $E$ . $$∫_{M}\mathrm{Pf}\Big(\frac{−F^{∇}}{2\pi}\Big)=χ(M^{2n}).$$ My questions are: How does this relate to counting (with multiplicities) the number of zeros of generic sections of the vector bundle? Also, are there other good references for learning this topic?
|
When $E\to M$ is an oriented vector bundle of rank $2n$ over a
compact manifold $M$, it has a well-defined de Rham Euler class $e(E)$
in $H^{2n}_{dR}(M)$, and a representative $2n$-form for $e(E)$
can be computed as follows: Fix a positive definite inner product $\langle,\rangle$ on $E$. (Since any two such inner products are equivalent under automorphisms of $E$, it doesn't matter which one.) Let $\nabla$ be a $\langle,\rangle$-orthogonal connection on $E$, and let $K^\nabla$ denote the curvature of $\nabla$, regarded as a $2$-form with values in ${\frak{so}}(E)$. Let $e(\nabla) = \mathrm{Pf}(K^\nabla/2\pi)$, which is a well-defined $2n$-form on $M$. (N.B.: The definition of $\mathrm{Pf}$ requires both the inner product and the orientation of $E$.) Then $e(E)=\bigl[e(\nabla)\bigr]\in H^{2n}_{dR}(M)$. (That $\bigl[e(\nabla)\bigr]$ is independent of the choice of $\nabla$
is one of the first results proved in Chern-Weil theory.) If $M$ is a compact, oriented $2n$-manifold, then the value of $e(E)$ on $[M]$, the fundamental class of $M$, can be computed as follows: Let $Y$ be a section of $E$ that has only a finite number of zeroes. (By Whitney transversality, the generic section of $E$ satisfies this condition.) Using the orientations of $E$ and $M$, one defines the index of an isolated zero $z$ of $Y$, which is an integer $\iota_Y(z)$. Then
$$
e(E)\bigl([M]\bigr) = \int_M e(\nabla) = \sum_{z\in Z}\ \iota_Y(z).
$$
By the Poincaré-Hopf Theorem, when $E = TM$ (as oriented bundles),
this sum is equal to the Euler characteristic of $M$, which explains
why $e(E)$ is called the Euler class of $E$. To prove this result, one chooses an orthogonal connection $\nabla$ on $E$ that depends on $Y$ and whose Euler form $e(\nabla)$ is easy to evaluate explicitly. This can be done as follows: Let $Z\subset M$ be the (finite) zero set of $Y$ and let $U\subset M$ be an open neighborhood of $Z$ that consists of disjoint, smoothly embedded $2n$-balls, one around each element of $Z$. Let $\phi$ be a smooth function on $M$ that is identically equal to $1$ on an open neighborhood of $Z$ and whose support $K\subset U$ is a disjoint union of closed, smoothly embedded balls,
one around each element of $Z$. Now, $E$ is trivial over $U$,
so choose a positively oriented $\langle,\rangle$-orthonormal basis of sections
$s_1,\ldots, s_{2n}$ of $E$ over $U$ and let $\nabla_1$ be the (flat) connection on $E_U$ for which these sections are parallel. Let $\bar Y = Y/|Y|$ be the normalized unit section defined on $M\setminus Z$, and define a second connection on $U$ by
$$
\nabla_2 s = \nabla_1 s
+ (1{-}\phi)\bigl( \bar Y\otimes \langle s,\nabla_1 \bar Y\ \rangle
- \langle s,\bar Y\ \rangle\ \nabla_1\bar Y\ \bigr).
$$
It is easily verified that this formula does define a connection on $U$,
that $\nabla_2$ is $\langle,\rangle$-orthogonal, and that,
outside of $K$ (i.e., where $\phi\equiv0$), the vector field $\bar Y$ is
$\nabla_2$-parallel. (Note that, because $\phi\equiv1$ near $Z$, where $Y$ is
not defined, this formula does extend smoothly across $Z$, agreeing with $\nabla_1$
on a neighborhood of $Z$.) Meanwhile, on $M\setminus K$, write $E$ as a $\langle,\rangle$-orthogonal direct sum $E = \mathbb{R}\cdot Y \oplus E'$ and choose a $\langle,\rangle$-compatible connection $\nabla_3$ on $E$ over $M\setminus K$ that preserves this splitting. Using a partition of unity subordinate to the open cover of $M$ defined by $U$ and $M\setminus K$, construct a $\langle,\rangle$-orthogonal connection $\nabla$ that agrees with $\nabla_2$ on $K$ and with $\nabla_3$ on $M\setminus U$, and for which the line bundle $\mathbb{R}\cdot Y\subset E$
is parallel on $M\setminus K$. Now, on $M\setminus K$, the curvature of $\nabla$ takes values in ${\frak{so}}(E')\subset{\frak{so}}(E)$, so $e(\nabla) = \textrm{Pf}\bigl(K^{\nabla}/(2\pi)\bigr)$ vanishes identically outside of $K$. It suffices now to evaluate the
integral of $e(\nabla)$ over a single component $B$ of $K$, which may be
assumed to be the unit ball in $\mathbb{R}^{2n}$, so restrict attention to
this case. Write $\bar Y = s_1 u_1 +\cdots + s_{2n}\ u_{2n}$, and note that,
on $B$, one has, by definition,
$$
\nabla s_i = \nabla_2 s_i
= \sum_{j=1}^{2n} s_j\otimes (1{-}\phi)(u_j\ du_i - u_i\ du_j),
$$
i.e., the connection $1$-forms of $\nabla$ in this basis are
$\omega_{ij} = (1{-}\phi)(u_i\ du_j - u_j\ du_i)$. Using the identity ${u_1}^2 +\cdots + {u_{2n}}^2 = 1$,
the curvature forms are easily computed to satisfy
$$
\Omega_{ij} = d\omega_{ij} + \omega_{ik}\wedge\omega_{kj}
= (1{-}\phi^2)\ du_i\wedge du_j - d\phi\wedge(u_i\ du_j - u_j\ du_i).
$$
At this point, you have to know the definition of the Pfaffian. (I'll wait while you look it up.) Using the fact that the result has to be invariant under $SO(2n)$-rotations, it is easy to show that, on $B$, one has
$$
e(\nabla) = \textrm{Pf}\left(\frac{\Omega}{2\pi}\right)
= c_n (1{-}\phi^2)^{n-1} d\phi\wedge u^*(\Upsilon)
$$
for some universal constant $c_n$ and where $\Upsilon$ is the $SO(2n)$-invariant $2n$-form
of unit volume on $S^{2n-1}\subset\mathbb{R}^{2n}$ and $u: B\setminus z\to S^{2n-1}$
is $u = (u_1,\ldots, u_{2n})$. (I'll let you evaluate the constant $c_n$. This can be done in a number of ways, but it's essentially a combinatorial exercise.) In particular,
$$
e(\nabla) = d\bigl(P_n(1{-}\phi)\ u^*(\Upsilon)\bigr)
$$
where $P_n(t)$ is the polynomial in $t$ (of degree $2n{-}1)$) that vanishes at $t=0$ and
satisfies $P'_n(1{-}t) = -c_n(1{-}t^2)^{n-1}$. By Stokes' Theorem, one has
$$
\int_B e(\nabla) = P_n(1) \int_{\partial B}u^*(\Upsilon)
= P_n(1)\ \textrm{deg}(u:\partial B\to S^{2n-1}) = P_n(1)\ \iota_Y(z)
$$
Thus, it follows that there is a universal constant $C_n = P_n(1)$ such that
$$
e(E)\bigl([M]\bigr) = \int_M e(\nabla) = C_n \ \sum_{z\in Z}\ \iota_Y(z).
$$
Now, $C_1 = 1$, as one can show by computing with the constant curvature metric on the $2$-sphere and noting that, for $Y$ the gradient of the height function on $S^2$, the sum of the indices is $2$. Finally, by properties of the Pfaffian and the index, one sees that,
if such a formula as above is to hold, then one must have $C_{n+m}=C_nC_m$. Thus, $C_n=1$ for all $n$, and the formula is proved. (Of course, one can avoid these final tricks for evaluating the constants by carefully doing the combinatorial exercise, but I'll leave that to the curious.)
|
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|
84,705 |
There are various frameworks around which enlarge the category of rings to include more exotic objects such as the 'field with one element,' $\mathbb{F}_1$. While these frameworks differ in their details, there are certain things this should be true of any object that deserves to be called $\mathbb{F}_1$. For example, The algebraic K-theory of $\mathbb{F}_1$ should be sphere spectrum, and the theory of toric varieties should be defined over $\mathbb{F}_1$. Question 1: Is there a moral reason why the moduli space of curves should (or should not) be defined over Spec $\mathbb{F}_1$? EDIT: For anyone who would like to be more concrete, I'm happy to take the Toen-Vaquie definition of schemes over $\mathbb{F}_1$. (see arXiv:math/0509684 ). In this setup (and most of the other frameworks I know) an affine scheme over $\mathbb{F}_1$ is just a commutative monoid $M$. After base change to $\mathbb{Z}$ this becomes the monoid ring $\mathbb{Z}[M]$. So here is a more precise question: Question 2: Does the moduli space of curves $\mathcal{M}_{g,n}$ (over $\mathbb{Z}$, say) admit a covering by affine charts of the form spec $\mathbb{Z}[M_i]$ for commutative monoids $M_i$? If so, can this covering be chosen so that (as in the case of toric varieties) the gluing is entirely determined by maps of monoids?
|
This "answer" will basically restate the comments of Marty and Jason Starr. Any variety covered by schemes of the form $\mathrm{Spec}(\mathbf Z[M_i])$, or any torified variety, is rational. And indeed Severi conjectured at one point that $M_g$ is rational for any $g$! But we know a lot about the Kodaira dimension of $M_g$ by work of Harris--Mumford, Farkas, Eisenbud, Verra, ... in particular we know that Severi's conjecture is maximally false. We have $\kappa(M_g) = -\infty$ for $g \leq 16$, we don't know anything for $17 \leq g \leq 21$, and $\kappa(M_g) \geq 0$ for $g \geq 22$. In fact we know that $M_{g}$ is of general type for $g = 22$ and $g \geq 24$. But if one only wants an example of a $g$ for which $M_g$ is not rational , then I think one can find examples much earlier (maybe $g \approx 6,7$?). See http://arxiv.org/abs/0810.0702 for a survey by Farkas. Moreover, the cohomology of a torified scheme can only contain mixed Tate motives. In particular, this implies properties like: the number of $\mathbf F_q$-points is a polynomial function of $q$. As Marty says there are no problems in genus zero; these might be honest-to-God $\mathbf F_1$ schemes. The cohomology of $M_{1,n}$ contains only mixed Tate motives when $n \leq 10$, but for $n \geq 11$ one finds motives associated to cusp forms for $\mathrm{SL}(2,\mathbf Z)$. (The number 11 arises as one less than the smallest weight of a nonzero cusp form; the discriminant form $\Delta$.) This implies that the polynomiality behaviour changes drastically -- the number of $\mathbf F_q$-points is now given by Fourier coefficients of modular forms, which are far more complicated and contain lots of arithmetic information. The connection with birational geometry is also very visible here. The cohomology classes on $M_{1,n}$ associated to the cusp forms of weight $n+1$ are of type $(n,0)$ and $(0,n)$ in the Hodge realization, since given such a cusp form one can explicitly write down a corresponding differential form in coordinates. So by the very definition of Kodaira dimension we can not have $\kappa = -\infty$ anymore. Bergström computed the Euler characteristic of $M_{2,n}$ in the Grothendieck group of $\ell$-adic Galois representations by point counting techniques. That is, he found formulas for the number of $\mathbf F_q$-points of $M_{2,n}$ by working really explicitly with normal forms for hyperelliptic curves. These formulas turned out to be polynomial in $q$ for $n \leq 7$ (and conjecturally for $n \leq 9$), just as they would be if we knew that the cohomology of $M_{2,n}$ contained only mixed Tate motives. By thinking a bit about the stratification of topological type, one concludes that this holds also for $\overline M_{2,n}$ when $n \leq 7$, which is smooth and proper over the integers. Then a theorem of van den Bogaart and Edixhoven implies that the cohomology of $\overline M_{2,n}$ is all of Tate type and with Betti numbers given by the coefficients of the polynomials. There are similar results by Bergström for $M_{3,n}$ when $n \leq 5$ and Bergström--Tommasi for $M_4$. But the general phenomenon is that increasing either $g$ or $n$ will rapidly take you out of the world of $\mathbf F_1$-schemes, at least if this is taken to mean "commutative monoids". However, I don't know enough $\mathbf F_1$--geometry to say what the answer is if one takes $\mathbf F_1$-schemes in the sense of Borger. The first nontrivial case to answer would be: are the motives associated to cusp forms for $\mathrm{SL}(2,\mathbf Z)$ defined over $\mathbf F_1$ in his set-up? To be clear, I don't believe that this is true, but I know almost nothing about $\lambda$-schemes. Let me also make two small remarks: (i) Everything I have written above is necessary conditions for being defined over $\mathbf F_1$. (ii) The fact that $M_{g,n}$ is smooth over the integers says that the cohomology of $M_{g,n}$ is at least quite "special", even though it is not defined over $\mathbf F_1$. Smoothness is a strong restriction on the Galois representations that can occur in the cohomology.
|
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|
84,726 |
I have several questions on Lindelöf property. If every point countable open cover of $X$ has a countable subcover ( Condition A ), does $X$ have Lindelöf property? How far is having Condition A from Lindelöf property? A space $X$ is called $\omega_1$-Lindelöf if every $\omega_1$-sized open cover of $X$ contains a countable subcover. Can every $\omega_1$-Lindelöf space with Condition A be Lindelöf? A space $X$ is called discretely Lindelöf if the closure of every discrete subspace of $X$ is Lindelöf. Can every discretely Lindelöf space with Condition A be Lindelöf?
|
This "answer" will basically restate the comments of Marty and Jason Starr. Any variety covered by schemes of the form $\mathrm{Spec}(\mathbf Z[M_i])$, or any torified variety, is rational. And indeed Severi conjectured at one point that $M_g$ is rational for any $g$! But we know a lot about the Kodaira dimension of $M_g$ by work of Harris--Mumford, Farkas, Eisenbud, Verra, ... in particular we know that Severi's conjecture is maximally false. We have $\kappa(M_g) = -\infty$ for $g \leq 16$, we don't know anything for $17 \leq g \leq 21$, and $\kappa(M_g) \geq 0$ for $g \geq 22$. In fact we know that $M_{g}$ is of general type for $g = 22$ and $g \geq 24$. But if one only wants an example of a $g$ for which $M_g$ is not rational , then I think one can find examples much earlier (maybe $g \approx 6,7$?). See http://arxiv.org/abs/0810.0702 for a survey by Farkas. Moreover, the cohomology of a torified scheme can only contain mixed Tate motives. In particular, this implies properties like: the number of $\mathbf F_q$-points is a polynomial function of $q$. As Marty says there are no problems in genus zero; these might be honest-to-God $\mathbf F_1$ schemes. The cohomology of $M_{1,n}$ contains only mixed Tate motives when $n \leq 10$, but for $n \geq 11$ one finds motives associated to cusp forms for $\mathrm{SL}(2,\mathbf Z)$. (The number 11 arises as one less than the smallest weight of a nonzero cusp form; the discriminant form $\Delta$.) This implies that the polynomiality behaviour changes drastically -- the number of $\mathbf F_q$-points is now given by Fourier coefficients of modular forms, which are far more complicated and contain lots of arithmetic information. The connection with birational geometry is also very visible here. The cohomology classes on $M_{1,n}$ associated to the cusp forms of weight $n+1$ are of type $(n,0)$ and $(0,n)$ in the Hodge realization, since given such a cusp form one can explicitly write down a corresponding differential form in coordinates. So by the very definition of Kodaira dimension we can not have $\kappa = -\infty$ anymore. Bergström computed the Euler characteristic of $M_{2,n}$ in the Grothendieck group of $\ell$-adic Galois representations by point counting techniques. That is, he found formulas for the number of $\mathbf F_q$-points of $M_{2,n}$ by working really explicitly with normal forms for hyperelliptic curves. These formulas turned out to be polynomial in $q$ for $n \leq 7$ (and conjecturally for $n \leq 9$), just as they would be if we knew that the cohomology of $M_{2,n}$ contained only mixed Tate motives. By thinking a bit about the stratification of topological type, one concludes that this holds also for $\overline M_{2,n}$ when $n \leq 7$, which is smooth and proper over the integers. Then a theorem of van den Bogaart and Edixhoven implies that the cohomology of $\overline M_{2,n}$ is all of Tate type and with Betti numbers given by the coefficients of the polynomials. There are similar results by Bergström for $M_{3,n}$ when $n \leq 5$ and Bergström--Tommasi for $M_4$. But the general phenomenon is that increasing either $g$ or $n$ will rapidly take you out of the world of $\mathbf F_1$-schemes, at least if this is taken to mean "commutative monoids". However, I don't know enough $\mathbf F_1$--geometry to say what the answer is if one takes $\mathbf F_1$-schemes in the sense of Borger. The first nontrivial case to answer would be: are the motives associated to cusp forms for $\mathrm{SL}(2,\mathbf Z)$ defined over $\mathbf F_1$ in his set-up? To be clear, I don't believe that this is true, but I know almost nothing about $\lambda$-schemes. Let me also make two small remarks: (i) Everything I have written above is necessary conditions for being defined over $\mathbf F_1$. (ii) The fact that $M_{g,n}$ is smooth over the integers says that the cohomology of $M_{g,n}$ is at least quite "special", even though it is not defined over $\mathbf F_1$. Smoothness is a strong restriction on the Galois representations that can occur in the cohomology.
|
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|
84,730 |
Greetings to all ! Let me first confess that this question was mentionned to me by Bernard Dacorogna, who doesn't sail on MO. Let $A\in M_{2n}(k)$ be an alternate matrix. Say that $A$ is non-singular. It is well-known that there exists an $M\in GL_{2n}(k)$ such that $A=M^TJM$, where
$$J=\begin{pmatrix} 0_n & I_n \\\\ - I_n & 0_n \end{pmatrix}.$$
Of course, $M$ is not unique. Every product $M'=QM$ with $Q\in Sp_n(k)$ ($Q$ symplectic) works as well. Is it always possible to choose $M$ symmetric ? In this case, we have $A=MJM$, but an identity $A=RJR$ does not imply that $R$ be symmetric. Equivalently, Let $M\in GL_{2n}(k)$ be given. Is it true that $Sp_n(k)\cdot M$ meets $Sym_{2n}(k)$ non-trivially? Notice that we must have $\det M=Pf(A)$. Therefore, if $k=\mathbb R$, the symmetric $M$ that we are looking for cannot always be positive definite.
|
I feel that framing this question in terms of matrices rather than bilinear forms on a vector space obscures what is actually going on and makes it harder to understand what needs to be proved. Here is how I would describe the problem and the partial answer that results from this description: Let $V$ be a finite dimensional vector space over a field $k$, and let $\mathsf{B}(V)$ denote the vector space over $k$ consisting of bilinear forms on $V$, i.e., an element $\beta\in \mathsf{B}(V)$ is a bilinear mapping $\beta: V\times V\to k$. An element $\alpha\in \mathsf{B}(V)$ is said to be alternating if $\alpha(x,x) = 0$ for all $x\in V$, and an element $\sigma\in \mathsf{B}(V)$ is said to be symmetric if $\sigma(x,y)=\sigma(y,x)$ for all $x,y\in V$. The subset $\mathsf{A}(V)\subset\mathsf{B}(V)$ of alternating forms is a subspace, as is the subset $\mathsf{S}(V)\subset\mathsf{B}(V)$ of symmetric forms. When the characteristic of $k$ is not $2$, there is a $GL(V)$-invariant direct sum decomposition $\mathsf{B}(V) = \mathsf{A}(V)\oplus\mathsf{S}(V)$. When the characteristic of $k$ is $2$, one has, instead, $\mathsf{A}(V)\subset \mathsf{S}(V)\subset\mathsf{B}(V)$, and, apparently, these inclusions have no $GL(V)$-invariant splittings. An element $\beta\in \mathsf{B}(V)$ is nondegenerate if, for each $x\not=0$ in $V$, there exists a $y\in V$ such that $\beta(x,y)\not=0$. If $\alpha\in \mathsf{A}(V)$ is nondegenerate, then the dimension of $V$ over $k$ must be even. Conversely, if the dimension of $V$ over $k$ is even, then there exists a nondegenerate $\alpha\in\mathsf{A}(V)$, and, moreover, any other nondegenerate $\overline\alpha\in\mathsf{A}(V)$ is of the form $\overline\alpha = m^\ast(\alpha)$ for some $m\in GL(V)$, where, by definition,
$$
\bigl(m^\ast(\alpha)\bigr)(x,y) := \alpha(mx,my)
$$
for any $m:V\to V$. When $\dim_k(V) = 2n$, let $\mathsf{K}(V) = \Lambda^{2n}(V^\ast)$ denote the $1$-dimensional vector space consisting of $2n$-multilinear alternating functions on $V$. There exists a canonical polynomial mapping $\mathrm{Pf}:\mathsf{A}(V)\to \mathsf{K}(V)$ of degree $n$ that satisfies $\alpha^n = n!\ \mathrm{Pf}(\alpha)$ (a property that defines $\mathrm{Pf}$ when $n$ is less than the characteristic of $k$). This Pfaffian vanishes if and only if $\alpha$ is degenerate, and it satisfies $\mathrm{Pf}\bigl(m^\ast(\alpha)\bigr) = \det(m)\ \mathrm{Pf}(\alpha)$. When $\alpha\in\mathsf{A}(V)$ is nondegenerate, let $Sp(\alpha)\subset GL(V)$ denote the subgroup consisting of those $m\in GL(V)$ such that $\alpha(mx,my)=\alpha(x,y)$ for all $x,y\in V$. Define two subspaces ${\frak{s}}(\alpha)\subset \mathrm{End}(V) \simeq V\otimes V^\ast$ and ${\frak{a}}(\alpha)\subset \mathrm{End}(V)$, by saying that $s\in{\frak{s}}(\alpha)$ if $s^\flat(x,y) := \alpha(x,sy)$ is symmetric, while $a\in{\frak{a}}(\alpha)$ if $a^\flat(x,y) := \alpha(x,ay)$ is alternating. Note that ${\frak{s}}(\alpha)$ is a subalgebra of $V\otimes V^*$ under the commutator bracket; in fact, it is the Lie algebra of $Sp(\alpha)$. The subspaces ${\frak{s}}(\alpha)$ and ${\frak{a}}(\alpha)$ are invariant under conjugation by elements of $Sp(\alpha)$. Now, there is a natural map $S:{\frak{s}}(\alpha)\to {\frak{a}}(\alpha)$, given by $S(s) = s^2$. In other words, if $\alpha(x,sy)$ is symmetric, then $\alpha(x,s^2x) = \alpha(sx,sx) = 0$, so $\alpha(x,s^2y)$ is alternating. Here, then, is the question: What is the image of $S$? (The OP is actually asking whether the image of $S$ contains the invertible elements of ${\frak{a}}(\alpha)$.) Note that the dimension of ${\frak{s}}(\alpha)$ is $2n^2{+}n$, while the dimension of ${\frak{a}}(\alpha)$ is $2n^2{-}n$, so it's conceivable that $S$ is actually surjective.
Also, the map $S$ is $Sp(\alpha)$-equivariant, so it's a question that can be studied by looking at the orbits of this group acting on ${\frak{a}}(\alpha)$. Remark: It took me a while to recognize that this is what is going on because the question, as asked, sneaks in an extraneous quadratic form that breaks the symplectic symmetry. A 'reference' alternating form on $k^{2n}$ has been specified by the formula $\alpha_0(x,y) = x^TJy$ for $x,y\in k^{2n}$. Note that the matrix $J$ satisfies $J^2 = -I$, an identity that has no meaning for an alternating form. The only way one can interpret an alternating form as a linear transformation (so that squaring makes sense) is to have some other way of identifying $V$ with $V^\ast$. Of course, this is supplied by the linear map $x\mapsto x^T$ in the formula. In other words, a (symmetric) bilinear form $\beta(x,y) = x^Ty$ has been introduced into the picture, and it breaks the symplectic symmetry. Anyway, writing $\alpha(x,y) = x^TAy$ and asking whether one can write $A = MJM$ for $M$ symmetric can be re-interpreted as follows: Note that $M=Js$ where $s\in{\frak{s}}(\alpha_0)$ and that $\alpha(x,y) = x^TAy = x^TJJ^{-1}Ay = \alpha_0(x,J^{-1}Ay) = \alpha_0(x,ay)$ where $a = J^{-1}A$ lies in ${\frak{a}}(\alpha_0)$. Putting this together says that
$$
Ja = A = MJM = (Js)J(Js) = -Js^2,
$$
so showing that the equation $A = MJM$ can be solved is equivalent to showing that the equation $a = -s^2$ for a given $a\in{\frak{a}}(\alpha_0)$ can be solved for some $s\in{\frak{s}}(\alpha_0)$. (It's off by a minus sign, but that's OK because the goal is to characterize the image of $S$, so characterizing its negative is just as good.) Anyway, back to the question: One approach is to look at the orbits of $Sp(\alpha)$ acting on ${\frak{s}}(\alpha)$ and see what their squares look like. This may be easier because the adjoint orbits of $Sp(\alpha)$ on its Lie algebra have been much studied. As an example, suppose that $k$ is algebraically closed and (for my comfort) that it has characteristic zero. Say that a pair of nondegenerate alternating $2$-forms $(\alpha_0,\alpha)$ on $k^{2n}$ is generic if the $n$ roots of the equation $\textrm{Pf}(\alpha - \lambda\ \alpha_0) = 0$ are all distinct. Then one can prove (see below) that a basis of $1$-forms on $k^{2n}$ exists so that
$$
\alpha_0 =\theta^1\wedge\theta^2+\theta^3\wedge\theta^4+\cdots
+\theta^{2n-1}\wedge\theta^{2n}
$$
while
$$
\alpha =\lambda_1\ \theta^1\wedge\theta^2+\lambda_2\ \theta^3\wedge\theta^4+\cdots
+\lambda_n\ \theta^{2n-1}\wedge\theta^{2n}.
$$
Thus, the problem uncouples into $n$ separate problems that are each trivially solvable, so, the problem is solvable for the generic pair in this case. As another example, in the case $n=2$, for an arbitrary field (even one of characteristic $2$), one can, by hand, classify the pairs $(\alpha_0,\alpha)$ with $\alpha_0$ nondegenerate and show that $S:{\frak{s}}(\alpha_0)\to{\frak{a}}(\alpha_0)$ is surjective. (I'll put in the details if someone asks. Note, by the way, that the claimed counterexample when $n=2$ in the 'answer' below does not actually work.) To prove surjectivity for all $n$, one may need to understand the orbits of $Sp(\alpha)$ acting on ${\frak{a}}(\alpha)$. I think that this is a classical problem (I'm not an algebraist, so I'm not completely sure), so maybe it's time to look at the literature. The classification of the possible $Sp(\alpha)$-orbit types in ${\frak{a}}(\alpha)$ gets more complicated as $n$ increases, so maybe some other approach needs to be tried. One would expect the orbits of $Sp(\alpha)$ in ${\frak{a}}(\alpha)$ to be somewhat simpler than the orbits of $Sp(\alpha)$ in ${\frak{s}}(\alpha)$, just because the dimension is lower. However, I note that the rings of $Sp(\alpha)$-invariant polynomials on each of the vector spaces ${\frak{s}}(\alpha)$ and ${\frak{a}}(\alpha)$ are each free polynomial rings on $n$ generators, so it may be that the complexity of the orbit structures are (at least roughly) comparable in the two cases. The uncoupling step: In the general case, for $a\in{\frak{a}}(\alpha_0)$, one has $\textrm{Pf}(a^\flat - \lambda\ \alpha_0) = p_a(\lambda)\ \textrm{Pf}(\alpha_0)$, where $\det(a - \lambda I) = p_a(\lambda)^2$. Letting $f_1(\lambda),\ldots,f_k(\lambda)$ denote the distinct irreducible factors of $p_a(\lambda)$, one has
$$
p_a(\lambda) = f_1(\lambda)^{d_1}\cdots f_k(\lambda)^{d_k}.
$$
There is a direct sum decomposition $V = V_1\oplus\cdots\oplus V_k$ into the corresponding generalized eigenspaces of $a$, and one sees without difficulty (using the identity $\alpha_0(ax,y) = \alpha_0(x,ay)$) that $\alpha_0(V_i,V_j) = 0$ for $i\not=j$. Moreover, any solution $s\in{\frak{s}}(\alpha_0)$ of $a = \pm s^2$ must commute with $a$ and therefore preserve its generalized eigenspaces. Thus, generalizing the 'uncoupled' situation given in the first example, one sees that the problem reduces to the case in which $p_a(\lambda)$ is a power of a single irreducible polynomial. Unfortunately, it turns out that, even in this uncoupled case, the minimal polynomial of $a$ can fail to be irreducible (i.e., $a$ need not be semi-simple), and I do not know a simple way to handle all of these cases. When $n=2$, this can be handled 'by hand', but even for $n=3$, it seems to be a little tricky (although, I think that I have correctly handled the cases there and shown surjectivity in that case as well). The uncoupled semi-simple case: Here is how one can complete the proof of solvability in the uncoupled, semi-simple case, i.e., when the minimal polynomial of $a$ is irreducible. Let $p(\lambda)=0$ be the (irreducible) minimal polynomial of $a$, say of degree $m$. We can assume that $p(0)\not=0$, since, otherwise, $a=0$, and there is nothing to prove (i.e., one can just take $s=0$, and the problem is solved). Let $K\subset End(V)$ denote the field generated by $a$, so that $[K:k]=m$. Now, for any nonzero $x\in V$, one has $\alpha_0(K{\cdot}x, K{\cdot}x) = 0$, since $\alpha_0(a^ix,a^jx)=0$ for all $i$ and $j$. If one chooses $x,y\in V$ such that $\alpha_0(x,y)\not=0$, then it is easy to see (using $p(0)\not=0$) that $\alpha_0$ is nondegenerate on the $a$-invariant subspace $W = K{\cdot}x\oplus K{\cdot}y$. One then can write $V = W\oplus W^\perp$ (where the $\perp$ is taken with respect to $\alpha_0$) and see that the problem uncouples into separate problems on $W$ and $W^\perp$. By induction, it then suffices to consider the case $W = V$ (and, hence, $n=m$). In this case, the general element of $V = W = K{\cdot}x\oplus K{\cdot}y$ can be written uniquely in the form $z = f_1(a)x + f_2(a)y$ where $f_1$ and $f_2$ are polynomials (with coefficients in $k$) of degree at most $m{-}1$. Let $q(a)$ and $r(a)$ be any polynomials in $a$ (i.e., elements of $K$) and define a map $s:V\to V$ by
$$
s\bigl(f_1(a)x + f_2(a)y\bigr) = f_1(a)q(a)y + f_2(a)r(a)x.
$$
One checks that $s\in{\frak{s}}(\alpha_0)$ and notes that, by construction, one has the identity $s^2 = q(a)r(a)$. By setting $q(a) = a$ and $r(a) = \pm1$, one sees that one may arrange $s^2 = \pm a$. This $s$ solves the problem.
|
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|
84,742 |
Let $f$ and $g$ be complex rational functions (of degree $\geq 2$ if that helps). What can be said about the relationship between $J(fg)$ and $J(gf)$, the Julia sets of the composite functions $f \circ g$ and $g \circ f$? If I'm not mistaken, $f$ restricts to a map $J(gf) \to J(fg)$, and $g$ restricts to a map $J(fg) \to J(gf)$. So $J(fg)$ and $J(gf)$ surject onto each other in a particular way (and, indeed, in a way that commutes with the actions of $fg$ on $J(fg)$ and $gf$ on $J(gf)$). Since Julia sets are completely invariant, the restricted map $f: J(gf) \to J(fg)$ is $deg(f)$-to-one, and similarly the other way round. So there's some kind of relationship between the two sets. If $f$ or $g$ has degree one then $J(fg)$ and $J(gf)$ are "isomorphic", in the sense that there's a Möbius transformation carrying one onto the other. Thus, the simplest nontrivial example would be to take $f$ and $g$ to be of degree 2. I don't know a way of computing, say, the example $f(z) = z^2$ and $g(z) = z^2 + 1$. That would mean computing the Julia sets of $gf(z) = z^4 + 1$ and $fg(z) = z^4 + 2z^2 + 1$. My question isn't completely precise, I'm afraid. But here are some of the things that I would value in an answer: theorems on what $J(fg)$ and $J(gf)$ have in common, examples showing how different they can be, pictures of $J(fg)$ and $J(gf)$ for particular functions $f$ and $g$, and references to where I can find out more (especially those accessible to a non-specialist). Thanks.
|
I'm not sure if this is helpful, but here is an example. The following picture shows the filled Julia set for $z^6 - 1$. and the following picture shows the filled Julia set for $(z^2-1)^3$: This is the case where $f(z) = z^2 - 1$ and $g(z) = z^3$. Note that the bottom image is a double cover of the top, while the top image is a triple cover of the bottom. (These images were produced using Mathematica .)
|
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|
84,958 |
Is $$ \sum_{n=0}^\infty {x^n \over \sqrt{n!}} > 0 $$ for all real $x$?
(I think it is.) If so, how would one prove this? (To confirm: This is the power
series for $e^x$, except with the denominator replaced by $\sqrt{n!}$.)
|
Looks like the computers really spoiled us :) GH gave a perfectly valid answer already but the cheapest way to prove positivity is to write $\int_0^1(1-t^n)\log(\frac 1t)^{-3/2}\,\frac{dt}t=c\sqrt n$ with some positive $c$ (just note that the integral converges and the integrand is positive, and make the change of variable $t^n\to t$). Hence $\int_0^1 (f(x)-f(xt))\log(\frac 1t)^{-3/2}\,\frac{dt}t=cxf(x)$. If $x$ is the largest zero of $f$ (which must be negative), then plugging it in, we get $0$ on the right and a negative number on the left, which is a clear contradiction. Thus, crossing the $x$-axis is impossible. Of course, there is nothing sacred about $1/2$. Any power between $0$ and $1$ works just as well.
|
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|
85,013 |
Let
$$ c_n = \sum_{r=0}^n (-1)^r \sqrt{\binom{n}{r}}. $$
It is clear that $c_n = 0$ if $n$ is odd. Remarkably, it appears that despite the huge positive and negative contributions in the sum defining $c_{2m}$, the sequence $(c_{2m})$ may be very well behaved. Is $c_n > 0$ for all even $n$? An affirmative answer will imply that the function $F(x) = \sum_{n=0}^\infty x^n/\sqrt{n!}$ is always strictly positive, thereby answering this earlier question . Numerical computation using Magma shows that $c_n > 0$ if $n$ is even and $n \le 2000$. To give some illustrative values, $c_{100} = 0.077737 \ldots$, $c_{1000} = 0.019880 \ldots $ and $c_{2000} = 0.013317 \ldots$. A comment by Mark Sapir on the earlier question suggests a stronger result might hold. Is $c_{n} > c_{n+2} > 0$ for all even $n$? I have checked that this is the case for all even $n \le 2000$. It is very natural to ask what happens if we replace $\sqrt{\binom{n}{r}}$ with $\binom{n}{r}^\alpha$ for $\alpha \in (0,1)$. For $n\le 250$ the generalized version of the conjecture continues to hold if $\alpha = k/10$ where $k \in \mathbf{N}$ and $k \le 9$. Of course when $\alpha = 1$ we have $c_n = 0$ for all $n$, so, as David Speyer remarked in a comment on the earlier question, there is a good reason for the cancellation in this case.
|
Here's a proof of the positivity of
$$
c_n(\alpha) := \sum_{r=0}^n (-1)^r {n\choose r}^\alpha
$$
for all even $n$ and real $\alpha < 1$. It follows
(via M.Wildon's clever $F(x) F(-x)$ trick at mo.84958) that
$\sum_{n=0}^\infty \phantom. x^n / n!^{\alpha} > 0$ for all $x \in\bf R$.
[ EDIT fedja has meanwhile provided a very nice direct proof of
the positivity of $\sum_{n=0}^\infty \phantom. x^n / n!^{\alpha}$.] The key is to write $c_n(\alpha)$ as a finite difference
$$
\sum_{r=0}^n \phantom. (-1)^r {n\choose r} \cdot {n\choose r}^{\alpha - 1}
$$
and show that the Gamma interpolation
$$
\bigl(\Gamma(r+1)\Gamma(n-r+1) / n!\bigr)^{1-\alpha}
= n!^{\alpha-1} \exp\bigl((1-\alpha) (\log\Gamma(r+1) + \Gamma(n-r+1)\bigr)
$$
of ${n\choose r}^{\alpha - 1}$ has a positive $n$-th derivative
for all $r \in [0,n]$. This in turn follows from the fact that the expansion of
$\log\Gamma(r+1) + \log\Gamma(n-r+1)$ in a Taylor series about $r = n/2$
has positive $(r - (n/2))^k$ coefficient for each $k=2,4,6,\ldots$.
[The coefficient vanishes for odd $k$ because
$\log\Gamma(r+1) + \log\Gamma(n-r+1)$ is an even function of $r-(n/2)$.]
Indeed the well-known formula
$$
\log \Gamma(x) = -\gamma x - \log x + \sum_{j=1}^\infty
\left[ \frac{x}{j} - \log \left( 1 + \frac{x}{j} \right) \right]
$$
shows that the $k$-th derivative of $\log\Gamma(x)$ is positive
for all $x>0$ and $k=2,4,6,\ldots$, because this is true for
$-\gamma x - \log x$ and for each term in the sum; explicitly
the derivative is $k! \phantom. \sum_{j=0}^\infty (x+j)^{-k}$ which is
positive termwise. Therefore in the Taylor expansion
$$
\log \Gamma(r+1) = \log(n/2)! + \sum_{k=1}^\infty \phantom. g_k (r-(n/2))^k
$$
each of $g_2,g_4,g_6,\ldots$ is even.
Since $\log\Gamma(r+1) + \log\Gamma(n-r+1)$ is
$$
2\log(n/2)!
+ 2 \Bigl( g_2 (r-(n/2))^2 + g_4 (r-(n/2))^4 + g_6 (r-(n/2))^6 + \cdots\Bigr),
$$
the claim follows. [ EDIT David Speyer notes that the convergence
of the Taylor series on $|r-(n/2)| \leq n/2$ requires justification,
and that the justification is easy because the $\Gamma(z)$ has no zeros
and poles only at $0,-1,-2,\ldots$ so the radius of convergence is $(n/2)+1$.]
Multiplying by $1 - \alpha$ and substituting into the exponential series,
we deduce that $(\Gamma(r+1) \Gamma(n-r+1))^{1-\alpha}$, too,
is a positive combination of even powers of $r-(n/2)$. Now if a function $g$ has positive $n$-th derivative, then its
first finite difference
$$
g(x+1) - g(x) = \int_x^{x+1} g'(y) dy
$$
has positive $(n-1)$-st derivative; repeating this argument $n$ times,
we find that the $n$-th finite difference is positive, and we're done.
|
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|
85,309 |
Let A and B be positive semidefinite matrices. It is not hard to see that $(A-B)^2 \leq 2A^2 + 2B^2$. In fact, $2A^2 + 2B^2 - (A-B)^2 = (A+B)^2$ is positive semidefinite. My question is: Is there a constant c (independent of A and B and the dimension) such that $$(A-B)^2 \leq c (A+B)^2?$$ Thanks.
|
There is no such $c$ even if we use only $2 \times 2$ matrices.
For any $c \geq 1$ let $A,B$ be the positive-semidefinite matrices
$$
A = \left( \begin{array}{lc} c^2 & c \cr c & 1 \end{array} \right),
\phantom\infty
B = \left( \begin{array}{cc} 1 & 0 \cr 0 & 0 \end{array} \right).
$$
of rank $1$. Then we calculate that the difference
$$
D := c(A+B)^2 - (A-B)^2
$$
has determinant $(c-1)^2 - 4c^3 < 0$, and is thus not positive semidefinite. In fact this counterexample works for all $c \in \bf R$:
looking around $\ker A = {\rm span} \lbrace(-1,c)\rbrace$
we find the negative vector $v = (-c, c^2+1)$.
To verify that $\langle v, Dv \rangle < 0$, recall that
for any vector $x$ and any symmetric matrix $M$ of the same order we have
$$
\langle x, M^2 x \rangle = \langle Mx, Mx \rangle = |Mx|^2.
$$
Here we compute $v(A+B) = (0,1)$ and $v(A-B) = (2c,1)$, so
$$
\langle v, Dv \rangle = c \phantom. |(0,1)|^2 - |(2c,1)|^2 = -4c^2 + c - 1,
$$
which is negative for all real $c$. If $c$ is small enough that
$\det(D) > 0$ then $D$ is negative definite.
|
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|
85,540 |
What are the groups $X$ for which there exists a group $G$ such that $G' \cong X$? My considerations: $\bullet$ If $X$ is perfect we are happy with $G=X$. $\bullet$ If $X$ is abelian then $G := X \wr C_2$ verifies $G'=\{(x,x^{-1}): x \in X\} \cong X$. $\bullet$ If $X$ satisfies the following properties: (1) $X \neq X'$, (2) The conjugation action $X \to \text{Aut}(X)$ is an isomorphism, then there is no $G$ such that $G' = X$ (consider the composition $G \to \text{Aut}(X) \cong X \to X/X'$, it is surjective so its kernel contains $X$, contradiction). For instance, the symmetric group $S_n$ verifies (1) and (2) if $n \neq 2,6$. I have been looking for this problem on the web but I didn't find anything.
Do you have any reference and/or suggestion on how to solve this problem?
|
A complete answer seems not to be known. Let me give you the following two nearly-contemporaneous references from the mid-70s: Robert Guralnick, On groups with decomposable commutator subgroups Michael Miller, Existence of Finite Groups with Classical Commutator Subgroup Both Guralnick and Miller call groups which are commutator subgroups $C$-groups (though I don't know who, if either, originated the term) and give partial answers to your general question. For example, Theorem 4 from Miller gives the following: Let $G$ be a subgroup of $\operatorname{GL}_n(K)$ containing $\operatorname{SL}_n(K)$ for $K$ a finite field of characteristic not equal to 2. Then $G$ is the commutator subgroup of some group unless it is of odd index and $n$ is even. The groupprops-wiki calls such groups commutator-realizable, and give a basic result on such groups, but mention that this terminology is not standard (though is probably safer than the overloaded term $C$-group.) Edit: Some googling around led to the following slick argument of Schoof (from his Semistable abelian varieties with good reduction outside 15 ), which is closely related to your observation in bullet (3), and also serves to eliminate the symmetric groups. I'll quote verbatim except for change of variable names: Let $G$ be a group and let $G'$ be its commutator subgroup. Conjugation gives rise to
a homomorphism $G \to \operatorname{Aut}(G')$. On the one hand it maps $G'$ to the commutator subgroup of $\operatorname{Aut}(G')$. On the other hand the image of $G'$ is the group $\operatorname{Inn}(G')$ of inner automorphisms of $G'$. Therefore, if a group $X$ is the commutator subgroup of some group, we must have $\operatorname{Inn}(X)\subset \operatorname{Aut}(X)'$.
|
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|
85,678 |
Much of the theory of continued fractions has been developped by Euler in the 18th century. The little survey "Euler: continued fractions and divergent series (and Nicholas Bernoulli)" , mentions towards the end the continued fraction $$f(x)=\dfrac{1}{1+\dfrac{x}{1+\dfrac{x}{1+\dfrac{2x}{1+\dfrac{2x}{1+\dfrac{3x}{1+\dots}}}}}}$$ which Euler has "derived" from the divergent power series $1-x+2x^2-6x^3+24x^4-+...$ . For $x=1$, the continued fraction converges to a limit $f(1)\approx 0.5963475922$. I was wondering: what is known about the values $f(x)$, in particular $f(1)$? Are they known to be transcendental for $x\in\mathbb N$? Can they be expressed by other known constants?
|
The sequence $a_n=(-1)^n n!$ satisfies $a_{n+1}+(n+1)a_n = 0$ (with $a(0)=1$). Thus the generating function satisfies the differential equation $x^2y'+(x+1)y=1$ (where $y(0)=1$). The unique solution is
$$\frac{e^{\frac{1}{x}}Ei\left(1,\frac{1}{x}\right)}{x}$$
For $x=1$, the constant is $e Ei(1,1) \approx .5963473623231940743410785$. The reason this is justified is because the sequence is Gevrey , i.e. it does not diverge 'too fast', so associated to it is a unique (generating) function which has that sequence as coefficients (asymptotically) at 0, when approached along the real line. The modern theory that 'this all works' is essentially due to Écalle, although Lindelof had worked out quite a bit already $100$ years before.
|
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|
85,881 |
I'm betting `yes, sure!', but don't see it. Could someone please point me toward,
or construct for me, a Lagrangian submanifold immersed in
standard symplectic ${\mathbb R}^{2n}$ for $n > 1$, whose closure is all of ${\mathbb R}^{2n}$? (For an $n =1$ example, one can use the leaves arising from this modification by Panov of irrational flow on the two-torus.) Strong preference given to analytic immersions of ${\mathbb R}^n$. Holomorphically immersed complex lines which are dense in complex 2-space -- i.e. dense ${\mathbb C}$'s in ${\mathbb C}^2$ -- are well-known. Ilyashenko in 1968 showed that the typical solution of the typical polynomial ODE (in complex time) yields such a curve. Following his line
of thought, it might be easier to construct an entire singular Lagrangian foliation of ${\mathbb R}^{2n}$ whose typical leaf is dense, rather than the one submanifold. Motivation: I have a certain unstable manifold related to a Hamiltonian system.
It is Lagrangian. I would like to be ``as dense as can be'', so I'd like to know
how dense can that be.
|
Your question already has the answer in it for $n=2$. Take a connected complex curve $L\subset\mathbb{C}^2$ that is dense in $\mathbb{C}^2$. Then $L$ is Lagrangian for the real part of the holomorphic $2$-form $\Upsilon = dz^1\wedge dz^2$. This real part of $\Upsilon$ is equivalent to the standard symplectic structure on $\mathbb{R}^4$ by a linear change of variables. Added comment about injectivity: Note, by the way, that one can easily arrange for such an $L$ to be a submanifold, not just the image of an immersion (i.e., the immersion is injective ). One simple explicit way to do this is to select constants $\lambda_1,\ldots,\lambda_k$ such that the subgroup in $\mathbb{C}^\times$ generated by the numbers $\mathrm{e}^{2\pi i\lambda_1},\ldots,\mathrm{e}^{2\pi i\lambda_k}$ is dense in $\mathbb{C}^\times$ and consider the linear differential equation
$$
\frac{dy}{dx} = \left(\frac{\lambda_1}{x-x_1}+\cdots + \frac{\lambda_k}{x-x_k}\right)\ y
$$
where $x_1,\ldots,x_k\in \mathbb{C}$ are distinct. The graph of any nonzero multi-valued solution $y(x)$ over $\mathbb{C}\setminus\{x_1,\ldots,x_k\}$ will then be dense in $\mathbb{C}^2$. (Consider the holonomy around the punctures $x_j$.) Of course, these graphs are the Riemann surfaces associated to the multivalued functions
$$
y = y_0 (x{-}x_1)^{\lambda_1}\cdots(x{-}x_k)^{\lambda_k}
$$
(when $y_0\not=0$). These are obviously integral curves (leaves) of the polynomial $1$-form
$$
\omega = (x{-}x_1)\cdots(x{-}x_k)\ dy - q(x) y\ dx
$$
for some polynomial $q$ of degree at most $k{-}1$ in $x$. Aside from the obvious closed leaves $x-x_j=0$ and $y=0$, the rest of the leaves are dense submanifolds. (This just gives a simple, explicit example of the general theorem that Richard quoted.) Dense analytic curves in $\mathbb{R}^2$: It is not hard to construct dense, connected analytic curves in $\mathbb{R}^2$: There exist analytic metrics on the $2$-sphere that have geodesics that wander densely over the surface. Now take such a geodesic and remove a point from $S^2$ through which the geodesic doesn't pass. What's left is a dense analytic curve in $\mathbb{R}^2$. If you are willing to use Finsler metrics, you can even do this with a rotationally invariant real analytic Finsler metric on the $2$-sphere (eg. Katok's examples), so that you can write down the dense analytic curve very explicitly. I'll think about the case $n>2$. I don't see it yet either, but maybe it's not too hard.
|
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|
85,904 |
My brother asked me a question which I didn't know the answer to. Are there theorems about existence, uniqueness and stability of solutions of ODEs of the followin type $$
\frac{d^2 y}{dt^2} = f(t,y,\frac{dy}{dt})H(g(y)),
$$ where $f$ and $g$ are Lipschitz functions and $H$ is the Heaviside function? If there are no general theorems that could apply, how does one analyze these problems for given $f,g$?
|
Your question already has the answer in it for $n=2$. Take a connected complex curve $L\subset\mathbb{C}^2$ that is dense in $\mathbb{C}^2$. Then $L$ is Lagrangian for the real part of the holomorphic $2$-form $\Upsilon = dz^1\wedge dz^2$. This real part of $\Upsilon$ is equivalent to the standard symplectic structure on $\mathbb{R}^4$ by a linear change of variables. Added comment about injectivity: Note, by the way, that one can easily arrange for such an $L$ to be a submanifold, not just the image of an immersion (i.e., the immersion is injective ). One simple explicit way to do this is to select constants $\lambda_1,\ldots,\lambda_k$ such that the subgroup in $\mathbb{C}^\times$ generated by the numbers $\mathrm{e}^{2\pi i\lambda_1},\ldots,\mathrm{e}^{2\pi i\lambda_k}$ is dense in $\mathbb{C}^\times$ and consider the linear differential equation
$$
\frac{dy}{dx} = \left(\frac{\lambda_1}{x-x_1}+\cdots + \frac{\lambda_k}{x-x_k}\right)\ y
$$
where $x_1,\ldots,x_k\in \mathbb{C}$ are distinct. The graph of any nonzero multi-valued solution $y(x)$ over $\mathbb{C}\setminus\{x_1,\ldots,x_k\}$ will then be dense in $\mathbb{C}^2$. (Consider the holonomy around the punctures $x_j$.) Of course, these graphs are the Riemann surfaces associated to the multivalued functions
$$
y = y_0 (x{-}x_1)^{\lambda_1}\cdots(x{-}x_k)^{\lambda_k}
$$
(when $y_0\not=0$). These are obviously integral curves (leaves) of the polynomial $1$-form
$$
\omega = (x{-}x_1)\cdots(x{-}x_k)\ dy - q(x) y\ dx
$$
for some polynomial $q$ of degree at most $k{-}1$ in $x$. Aside from the obvious closed leaves $x-x_j=0$ and $y=0$, the rest of the leaves are dense submanifolds. (This just gives a simple, explicit example of the general theorem that Richard quoted.) Dense analytic curves in $\mathbb{R}^2$: It is not hard to construct dense, connected analytic curves in $\mathbb{R}^2$: There exist analytic metrics on the $2$-sphere that have geodesics that wander densely over the surface. Now take such a geodesic and remove a point from $S^2$ through which the geodesic doesn't pass. What's left is a dense analytic curve in $\mathbb{R}^2$. If you are willing to use Finsler metrics, you can even do this with a rotationally invariant real analytic Finsler metric on the $2$-sphere (eg. Katok's examples), so that you can write down the dense analytic curve very explicitly. I'll think about the case $n>2$. I don't see it yet either, but maybe it's not too hard.
|
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"https://mathoverflow.net/questions/85904",
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|
86,118 |
Recall that a derangement is a permutation $\pi: \{1,\ldots,n\} \to \{1,\ldots,n\}$ with no fixed points: $\pi(j) \neq j$ for all $j$. A classical application of the inclusion-exclusion principle tells us that out of all the $n!$ permutations, a proportion $1/e + o(1)$ of them will be derangements. Indeed, by computing moments (or factorial moments) or using generating function methods, one can establish the stronger result that the number of fixed points in a random permutation is asymptotically distributed according to a Poisson process of intensity 1. In particular, we have: Corollary : the proportion of permutations that are derangements is bounded away from zero in the limit $n \to \infty$. My (somewhat vague) question is whether there is a "non-enumerative" proof of this corollary that does not rely so much on exact combinatorial formulae. For instance, a proof using the Lovasz Local Lemma would qualify, although after playing with that lemma for a while I concluded that there was not quite enough independence in the problem to make that lemma useful for this problem. Ideally, the non-enumerative proof should have a robust, "analytic" nature to it, so that it would be applicable to other situations in which one wants to lower bound the probability that a large number of weakly correlated, individually unlikely events do not happen (much in the spirit of the local lemma). My original motivation, actually, was to find a non-enumerative proof of a strengthening of the above corollary, namely that given $l$ permutations $\pi_1,\ldots,\pi_l: \{1,\ldots,n\} \to \{1,\ldots,n\}$ chosen uniformly and independently at random, where $l$ is fixed and $n$ is large, the probability
that these $l$ permutations form a $2l$-regular graph is bounded away from zero in the limit $n \to \infty$. There is a standard argument (which I found in Bollobas's book) that establishes this fact by the moment method (basically, showing that the number of repeated edges or loops is distributed according to a Poisson process), but I consider this an enumerative proof as it requires a precise computation of the main term in the moment expansion.
|
The mean number of fixed points is 1. This is very elementary. Consider the operation of rotating three values around: $p(i)\to p(j)\to p(k)\to p(i)$. Given a permutation with no fixed points, there are $n^2-O(n)$ rotations that create from it a permutation with exactly one fixed point. Given a permutation with exactly one fixed point, there are $n^2-O(n)$ rotations that create from it a permutation with no fixed points. From (2), the numbers of permutations with 0 and 1 fixed points are the same to $O(1/n)$ relative error. Combining this with (1) shows that the fraction of permutations with no fixed points is at least $\frac13-O(1/n)$. This is the switching method and it can be used in extremely many circumstances. By continuing to increase the number of fixed points by further switchings, step 1 could be avoided. Incidentally, congratulations to Terry and Jean Bourgain .
|
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|
86,742 |
In two-dimensional case one can generalize figures of constant width as figures which can rotate in a convex polygon.
Here is one example which can be used to drill triangular holes: I would like to know what happens with this generalization in dimension $3$ and maybe higher.
Obviously a body of constant width $1$ can rotate arbitrary in a unit cube.
More formally, given a body $B$ of constant width $1$ and $A\in SO(3)$ there is $v\in \mathbb R^3$ such that $$A(B)+v\subset\square,$$ where $\square$ is unit cube.
On the other hand, except for the cube, I do not see any other examples of convex polyhedron which have nontrivial rotating bodies (i.e. distinct from the inscribed ball). I hope that the answer is known.
(= I hope I should wait for the answer and I do not have to think.) The question is inspired by this one: " Local minimum from directional derivatives in the space of convex bodies ."
|
I found the reference I was looking for. The full list of cases under which $K$ is a rotor in a cavity shaped like the polytope $P$ is available on page 27 of the notes titled " The use of spherical harmonics in convex geometry " by Rolf Schneider. They are available under "Course Materials" on his website . As I recall, there is one more non-trivial case in $d=3$ if the cavity is allowed to be unbounded (e.g. a cone), and this case appears in the more complete list in Groemer's book.
|
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|
86,792 |
The two standard approaches to the quantization of Chern-Simons theory are geometric quantization of character varieties, and quantum groups plus skein theory. These two approaches were both first published in 1991 (the geometric quantization picture here and the skein theoretic approach here ), and despite a tremendous amount of development since then, it is still not known whether they are equivalent! I guess that it is reasonable to say that the problem of their equivalence has been around for 20 years. There is (at least) one important theorem, namely the asymptotic faithfulness of the mapping class group representations produced by these two quantizations, which has proofs in both settings. The two proofs are of completely different character, and are of course logically independent, since the two representations are not known to be the same (this was proved for the quantum group skein representation by Freedman, Walker, and Wang , and for the geometric quantization representation by Andersen ). Is there a good reason why the equivalence of these two viewpoints is not yet a theorem? Is there an idea for a proof, which hasn't been completed because "it's just a long calculation" or "everyone knows it's true" or "it's nice to know, but it wouldn't actually help us prove theorems"? Or is it that it's actually a hard problem that no one knows how to approach? Is it an "important" problem whose solution would have lots of consequences and applications, or at least advance our understanding of "quantization"?
|
The equivalence of these two construction is actually known now. It follows by combining the main result of: Yves Laszlo, Hitchin's and WZW connections are the same ., J. Differential Geom. 49 (1998), no. 3, 547–576, doi: 10.4310/jdg/1214461110 with my joint work with Kenji Ueno presented in a series of four papers: J.E. Andersen & K. Ueno, Abelian Conformal Field theories and Determinant Bundles , International Journal of Mathematics, 18 (2007) 919–993 doi: 10.1142/S0129167X07004369 , arXiv: math/0304135 . J.E. Andersen & K. Ueno, Geometric Construction of Modular
Functors from Conformal Field Theory , Journal of Knot theory and its
Ramifications, 16 (2007) 127–202 doi: 10.1142/S0218216507005233 , arXiv: math/0306235 . J.E. Andersen & K. Ueno, Modular functors are determined by
their genus zero data , Quantum Topol. 3 (2012), 255–291, doi: 10.4171/QT/29 , arXiv: math.QA/0611087 . J.E. Andersen & K. Ueno, Construction of the Reshetikhin-Turaev TQFT via
Conformal Field Theory , Invent. Math. 201 (2015) 519–559, doi: 10.1007/s00222-014-0555-7 , arXiv: 1110.5027 The work with Ueno establishes the isomorphism between the Reshetikhin-Turaev TQFT for $SU(n)$ using Quantum groups (We work with the skein theory model due to Blanchet, Habegger, Masbaum and Vogel) with the one coming from Conformal Field Theory as described in the above mentioned papers. We identify the underlying modular functors of the two theories. The natural identification of the vector space associated to a closed surface by the modular functor constructed from Conformal Field Theory with the space of covariant constant sections over Teichmüller space of the Hitchin connection, i.e. the space coming from the geometric quantization of the $SU(n)$ character variety is provided in the above mentioned paper with Laszlo. The composition of these isomorphisms gives the desired isomorphism.
|
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|
86,947 |
For a (bounded) double complex (of abelian groups or vector spaces) one can consider two spectral sequences that converge to the cohomology of the totalization: one can first compute either the cohomology of rows, or the cohomology of columns. Suppose that one of these spectral sequences degenerates at $E_1$ (i.e. the cohomology of rows yields the factors of the induced filtration of the limit). Do any 'nice' properties of the second spectral sequence follow?
|
There is a basic way to see whether things like this should be true. Any bounded double complex of vector spaces over a field $k$ is (noncanonically) the direct sum of complexes of the following two sorts: Squares: $$\begin{matrix} k & \rightarrow & k \\ \uparrow & & \uparrow \\ k & \rightarrow & k \end{matrix}$$ Staircases: $$\begin{matrix}
k & \rightarrow & k & & & & \\
& & \uparrow & & & & \\
& & k & \rightarrow & k & & \\
& & & & \uparrow & & \\
& & & & k & \rightarrow & k\\
\end{matrix}$$
We'll say that the "length" of a staircase is the number of nonzero entries in it, so the above stair case has length $6$. Staircases may have even odd or even length, and may start and end either with vertical or horizontal maps. The operation of "forming the spectral sequence" commutes with direct sum, so it is enough to know what the spectral sequences of each of these look like. The spectral sequence of a square is zero on every page except for the square itself; in particular, it converges in one step. The spectral sequence of an odd length staircase is one dimensional on every page except for the stair case itself. The one nonzero term is at one end of the staircase for the horizontal spectral sequence and the other for the vertical spectral sequence. So it also converges in one page. The spectral sequence of an even staircase is the one which violates your claim. In one direction, it is zero on every page after the staircase itself. In the other direction, it there are $m$ pages with two nonzero terms, where the length of the staircase was $2m$. These terms are at the two ends of the stair case, and they annihilate each other on the $(m+1)$st page. In particular, a double complex which consists simply of a length $2m$ staircase will die on the first page in one direction, but will survive for $m$ pages in the other.
|
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|
87,070 |
Suppose a finite-dimensional Lie group $G$ is given. Does there exist a connected manifold $M$ and a Riemannian metric $g$ , such that $G$ is the full isometry group of $(M,g)$ ? For example if I try to do this for a connected $G$ , then I often get a bigger group as the full isometry group, which includes e.g. the orientation reversing isometries. (Maybe one has to take a non--orientable space for that?) Even if I try to realize $\mathbb R$ as a full isometry group, I fail. (One could take the full isometry group of $\mathbb R$ with the standard metric, which is given by $\mathbb R \rtimes \mathbb Z_2$ and divide out the $\mathbb Z_2$ action. But this leads to a fixpoint and the quotient is therefore not a manifold any more.) There is an article of J. de Groot 1 which proves that every abstract group can be realized as an isometry group of some metric space, but it is not clear to me, if this is true in the category of Lie groups and Riemannian manifolds. 1 de Groot, J. "Groups represented by homeomorphism groups."
Math. Ann. 138 (1959) 80–102. MR119193 doi:10.1007/BF01369667</a
|
The article of de Groot is the one cited here: What kind group can be realized as a Isometry group of some space? That every compact group is the full isometry group of a compact Riemannian manifold is shown in: Saerens, Rita; Zame, William R. , The isometry groups of manifolds and the automorphism groups of domains , Trans. Am. Math. Soc. 301, 413-429 (1987). ZBL0621.32025 . (such an isometry group must be compact priori). Winkelmann, Jörg , Realizing connected Lie groups as automorphism groups of complex manifolds , Comment. Math. Helv. 79, No. 2, 285-299 (2004). ZBL1056.32022 . shows that every connected real lie group is the full automorphism group of a complete, hyperbolic (in the sense of Kobayashi) complex manifold.
|
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|
87,238 |
I've come across several references to MK (Morse-Kelley set theory), which includes the idea of a proper class, a limitation of size, includes the axiom schema of comprehension across class variables (so for any $\phi(x,\overline y)$ with $x$ restricted to sets, there a class $X=(x : \phi(x,\overline y))$). I have seen various statements about MK and how it proves the consistency of various things, including $Con(ZF)$, $Con(ZFC)$, $Con(NBG)$, and in fact, for any $T\subset MK$ finitely axiomatized, it proves $Con(T)$. However, and quite frustratingly, I don't see any references to back up these claims, except occasionally links to other places where the claim was made, but not proven (or even proof-sketched). I would really appreciate a reference where I can see a proof of these claims, or (if it's easier) a quick sketch of why it should be true. It's not obvious to me at all why quantifying across proper classes should allow this sort of thing, since all relevant sets (sets of proofs, or sets of statements, or whatever) should be contained in some subset of $\omega$, so should be able to be constructed in $ZF$.
|
Let me give an easier (sketch of an) answer to the part of the question about proving Con(ZFC) in MK. Unlike Emil's answer, the following does not cover the case of arbitrary finitely axiomatized subtheories of MK. Intuitively, there's an "obvious" argument for the consistency of ZFC: All its axioms are true when the variables are interpreted as ranging over arbitrary sets. (The universe is a model of ZFC, except that it isn't a set.) And anything deducible from true axioms is true, so you can't deduce contradictions from ZFC. The trouble with this argument is that it relies on a notion of "truth in the universe" that can't be defined in ZFC. What goes wrong if you try to define, in the language of ZFC, this notion of truth (or satisfaction) in the universe? Just as in the definition of truth in a (set-sized) model, you'd proceed by induction on formulas, and there's no problem with atomic formulas and propositional connectives. Quantifiers, though, give the following problem: The truth value of $\exists x\ \phi(x)$ depends on the truth values of all the instances $\phi(a)$, and there are a proper class of these. In showing that definitions by recursion actually define functions, one has to reformulate the recursion in terms of partial functions that give enough evidence for particular values of the function being defined. (For example, the usual definition of the factorial can be made into an explicit definition by saying $n!=z$ iff there is a sequence $s$ of length $n$ with $s_1=1$ and $s_k=ks_{k-1}$ for $2\leq k\leq n$ and $s_n=z$.) If you use the same method to make the definition of "truth in the universe" explicit, you find that the "evidence" (analogous to $s$ for the factorial) needs to be a proper class. So ZFC can't handle that (and it's a good thing it can't, because otherwise it would prove its own consistency). But MK can; it's designed to deal nicely with quantification of proper classes. So in MK, one can define what it means for a formula to be true in the ZFC universe. Then one can prove that all the ZFC axioms are true in this sense and truth is preserved by logical deduction (here one uses induction over the number of steps in the deduction). Therefore deduction from ZFC axioms can never lead to contradictions.
|
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|
87,242 |
In the proof, the author consider the normalization $\tilde{A}$ of $A$ and show $\tilde{A}/t \tilde{A}$ is a integral domain. He showed that the localizations at points of Spec $A$ are domains, but we know a non-domain ring can have integral localizations. How should I understand the proof? Thanks a lot.
|
Let me give an easier (sketch of an) answer to the part of the question about proving Con(ZFC) in MK. Unlike Emil's answer, the following does not cover the case of arbitrary finitely axiomatized subtheories of MK. Intuitively, there's an "obvious" argument for the consistency of ZFC: All its axioms are true when the variables are interpreted as ranging over arbitrary sets. (The universe is a model of ZFC, except that it isn't a set.) And anything deducible from true axioms is true, so you can't deduce contradictions from ZFC. The trouble with this argument is that it relies on a notion of "truth in the universe" that can't be defined in ZFC. What goes wrong if you try to define, in the language of ZFC, this notion of truth (or satisfaction) in the universe? Just as in the definition of truth in a (set-sized) model, you'd proceed by induction on formulas, and there's no problem with atomic formulas and propositional connectives. Quantifiers, though, give the following problem: The truth value of $\exists x\ \phi(x)$ depends on the truth values of all the instances $\phi(a)$, and there are a proper class of these. In showing that definitions by recursion actually define functions, one has to reformulate the recursion in terms of partial functions that give enough evidence for particular values of the function being defined. (For example, the usual definition of the factorial can be made into an explicit definition by saying $n!=z$ iff there is a sequence $s$ of length $n$ with $s_1=1$ and $s_k=ks_{k-1}$ for $2\leq k\leq n$ and $s_n=z$.) If you use the same method to make the definition of "truth in the universe" explicit, you find that the "evidence" (analogous to $s$ for the factorial) needs to be a proper class. So ZFC can't handle that (and it's a good thing it can't, because otherwise it would prove its own consistency). But MK can; it's designed to deal nicely with quantification of proper classes. So in MK, one can define what it means for a formula to be true in the ZFC universe. Then one can prove that all the ZFC axioms are true in this sense and truth is preserved by logical deduction (here one uses induction over the number of steps in the deduction). Therefore deduction from ZFC axioms can never lead to contradictions.
|
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