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67,384 |
Possibly the most striking proof of Archimedes's inequality $\pi < 22/7$ is an integral formula for the difference:
$$
\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 \frac{dx}{1+x^2},
$$
where the integrand is manifestly positive. This formula is "well-known" but its origin remains somewhat mysterious. I ask: Who discovered this integral, and in what context? The earliest reference I know of is Problem A-1 on the 29th Putnam Exam (1968). According to J.H.McKay's report in the American Math. Monthly (Vol.76 (1969) #8, 909-915), the Questions Committee consisted of N.D.Kazarinoff, Leo Moser, and Albert Wilansky. Is one of them the discoverer, and if so which one? The printed solution, both in the Monthly article and in the book by Klosinski, Alexanderson, and Larson, says only "The standard approach, from elementary calculus, applies. By division, rewrite the integrand as a polynomial plus a rational function with numerator of degree less than 2. The solution follows easily." But surely there's more to be said, because this integral is a minor miracle of mathematics: $\bullet$ Not only is the integrand manifestly positive, but it is always small: $x-x^2 \in [0,1/4]$ for $x \in [0,1]$, and the denominator $1+x^2$ is at least 1, so $(x-x^2)^4/(x^2+1) < 1/4^4 = 1/256$. A better upper bound on the integral is $\int_0^1 (x-x^2)^4 dx$, which comes to $1/630$ either by direct expansion or by recognizing the Beta integral $B(5,5)=4!^2/9!$. Hence $\frac{22}{7} - \pi < 1/630$, which also yields Archimedes's lower bound $\pi > 3\frac{10}{71}$. $\bullet$ The "standard approach" explains how to evaluate the integral, but not why the answer is so simple. When we expand
$$
\frac{(x-x^2)^4}{1+x^2} = x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac4{x^2+1},
$$
the coefficient of $x/(x^2+1)$ vanishes, so there's no $\log 2$ term in the integral. [This much I can understand: the numerator $(x-x^2)^4$ takes the same value $(1\pm i)^4 = -4$ at both roots of the denominator $x^2+1$.] When we integrate the polynomial part, we might
expect to combine fractions with denominators of 2, 3, 4, 5, 6, and 7, obtaining a complicated rational number. But only 7 appears: there's no $x$ or $x^3$ term; the $x^4$ coefficient 5 kills the denominator of 5; and the terms $-4x^5-4x^2$ might have contributed denominators of 6 and 3 combine to yield the integer $-2$. Compare this with the next such integrals
$$
\int_0^1 (x-x^2)^6 \frac{dx}{1+x^2} = \frac{38429}{13860} - 4 \log 2
$$
and
$$
\int_0^1 (x-x^2)^8 \frac{dx}{1+x^2} = 4\pi - \frac{188684}{15015},
$$
which yield better but much more complicated approximations to $\log 2$ and $\pi$... This suggests a refinement of the "in what context" part of the question: Does that integral for $(22/7)-\pi$ generalize to give further approximations to $\pi$ (or $\log 2$ or similar constants) that are useful for the study of Diophantine properties of $\pi$ (or $\log 2$ etc.)?
|
Beukers mentions ( http://www.staff.science.uu.nl/~beuke106/Pi-artikel.ps ) that the integrals:
$$\int^{1}_{0} \frac{(x - x^2)^{4n}}{(1 + x^2)} dx$$
give approximations to $\pi$ of the form $p/q$ with $\displaystyle{\left| \pi - \frac{p}{q} \right| < \frac{1}{q^{\theta}}}$ with $\theta \rightarrow \log(4)/\log(2 e^8) = 0.738\ldots$
as $n$ goes to infinity. So these are not really arithmetically significant.
He also mentions that the integrals:
$$J_n = \int^{1}_{0} \frac{(x - x^2)^n}{(1+x^2)^{n+1}} dx$$
give approximations with $\theta \rightarrow 0.9058\ldots$ as $n$ goes to infinity. However, a further variation by Hata gives the integrals:
$$I_n = \int^{1}_{-1} \frac{x^{2n} (1 - x^2)^{2n}}{(1 + i x)^{3n+1}} dx,$$
with $\theta \rightarrow 1.0449\ldots$, giving an irrationality measure for $\pi$
and providing "explicit" rational approximations. Notes that
$I_1 = 14 \pi - 44$ gives the approximation $22/7$. (Beukers own integral proofs of the irrationality of $\pi^2$ and $\zeta(3)$ use somewhat different integrals.)
|
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67,387 |
Let $X$ and $Y$ be varieties. Let $E$ be a locally free sheaf over $X$. Let $f: X \to Y$. Is there some nice criteria which ensures that $f_\ast E$ is still locally free? Sorry, if this is a very standard question.
|
Under reasonable hypotheses on $X$ , $Y$ and $f$ , the answer is that $f_*E$ is locally free if and only if $\dim H^0(X_y, \, E_y)$ is a constant function, where $$X_y:=f^{-1}(y), \quad E_y:=E|_{X_y}.$$ More precisely, there is the following result, whose proof can be found in [Mumford, Abelian Varieties , Chapter II]: Theorem (Base Change). Let $f \colon X \to Y$ be a proper morphism of Noetherian schemes, with $Y$ reduced and connected, and $E$ a coherent sheaf on $X$ , flat over $Y$ . Then for all integers $p \geq 0$ the following conditions are equivalent: $\boldsymbol{(i)}$ $y \to \dim H^p(X_y, E_y)$ is a constant function; $\boldsymbol{(ii)}$ $F:=R^pf_*E$ is a locally free sheaf on $Y$ and, for all $y \in Y$ , the natural map $$F \otimes_{\mathcal{O}_Y} k(y) \to H^p(X_y, E_y)$$ is an isomorphism. For instance, if $f$ is a finite map and $E$ is a line bundle we obtain that $f_*E$ is a vector bundle with $\operatorname{rank} f_*E=\deg f$ , whereas $R^p f_*E=0$ for $p >0$ .
|
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|
67,436 |
Is there any sequence $a_n$ of nonnegative numbers for which $\displaystyle\sum_{n \geq 1}a_n^2 <\infty$ and $$\sum_{n \geq 1}\left(\sum_{k \geq 1}\frac{a_{kn}}{k}\right)^2=\infty\quad?$$ See also https://math.stackexchange.com/questions/42624/double-sum-miklos-schweitzer-2010
|
Yes, such sequences exist. In effect the problem concerns the linear operator, call it $T$, that maps any sequence $(a_n)$ to the sequence whose $n$-th term is $\sum_{k\geq1} a_{kn}/k$. The problem asks whether there exists $a$ such that the $l^2$ norm $\|a\|_2$ is finite but $\|Ta\|_2 = \infty$. We show that such $a$ exist for each $l^q$ norm ($q \geq 1$). [I can't call it $l^p$ because I'll soon want to use $p$ for a generic prime number.] The key is that $T$ can be regarded as an infinite tensor product (Kronecker product) of operators $T_p$, where for each prime $p$ the operator $T_p$ is convolution with $(1,p^{-1},p^{-2},p^{-3},\ldots)$ on $(a_1,a_p,a_{p^2},a_{p^3},\ldots)$. The operator $T_p$ has norm $\sum_{i=0}^\infty p^{-i} = p/(p-1)$, so $T = \otimes_p T_p$ should have norm $\prod_p p/(p-1) = \infty$. To get a construction from this, we first make for each $M>1$ a nonnegative sequence $\alpha = (\alpha_n)$ such that $\|\alpha\|_q < \infty$ and $T\alpha = M\alpha$. Since $\prod_p p/(p-1)$ diverges, we can find $x$ large enough that $\prod_{p\leq x} p/(p-1) > M$. Choose for each prime $p\leq x$ a positive number $z_p$ smaller than 1 but close enough to 1 that $\prod_{p\leq x} p/(p-z_p) = M$. Now let $\alpha$ be the totally multiplicative function taking each $p\leq x$ to $z_p$ and each $p>x$ to $0$. (That is, if $n=\prod_{p\leq x} p^{e_p}$ then $\alpha_n = \prod_{p \leq x} z_p^{e_p}$, while if $n$ has a prime factor bigger than $x$ than $\alpha_n = 0$.) Then the $n$-th term of $T\alpha$ is
$$
\sum_{k\geq 1} \frac{\alpha_{kn}}{k} = \sum_{k\geq 1} \frac{\alpha_k \alpha_n}{k}
= \alpha_n \sum_{k\geq 1} \frac{\alpha_k}{k}
$$
because $\alpha$ is multiplicative, and $\sum_{k\geq 1} \alpha_k/k$ is the product over $p\leq x$ of geometric series
$$
\sum_{e=0}^\infty \frac{\alpha_{p^e}}{p^e} = \sum_{e=0}^\infty (z_p/p)^e = \frac{p}{p-z_p},
$$
so
$$
T\alpha = \prod_{p\leq x} \frac{p}{p-z_p} \cdot \alpha = M \alpha;
$$
and $\|\alpha\|_q < \infty$ because $\|\alpha\|_q^q$ is the product of $\pi(x)$ convergent geometric series. The conclusion is a standard argument (as already noted in the parallel stackexchange thread). For example, let $a^{(M)} = \alpha / \|\alpha\|$ where $\alpha$ is a sequence with $T\alpha = M\alpha$ as in the last paragraph. Then $\|a^{(M)}\| = 1$ and $Ta^{(M)} = M a^{(M)}$. Set
$$
a = \sum_{m=1}^\infty \frac{a^{(4^m)}}{2^m} = \frac{a^{(4)}}{2} + \frac{a^{(16)}}{4} + \frac{a^{(64)}}{8} + \cdots .
$$
The $m$-th term has norm $1/2^m$, so the sum converges in $l^q$ to a vector of norm at most $\sum_{m=1}^\infty 1/2^m = 1$. But since each term $a^{(4^m)} / 2^m$ is a nonnegative sequence, and $T$ has nonnegative coefficients, we have for each $m \geq 1$
$$
Ta > T\frac{a^{(4^m)}}{2^m} = 2^m a^{(4^m)};
$$
and $2^m a^{(4^m)}$ is a vector of norm $2^m$. Hence $\|Ta\|_q = \infty$, QED .
|
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67,458 |
In an appendix to his notes on intersection homology and perverse sheaves, MacPherson writes Why do we want to consider only spaces $V$ that admit a decomposition into manifolds? The intuitive answer is found by considering the group of all self homeomorphisms of $V$. Certainly if $V$ is to be of "finite topological type", then this group should have finitely many orbits. It is these orbits that should be the natural strata of $V$. That the orbits of this group should be manifolds results from the meta-mathematical pricinple that a space of "finite topological type" whose group of self-homeomorphisms acts transitively must be a manifold. I don't know a precise mathematical statement that realizes this meta-mathematical principle, but I expect that there is one. As these notes are dated from 1990, I was wondering if the past twenty years have seen any work done towards a precise formulation of this meta-mathematical principle.
|
A precise version of this statement is the Bing-Borsuk conjecture
that a homogeneous ENR is a manifold. Here is a recent survey,
generally in the direction of my answer.
There is a candidate counter-example, due to Bryant, Ferry,
Mio, and Weinberger, but they can't show it is homogeneous.
Some people think that these generalized
manifolds are very nice (especially if it does turn out that they're homogeneous) and
one should just weaken statements like MacPherson's to allow them.
If you throw on enough hypotheses, the BFMW machinery applies.
This uses surgery theory
and so is limited to high dimensions.
The question is wide open in dimension 3, where
it implies the Poincare conjecture. I think of the Euclidean neighborhood retract (ENR) condition as a
finiteness hypothesis. It is that the space is an
absolute neighborhood retract that embeds in Euclidean space;
equivalently a retract of an open subset of Euclidean space.
This rules out the Cantor set because a neighborhood can have only
countably many components, while the Cantor set has uncountably
many.
This condition implies that the space is homotopy equivalent to a
finite dimensional CW complex, but it imposes a lot of
tameness on the topology as well.
This is a local condition and does not rule out the integers or
the infinite genus surface.
If one wants to impose such a global finiteness, one can require
that the one-point compactification also be an ENR. Following things like Bing's proof of Kline's characterization of
the 2-sphere ,
by being disconnected by all embedded circles, but no pairs of
points, there were attempts to characterize $n$ -manifolds by
separation conditions, such as homology.
A manifold is locally a disk, which is the cone on a sphere.
Given a space and a point, one can define the homotopy link of
that point, which would be the base of the cone, if the space
were locally a cone. The local homology $H_k(X,X-\{x\})$ is the
homology of the link. If these groups are the homology of spheres
and they form a local system, the space is called a homology
manifold.
To be a manifold, the links must be simply connected.
The disjoint disks property implies this and it was conjectured
that a homology manifold with the disjoint disks property is a
manifold.
This implies the shocking Cannon-Edwards theorem that the double suspension of the Poincare homology three-sphere
is a topological manifold, even though it is not homogeneous
under piecewise linear maps (or bi-Lipschitz maps). Bryant, Ferry, Mio, and Weinberger produced counterexamples that
were not manifolds, but they showed that the obstruction was a
local invariant and that these spaces were amenable to surgery
theory and classified them up to s-cobordism.
They conjectured that these generalized manifolds are homogeneous
and that s-cobordism implies homeomorphism, as with manifolds. Bredon and later Bryant showed that if the local homologies of a homogeneous ENR are
finitely generated, the space is a homology manifold. This sounds like a pretty tame finiteness assumption. More recently , Bryant
achieved the homology manifold conclusion by strengthening the
hypothesis from homogeneity to arc-homogeneity.
To get from homology manifold to the BFMW generalized manifolds,
one needs a hypothesis like the disjoint disks property, but I don't think
anyone knows how to get this from homogeneity.
|
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|
67,903 |
Let $X$ be a complex manifold and $g$ a hermitian metric on $X$. Consider the Riemannian exponential $\exp_p: T_p X \to X$. If $\exp_p$ is holomorphic for every $p \in X$, then $(\exp_p)^{-1}$, suitably restricted, provide holomorphic normal coordinates near $p$, with respect to which the metric osculates to order 2 to the standard metric at the origin. This shows that $g$ is a Kähler metric. However, Kähler is not sufficient to ensure that $\exp_p$ is holomorphic: take $X$ a curve of genus $g \geq 2$. If $\exp_p:T_pX \to X$ is holomorphic, then it lifts to a holomorphic map from $T_pX$ to the universal cover $\widetilde{X} = \Delta$, giving a holomorphic map $T_pX \simeq \mathbb{C} \to \Delta$, which must be constant by Liouville's theorem. In fact, one can see that $\exp$ cannot be holomorphic if $X$ is Kobayashi hyperbolic. This leaves the question: What are the hermitian manifolds/metrics whose exponential map is holomorphic?
|
NB: I've had a little time to think about this and can now improve my answer, in particular, removing the real-analytic assumption, which, as I suspected, was not necessary. Here is the improved answer: If the metric $g$ is Kähler, then having the exponential map from a point $p\in M$ be holomorphic makes it flat in a neighborhood of $p$. Suppose that $\exp_p:T_pM\to M$ is holomorphic near $0_p\in T_pM$ (where we use the natural holomorphic structure on the complex vector space $T_pM$). Let $z:T_pM\to\mathbb{C}^n$ be a complex linear isometry, so that the hermitian metric on $T_pM$ is just $|z|^2$ in the usual sense. Let $Z$ be the holomorphic 'radial' vector field on $\mathbb{C}^n$, whose real part is the standard radial vector field on $\mathbb{C}^n$. Then
$$
{\exp_p}^*g = g_{i\bar j}(z)\ dz^i\ d\overline{z}^j
$$
for some functions $g_{i\bar j}$ on a neighborhood of $0\in\mathbb{C}^n$. Since $g$ is Kähler, there is a function $f$ defined on a neighborhood of $0\in\mathbb{C}^n$ such that
$$
g_{i\bar j} = \frac{\partial^2f}{\partial z^i\ \partial\overline{z}^j}.
$$ Now, the condition that $z$ furnish Gauss normal coordinates for ${\exp_p}^*g$ is easily seen to be that $$
\mathcal{L}_Z\bigl(\bar\partial f\bigr) = \bar\partial\bigl(|z|^2\bigr).
$$
In particular, $ \bar\partial\bigl(\mathcal{L}_Z(f - |z|^2)\bigr) = 0$,
so $\mathcal{L}_Z(f - |z|^2) = h$ for some holomorphic function $h$ on a neighborhood of $0$. This $h$ must vanish at $0$, so it is easy, by adding the real part of the appropriate holomorphic function to $f$ (which won't change $g$) to arrange that $h\equiv0$ and, moreover, that $f(0) = 0$. But this now implies that the real-valued function $f-|z|^2$ vanishes at the origin and also is constant along the radial vector field. Thus, $f = |z|^2$, and the metric $g$ is flat in these coordinates.
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|
68,081 |
I know this is not the forum for this particular question, but the majority of users are immersed in the environment. I want to study differential geometry, but I need to know what courses would help guide me in that direction. I enjoy and/or also want to study the following as well: abstract algebra, real analysis, partial differential equations, complex analysis, functional analysis, and topology. Which of these topics should I study when I start next year? Thanks
|
The sociological and metamathematical aspects of this question are too often overlooked, I think. First, undergrad or grad students' discussions with their peers too often subtly veers into a "Lord of the Flies" scenario. Second, many "advisors" (whether undergrad or grad), have some weaknesses in communications skills and in perception of others' non-verbal expression (e.g., affect). Nevertheless, yes, one should talk to faculty quite a lot (even if taking remarks with a grain of salt). Now some objections: the labels of "subjects" or "specialties", while seemingly sanctioned or even mandated by the AMS subject classification, by faculty "research descriptions", and grad students' desires to taxonify ambient activity, are innately misleading. There are no clear separators... except those artificially imposed. True, the "requirements" have such labels, and "everyone" speaks in terms of them, and... yes... one can live one's whole professional life speaking in those terms, ... but this partitioning is fundamentally invidious. The next objection is that it is usually very difficulty to understand the significance of things until one sees how they're used "in the sequel". Thus, a misguided fixation on "mastery" at an entry level really is misguided, in that one is doing exercises without a notion of real-life activity. The utility of things is not well-illustrated by contrived (a.k.a., "textbook") exercises. These remarks are all cliches, but perhaps bear repeating... Edits: Seeing the responses, I'd like to add clarifications. First, one should not depend on coursework for learning mathematics, especially not for seeing how it is done in real life. One must learn more-and-different things than what the traditional curriculum promotes, no matter which courses one signs up for. One special corruption is the usual convention of assigning piles of weekly homework, exams, grades... leaving people little time or energy to think critically about anything, and confusing compliance with scholarship. (Observe: in mathematics, apparently one is not permitted to question the goodness of course content, insofar as grading systems reward obedient technical responses rather than critiques.) Far more important is awareness of things, of their utility, of their interactions with other things. Earliest-possible awareness of as many ideas as possible is highly desirable, whether or not piles of exercises are completed. Thus, it is desirable to "look at everything", and obviously this can't happen via coursework. It is desirable to witness the actual practice of mathematics, thereby to be aware how different doing mathematics is from doing homework or exams or contest problems. Seminars sometimes represent this, although often they amount to reports or job talks. Regular conversations with faculty about mathematics, not about coursework, surely cultivates a more useful outlook than any amount of coursework.
|
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68,145 |
All the statements below are considered over local rings, so by regular, I mean a regular local ring and so on; It is well-known that every regular ring is Gorenstein and every Gorenstein ring is Cohen-Macaulay. There are some examples to demonstrate that the converse of the above statements do not hold. For example, $A=k[[x,y,z]]/(x^2-y^2, y^2-z^2, xy, yz, xz)$ where $k$ is a field, is Gorenstein but not regular, or $k[[x^3, x^5, x^7]]$ is C.M. but not Gorenstein.
Now, here is my question: I want to know where these examples have come from, I mean, have they been created by the existence of some logical translations to the Algebraic combinatorics (like Stanley did), or even algebraic geometry, or they are as they are and they are some kind of lights that have been descended from heaven to their creators by any reason!
|
I will argue that the examples you gave are "simplest" in some strong sense, so although they look unnatural, if Martians study commutative algebra they will have to come up with them at some point. Let's look at the first one $A=k[[x,y,z]]/(x^2-y^2, y^2-z^2, xy,yz,zx)$. Suppose you want a $0$-dimensional Gorenstein ring which is not a complete intersection (complete intersections are the cheapest way to get Gorenstein but non-regular, for example $k[[x]]/(x^2)$). Then it would look like $A=R/I$, where $R$ is regular and $I$ is of height equals to $\dim R$. If $\dim R=1$ or $2$, then $I$ would have to be a complete intersection, no good (now, the poor Martian may not know this at the begining, but after trying for so long she will have to give up and move on to higher dimension, or prove that result for herself). Thus $\dim R=3$ at least, and we may assume $R=k[[x,y,z]]$ (let's suppose everything are complete and contains a fied). Since $I$ is not a complete intersection, it must have more than $3$ minimal generators. If it has $4$ then it would be an almost complete intersection, and by an amazing result by Kunz (see this answer ), those are never Gorenstein. So, in summary, a simplest $0$-dimensional Gorenstein but not complete intersection would have to be $k[[x,y,z]]/I$, where $I$ is generated by at least $5$ generators. At this point our Martian would just play with the simplest non-degenerate generators: quadrics, and got lucky! (You can look at this from other point of view, Macaulay inverse system or Pfaffians of alternating matrices, see Bruns-Herzog, but because of the above reasons all the simplest examples would be more or less the same, up to some linear change of variables) On to your second example, $k[[t^3,t^5,t^7]]$. Now very reasonably, our Martian wants a one dimensional Cohen-Macaulay ring B that is not Gorenstein. Since $\dim B=1$, being a domain would automatically make it Cohen-Macaulay. The most natural way to make one dimensional domains is to use monomial curves, so $B=k[[t^{a_1},...,t^{a_n}]]$. But if $n=2$ or $n=3$ and $a_1=2$ you will run into complete intersections, so $a_1$ would have to be $3$ and $n=3$ at least, and on our Martian goes... (Again, you can arrive at this example by looking at things like non-symmetric numerical semi-groups, but again you would end up at the same simplest thing). Reference: I would recommend this survey for a nice reading on Gorenstein rings.
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68,208 |
My question is somewhat similar to this previous question , but from a slightly different perspective. Is there any textbook on elementary number theory that develops the properties of $\mathbb{Z}$ as, say, the initial object in the category of commutative rings with identity? I am looking for something that presupposes a knowledge of category theory at the level of Categories for the Working Mathematician. Edit: I had no idea that this question would provoke the storm of criticism that is has. My intention was not to imply that number theory is best learned from a categorical perspective, or that number theory should be subsumed by category theory. I was simply wondering what sort of interesting things one could say about $\mathbb{Z}$ from a category-theoretic perspective. So, I'll narrow the question: "Are there any good sources for learning about the properties of a natural numbers object in an arbitrary topos (possibly well-pointed and satisfying the axiom of choice)?"
|
I have to admit that this is not really an answer, but rather some sort of meta-answer with some very general remarks which I hope do not bore everyone reading this; it just seems to me that this is necessary to indicate that it is rather misguided, as Yemon already says in the comments and I strongly agree with, to ask such a question if some book introduces elementary number by means of category theory. Mathematics is all about the nontrivial, unexpected relationships. Category Theory is not really about finding such relationships, but rather about the correct setting, language and color some theory is developed. This point of view does not really contradict the hitherto development of category theory into a huge area of mathematics in its own right, full of nontrivial deep theorems; namely because often there is some geometric or whatever background which is our real motiviation. There are ubiquitous examples (model categories, topoi, stacks, $\infty$-categories, ...) which I don't want to elaborate here. Anyway, as I said, mathematics really starts when something unexpected happens, which does not follow from general category theory. For example, the covariant functor $\hom(X,-)$ is always continuous, but when is it also cocontinuous, or respects at least filtered colimits? It turns out that this leads to a natural finiteness condition on $X$, namely we call $X$ then finitely presented. But finally to arrive at the question, $\mathbb{Z}$ is easily seen to be a inital object in the category of rings, but what theorems from category theory are known about initial objects? Well there is nothing to say, expect that every two initial objects are canonical isomorphic, which is just a trivial consequence of the definition. So $\hom(\mathbb{Z},-)$ is easy to describe, but what about the contravariant functor $\hom(-,\mathbb{Z})$? What happens when you plug in $\mathbb{Z}[x,y,z]/(x^n+y^n=z^n)$ for some fixed $n>2$? Does category theory help you to understand this? This example also shows that although the Yoneda-Lemma says that an object $X$ of a category is determined by its functor $\hom(X,-)$, it does not say you anything about the relationship of $X$ with other objects, for example when we just reverse the arrows. Instead, we have to use a specific incarnation of the category and its objects in order derive something which was not there just by abstract nonsense. Perhaps related questions are more interesting: Which investigations in elementary number theory have led to some category theory (for example, via categorification), which was then applied to other categories as well, thus establishing nontrivial analogies? Or for the other direction, which general concepts become interesting in elementary number theory after some process of decategorification? But in any case, it should be understood that you have to digest elementary number theory before that ...
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68,335 |
There is a theorem of Deligne in SGA4 that a "coherent" topos (e.g. one on a site where all objects are quasi-compact and quasi-separated) has enough points (i.e. isomorphisms can be detected via geometric morphisms to the topos of sets). I've heard it said that this is a form of Goedel's completeness theorem for first-order logic. Why is that? I get the intuition that having a model for a formula is supposed to be analogous to a point of a suitable topos, but this is very vague. I'm sorry for not providing more motivation, but I don't know enough about this connection to do so! (This was first posted on math.SE here , where it did not (yet) receive a response.)
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They are indeed formally equivalent. See for instance Johnstone: Topos theory, p. 243 but here is a quick explanation. Given a topos $T$ one may define a geometric theory associated to it consisting of formulas describing essentially the topos. More specifically a geometric morphism from another topos $S$ to $T$ is the same thing as a model for the theory in $S$. In particular if $S$ is the category of sets this a set-theoretical model fo the theory. In general the language of the theory has arbitrary disjunctions leading to theories that in general do not have models. However, if the topos is coherent only finite disjunctions are needed and we are in the realm of the Gödel completeness theorem which then can be interpreted as saying that a coherent topos has enough points. Conversely, given a geometric theory one can associate to it a syntactic site whose objects are the formulas. An implication from a disjunction of formulas to a formula is a covering. The topos of sheaves on this site will then be a classifying topos for the theory (i.e., geometric morphisms to it are the same as models). If the theory is finitary (i.e., uses only finite disjunctions) then the topology is coherent and there are models for the theory by Deligne's theorem. It is amusing that Deligne's fairly natural example of a topos without points, "measure sheaves" on a measure space for which all points have measure zero thus gives an example of a consistent geometric theory (with countable disjunctions) that doesn't have a model.
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68,339 |
Let $R$ be a commutative ring. For awhile I have been trying to motivate to myself more fully the definition of and various structures on the category $\text{Ch}(R)$ of chain complexes of $R$-modules (and various subcategories thereof). One significant piece of motivation is the Dold-Kan correspondence , at least when $R = \mathbb{Z}$, which tells us that studying connective chain complexes is like studying linearized homotopy theory (or linearized higher category theory). This is a great idea, but I don't have much intuition for what's going on in the proof of Dold-Kan, and I don't see how one could have predicted in advance that something like Dold-Kan might be true just by looking at all the definitions in the right way. I like the idea of linearized homotopy but I don't know what the conceptual path is from linearized homotopy to, for example, the braiding $a \otimes b \mapsto (-1)^{|a| |b|} b \otimes a$. Consider also the differential. I can think of various ways to motivate $d^2 = 0$, and I don't quite know how they fit together. For example, one can talk about boundaries of manifolds with boundary, the exterior derivative, and Stokes' theorem. If one starts from the simplicial / higher-categorical perspective, the differential encodes something like the generalized source / target of a higher morphism, and somehow the fact that this generalized source / target ought to satisfy a natural "gluing law" (for example if $a \to b$ is a $1$-morphism then $d(a \to b + b \to c)$ ought to equal $d(a \to c)$) is equivalent to it squaring to zero. I can sort of picture how this works in low dimensions but I don't completely grasp what the exact relationship between these two ideas is. Keeping in mind the symmetric monoidal structure, the differential behaves like an element of a super Lie algebra, concentrated in degree $-1$, acting on a representation (see for example Theo Johnson-Freyd's MO answer here ). The action of a super Lie algebra should be related to infinitesimal symmetry coming from a super Lie group, but I don't have a clear idea of what this super Lie group is or what it has to do with homotopy theory. This seems to have something to do with the supergeometric definition of differential forms, but I don't really know anything about this. Algebraically, the relation $d^2 = 0$ seems to come from at least two different ideas: first-order approximation, and odd things anti-commuting with themselves. Both of these ideas seem relevant to what I'm confused about, but I can't put them together into a cohesive story. So what is that cohesive story? Edit: If the bulk of the question seems sort of silly to you, feel free to focus on that last bit about super Lie algebras. I remember hearing that this has something to do with the action of the automorphism group of the odd real line; I would appreciate if someone could clarify that for me.
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$ d^2 $ and homology are almost the most basic and natural quests in mathematics. In general, the image of a map can be thought of as "things that can be constructed". In general, the kernel of a map is "things we can test". So $ \ker d = \operatorname{im }d$ is one of the basic quests in mathematics, linearized: "let us find a construction for all the things that satisfy a certain criterion". Likewise, given a construction for things that satisfy a certain property (name, given $A\stackrel{d_1}{\longrightarrow}B\stackrel{d_2}{\longrightarrow}C$, where $d_1$ is the construction and $d_2=0$ is the property), the homology $\ker d_2/\operatorname{im }d_1$ measures how successful you had been - to what extent you were able to construct all the things you wanted to construct. Again a very basic mathematical quest. So two-step complexes, $d^2=0$, and homology are as natural as anything. What puzzles me is that way too often two-step complexes have a natural extension to become many-step complexes. I have no good philosophical explanation for why that should be the case.
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68,378 |
This should be a very easy question, but the proof in Lawson/Michelson (Spin geometry) is wrong and I do not find a really correct and complete argument: Let V be a nonzero real vector space with scalar product: Why is the Clifford-algebra (constructed from the tensor algebra by quotiening out an ideal) non-zero?
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To show that an algebra constructed as a quotient of the tensor algebra of a vector space is nonzero, one of the main ways to go is to construct representations. We can do this for the Clifford algebra as follows. Let $V$ be a vector space over a field $k$ and $(,):V\times V \to k$ a symmetric bilinear form on $V$. The Clifford algebra (for this form) is given by
$$Cl(V)= T(V)/\langle v \otimes v - (v,v)\rangle.$$
We will construct a representation of the Clifford algebra on the exterior algebra $\bigwedge (V)$. For $v \in V$, define two $k$-endomorphisms of $\bigwedge(V)$ by
$$ l_v(x) = v \wedge x$$
and
$$ \delta_v(x) = \sum_{j=1}^k (-1)^{j-1}(v,x_j) x_1 \wedge \dots \wedge \widehat{x_j} \wedge \dots \wedge x_k$$
if $x = x_1 \wedge \dots \wedge x_k$. Then check that $l_v^2 = \delta_v^2 = 0$, and moreover that $l_v \delta_v + \delta_v l_v = (v,v) \cdot \mathrm{id}$. Extend the map linear $v \mapsto l_v + \delta_v$ to an algebra homomorphism from the tensor algebra $T(V)$ to $\mathrm{End}_k(\bigwedge(V))$. By the previous remark, this descends to a map, let's call it $\phi$, from the Clifford algebra to $\mathrm{End}_k(\bigwedge(V))$. In particular, $\phi(v)1 = v$, so $V$ injects into the Clifford algebra. Edit: I believe also that the map
$$ x \mapsto \phi(x)1$$
gives a linear isomorphism of the Clifford algebra with the exterior algebra. A great reference for this stuff is Chevalley's monograph, The Algebraic Theory of Clifford Algebras and Spinors.
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68,421 |
If they are not proper, two complex algebraic varieties can be nonisomorphic yet have isomorphic analytifications. I've heard informal examples (often involving moduli spaces), but am not sure of the references. What are the simplest examples of nonisomorphic complex algebraic varieties with isomorphic analytificaitons? By "simplest", I mean by one of the following measures. (best) An example whose proof is as elementary as possible, and ideally short. This of course requires proof that the complex algebraic varieties are nonisomorphic, and that the analytifications are isomorphic. A known example that is simple to state, but may have a complicated proof. (Ideally there should be a reference.) An expected, folklore, or conjectured example.
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I believe the following is an elementary example: Let $X$ be an affine smooth curve of geometric genus at least one. Let $L$ be a non-trivial algebraic line bundle
on $X$ (easy to produce such things). Then $L$ is analytically trivial because $X$ is a Stein space ($H^1(X, O_X)=0$) with trivial integral $H^2$. Hence, there is an
analytic isomorphism $L\simeq X\times A^1$. We see that there is no algebraic isomorphism
by noting: Suppose there is an isomorphism $$f: L\simeq X\times A^1$$
of algebraic varieties. Then $L$ and $X\times A^1$ are isomorphic as algebraic line bundles. Proof: For any fiber $L_y$ of $L$, if we consider the composite $$p\circ f: L_y\rightarrow X\times A^1 \stackrel{p}{\rightarrow} X$$
of $f$ with the projection, it must be constant, since $X$ is not rational. Hence, we have
$$f(L_y)\subset z\times A^1$$ for some point $z$. Since the map $L_y\rightarrow A^1$ thus obtained is injective,
it must be of the form $ax+b$ for non-zero $a$, that is, $f$ induces an isomorphism
$$L_y\simeq z\times A^1.$$ Also by injectivity, we see that $y\neq y'$ implies
$f(L_y)=z\times A^1$ and $f(L_{y'})=z'\times A^1$ for $z\neq z'$.
Now let $s:X\rightarrow L$ be the zero section. Then $$\phi=p\circ f\circ s: X\rightarrow X$$ is an injective map, and hence,
an automorphism. Thus, $$(\phi^{-1}\times 1)\circ f:L\rightarrow X\times A^1 \stackrel{\phi^{-1}\times 1}{\rightarrow} X\times A^1$$ is a map preserving the fibers of the projections to $X$. Now let $$q:X\times A^1 \rightarrow A^1$$
be the other projection, and $$h=q\circ (\phi^{-1}\times 1)\circ f\circ s.$$ So $h(y)$ is the image in $A^1$ under $(\phi^{-1}\times 1)\circ f$ of the origin of
$L_y$. We use this function to get a fiber-preserving isomorphism $g: X\times A^1\simeq X\times A^1$ that sends
$(y,\lambda )$ to $(y, \lambda-h(y))$. So finally, $$g\circ (\phi^{-1}\times 1)\circ f: L\simeq X\times A^1$$ is a fiber-preserving isomorphism of
varieties that furthermore preserves the origins of each fiber. It must then be fiber-wise
an isomorphism of vector spaces. Thus, it is an isomorphism of line bundles. Added: The argument above can be easily modified to show that if $X$ is an irrational smooth curve and $L$ and $M$ are line bundles on $X$, then any isomorphism of algebraic varieties $$f:L\simeq M$$ is of the form
$$f=T_s\circ \tilde{\phi} \circ g$$
where $$\tilde{\phi}:\phi^*M\rightarrow M$$ is the base-change map for an automorphism $\phi$ of $X$, $$g:L\simeq \phi^*M$$ is an isomorphism of line bundles, and $$T_s:M\rightarrow M$$ is translation by a section $s:X\rightarrow M$ of $M$. Since the algebraic automorphism group of an affine irrational curve is finite, we see, by varying $L$, that for $X$ as above, there is in fact a continuum of distinct algebraic structures on the analytic space $X\times A^1$.
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68,436 |
On page 98 of Weibel's An Introduction to Homological Algebra he mentions that the ring $R = \prod_{i=1}^\infty \mathbb{C}$ has global dimension $\geq 2$ with equality iff the continuum hypothesis holds. He doesn't give any clue as to the proof of this fact or why the continuum hypothesis got involved. On page 92 he mentions some examples of Osofsky and says the continuum hypothesis gets involved there because of non-constructible ideals over uncountable rings. I think this explains at least the "why" of the appearance of the continuum hypothesis (though I would welcome more details on this!), but it leaves me with some other questions: How is the continuum hypothesis used in this proof? Why wouldn't the proof work without the continuum hypothesis? I will understand if the above have to do with some work of Osofsky that is not widely known. If I can't get answers for those questions, perhaps I can still get help on the below. I got involved with this because I wanted to understand an example of a ring that is von Neumann regular but not semisimple (and an infinite product of fields is such an example). I had hoped all such examples would have weak dimension zero (to be VNR) and right global dimension 1. In particular, I wanted to know that the global dimension of $A = \prod_{i=1}^\infty \mathbb{F}_2$ was $1$ . According to this MO answer and its comments, $Spec A$ is the Stone-Cech compactification of $\mathbb{N}$ . Now I'm concerned that things from set theory which I try to avoid thinking about will come into play in this example as well as in the above ring $R$ . What is the global dimension of $\prod_{i=1}^\infty \mathbb{F}_2$ ? Do we need to assume the continuum hypothesis at any point? What about an uncountable product of $\mathbb{F}_2$ ?
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In [Osofsky, B. L. Homological dimension and cardinality. Trans. Amer. Math. Soc. 151 1970 641--649. MR0265411 (42 #321)] she proved that the global dimension of a countable product of fields is $k+1$ iff $2^{\aleph_0}=\aleph_{k}$. In particular, if the continuum hypothesis holds, so that $2^{\aleph_0}=\aleph_1$, the global dimension of such a product is exactly $2$. Because the AMS is nice, you can see the paper here .
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68,437 |
The divisor bound asserts that for a large (rational) integer $n \in {\bf Z}$, the number of divisors of $n$ is at most $n^{o(1)}$ as $n \to \infty$. It is not difficult to prove this bound using the fundamental theorem of arithmetic and some elementary analysis. My question regards what happens if ${\bf Z}$ is replaced by the ring of integers in some other number field. For sake of concreteness let us work with the simple extension ${\bf Z}[\alpha]$, where $\alpha$ is some fixed algebraic integer. Of course, one may now have infinitely many units in this ring, but if we restrict the height then it appears that we have a meaningful question, namely: Question: Let $n \in {\bf Z}[\alpha]$ be of height $O(H)$ (by which I mean that $n$ is a polynomial in $\alpha$ with rational integer coefficients of size $O(H)$ and degree $O(1)$). Is it true that the number of elements of ${\bf Z}[\alpha]$ of height $O(H)$ that divide $n$ is at most $H^{o(1)}$? Here $o(1)$ denotes a quantity that goes to zero as $H \to \infty$, holding $\alpha$ fixed. (Actually, for my applications I would like $\alpha$ to not be fixed, but to have a minimal polynomial of bounded degree and coefficients of polynomial size in $H$, but for simplicity let me stick to the fixed $\alpha$ question first.) It is tempting to take norms and apply the divisor bound to the norm, but then I end up needing to bound the number of elements in ${\bf Z}[\alpha]$ with a given norm and of controlled height, and I don't know how to do that except for quadratic extensions. A related problem comes up if one tries to exploit unique factorization of ideals to answer this problem. (On the other hand, it appears to me from the Dirichlet unit theorem that the number of units of height $O(H)$ is at most polylogarithmic in H, so the unit problem at least should go away.)
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As long as you allow a fixed number field $F = {\bf Q}(\alpha)$ you can prove $H^{o(1)}$ as you more-or-less suggest towards the end, by first showing that the number of ideals of $F$ that divide $n$ is $H^{o(1)}$ and then proving that any ideal has $O(\log^r H)$ generators of height at most $H$, where $r = r_1 + r_2 - 1$ is the rank of the unit group $U_F$ of $F$. The first part is basically the same as the argument over $\bf Z$. If the ideal $(n)$ factors into prime powers as $\prod_i \wp_i^{e_i}$ then there are $\prod_i (e_i + 1)$ ideals that divide $n$. Given $\epsilon > 0$ there are finitely many choices of rational prime $p$ and integer $e>0$ such that $e+1 > (p^e)^\epsilon$, and therefore only finitely many choices of a prime $\wp$ of $F$ and $e>0$ such that $e+1$ exceeds the $\epsilon$ power of the norm of $\wp^e$. Therefore $\prod_i \wp_i^{e_i} \ll_\epsilon N^\epsilon$ where $N$ is the absolute value of the norm of $n$. But $N \ll H^{[F : {\bf Q}]}$, and $\epsilon$ was arbitrary, so we've proved the $H^{o(1)}$ bound. For the second part, Dirichlet gives a logarithm map $U_F \rightarrow {\bf R}^r$ whose kernel is finite (the roots of unity in $F$) and whose image is a lattice $L$. A unit of height at most $H$ has all its conjugates of size $O(H)$, so is mapped to a ball of radius $\log(H) + O(1)$. Therefore there are $O(\log^r(H))$ units of height at most $H$. Much the same argument (involving a translate of $L$) shows that he same bound applies also to generators of any ideal $I$, since the ratio of any two generators of $I$ is a unit. [I see that "Pitt the Elder" just gave much the same answer.]
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68,442 |
This is a soft question but it's not meant as a big-list question. I have recently been asked whether I want to provide feedback at the pre-beta stage on a forthcoming website that will provide a platform for data sharing, and rather than giving just my personal opinion I'd rather consult other mathematicians first. I was going to write a blog post but then I thought that Mathoverflow was a more suitable place since I have a question and I'm looking for answers of a certain type rather than general comments. The website seems to be aimed mostly at scientists who want to share raw data, so at first I thought it probably wouldn't be much use to mathematicians since our data is (or are if you prefer) mostly highly interlinked -- the connections are often more interesting than what they connect. But on further reflection, it seems to me that a good data sharing site could be a valuable resource, even if it doesn't do absolutely everything any mathematician would ever want. For instance, Sloane's database is fantastically useful. A rather different sort of database that is also useful is Scott Aaronson's Complexity Zoo. So useful databases exist already. Is this an aspect of mathematical life that could be greatly expanded given the right platform? And if so, what should the platform be like? I don't know anything about the design of the site, but if I'm going to comment intelligently on what features it would need to have to be useful to mathematicians, I'd like to be armed with some examples of the kind of data sharing we might actually go in for. Here are a few ideas off the top of my head. Diophantine equations: one could have a list of what is known about various different ones. Mathematical problems: listed in some nice categorized way, each problem accompanied by a description, complete with reading list, of what you really ought to know before thinking about the problem. (As an example, if you are thinking about the P versus NP problem, then you really ought to know about the Razborov/Rudich natural proofs paper.) Key examples in various different areas and subareas of mathematics. Sometimes you have a whole lot of related mathematical properties with a complicated pattern of implications between them. Under such circumstances, it could be nice to have this information presented in a nice graphical way (something I think this site may be able to do well -- they seem to be keen on visualization) with links to proofs of the implications or counterexamples that demonstrate when the implications do not hold. (The example I'm thinking of while writing this is different forms of the approximation property for Banach spaces, but there are presumably several others.) List of special functions and the facts about each one that are the main facts one uses to prove things about them. List of integrals that can be evaluated, with descriptions of how they can be evaluated. List of important irrational numbers with their decimal expansions to vast numbers of places. (I'm not sure why this would be useful but it might be amusing.) These are supposed to be examples where people could usefully pool the background knowledge that they pick up while doing research. I'm not particularly pleased with them: they should be thought of as a challenge to come up with better ones, which almost certainly exist. If you've ever thought, "Wouldn't it be nice if there's somewhere where I could look up X," then X would make a great answer. I think the most interesting answers would be research-level answers (unlike some of the suggestions above). If there were a site with a lot of databases, it would make a great place to browse: it would be much easier to find useful data there than if it was scattered all round the internet. One constraint on answers: there should be something about a suggested database that makes it unsuitable for Wikipedia, since otherwise putting it on Wikipedia would appear to be more sensible.
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It would be really useful to have a database containing various computations of (co)homology/homotopy groups of various spaces that arise in algebraic topology... Note: There is so much known out there that one would have to first think really hard about how to organize it all. Here's an example: I could imagine that, for certain users, listing the first 30 integral cohomology groups of the spaces $K(\mathbb Z,1)$, $K(\mathbb Z,2)$, $K(\mathbb Z,3)$, and $K(\mathbb Z,4)$
could be more useful¹ than listing all the cohomology groups of all the $K(\mathbb Z,n)$'s. The reason is that, in order to do the latter, the information has to be packaged in a certain way that might be hard to understand: the user would need to unpack that information before she can access it. ¹ Of course, it's even better to have both pieces of information available.
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68,465 |
Early this year, I started to learn about p-adic modular forms. Very recently, a mathematician tells me Emerton constructed an object called completed cohomology group with very rich structure, and the author could use it to prove fantastic results about Galois representations. (see Emerton's paper "Local-Global Compatibility In The p-adic Langlands Programme For $GL_{2/\Bbb{Q}}$". In page 45, he defines the huge $\hat{H}^{1}_{A}$) With little understanding of it, can I ask if we can have a "natural" embedding of the space of Katz generalized p-adic modular functions (as defined in Section I.3 of Gouvea's Arithmetic of p-adic Modular Forms) into Emerton's completed cohomology group, assuming the base ring is $\Bbb {Z}_p$? If so, can we describe the image? This would be some type of Eichler-Shimura theorem for Katz generalized p-adic modular functions. Could this be one of the motivations when he defined such completed cohomology group?
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You already have two helpful answers related to general aspects of Eichler--Shimura isomorphisms in a $p$-adic context. Here is an answer that more directly addresses your original question. I will begin by recalling/stating some facts on the $p$-adic modular form side: Fix a tame (i.e. prime-to-$p$) level $N$, and let $\mathbb T(N)$ be the completed Hecke algebra generated by the $S_{\ell}$ and $T_{\ell}$ for $\ell \nmid N p$ acting on Katz's space $V(N)$ of generalized $p$-adic modular fuctions of level $N$. Fix a maximal ideal $\mathfrak m$ in $\mathbb T(N)$, and let $V(N)_{\mathfrak m}$ be the localization of Katz's space at the maximal ideal $\mathfrak m$. In fact, in order to deal sensibly with oldforms at $N$, I find it helpful to do the
following: fix the finite set of primes $\ell_1,\ldots,\ell_n$ dividing $N$, and take
a direct limit of $V(N)_{\mathfrak m}$ as $N$ ranges over all levels divisible by
just $\ell_1,\ldots,\ell_n$. Note that $\mathbb T_{\mathfrak m}$ may grow as $N$ increases (if $N$ divides $N'$ then $\mathbb T(N)\_{\mathfrak m}$ is a quotient of $\mathbb
T(N')\_{\mathfrak m}$), but eventually stabilizes (even thought the $V(N)_{\mathfrak m}$ don't stabilize), because if $\rho$ is any lift of
the Galois representation $\overline{\rho}$ attached to $\mathfrak m$ then the difference between the prime-to-$p$ conductor of $\rho$ and $\overline{\rho}$ is bounded. So now let $V_{\mathfrak m}$ be this direct limit of $V(N)\_{\mathfrak m}$, and let $\mathbb T_{\mathfrak m}$ be the Katz Hecke algebra acting on it. Note that $V_{\mathfrak m}$ is a smooth representation of the product $\prod_{i = 1}^n GL_2(\mathbb Q_{\ell_i}).$ Note also that there is an action of $U_p$ on $V_{\mathfrak m}$. (Let me reiterate that I did not include $U_p$ in my Hecke algebra $\mathbb T_{\mathfrak m}$!) Now completed cohomology: If we take completed cohomology at tame level $N$, we get a $p$-adically complete
$\mathbb Z_p$-module $\widetilde{H}^1(N)$, with an action of $\mathbb T_{\mathfrak m}$,
as well as of $G_{\mathbb Q}$ and $GL_2(\mathbb Q_p)$. We can complete it at $\mathfrak m$
to get $\widetilde{H}^1(N)\_{\mathfrak m}$. If we then take the direct limit over all $N$
which are divisible exactly by $\ell_1,\ldots,\ell_n$, we get a module I'll denote
$\widetilde{H}^1_{\mathfrak m}$, which has an action of $\mathbb T_{\mathfrak m}$, of $G_{\mathbb Q}$
and $GL_2(\mathbb Q_p)$, and also of the product $\prod_{i=1}^n GL_2(\mathbb Q_{\ell_i})$. Now suppose that $\overline{\rho}$ satisfies some technical conditions, irreducibility
being the most significant one. (The precise conditions are in the local-global compatibility paper that you mention. Note also that the irreducibility assumption eliminates the distinction between cohomology and cohomology with compact support, and --- more or
less equivalently --- the distinction between working on closed vs. open modular curves.) Then you can show that there is an isomorphism of $\mathbb T_{\mathfrak m}[G_{\mathbb Q}\times GL_2(\mathbb Q_p) \times \prod_{i = 1}^n GL_2(\mathbb Q_{\ell_i})]$-modules $$\widetilde{H}^1\_{\mathfrak m} = \rho^u \otimes_{\mathbb T\_{\mathfrak m}} \pi^u
\hat{\otimes}_{\mathbb T\_{\mathfrak m}} V\_{\mathfrak m}^{U_p =0}.
$$
Here $\rho^u$ is the univeral modular deformation of $\overline{\rho}$ over $\mathbb T_{\mathfrak m}$, $\pi^u$ is a representation of $GL_2(\mathbb Q_p)$ on an orthonormalizable $\mathbb T_{\mathfrak m}$-Banach module constructed from $\rho^u_{| G_{\mathbb Q_p}}$ via the $p$-adic local Langlands, and $V_{\mathfrak m}^{U_p = 0}$ is, as indicated, the kernel
of $U_p$ on $V\_{\mathfrak m}$. Also, the completed tensor product $\hat{\otimes}$ has to be suitably interpreted. (One should cut back to a fixed tame level $N$, then form the completed tensor product, and then take a direct limit over all $N$.) So this is a kind of $p$-adic Eichler--Shimura isomorphism relating $p$-adically completed cohomology and $p$-adic modular forms. It brings out the difference between the two sides quite clearly: on the completed cohomology side, we have a Galois action, which is encoded in the appearance of $\rho^u$.
(This reflects the classical fact that every cuspform appears "twice" in cohomology.) Also, completed cohomology has an action of $GL_2(\mathbb Q_p)$, while $p$-adic modular forms just have an action of $U_p$. So to compare the two, we have to first get rid of the $U_p$-action on $p$-adic modular forms (which we do by passing to $U_p = 0$), and then add in a $GL_2(\mathbb Q_p)$-action, which we do by tensoring with $\pi^u$. Note also that this isomorphism is not canonical. In this sense, it is analogous
to looking at classical cohomology of modular forms with say $\mathbb Q$-coefficients,
and modular forms with $\mathbb Q$-coefficients. These will be isomorphic as Hecke modules
--- up to the issue of cuspforms appearing twice in cohomology --- but not canonically so. In order to make the Eichler--Shimura isomorphism canonical, one has to extend scalars to an appropriate period ring. Whether this is possible with completed cohomology I'm not sure about at the moment. One more remark: trading in a $U_p$-action for a $GL_2(\mathbb Q_p)$-action is a fairly significant upgrading of structure, and this is why completed cohomology provides a useful tool for proving modularity theorems for Galois representations, over and above the already-existing theories of $p$-adic modular forms and $p$-adic modular symbols. Added: You asked about motivation. The original motivation for defining completed cohomology was to construct eigenvarieties. Later it became clear that it was an important object in its own right, providing a global counterpart to the representations of $p$-adic groups that were beginning to appear as part of $p$-adic local Langlands. E.g. the theorem that locally algebraic vectors in cohomology are classical was first proved as an ingredient in the proof of an analogue of Coleman's "small slope implies classical" result for the eigenvariety constructed from completed cohomology. Only later was it realized that this could be combined with a local-global compatiblity result to prove modularity theorems for Galois representations. Note that the rough relation with $p$-adic modular forms, namely that one gets the same Hecke algebra via either approach, was clear from the beginning, even though the more precise Eichler--Shimura-like statement above was not. Since eigenvarieties (as their name indicates) only care about Hecke eigenvalues, this meant that completed cohomology was good enough for constructing them.
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68,680 |
Let us compile a list of counterexamples in PDE, similar in spirit to the books Counterexamples in topology and Counterexamples in analysis . Eventually I plan to type up the examples with their detailed derivations. Please give one example per answer, preferably with clear descriptions and pointers to literature. A related question :
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Scheffer has shown that there is a nontrivial weak solution $u(x,t)\in L^2(\mathbb R^2\times\mathbb R)$ to the incompressible Euler equations in 2D $$\begin{cases} \frac{\partial u}{\partial t}+\nabla\cdot(u\otimes u) +\nabla p=0,
\\[5pt]
\nabla\cdot u=0 . \end{cases}$$ such that $u(x,t)\equiv 0$ for $|x|^2+|t|^2>1$ . In other words, the solution is identically zero for $t<-1$ , then "something happens" and the solution becomes non-zero, and for all $t>1$ the solution vanishes again. In the real world, this would look like if the water suddenly started to move in a cup that stands firmly on a table. See V. Scheffer, "An inviscid flow with compact support in space-time" , Journal of Geometric Analysis , vol. 3 (1993), pp. 343-401.
|
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|
68,842 |
I have two questions that are inspired by a couple of questions here on MO (referenced below), as well as by a conversation with some other grad students at a summer school. Caveat : I'm not a symplectic geometer, nor a differential topologist in the 'classical' sense, so my questions might have a well-known answer, or be open. It's known that exotic $\mathbb{R}^4$'s have diffeomorphic cotangent bundles, as it's known (see here ) that the Milnor spheres have isomorphic (topological) cotangent bundles. 1 . Can the smooth structure of cotangent bundles distinguish exotic smooth structures on the base? Phrased differently, does it exist a pair of smooth manifolds $M,M'$ that are homeomorphic, such that $T^*M$ and $T^*M'$ are isomorphic (see footnote) as vector bundles, but not diffeomorphic as smooth manifolds? As said before, exotic smooth structures on $\mathbb{R}^4$ are not detected by the cotangent bundle as a smooth manifold. But the cotangent bundle admits a canonical symplectic structure, so... 2 . Can the symplectic structure of cotangent bundles see the base? I.e. Does it exist a pair $M,M'$ of smooth manifolds that have diffeomorphic tangent bundles, but such that their symplectic cotangent bundles are/are not symplectomorphic? In this phrasing, I'm including also pairs like $\mathbb{R}^3$ and the Whitehead manifold (see this question), but we can impose some further restrictions: what if $M$ and $M'$ are homeomorphic? UPDATE : on this page, Igor Belegradek gives a reference to a paper where it's proved that homeomorphic $n$-spheres have diffeomorphic cotangent bundles. As Andy Putman points out in his answer, Mohammed Abouzaid found examples of spheres with non-symplectomorphic cotangent bundles, so the answer is YES : cotangent bundles can (at least sometimes) see the base. UPDATE (16/07/2012): regarding question 2, Tobias Ekholm and Ivan Smith have an interesting preprint : Corollary 1.4 says that the symplectic topology on $T^*(S^1\times S^{8k-1})$ detects the smooth topology of the underlying manifold. Their paper is about double points of Lagrangian immersions (mentioned in Tim Perutz's answer, below). As Igor Belegradek remarked, the word "isomorphic" might be a bit confusing, in this context. What I mean is that there is a homeomorphism $M\to M'$ that pulls back $T^*M'$ to $T^*M$. This can be strengthened/weakened by asking that this homeomorphism is topologically isotopic to the identity (since $M$ and $M'$ are homeomorphic, this makes sense). For example, I might want $M$ and $M'$ to be both parallelisable.
|
I wrote a little expository piece about this and related matters in the Newsletter of the European Mathematical Society: http://www.ems-ph.org/journals/newsletter/pdf/2010-03-75.pdf The classical topology of $X:=T^\ast L$ can be taken to include a little more than its diffeomorphism type: there's also an almost complex structure $J$ on $X$, canonical up to homotopy. The pair $(X,J)$ knows the Pontryagin classes of $L$, because $$c_{2k}(TX,J)|_{L}=c_{2k}(TL\otimes\mathbb{C})=(-1)^k p_k(TL),$$
so $(-1)^k p_k(TL)$ pulls back to $c_{2k}(L)$ under projection $X\to L$. However, even with this embellishment, the smooth topology of $X$ doesn't determine $L$. Faithfulness conjecture: the exact symplectomorphism type of the cotangent bundle $(X,\omega=d\lambda_L)$ of a compact manifold $L$ determines $L$. An exact symplectomorphism $T^\ast L \to T^\ast L'$ is a diffeomorphism $f$ such that $f^*\lambda_{L'}-\lambda_L= dh$ for $h$ a compactly supported function. The conjecture (but not the name) is standard. Attempts to use symplectic invariants of $X$ to distinguish smooth structures on $L$ have so far been a complete failure. For example, the symplectic cohomology ring $SH^*(X)$ is isomorphic to loopspace homology $H_{-*}(\mathcal{L}L; w)$ (the coefficients are the local system $w$ of $\mathbb{Z}$-modules determined by $w_2$), with the string product. This invariant is determined by the homotopy type of $L$. "Arnol'd's conjecture" (scare quotes because Arnol'd really made a much more circumspect conjecture). Any exact Lagrangian embedding $\Lambda \to X$ (with $\Lambda$ compact) is exact-isotopic to the embedding of the zero-section. This would immediately imply the faithfulness conjecture. There has been progress towards Arnol'd's conjecture of three kinds: (1) It's true for $L=S^2$ (Hind, http://arxiv.org/abs/math/0311092 ). (2) The work of several authors (Fukaya-Seidel-Smith http://arxiv.org/abs/0705.3450 , Nadler http://arxiv.org/abs/math/0612399 , Abouzaid http://arxiv.org/abs/1005.0358 , and Kragh's work in progress) cumulatively shows that the projection from an exact Lagrangian to the zero-section is a homotopy equivalence. This is good evidence for the truth of the conjecture, but for the application to faithfulness one might as well make homotopy-equivalence an assumption. (3) As Andy mentioned, Abouzaid http://arxiv.org/abs/0812.4781 has shown that a homotopy $(4n+1)$-sphere $S$, such that $T^*S$ contains an exact embedded Lagrangian $S^{4n+1}$, bounds a parallelizable manifold. This is proved by a stunning analysis of the geometry of a space of pseudo-holomorphic discs. The existence of exact Lagrangian immersions is governed by homotopy theory (there is an h-principle which finds such an immersion given suitable homotopical data). Just as the subtleties of 4-manifold topology can be located at the impossibility of removing double points of immersed surfaces (the failure of the Whitney trick), so the subtlety of the symplectic structure of cotangent bundles comes down to the question of removability of double points of Lagrangian immersions.
|
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|
68,936 |
I want to understand algebraic geometry from the functorial viewpoint. I've found a set of notes (linked below) that develop algebraic geometry from the elementary beginnings in this framework. They go under the name "Introduction to Functorial Algebraic Geometry" (following a summer course held by Grothendieck), and are in parts in an almost unreadable shape. Is there an available, elementary, and readable source which takes the functorial viewpoint? http://www.math.jussieu.fr/~leila/grothendieckcircle/FuncAlg.pdf EDIT: Great answers so far, thanks! I would like to add that if anyone were to have a better scan of the document linked to above, or if anyone has vol. 2 of the notes I would be very interested in aquiring a copy.
|
Since your question might interest other readers, allow me to expand it. Given a scheme $T$ , you can associate to it the contravariant functor $h_T: \mathcal{ Schemes}^\text{opp} \to \mathcal{Sets}$ . In a nutshell, Eivind's request is for documents showing how you can study the scheme $T$ by studying the functor $h_T$ . Not all functors $$h: \mathcal{ Schemes}^\text{opp} \to \mathcal{Sets}$$ come from a scheme $T$ and you have to characterize those that do. A pleasant surprise is that it is enough to look at what your functor does on affine rings, so that in effect you study a functor $$k: \mathcal{ Rings} \to \mathcal{Sets}$$ The characterization is then relatively easy (once Grothendieck has shown us what to do!): the functor must be a sheaf in the Zariski topology and satisfy a condition which translates that a scheme is covered by affines. There is a real aesthetic appeal to the realization that concepts like closed or open immersions, tangent spaces,… can be expressed purely in terms of functors. The appeal is not only aesthetic, but also technical: it is with the functorial method that parameter spaces, like Grassmannians, Hilbert schemes,… are constructed.
And where is all this to be found? a) In Mumford's Introduction to Algebraic Geometry , affectionately called The Red Book ,
Chapter Ii, §6: The functor of points of a prescheme . The book has been reprinted by Springer . b) Unfortunately Mumford didn't publish the rest of his course. However there is a set of notes (more than 300 pages) coauthored by Oda, corresponding to Chapters I to VIII, which can be found on-line here . c) Eisenbud and Harris wrote a book on schemes which came out of a common project with Mumford. The last chapter (chapter VI) is called Schemes and Functors and, as the name says, is dedicated to the point of view we are discussing. d) Brian Osserman has a very pleasant short hand-out here on the use of the functorial point of view to illuminate the construction of the product of two schemes. Edit Mumford and Oda have published in 2015 the rest of Mumford's course mentioned in b) as the book Algebraic Geometry II at the Hindustan Book Agency. The link I gave in b) fortunately still works, and we should be grateful to the authors for this act of generosity.
|
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|
68,952 |
Is there an elliptic curve in CP^2 whose induced Remannian metric ( induced from the Fubini-Sudy metric on CP^2) is Euclidian flat?
|
According to this paper by Linda Ness the Gaussian curvature of a curve $C\subset \mathbb P^2$ defined by the
zeros of a degree $d>1$ homogeneous polynomial $F \in \mathbb C[x,y,z]$ at a smooth
point $p$ is given by
$$
K(p) = 2- \frac{\|p\|^6 \cdot | \rm{Hessian}(F)(p)|^2}{ (d-1)^4 \cdot \| \nabla F(p) \|^6} ,
$$
where $\| \cdot \|$ stands for the usual norm in $\mathbb C^3$, and $\nabla F$ is the gradient of $F$. In particular, if $p$ is a smooth inflection point of $C$ then $K(p) = 2$. Thus, there are no smooth cubics in $\mathbb P^2$ which are Euclidean flat, since these have $9$ inflection points. N.B. : Ness normalizes the Fubiny-Study metric to have sectional curvature $2$. After googling a bit I've found the paper The Riemannian geometry of holomorphic curves by Blaine Lawson which is strictly related to the subject. There he says that Eugenio Calabi proved, in Isometric imbedding of complex manifolds , that ($\ldots$) modulo holomorphic congruences, there is only
one curve $C_n$ of
constant Gauss curvature in $\mathbb
> C P^n$ which does not lie in any linear
subspace. This curve has curvature $1/n$ and is given by the following embedding of $\mathbb C P^1\to \mathbb C P^n$:
$$(z_0,z_1) \mapsto \left(z_0^n, \sqrt{n} z_0^{n-1} z_1, \ldots, \sqrt{\binom{n}{k}}z_0^{n-k}z_1^k, \ldots, z_1^n \right). $$ I could not find this statement in Calabi's paper, but this does not exclude the possibility that it is indeed there. The paper is the published version of Calabi's Phd thesis, so another possibility is that the statement is in the thesis but did not make its way into the paper. N.B. : Lawson normalizes the Fubiny-Study metric to have sectional curvature $1$.
|
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|
69,086 |
A Lawvere theory is a small category with finite products such that every object is isomorphic to a finite product of copies of a distinguished object x. A model of the theory in a category with finite products is a product preserving functor from the theory to that category. This notion is supposed to be the right categorical viewpoint on universal algebra. Given a classical variety of universal algebra, one gets a theory by considering the opposite of the category of finitely generated free algebras. My question is, what are the advantages/disadvantages of theories versus classical universal algebra? I believe that theories are more general. What are natural examples of theories that are not covered by classical universal algebra? Does the classical equational theory of universal algebra have a nice analog for theories?
|
My own experience is that Lawvere theories help one "think outside the box" in ways that I really don't think are too likely with classical universal algebra. Qiaochu has already pointed to what is the key idea: that they enable one to consider models other than in $Set$. Actually, you could put it more strongly. Namely, in Lawvere's formulation, the theory is a model in this extended sense: it is the universal model. For example, the Lawvere theory of groups is the category with finite products where all you know about it is that it has a group object, and nothing more. You could take this philosophy further still: any category with finite products $C$ could be considered an algebraic theory in its own right, where a model of $C$ in a category with finite products $D$ is a product-preserving functor $C \to D$. And this conceptual expansion pays off. Let me give a concrete example of this: consider the concept of Boolean algebra. Here the Lawvere theory is the category of finite sets of cardinality $2^n$, denoted $Fin_{2^\bullet}$. This is already a neat way to define a Boolean algebra: as a product-preserving functor $$Fin_{2^\bullet} \to Set.$$ But it's even nicer when you complete $Fin_{2^\bullet}$ to its Karoubi envelope. It's not hard to see that if $C$ is a category with finite products, then the Karoubi envelope $\bar{C}$ also has finite products, and the models of $C$ are equivalent to models of $\bar{C}$. In the present case, the Karoubi envelope of $Fin_{2^\bullet}$ is $Fin_+$, the category of finite nonempty sets. Thus a Boolean algebra is equivalently a product-preserving functor $$Fin_+ \to Set$$ This not only looks prettier, but it frees us from tendencies of thinking of Boolean algebras as having intrinsically to do with things like power sets and powers of 2: there's no power of 2 sticking out in this description. I call this an unbiased Boolean algebra -- we have removed the bias toward 2. It also allows one to see that one could be biased in different ways, and see Boolean algebras, if we want, as having to do with powers of 3 instead. In other words, the Lawvere theory $Fin_{3^\bullet}$ has, by the same Karoubi envelope reasoning, the same models as $Fin_+$, so we could just as well think of a Boolean algebra as a product-preserving functor $$Fin_{3^\bullet} \to Set$$ which gives a totally different way of seeing Boolean algebras as monadic over $Set$ -- here the "underlying sets" of such algebras have a tendency to look like structures on sets of cardinality $3^n$. If you are wondering what good that is for, I might recommend looking at the nLab article on Boolean algebras , and on the multiplicity of different ways of understanding the ultrafilter monad which this suggests. I personally have found this quite eye-opening. Historically, the heyday of Lawvere theories was in the sixties, when the close connections with monads were ironed out. Lawvere theories are essentially the same things as finitary monads; one advantage of Lawvere theories is that it enabled one to more clearly see this connection, which in turn leads to a clear view of the story about infinitary theories and their equivalence with monads on $Set$ (at least for locally small theories). Nowadays these things are under good control, and they allow us to define things like "compact Hausdorff objects" as models of the infinitary Lawvere theory attached to the ultrafilter monad -- these too are useful. I don't think any of this would have been at all easy to see from the point of view of classical universal algebra. The connection between theories and monads is worked out in this nLab article . Cf. also this MO answer , which goes into some more detail on the discussion about Boolean algebras. Edit: In a comment below, Gerhard Paseman very kindly called my attention to the example of $n$-valued Post algebras, which I had not heard of prior to this discussion. Apparently these were introduced by the Polish logician Emil Post based on his studies of $n$-valued logic as an extension of ordinary 2-valued logic; the $n$-valued Post algebras are to $n$-valued logic as Boolean algebras are to 2-valued logic. There are a number of equational presentations of Post algebras; see for example this article by George Epstein from the Transactions of the AMS. Perhaps I can use this very example to follow Gerhard's advice, and deliver an attempted sales pitch specifically to him. :-) I hope it's not considered off the topic; it is meant to illustrate the basic conceptual simplicity of the Lawvere-theory way of thinking. If my suspicions are right, the Lawvere theory of $n$-valued Post algebras is nothing but $Fin_{n^\bullet}$, the category of finite sets whose cardinality is a power of $n$. Or, in other words, that such Post algebras can be identified with functors $$Fin_{n^\bullet} \to Set$$ that preserve finite products. Certainly for anyone used to Lawvere theories, this gives a very tidy description, and this description makes it easy to see the essential equivalence between $n$-valued Post algebras and Boolean algebras, as in the Karoubi envelope analysis above (which uses nothing more complicated than idempotent functions between finite sets). For those not used to Lawvere theories (and of course this is assuming my suspicions are correct), the more traditional syntactic descriptions given by Post and his followers can be extracted from this categorical description. The rough idea is this. If $X$ is a Boolean algebra, and if $P(n)$ is the power set of a set with $n$ elements, then the elements of the corresponding $n$-valued Post algebra are simply Boolean algebra homomorphisms $P(n) \to X$. Let $X(n)$ denote this set. The $m$-fold cartesian product $X(n)^m$ is naturally identified with $X(n^m)$, the set of Boolean algebra homomorphisms $P(n^m) \to X$. Then, we may describe the clone of Post algebra operations: the $m$-ary operations $X(n)^m \to X(n)$ in the clone are in one-to-one correspondence with functions $f: [n^m] \to [n]$ from an $n^m$-element set to an $n$-element set. Namely, each function $f: [n^m] \to [n]$ induces a Boolean algebra map $f^{-1}: P(n) \to P(n^m)$ (the inverse image $f^{-1}$ takes a subset $S \subseteq [n]$ to $f^{-1}(S) \subseteq [n^m]$); then the corresponding Post algebra operation $X(n^m) \to X(n)$ sends a Boolean algebra map $\phi: P(n^m) \to X$ to the composite $$P(n) \stackrel{f^{-1}}{\to} P(n^m) \stackrel{\phi}{\to} X.$$ This gives the clone; giving an explicit description of a set of operations and identities for the theory in a universal algebra sense is pretty much the same thing as giving a combinatorial analysis of functions between sets of cardinality a power of $n$, in other words how to generate such functions from a smaller class using cartesian products and composition. I think with that clue, we can figure out what is going on in Epstein's paper. The constants of the theory are given by functions $[1] \to [n]$, so there are $n$ of them. These are the $e_i$ of his paper. Next, unary operations of the clone are given by functions $[n] \to [n]$; for his purposes, Epstein selects $n$ of them, and calls them $C_i$ ($i = 0, \ldots, n-1$). From our point of view, they are uniquely specified by how they act on the constants, since a function $[n] \to [n]$ is uniquely determined by how it acts on functions $[1] \to [n]$. Epstein's $C_i$ are just the indicator or characteristic functions given by $C_i(e_j) = \delta_{ij}$ (returning the "bottom" constant $e_0 = \bot$ if $i \neq j$, and the "top" constant $e_{n-1} = \top$ if $i = j$). These together with the lattice operations $\wedge: [n]^2 = [n^2] \to [n]$ and $\vee: [n]^2 = [n^2] \to [n]$, which are again uniquely specified by how they act on constants (and defined by the expected rules if the $e_i$ are ordered by $e_i \leq e_j$ iff $i \leq j$), are shown by Epstein to generate the theory of Post algebras, but seems pretty intuitive that they generate the clone as described here by the Lawvere theory $Fin_{n^\bullet}$: it says that any finite function $[n^m] \to [n]$ can be built from a combination of meets and joins applied to the constants $e_i$ and their characteristic functions $C_j$. Part of my point above was that all the Post algebra theories can be united under a single simple umbrella given by the category of product-preserving functors $Fin_+ \to Set$.
|
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|
69,099 |
Let $S$ be a sphere in $\mathbb{R}^3$ . Let $C$ be a closed curve in $\mathbb{R}^3$ disjoint from and
exterior to $S$ which has the property that every point $x$ on $S$ is visible to some point $y$ of $C$ ,
in the sense that the segment $xy$ intersects $S$ at precisely the one point $x$ . I am interested
in the shortest $C$ with this property. In computational geometry, such paths are called watchman tours , and there are many results concerning polygons in the plane finding such tours. This question arose at a conference I'm attending, and I was
pointed to a paper by
V. A. Zalgaller: "Shortest Inspection Curves for the Sphere"
( Journal of Mathematical Sciences ,
Volume 131, Number 1, 5307-5320 ;
Translated from Zapiski Nauchnykh Seminarov POMI , Vol. 299, 2003, pp. 87–108.) I cannot access the paper from the conference, but from the abstract it
appears he focused on open rather than closed curves. Has anyone heard of this natural question?
Can you point me to relevant literature? Thanks! Addendum. Here is the $4\pi$ saddle / baseball-stitches curve suggested by Gjergji Zaimi:
|
James Wenk and I just finished a paper proving Zalgaller's sphere inspection conjecture for closed curves: Shortest closed curve to inspect a sphere . We show that in $R^3$ any closed curve $\gamma$ which inspects the unit sphere $S^2$ , i.e., lies outside $S^2$ and contains $S^2$ within its convex hull, has length $L(\gamma)\geq 4\pi$ . Equality holds only when $\gamma$ is composed of 4 semicircles of length $\pi$ , arranged in the shape of a baseball seam, as conjectured by Zalgaller in 1996. The proof, which runs 38 pages, uses some notions from the earlier work on this problem which I had mentioned in my last post below, together with other ideas from integral geometry, convex analysis, geometric measure theory, and geometric knot theory. Furthermore, we derive a number of formulas for inspection efficiency of curves, which can be verified with the aid of a Mathematica notebook which we have provided.
The basic approach is as follows. Let $\gamma\colon[a,b]\to R^3$ be a closed rectifiable curve inspecting $S^2$ . As I had mentioned in my last post, the horizon of $\gamma$ is defined as $$
H(\gamma):=\int_{p\in S^2} \#\big(\gamma^{-1}(T_pS^2)\big)\, dp,
$$ i.e., the measure in $S^2$ of all points $p$ counted with multiplicity where the tangent plane $T_p S^2$ intersects $\gamma$ .
The horizon of a line segment, for instance, is the area of the region illustrated in the picture above. In the current paper we define the (inspection) efficiency of $\gamma$ as $$
E(\gamma):=\frac{H(\gamma)}{L(\gamma)}.
$$ We want to show that $L(\gamma)\geq 4\pi$ . Since $\gamma$ is closed, $H(\gamma)\geq 8\pi$ . So it suffices to show that $E(\gamma)\leq 2$ . To this end we note that, since $H$ is additive, for any partition of $\gamma$ into subsets $\gamma_i$ , $i\in I$ , $$
E(\gamma)= \sum_i \frac{H(\gamma_i)}{L(\gamma)}=\sum_i \frac{L(\gamma_i)}{L(\gamma)} E(\gamma_i)\leq \sup_i E(\gamma_i) .
$$ So the key to solving the problem would be to construct a partition with $E(\gamma_i)\leq 2$ for all $i\in I$ . To construct the desired partition we may assume that $\gamma$ has minimal length. Then we connect all points of $\gamma$ to the origin $o$ of $R^3$ to obtain a conical surface. This surface can be isometrically developed into the plane to yield a curve $\tilde\gamma\colon[a,b]\to R^2$ which is called the (cone) unfolding of $\gamma$ . This operation goes back to a paper of Cantarella, Kusner, and Sullivan on thickness of knots. It turns out that $$
E(\gamma)=E(\tilde\gamma).
$$ Furthermore, since $\gamma$ is minimal, it follows that $\tilde\gamma$ is locally convex with respect to $o$ . Consequently it admits a partition into segments we call spirals , which are maximal subsets of $\tilde\gamma$ with monotone distance from $o$ . We prove that the efficiency of any spiral is at most $2$ which yields the desired inequality $L(\gamma)\leq 4\pi$ . The proof is based on a computation of $E$ for line segments, polygonal approximations, and a variational procedure. The above arguments fill the first half of the paper. The second half is devoted to characterizing the case where $L(\gamma)=4\pi$ . This part of the paper gets quite a bit more analytical, and requires more precise estimates for efficiency. The main idea here is that if $L(\gamma)=4\pi$ , then $E(\gamma)=2$ which in turn yields that $E(\tilde\gamma_i)=2$ for all the spirals in the unfolding of $\gamma$ . We show that $E(\tilde\gamma_i)=2$ only when $\tilde\gamma_i$ has constant distance $\sqrt2$ from $o$ , which in turn yields that $\gamma$ must have constant distance $\sqrt2$ from $o$ as well, or lie on a sphere of radius $\sqrt2$ . Finally we show that $\gamma$ must be composed of $4$ semicircles using a Crofton type formula of Blaschke-Santalo for length of spherical curves, and a technique from the proofs of the classical $4$ vertex theorem, which has been developed further by Umehara and Thorbergsson. In short, as had been anticipated at the end of my last post, the proof of the full result did require substantial effort beyond the estimates obtained in the earlier work. Zalgaller's conjectures for open inspection curves still remain unresolved. It is also natural to consider higher dimensional versions of the problem. Some of the techniques discussed above apply in all dimensions, but more ideas would be needed to extend the result. Update (July 2021): The proof has been simplified and the paper, which has been shortened to 25 pages, has been accepted for publication in Crelle's journal .
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|
69,222 |
The Monster group $M$ acts on the moonshine vertex algebra $V^\natural$. Because $V^\natural$ is a holomorphic vertex algebra (i.e., it has a unique irreducible module), there is a corresponding cohomology class $c\in H^3(M;S^1)=H^4(M;\mathbb Z)$
associated to this action. Roughly speaking, the construction of that class goes as follows: For every $g\in M$, pick an irreducible twisted module $V_g$ (there is only one up to isomorphism). For every pair $g,h\in M$, pick an isomorphism $V_g\boxtimes V_h \to V_{gh}$, where $\boxtimes$ denotes the fusion of twisted reps. Given three elements $g,h,k\in M$, the cocycle $c(g,h,k)\in S^1$ is the discrepancy between
$$
(V_g\boxtimes V_h)\boxtimes V_k \to V_{gh}\boxtimes V_k \to V_{ghk}\qquad\text{and}\qquad
V_g\boxtimes (V_h\boxtimes V_k) \to V_g\boxtimes V_{hk} \to V_{ghk}
$$ I think that not much known about $H^4(M,\mathbb Z)$. But is anything maybe known about that cohomology class? Is it non-zero? Assuming it is non-zero, would that have any implications? More importantly: what is the meaning of that class?
|
There is some evidence from characters that $H^4(M,\mathbb{Z})$ contains $\mathbb{Z}/12\mathbb{Z}$. In particular, the conjugacy class 24J (made from certain elements of order 24) has a character of level 288, and the corresponding irreducible twisted modules have a character whose expansion is in powers of $q^{1/288}$. Fusion in a cyclic group generated by a 24J element then yields a $1/12$ discrepancy in $L_0$-eigenvalues, meaning you will pick up 12th roots of unity from the associator. If you pull back along a pointed map $B(\mathbb{Z}/24\mathbb{Z}) \to BM$ corresponding to an element in class 24J (i.e., if you forget about twisted modules outside this cyclic group) you get a cocycle of order 12. This is the largest order you can get by this method - everything else divides 12. I don't know how the cocycles corresponding to different cyclic groups fit together. I don't know if you've seen Mason's paper, Orbifold conformal field theory and cohomology of the monster , but it is about related stuff. I don't understand how he got his meta-theorem with the number 48 at the end, though. As far as implications or meaning of the cocycle, all I can say is that the automorphism 2-group of the category of twisted modules of $V^\natural$ has the monster as its truncation, and its 2-group structure is nontrivial. I've heard some speculation about twisting monster-equivariant elliptic cohomology, but I don't understand it. If you believe in AdS/CFT, this might say something about pure quantum gravity in 3 dimensions, but I have no idea what that would be. Update Nov 2, 2011: I was at a conference in September, where G. Mason pointed out to me that $H^4(M,\mathbb{Z})$ probably contains an element of order 8, and therefore also $\mathbb{Z}/24\mathbb{Z}$. I believe the argument was the following: there is an order 8 element $g$ whose centralizer in the monster acts projectively on the unique irreducible $g$-twisted module of the monster vertex algebra $V^\natural$, such that one needs to pass to a cyclic degree 8 central extension to get an honest action. Rather than just looking at $L_0$-eigenvalues, one needs to examine character tables to eliminate smaller central extensions here. Naturally, like the claims I described before, the validity of this argument depends on some standard conjectures about the structure of twisted modules. It seems that the relevant group-theoretic computation may have been known to S. Norton for quite some time. In his 2001 paper From moonshine to the monster that reconstructed information about the monster from a revised form of the generalized moonshine conjecture, he explicitly included a 24th root-of-unity trace ambiguity. I had thought perhaps he just liked the number 24 more than 12, but now I am leaning toward the possibility that he had a good reason.
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|
69,239 |
Consider a hypersurface $X=V(f) \subset \mathbb A^n_{\mathbb C}$, where $f(T_1, T_2,\ldots,T_n)\in \mathbb C[T_1,Y_2,\ldots, T_n]$ is a polynomial . Assume that $X$ is smooth, i.e. that $df(x)\neq 0 \;$ for all $x\in X$ . My question is simply whether $X $ is parallelizable i.e. whether its tangent bundle $T_X$ is algebraically trivial. I've asked a few friends and their answer was unanimously "no, why should it be?", but they couldn't provide a counter-example. Here are a some considerations which might show that the question is not so ridiculous as it looks. We have the exact sequence of vector bundles on $X$ $$0 \to T_X\to T_{A^n_{\mathbb C}}|X\to N(X/A^n_{\mathbb C})\to 0$$ Now, the normal bundle $N(X/A^n_{\mathbb C})$ is trivial (trivialized by $df$) and the restricted bundle $T_{A^n_{\mathbb C}}|X$ is trivial because already $T_{A^n_{\mathbb C}}$ is trivial. Moreover the displayed exact sequence of vector bundles splits, like all exact sequences of vector bundles, because we are on an affine variety. So we deduce (writing $\theta$ for the trivial bundle of rank one on $X$)
$$\theta^n=T_X\oplus \theta $$
In other words the tangent bundle is stably trivial, and this is already sufficient to deduce (by taking wedge product) that $\Lambda ^{n-1}T_X=\theta$ (hence the canonical bundle $K_X=\Lambda ^{n-1}T_X^\ast$ is also trivial). This suffices to prove that indeed for $n=2$ the question has an affirmative answer: every smooth curve in $A^2_{\mathbb C}$ is parallelizable. Another argument in favor of parallelizability is that there are no analytic obstructions: O. Forster has proved a result for complex analytic manifolds which implies that analytically (and of course differentiably) our hypersurface is parallelizable: $T_{X_{an}}=\theta _{an}^{n-1}$.This is why I choose $\mathbb C$ as the ground field: the question makes perfect sense over an arbitrary algebraically closed field but I wanted to be able to quote the related analytic result.[ As ulrich remarks, parallelizability can't be deduced over the non-algebraically closed field $\mathbb R$, as shown by a 2-sphere] Edit ulrich's great reference not only answers my question but seems to yield more results in the same direction. For example consider a smooth complete intersection: $X=\{ x\in \mathbb C^n|f_1(x)=f_2(x)=\ldots=f_k(x)=0 \} $ with the $f_i$'s polynomials and the $df_i(x)$'s linearly independent at each $x\in X$ . Then, just as above, the normal bundle is trivial and the tangent bundle is stably trivial: $\theta^n=T_X\oplus \theta ^{k} $ So Suslin's incredible theorem again allows us to conclude that $X$ is parallelizable. However not all affine smooth algebraic varieties are parallelizable: for example the complement of a smooth conic in $\mathbb P^2(\mathbb C)$ is a smooth affine variety (Veronese embedding !) but is not even differentiably parallelizable. I wonder if these differentiable obstructions are the only ones preventing algebraic parallelizability of smooth algebraic subvarieties of $\mathbb C^n$. Any thoughts, dear friends?
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Yes. Suslin has proved that every stably trivial vector bundle of rank $n$ on an affine variety of dimension $n$ over an algebraically closed field is trivial. See: Suslin, A. A. Stably free modules. Mat. Sb. (N.S.) 102(144) (1977), no. 4, 537–550, 632. Note that this is not true over arbitrary fields; for example, the tangent bundle of the $2$-sphere (given by $x^2 + y^2 + z^2 - 1 = 0$) over $\mathbb{R}$ is not trivial (since it is not so even topologically). However, it is true over finite fields and also over $C_1$ fields of characteristic $0$. See: Bhatwadekar, S. M. A cancellation theorem for projective modules over affine algebras over $C_1$-fields. J. Pure Appl. Algebra 183 (2003), no. 1-3, 17–26.
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69,253 |
Are there any positive integer solutions to $2^n-3^m=1$ with $n,m>2$ ? By way of justifying the question, I've found lots of info on what happens when $m=n$ (mostly FLT variations, Darmon + Merel,...), but I don't really know where to look for $m\not=n$ . Also it's pretty obvious that you can't have solutions to similar equations, e.g., $2^n-3^m=2$ . There are no solutions for $n,m<1000$ aside from $n=2, m=3$ . It seems pretty likely to me that it should happen for some large numbers at some point though. Are there any theorems I don't know about regarding primes $p,q$ and $p^n-q^m=k$ , $k \in \mathbb{N}$ that might rule out a solution or help me find one?
|
Here is the proof of Gersonides [Levi ben Gershon] (1343) for $2^n-3^m=1$. It uses nothing more that arithmetic modulo $8$. Case I: $m$ is even. Then $3^m$ is 1 mod 4, so $2^n$ is 2 mod 4, implying $n=1$ and $m=0$. Case II: $m$ is odd. Then $3^m$ is 3 mod 8, so $2^n$ is 4 mod 8, implying $n=2$ and $m=1$. The alternative equation $3^m-2^n=1$ follows similarly when $m$ is odd, but is a bit more tricky when $m$ is even (hint, factor $2^n=3^m-1=(3^{m/2}+1)(3^{m/2}-1)$ and argue from there).
|
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69,307 |
I assume it is partially because they are good generalizations of polynomial rings, but what makes this generalization better than graded algebras or other generalizations of polynomial rings?
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The best answer I've ever been able to come up with is that the class of noetherian rings contains the classical number rings $\mathbf{Z}$ and $\mathbf{R}$ and is closed under the formation of polynomial rings, localization, completion, and quotients. So it contains many of the rings you will come across in ordinary situations (whatever that means). It also has the advantage that the definition is tractable enough that if someone hands you an explicit ring, it's not out of the question to try to work out from scratch whether it's noetherian. If you're the kind of person who likes abstract fields, then they're also included. On the other hand, I don't think of it as a truly fundamental concept, like say finite presentation. But there is no denying its convenience. If you need to avoid some infinitary phenomena but you still want a broad class of rings, it's often hard to beat noetherianness. It's also quite good in situations where you're too lazy to work out exactly what finiteness conditions you care about.
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69,352 |
$U(1)$ is diffeomorphic to $S^1$ and $SU(2)$ is to $S^3$, but apparently it is not true that $SU(3)$ is diffeomorphic to $S^8$ (more bellow). Since $SU(3)$ appears in the standard model I would like to understand its topology. By one of the tables here $SU(3)$ is a compact, connected and simply connected 8-dimensional manifold. This MO post says that its $\pi_5$ is $\mathbb{Z}$ thus it can not be homeomorphic to $S^8$(e.g.: see this wiki article). Even if it was a homotopy sphere Poincaré conjecture would not be helpful (at least in the smooth category: there exists exotic 8-spheres, right?). I guess that this is what the author of this question was trying to know... Anyway, is it known any manifold diffeomorphic to $SU(3)$?
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Apart from jokes, an answer which may satisfy you is the following: $SU(3)$ is a $S^3$-bundle over $S^5$. To see this just consider the defining representation of $SU(3)$ on $\mathbb{C}^3$; this induces a transitive action of $SU(3)$ on the unit sphere of $\mathbb{C}^3$, which is $S^5$. Since the stabilizer of a point for this action is $SU(2)$ this exhibits $SU(3)$ as an $SU(2)$-bundle over $S^5$, and as you wrote $SU(2)$ is diffeomophic to $S^3$. Now, the next question is: which $SU(2)$-bundle over $S^5$ is $SU(3)$? to answer this, recall that isomorphic classes of principal $SU(2)$-bundles over (a not too wild) topological space $X$ are in bijection with the set $[X,BSU(2)]$ of homotopy classes of maps from $X$ to the classifying space of $SU(2)$. So in the case at hand you are interested in $[S^5,BSU(2)]= \pi_5(BSU(2))= \pi_4(SU(2))= \pi_4(S^3)= \mathbb{Z}/2\mathbb{Z}$. So there are only two $S^3$-bundles over $S^5$, the trivial one and the nontrivial one: $SU(3)$ is the nontrivial one (otherwise one would have $\pi_4(SU(3))=\mathbb{Z}/2\mathbb{Z}$, which is not the case: it is $\pi_4(SU(3))=\{0\}$).
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69,631 |
A smooth curve $X$ in $\mathbb{P}^n$ is strange if there is a point $p$ which lies on all the tangent lines of $X$. Examples are $\mathbb{P}^1$ is strange and so is $y=x^2$ in characteristic $2$. These are in fact all (see Hartshorne IV.3.9). For this reason, in any characteristic we do not use lines to get embeddings in projective space. But this actually has consequences. Many classical theorems in algebraic geometry do not use 'transcendental methods', i.e. the only result they use is that the base field is algebraically closed, and so they can be applied in finite characteristic. Or they cannot? Here is where characteristic $2$ breaks down the standard results. For instance, when embedding blow-ups of $\mathbb{P}^2$ in $\mathbb{P^n}$ we use linear systems of conics and cubics in $\mathbb{P}^2$ to separate the points, but this is not possible in characteristic $2$ (have a look at Beauville IV.4 if you do not know how linear systems can embed spaces). This means that in characteristic 2 we cannot interpret cubic surfaces ni $\mathbb{P}^3$ in terms of blow-ups of the projective plane in 6 points in general position and viceversa. OK, enough intro. My question is:
"Are there other examples of results which do not apply in characteristic 2 due to other reasons not involving embeddings in projective space?" Also, I am looking for results that do not hold in low characteristic. i.e. I know that vanishing theorems do not hold in characteristic p in general, but I am looking for pathologies for some but not all finite characteristic (usually 2, or 3). I suspect 'probably yes but not many', since I cannot come up with any, but if it turns out there are lots, maybe I'll make this question community wiki.
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I am not going to add any new examples but suggest a systematic way of looking at examples. If one looks at special phenomena in characteristic $2$ one can classify them as follows (though this division is far from clear cut): They are really special to positive characteristic and not only characteristic $2$. They are still really really positive characteristic phenomena but they only appear for some numerical invariants that depend on $p$ (normally increasing with $p$) and as $2$ is the smallest prime they appear "earlier" in characteristic $2$ and hence are encountered there first. They are really special to characteristic $2$. Some examples and their classification: Here one can look at failure of strong versions of Bertini, for instance a base point free linear system all of whose members are singular (take the $p$'th power of a very ample linear system).
This is uniform in $p$ (though if one starts with a characteristic free ample system, the degree will grow as $p$ grows so in that sense it could also be classified under 2). The existence of quasi-elliptic fibrations in characteristic $2$ (and $3$) is an example as the same phenomemena of a regular but non-smooth curve over a non-perfect field exists in all positive characteristics. However, by a result of Tate the genus of such an example is bounded from below by a linear function of $p$ so they appear later and later. However, there is one further complication in that the quasi-elliptic case is of Kodaira dimension $0$ which makes $2$ and $3$ special as all other examples are of general type. This gives an example of overlap between 2) and 3). Another such example is that of Enriques surfaces. On the one hand the Godeaux construction gives examples of smooth surfaces whose fundamental group scheme is any group scheme of order $p$ with various numerical invariants depending on $p$. However, only in characteristic $2$ (and $3$ I think) is it of Kodaira dimension $0$. Here the examples that come to my mind are mostly somehow related to quadratic forms. They themselves of course really behave differently in characteristic $2$ (even purely geometrically) as does the orthogonal group. However, their influence goes further, for instance that theta characteristics behave differently in characteristic $2$ can be traced back to quadratic forms. Addendum : To test my claim I went through the answers given so far and tried to classify as per above. Most of them are already mentioned above but two are not. First there is Jeremy's comment on torsion in $1+p\mathbb Z_p$ which on the face of it belongs to category 3). However, it is clearly related to $p$-adic radius of convergence of the logarithm and exponential series and that radius grows as $p$ grows. Hence, for absolutely ramified rings you can get the same phenomenon in all characteristics, what is special with $2$ is that it happens in the absolutely unramified. Note also that though the consequence mentioned by Jeremy is more arithmetic than algebro-geometric there are consequences of the latter type. Certainly, mixed characteristic ones such as the structure of finite group schemes but also for crystalline issues (technically the divided power structure on $2\mathbb Z_2$ is not nilpotent). Sándor's example of failure of Kodaira vanishing is mostly of type 1) but the examples usually have $p$ as a parameter in numerical characters making it partly of type 2). There is even the fact that there are a few minimal surfaces of general type in characteristic $2$ (but in no other characteristic) for which $H^1(X,\omega_X^{-1})\ne 0$ which technically is of type 3).
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69,937 |
I am doing my PhD in algebraic graph theory, for not much more reason than that was what was available. However, I love deep structure and theory in mathematics, and I do not particularly want to be a graph theorist for the rest of my life. I have heard of mathematicians changing from more theoretical subjects to broad ones like combinatorics, but not the other way round, and am concerned that this is because the time it takes to learn the requisite theory for deeper subjects is hard to come by after grad school. Has anyone done this, or known of someone who has? Is it at all likely or possible I will be able to get a post-doc position in a subject only marginally connected to my thesis? I have been using small amounts of algebraic number theory, and would ideally like to go on and specialise in something similar. Any advice much appreciated! EDIT (June '12) - I came across this old question of mine, and now feel rather silly for having asked it at all. For the record, and for anyone who might be having similar doubts to me: a year or so on I have realised that, as was mentioned below, I should not pigeonhole myself, and that the most interesting problems are often those that lead to other seemingly disparate subfields. Perhaps more to the point, I actually can't now think of a subject I'd rather do than algebraic graph theory! If you find yourself in the position I was in, I advise you to just look for those problems you find most interesting in your "chosen" field - they are bound to lead to other areas. And anyway, surely all that matters is that you are interested and inspired?! In my own research I have used number theory, galois theory, group theory, and even a bit of probability. I might even opine that graph theory is one of the subjects most likely to appear in intra-disciplinary work, which seems to be forming an ever higher proportion of mathematical research.
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Speaking as someone whose thesis was also in algebraic graph theory but who has later gone on to do research in other areas, I would say that it is definitely possible to switch fields. The main skills you need are management skills: the ability to manage your own time so that you can spend some time learning a new field while also producing something that others will value, and the ability to manage other people's view of your work. In this regard, I believe that Deane Yang's comment is right on the money. You probably can't afford to drop your initial specialty abruptly and produce no results while you retrain yourself. But if you manage things carefully then anything is possible, regardless of whether you have tenure or are switching to a field that requires a lot of background study. There are some things you can do to help you solve these management problems. If you can find overlap between your current field and your new field, that will obviously help. Mathematics is so interconnected that this is usually not too hard; in the specific case of algebraic graph theory versus algebraic number theory, the first topic to come to mind is the Ihara zeta function of a graph, but I'm sure there are others. In my case, I decided it would help to switch from academics to industry/government. I personally found it easier in industry/government to spend x% of my time producing results that pleased my employers and spending the remaining (100-x)% of my time training myself in a new area. Your mileage may vary, of course; in my case, I found that teaching drained me of too much of my energy, but this is not true of everybody. One last comment I have is that if you take this route, then you will need the ability to maintain a clear sense of your own identity and goals and not be unduly swayed by other people's categorizations. For example, when I switched out of academics, many others regarded me as "leaving mathematics." In fact I was only leaving academics and not leaving mathematics, and it was important for me to ignore other people's view of the matter. As another example, people will want to pigeonhole you as a "something-ist" (and it seems you have been influenced by this point of view, since you use the phrase "being a graph theorist for the rest of my life"); you should resist this pigeonholing, and instead think of yourself just as someone with certain abilities and interests. Thinking of yourself as either a graph theorist or a number theorist is unnecessarily limiting. (Of course you may need to bill yourself as one or the other for the purposes of managing other people's view of you while you're making a transition, but you should not necessarily believe your own propaganda.)
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69,967 |
I apologize for burdening MO with such a vapid, nonresearch question, but
I have been curious ever since
Suvrit's popular October 2010 Most memorable titles MO question if there were any " $E=mc^2$ -titles," as I think of them—how Einstein in retrospect might have entitled his 1905 paper
(instead of
"Zur Elektrodynamik bewegter Körper"!)—paper/book titles composed entirely of math symbols. There are two close misses in the responses to that MO question:
Connes et al.'s "Fun with $\mathbb{F}_{1}$ " ,
and Taubes's " ${\rm GR}={\rm SW}$ : Counting curves and connections." The only title entirely composed of math symbols with which I'm familiar is the delightful book A=B , by Marko Petkovsek, Herbert Wilf, and Doron Zeilberger.
Can you identify others? Please interpret this question in a weekend-recreational spirit! :-)
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$SL_2(\mathbf{R})$ ( link )
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70,031 |
Let us say that a mathematical structure of cardinality $\omega_1$ is Jonsson whenever every one of its proper substructures is countable. There are examples of Jonsson groups due to Shelah or Obratzsov. I am almost sure that there is no Jonsson Boolean algebra but I cannot (dis)prove it by hand. Am I right? PS. feel free to give any further examples of Jonsson structures or structures which are never Jonsson.
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Boolean algebras are never Jonsson. Suppose that $\mathbb{B}$ is a Boolean algebra of size $\omega_1$. Let $a$ be any element such that neither $a$ nor $\neg a$ is an atom. Note that every element $b\in\mathbb{B}$ is the join $b=(b\wedge a)\vee(b\wedge \neg a)$, and so there must be uncountably many elements either in the cone below $a$ or below $\neg a$. Assume without loss of generality that there are uncountably many elements below $a$. Let $\mathbb{C}$ be the subalgebra of $\mathbb{B}$ consisting of the elements below-or-equal $a$ or above-or-equal $\neg a$. This is closed under meets, joins and complements, and hence is a sub-Boolean algebra. And it has size $\omega_1$ by the choice of $a$. But it has no elements below $\neg a$ other than $0$, and so $\mathbb{C}$ is an uncountable proper subalgebra, as desired. QED It seems that the same idea generalizes to any uncountable cardinal.
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70,154 |
Hi all. Is there any explicit matrix expression for a general element of the special orthogonal group $SO(3)$? I have been searching texts and net both, but could not find it. Kindly provide any references.
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Here is the standard quaternion answer: Given $(a,b,c,d)$ such that $a^2+b^2+c^2+d^2=1$, the matrix
$$\begin{pmatrix}
a^2+b^2-c^2-d^2&2bc-2ad &2bd+2ac \\
2bc+2ad &a^2-b^2+c^2-d^2&2cd-2ab \\
2bd-2ac &2cd+2ab &a^2-b^2-c^2+d^2\\
\end{pmatrix}$$
is a rotation and every rotation matrix is of this form. Note that $(a,b,c,d)$ and $(-a, -b,-c,-d)$ give the same rotation.
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70,162 |
According to Wikipedia, a Lawvere theory consists of a small category $L$ with (strictly associative) finite products and a strict identity-on-objects functor $I:\aleph_0^\text{op}\rightarrow L$ preserving finite products. Why a Lawvere theory have n-products for any n finite?
For example, why isn't a Lawvere theory for monoids a category T with four elements: $1$, $T$, $T^2$ and $T^3$, and morphisms $e:1 \to T$ and $*:T^2 \to T$ making the appropiate diagrams commute and such that they are products of each other as expected? ($T^3$ is needed in order to state these diagrams) Also, why do they usually use the 'free' maps between the desired objects to model and not just the operators that the desired family of algebras has?
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I think you must have misunderstood how Lawvere theories are supposed to work. Let me try to motivate them from an algebraist point of view, and answer your question in passing. An algebraic structure such as a group or a ring is usually described in terms of operations and equations they satisfy. But why do we choose certain operations and certain equations? There are many ways of axiomatizing any given algebraic structure. For example, in a ring we could take the unary operation negation $-x$ as basic, or the binary operation subtraction $x - y$ as basic. And why not both? Why is multiplication taken as basic in the theory of groups, rather than division? Can we describe an algebraic structure in a canonical way, so that no preference is made about which operations and equations count as "basic"? If we are not allowed to prefer any particular choice of operations and axioms, then we must favor them all equally! So, a "canonical" description should include all operations and all equations. Let us consider the theory of groups to see how this works. We customarily start with three basic operations: a nullary operation (constant) unit $1$, a unary operation inverse ${}^{-1}$, and a binary operation multiplication $\cdot$. There are five axioms: $x \cdot (y \cdot z) = (x \cdot y) \cdot z$ $x \cdot 1 = x$ $1 \cdot x = x$ $x \cdot x^{-1} = 1$ $x^{-1} \cdot x = 1$ To get a canonical theory of groups which does not prefer the above operations and axioms, we should generate all possible operations and equations out of the basic ones, and then take them all as basic. Some generated operations will have names, e.g., division $x/y = x \cdot y^{-1}$, squaring $x^2 = x \cdot x$, etc., but others will not. An example of such an operation might be a ternary operation $p(x_1,x_2,x_3) = ((x_1 \cdot 1) \cdot x_2^{-1}) \cdot (x_3 \cdot x_2)$. In fact, for each $n$ there will be infinitely many $n$-ary operations, represented by expressions built from basic operations and variables $x_1, \ldots, x_n$. As new "axioms" we will simply take all equations that follow logically from the above five axioms. If we now "forget" that we started with the three basic operations and five axioms, we will get a somewhat unusual theory with infinitely many operations and infinitely many axioms, which nevertheless still describes what a group is. This all sounds very messy. Are we supposed to invent notation for infinitely many operations? And what use is there in having so many axioms that they already are closed under logical deduction? If we truly are algebraists at heart, we must free ourselves of the shackles of syntax. Let us make a category $\mathcal{T}$ out of the messy description of groups we generated above. Remember that we are not trying to construct the category of groups, but rather a category which nicely organizes the infinitely many operations and equations. The idea is simple enough. The morphisms of $\mathcal{T}$ should correspond to the operations of the theory. The equations should correspond to the fact that certain morphisms are equal. An $n$-ary operation can be thought of as a map $G^n \to G$ where $G$ is the carrier set of a group and $G^n$ is the $n$-fold product of $G$'s. Thus, for each $n$ there should be an object $\mathtt{G}^n$ in $\mathcal{T}$ and the morphisms $\mathtt{G}^n \to \mathtt{G}^1$ should correspond to the $n$-ary operations. We therefore posit that the objects of $\mathcal{T}$ are
$$\mathtt{G}^0, \mathtt{G}^1, \mathtt{G}^2, \ldots$$
where you must not think of $\mathtt{G}^n$ as any kind of set, or an $n$-fold product of anything. We formally write $\mathtt{G}^0, \mathtt{G}^1, \mathtt{G}^2, \ldots$, but we could have as well declared that the objects of $\mathcal{T}$ are the natural numbers and write them simply but confusingly as $0, 1, 2, \ldots$. The morphisms $\mathtt{G}^n \to \mathtt{G}^1$ are expressions built from $n$ variables $x_1, \ldots, x_n$ and the operations $1$, $\cdot$, and ${}^{-1}$. Two such expressions $p(x_1, \ldots, x_n)$ and $q(x_1, \ldots, x_n)$ are considered to be the same morphism if the theory of groups proves $p(x_1, \ldots, x_n) = q(x_1, \ldots, x_n)$. What about the morphisms $\mathtt{G}^n \to \mathtt{G}^m$? Since we expect that $\mathtt{G}^m$ will in fact end up being the $m$-fold product of $\mathtt{G}^1$'s, a morphism $\mathtt{G}^n \to \mathtt{G}^m$ corresponds uniquely to an $m$-tuple of morphisms $\mathtt{G}^n \to \mathtt{G}^1$. In other words, the morphisms $\mathtt{G}^n \to \mathtt{G}^m$ are $m$-tuples of expressions in variables $x_1, \ldots, x_n$, where again two such $m$-tuples represent the same morphism if the theory of groups proves them equal. Composition in $\mathcal{T}$ is performed by substitution. For example, the composition of $(x_1^{-1}, x_2 \cdot x_1) : \mathtt{G}^2 \to \mathtt{G}^2$ and $(x_1 \cdot x_1 \cdot x_2) : \mathtt{G}^2 \to \mathtt{G}^1$ is $(x_1^{-1} \cdot x_1^{-1} \cdot (x_2 \cdot x_1) : \mathtt{G}^2 \to \mathtt{G}^1$. The identity morphism from $\mathtt{G}^n \to \mathtt{G}^n$ is the $n$-tuple $(x_1, x_2, \ldots, x_n)$. You might think that a better $\mathcal{T}$ would contain just three objects $\mathtt{G}^0, \mathtt{G}^1, \mathtt{G}^2$ and morphisms corresponding to the basic operations $1$, ${}^{-1}$ and $\cdot$. But that would not even be a category, and if somehow you managed to make one, you would still have to make sure that $\mathtt{G}^2 = \mathtt{G}^1 \times \mathtt{G}^1$ and that $\mathtt{G}^0$ is the terminal object. It would not really be any prettier. We can now actually show that $\mathtt{G}^m$ is the $m$-fold product of $\mathtt{G}^{1}$'s. The $k$-th projection $\pi_k : \mathtt{G}^m \to \mathtt{G}^1$ is (represented by) the expression $x_k$. I leave the rest as an exercise. Another exercise is to show that $\mathcal{T}$ has finite products computed as $\mathtt{G}^n \times \mathtt{G}^m = \mathtt{G}^{n + m}$. We are still doing a lot of syntax disguised as category theory, but that is a necessary step that allows us to see what sort of category $\mathcal{T}$ is. We are ready to define when a category in general is the description of an algebraic theory. I am phrasing the following definition a bit imprecisely without the technical distraction of requiring a "strict identity on objects functor from $\aleph_0^\mathrm{op}$ ...": Definition [Lawvere]: An algebraic theory is a category with distinct objects $\mathtt{A}^0, \mathtt{A}^1, \mathtt{A}^2, \ldots$ such that $\mathtt{A}^n$ is the $n$-fold product of $\mathtt{A}^1$'s. It turns out that the models of such a theory/category $\mathcal{T}$ are precisely those functors $F : \mathcal{T} \to \mathsf{Set}$ which preserve finite products. Every such functor is already determined by how it maps $\mathtt{A}^1$ and morphisms $\mathtt{A}^n \to \mathtt{A}^1$, which of course correspond to the $n$-ary operations. In a particular case $F$ might be determined by even fewer pieces of information. For example, if $\mathcal{T}$ is the category which describes the theory of groups, $F$ will be determined already by how it maps the morphisms $1 : \mathtt{G}^0 \to \mathtt{G}^1$, $(x_1 \cdot x_2) : \mathtt{G}^2 \to \mathtt{G}^1$, and $(x_1^{-1}) : \mathtt{G}^1 \to \mathtt{G}^1$, because these generate all other morphisms (except projections and pairings, but $F$ preserves products). The whole point of the exercise was to arrive at a non-syntactic notion of "algebraic theory". The next step is to look for examples which really are non-syntactic in nature. Here is one: the category whose objects are Euclidean spaces $\mathbb{R}^n$ and the morphisms are smooth maps. This theory describes what is known as smooth algebras .
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|
70,347 |
While working on a project, I have run into a situation where I have integers x and n so that $x^n \equiv (x+1)^n \equiv (x+2)^n$ mod $p$ for a prime $p$. It seems to me that this an extremely restrictive condition, and I was wondering if there are any results about when (or if?) it can happen, but I couldn't figure out what to search. Any thoughts? What if I add the additional restriction that they are also congruent to $(x+3)^n$, etc. Thanks!
|
Write $a = (x+1)/x, b = (x+2)/x$, then your condition is equivalent to $a^n \equiv b^n \equiv 1 \mod p$ and $2a-b \equiv 1 \mod p$. Now, without loss of generality, you can assume $n|(p-1)$ (otherwise replace $n$ by the gcd of $n$ and $p-1$). Write $m = (p-1)/n$ and $a=u^m,b=v^m$. Finally, your conditions become $2u^m-v^m=1$. This defines a curve and, by the Weil bound, it will have points with $uv \ne 0$ if $p \gg m^4$. So, assuming $n|(p-1)$ what you need is $n \gg p^{3/4}$ to guarantee solutions. On the other hand, if $n$ is small, typically there won't be any solutions. If you add the $x+3$ condition, you still get a curve, but of higher genus, so a similar thing will happen with different bounds.
|
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|
70,421 |
Imagine a smooth curve $c$ sweeping out a unit-radius disk that is
orthogonal to the curve at every point.
Call the result a tube .
I want to restrict the radius of curvature of $c$ to be at most 1.
I am interested in the behavior of a light ray aimed directly down
the central axis of one end of the tube,
as it bounces with perfect reflection from the interior wall
of the tube.
This could be viewed as a model of an optic fiber,
although I want to treat the light ray as a billiard ball
and not a dispersing wave. Q1. Does the light ray always emerge from the other end? I believe the answer is Yes , although there can be close calls: I would be interested in a succinct, convincing proof
(or a counterexample!).
Perhaps I should stipulate that if the ray hits a
boundary singularity (as it nearly does above), it dies;
otherwise it could pass through the center of curvature
and reflect to its own reversal. Q1 Answer: Yes if $c$ is $C^\infty$ (Dimitri Panov), Not necessarily if $c$ is $C^2$ (Anton Petrunin). I explored one particular, maximally convoluted, snake-like tube,
composed of alternating semicircles of radius 2 (so the central
curve $c$ has curvature 1): The lightray behaves seemingly chaotically,
although when I look at the angles the rays make with the $(+x)$-axis,
there is a striking distribution.
Here is a histogram for a tube composed of 1000 semicircles: Q2. Can you offer an explanation for the observed distribution of ray orientations?
Why is the angle $\pm 17^\circ$ so prominent, and there are no
ray angles whose absolute value lies within
$[19^\circ,111^\circ]$?
There are approximately 1.55 ray bounces per semicircle: Why?
Is this approaching $\frac{3}{2}$? Or $\frac{\pi}{2}$? Ideas/insights welcomed—Thanks! Q2 Answer: Dimitri Panov's remarks and the phase portrait below show that likely the ray trajectory is quasiperiodic, which explains the angle histogram, which is, in a sense, a projection of the phase portrait.
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Edited 1. Some suggestions are added at the end concerning Q2 . Edited 2. An "explanation" of spikes at $17^0$ is added at the very end... Q1 I think that the answer to Q1 is positive provided the boundary of the tube is smooth. I'll consider the case of dimension $2$. So, by our assumptions the light is propagating in the strip bounded by two smooth curves $L_+$ and $L_-$ that are both equidistant from the central curve $L$ on distance $\frac{1}{2}$. It is important that the whole strip is foliated by unit intervals orthogonal to all three curves, we call this foliation $F$. Now consider our ray of light $R(t)$ propagating in the strip and introduce a function $\angle(t)$ that equals the angle between $R(t)$ and the orthogonal to $F$ in the direction of the tube. On the entrance of the tube the angle equals $0$. Claim 1. For any moment $t$ we have $\angle(t)<\frac{\pi}{2}$. Proof. Indeed, suppose that at some time $\angle(t)=\frac{\pi}{2}$. This means that $R(t)$ at this time goes in the direction of the foliation $F$. But since any segment of $F$ is a periodic ray in the strip, $R(t)$ must coincide with the segment, which is an absurd. END. So, we see now that $R(t)$ will always propagate in the strip in one direction. So the only possibility for the ray to stay forever in the strip is to accumulate at some point to a segment $F_0$ of the foliation $F$. Let me explain why this is impossible. The main idea is that this is impossible in the case the curve $L$ is a circle of radius $r>1$. In this case it is easy to check the statement. The statement for general $L$ roughly follows from the fact that $L$ can be approximated well by the circle at any point. To spell out the above in more details we can reduce the question to a question of billiards. Indeed, on the two-dimensional set of straight directed segments that join $L_+$ with $L_-$ there is a (partially defined) self map, consisting of two consequent reflection of the segment (first with respect $L_-$ then with respect to $L_+$). All the segments of $F$ are fixed points of the map. We need to show that for $F_0$ there is no a point that tends to it under the infinite iterations of the map. This self-map have three properties: 1) it preserves an area form 2) it fixes a segment (parametrizing the segments of $F$) 3) its linearisation is never identical on the fixed segment. These 3 properties are sufficient to deduce that everything roughly boils down to the following exercise: Exercise. Consider a sequence $a_n$, such that
$a_{n+1}=a_n(1-a_n)$, with $a_0$ positive and less than one . Then $\sum_i a_i=\infty$. PS . I think, that we can ask the curves $L_+$, $L_-$ and $L$ to be only $C^3$-smooth, but the proof uses the fact that the curvature of $L$ is strictly larger than $1$. It is not obvious if this condition can be relaxed. Q2 This is more of a suggestion rather than an answer. But this suggestion might help to get some clues to the answer. I would suggest you to make one more picture, namely, the picture of the Phase portrait - standard thing one does when dealing with a billiard. So, one only needs to consider the trajectory and for each reflection of the trajectory from the upper curve plot the point with two coordinates: (angle of the ray; $x$-coordinate modulo $2$) If you plot 1500 points, a certain shape will appear. Probably the points will fill a two-dimensional domain, but according to the histogram, the trajectory will avoid a large part of the phase portrait. This just reflects the fact that this billiard is not ergodic. I think, that to understand why there are no rays with angles in $[19^0, 111^0]$ one should analyse the boundary of the shape that will appear. This boundary might correspond to some "quasi-periodic" trajectory(ies) of the billiard. Further on Q2. I want to add a couple of remarks on Q2 , that are rather superficial. So, from the experiment of Joseph we see that with some probability it turns out that the original trajectory is quasi-periodic. I.e. the segments constituting the trajectory land on a one-dimensional curve in the 2-dimensional space of all segments. This at least explains the appearance of spikes in the first histogram. Indeed, when you project a measure evenly distributed on a curve on a plane to the $x$ axes - the projected measure will have singularities at points where the vertical lines $x=const$ are tangent to the curve. Now, I guess, that in order to really answer the question one can indeed try to prove that the initial trajectory is quasi-periodic. The billiard is rather simple of course, but I don't know how hard it will be. And before you prove this, you can not be sure that the trajectory is really quasi periodic...
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|
70,605 |
Let $C$ be a nonsingular projective curve of genus $g \geq 0$ over a finite field $\mathbb{F}_q$ with $q$ elements. From this curve, we define the zeta function
$$Z_{C/{\mathbb{F}}_q}(u) = \exp\left(\sum^{\infty}_{n = 1}{\frac{\# C(\mathbb{F}_{q^n})}{n} u^n}\right),$$
valid for all $|u| < q^{-1}$. This zeta function extends meromophicially to $\mathbb{C}$ via the equation
$$Z_{C / \mathbb{F}_q}(u) = \frac{P_{C / \mathbb{F}_q}(u)}{(1 - u) (1 - qu)}$$
for some polynomial with coefficients in $\mathbb{Z}$ that factorises as
$$P_{C/\mathbb{F}_q}(u) = \prod^{2g}_{j = 1}{(1 - \gamma_j u)}$$
with $|\gamma_j| = \sqrt{q}$ and $\gamma_{j + g} = \overline{\gamma_j}$ for all $1 \leq j \leq g$. This last point tells us that $Z_{C / \mathbb{F}_q}(u)$ has a functional equation and satisfies a version of the Riemann hypothesis. What happens if we run this construction in reverse? What if we start with a set of numbers $\gamma_j$, $1 \leq j \leq 2g$, such that $|\gamma_j| = \sqrt{q}$, $\gamma_{j + g} = \overline{\gamma_j}$ for all $1 \leq j \leq g$, and such that the polynomial
$$P(u) = \prod^{2g}_{j = 1}{(1 - \gamma_j u)}$$
has coefficients in $\mathbb{Z}$? Is there a way of telling whether the function
$$\frac{P(u)}{(1 - u) (1 - qu)}$$
is the zeta function of some curve $C$? Furthermore, what is this curve exactly? A simple case of this is if we look at the function
$$\frac{1 - au + qu^2}{(1 - u) (1 - qu)}$$
for some $a \in \mathbb{Z}$ with $|a| \leq 2 \sqrt{q}$. How do we determine whether this function is the zeta function $Z_{C / \mathbb{F}_q}(u)$ of an elliptic curve $C$ over $\mathbb{F}_q$? If it is indeed equal to $Z_{C / \mathbb{F}_q}(u)$, what is the Weierstrass equation for $C$ (assuming $\mathrm{char}(q) \geq 5$)?
|
To determine which potential zeta functions are actual zeta functions of curves is very difficult. The zeta function of a curve is determined by the zeta function of its Jacobian so one could instead ask which potential zeta functions are zeta functions of abelian varieties. This problem is solved (by Tate though as I recall Waterhouse was also involved in working out some details) and the answer is that essentially every $P(u)$ that could (i.e., having roots with the proper absolute values) occur with some extra restriction having to do with the endomorphism ring of the abelian variety having to be an order in a semi-simple algebra with some non-split factor. The conditions are anyway very explicit and reasonably easy to check. The next step is to pick out the zeta functions of principally polarised abelian varieties among all of them. This involves more arithmetic but is also also feasible. The tricky part is the Schottky problem, to pick out the Jacobians among the principally polarised abelian varieties. This is in principle solved (at least in characteristic zero, I am less sure about positive characteristic) but any of the existing solutions meshes very badly with the problem of zeta functions as the solution to the problem above is very non-explicit and only tells you about the existence of a p.p.a.v. with given zeta function not a description of all of them. Even where all of this can be done, for instance in genus $1$ where the Schottky problem is trivial, it is non-trivial to get actual equations. The reason for this is essentially that Tate's existence argument is through construction of a characteristic $0$ abelian variety with complex multiplication and then reducing it modulo $p$. Hence, the only way to be explicit would seem to be to first get the Weierstrass equation for the CM-curve. This is certainly possible (and fairly, for some definition of that term, efficiently) but it is far from easy. Of course, it still doesn't give all of the p.p.a.v.'s with given zeta function (though in some cases, for instance the ordinary case, all of them are the reduction of a CM abelian variety). Addendum : As Noam points out there are a (small) number of general results excluding some zeta functions for curves where the p.p.a.v. exists. These generally concern improvements of the Weil bounds exploiting the fact that given the zeta function we can compute the number of points of the curve over the field and extensions of it and that these number should be non-negative and increase as the field increases (though the actual proofs are sometimes quite sophisticated). To see if the bounds obtained are sharp a lot of effort has also been expended on constructing specific curves with many points. In some cases special arguments can be used to exclude some values. See this site for tables and these slides for a description of a particularly detailed analysis of some cases.
|
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|
70,692 |
Why is the Chebyshev function $\theta(x) = \sum_{p\le x}\log p$ useful in the proof of the prime number theorem. Does anyone have a conceptual argument to motivate why looking at $\sum_{p\le x} \log p$ is relevant and say something random like $\sum_{p\le x}\log\log p$ is not useful or for that matter any other random function $f$ and $\sum_{p\le x} f(p)$ is not relevant.
|
There are several ideas here, some mentioned in the other answers: One: When Gauss was a boy (by the dates found on his notes he was approximately 16) he noticed that the primes appear with density $ \frac{1}{\log x}$ around $x$. Then, instead of counting primes and looking at the function $\pi (x)$, lets weight by the natural density and look at $\sum_{p\leq x} \log p$. Since we are weighting by what we think is the density, we expect it to be asymptotic to be $x$. Two: Differentiation of Dirichlet series. If $$ A(s)=\sum_{n=1}^{\infty} a_{n} n^{-s} $$ then $$ (A(s))'=-\sum_{n=1}^{\infty} a_{n} log(n) n^{-s}$$
The $\log$ term appears naturally in the differentiation of Dirichlet series. Taking the convolution of $\log n$ with the Mobius function (that is multiplying by $\frac{1}{\zeta(s)}$) then gives the $\Lambda(n)$ mentioned above. The $\mu$ function is really the special thing here, not the logarithm. Expanding on this, there are other weightings besides $\log p$ which arise naturally from taking derivatives. Instead we can look at $\zeta^{''}(s)=\sum_{n=1}^\infty (\log n)^2 n^{-s}$, and then multiply by $\frac{1}{\zeta(s)}$ as before. This leads us to examine the sum $$\sum_{n\leq x} (\mu*\log^2 )(n)$$ (The $*$ is Dirichlet convolution) By looking at the above sum, Selberg was able to prove his famous identity which was at the center of the first elementary proof of the prime number theorem: $$\sum_{p \leq x} log^2 p +\sum_{pq\leq x}(\log p)(\log q) =2x\log x +O(x).$$ Three: The primes are intimately connected to the zeros of $\zeta(s)$, and contour integrals of $\frac{1}{\zeta(s)}$. (Notice it was featured everywhere here so far) We can actually prove that $$\sum_{p^k \leq x} \log p= x - \sum_{\rho :\ \zeta(\rho)=0} \frac{x^{\rho}}{\rho} +\frac{\zeta'(0)}{\zeta(0)} $$ Notice that the above is an equality , which is remarkable since the left hand side is a step function. (Somehow, at prime powers all of the zeros of zeta conspire and make the function jump.)
|
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|
70,710 |
Let $\mathcal{P}(S^{\infty})$ denote the set of probability measures on the unit sphere $S^{\infty} \subset \mathcal{H}$ in the Hilbert space of states of a quantum mechanical system. Measurement of an observable $\Omega$ corresponds to orthogonal projection, sending $\delta_{|\psi \rangle}$ to a particular measure supported on the eigenvectors of $\Omega$, thus inducing a map $T_{\Omega}: \mathcal{P}(S^{\infty}) \rightarrow \mathcal{P}(S^{\infty})$. If we think of $T_{\Omega}$ as the transition matrix of a Markov chain, we can say that a continuum $\lbrace \Omega_t \rbrace_{t \in \mathbb{R} \geq 0}$ of observables induces a stochastic process on $S^{\infty}$. If we let $\Omega_t = H(t)$, the time-dependent Hamiltonian of our system, is the associated stochastic process the deterministic one described by the Schrödinger equation? What does this construction have to do with the time-energy uncertainty relation?
|
There are several ideas here, some mentioned in the other answers: One: When Gauss was a boy (by the dates found on his notes he was approximately 16) he noticed that the primes appear with density $ \frac{1}{\log x}$ around $x$. Then, instead of counting primes and looking at the function $\pi (x)$, lets weight by the natural density and look at $\sum_{p\leq x} \log p$. Since we are weighting by what we think is the density, we expect it to be asymptotic to be $x$. Two: Differentiation of Dirichlet series. If $$ A(s)=\sum_{n=1}^{\infty} a_{n} n^{-s} $$ then $$ (A(s))'=-\sum_{n=1}^{\infty} a_{n} log(n) n^{-s}$$
The $\log$ term appears naturally in the differentiation of Dirichlet series. Taking the convolution of $\log n$ with the Mobius function (that is multiplying by $\frac{1}{\zeta(s)}$) then gives the $\Lambda(n)$ mentioned above. The $\mu$ function is really the special thing here, not the logarithm. Expanding on this, there are other weightings besides $\log p$ which arise naturally from taking derivatives. Instead we can look at $\zeta^{''}(s)=\sum_{n=1}^\infty (\log n)^2 n^{-s}$, and then multiply by $\frac{1}{\zeta(s)}$ as before. This leads us to examine the sum $$\sum_{n\leq x} (\mu*\log^2 )(n)$$ (The $*$ is Dirichlet convolution) By looking at the above sum, Selberg was able to prove his famous identity which was at the center of the first elementary proof of the prime number theorem: $$\sum_{p \leq x} log^2 p +\sum_{pq\leq x}(\log p)(\log q) =2x\log x +O(x).$$ Three: The primes are intimately connected to the zeros of $\zeta(s)$, and contour integrals of $\frac{1}{\zeta(s)}$. (Notice it was featured everywhere here so far) We can actually prove that $$\sum_{p^k \leq x} \log p= x - \sum_{\rho :\ \zeta(\rho)=0} \frac{x^{\rho}}{\rho} +\frac{\zeta'(0)}{\zeta(0)} $$ Notice that the above is an equality , which is remarkable since the left hand side is a step function. (Somehow, at prime powers all of the zeros of zeta conspire and make the function jump.)
|
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"https://mathoverflow.net/questions/70710",
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|
70,714 |
In the world of real algebraic geometry there are natural probabilistic questions one can ask: you can make sense of a random hypersurface of degree d in some projective space and ask about its expected topology where "expected" makes sense because there are sensible measures on the space of hypersurfaces. See Welschinger-Gayet http://arxiv.org/abs/1107.2288 and http://arxiv.org/abs/1005.3228 for recent progress on such questions (e.g. what is the expected Betti number of a random real hypersurface of degree d?). In geometry more generally you might want to make statements like "a general manifold is aspherical" or "a general manifold has positive simplicial volume". It seems difficult to construct sensible measures for which these questions have answers: to talk about probability you need some way of producing manifolds (and then distinguishing them) in a random way. However, Cheeger proved that for fixed L there is only a finite set $D_L$ of diffeomorphism classes of manifold admitting a Riemannian metric with sectional curvatures bounded above in norm by L, volume bounded below by 1/L and the diameter bounded above by L (see the first theorem in Peters "Cheeger's finiteness theorem for diffeomorphism classes of Riemannian manifolds" http://www.reference-global.com/doi/abs/10.1515/crll.1984.349.77 ). This means that you can ask questions like "what is the average total Betti number of a manifold in $D_L$" and "how does this increase with L?" (are there exponential upper bounds?), or one can try to make sense of the limit as $L\rightarrow\infty$ of the proportion of manifolds in $D_L$ with zero simplicial volume. Are there any known concrete answers to these questions, or other formulations of the questions which lead to answers?
|
To define a random $n$-manifold you typically need to define a complexity on the set $\mathcal M_n$ of all $n$-manifolds you want to consider, which satisfies a finiteness property: for every $k$, there are only a finite number of manifolds having complexity at most $k$. There are various ways to do this in a combinatorial framework. The minimum number of simplexes $t(M)$ in a triangulation of $M$ is a natural example. I don't think there is much known on random manifolds in this context. Since $t$ is roughly subadditive on connected sums, a random manifold might be a connected sum of many manifolds and would hence be far from being aspherical. In dimension 3 one may restrict to irreducible manifolds. It might seem reasonable to expect that in this conext a random 3-manifold is hyperbolic, but this is still unknown. The first segment $t(M)\leq 11$ shows up a huge number of graph manifolds, see the tables here and here . The hard point here is that it is very difficult to estimate such complexity from below, even for simple manifolds like lens spaces. As an example, the number of hyperbolic manifolds having complexity smaller than $t$ grows more than exponentially with $t$, and the number of lens spaces is conjecturally roughly $2^t$. However, I think we are not yet able to say that the number of lens spaces does not grow more than exponentially. Another fact that shows our ignorance in this conext is the following: we still don't know if the number of triangulation of the three-sphere grows exponentially with the number of tetrahedra, see Gromov's recent questions. As pointed by Jean-Marc Schlenker, there is another natural complexity in dimension 3 which is easier to treat and gives various nice results: we can consider the smaller set $\mathcal M_3^g$ of all 3-manifolds decomposing into two genus-$g$ handlebodies. A manifold there is determined by an element of the mapping class group MCG of the surface $\Sigma_g$ of genus $g$, and after fixing a set of generators for MCG we can define the complexity of one such element as the minimum lenght of a word which represents it. Many results have been obtained in this context by Nathan Dunfield (with D. and W. Thurston, and H. Wong), Juan Souto and others. In dimension 4 there are various combinatorial notions of complexity one may use, but they are difficult to treat. One can use Kirby diagrams to represent 4-manifolds and define a complexity by counting the number of crossings in the diagrams as I did here . It is very easy to produce doubles of 2-handlebodies in this context (draw a random diagram and add small 0-framed unknots encircling every component) and to perform blow-ups, so I suppose that in this context most manifolds are of this type: these manifolds are never aspherical and have simplicial norm zero. In this context, Auckly has proved that there is a big discrepancy between homeomorphic and diffeomorphic classes of manifolds: the number of simply connected manifolds of complexity up to $n$ seen up to homeomorphism grows like $n^2$, whereas the number of simply connected manifolds seen up to diffeomorphism grows more than polinomially.
|
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|
70,790 |
By Matiyasevich, for every recursively enumerable set $A$ of natural numbers there exists a polynomial $f(x_1,...,x_n)$ with integer coefficients such that for every $p\ge 0$, $f(x_1,...,x_n)=p$ has integer solutions if and only if $p\in A$. Now suppose that $A$ is a set of natural numbers with membership problem in $NP$. Is there a polynomial $f$ with integer coefficients such that $f(x_1,...,x_n)=p$ has integer solutions if and only if $p\in A$ and there exists a solution with $||x_i||\le Cp^s$ for some fixed $s, C$, where $||x_i||$ is the length of $x_i$ in binary (i.e. $\sim \log |x_i|$)? Clearly the converse is true: if such a polynomial exists, then the membership problem for $A$ is in NP.
|
I don’t know about the particular form of the polynomial you are using, but in general, it is a well-known open problem whether every NP set can be represented by a Diophantine equation with a polynomial bound on the length of the solutions. Adleman and Manders proved that the set $\{\langle a,b,c\rangle\in\mathbb N^3:(\exists x,y\in\mathbb N)(ax^2+by=c)\}$ is NP-complete, hence the answer is positive iff the class of such representable sets is closed under polynomial-time reductions, but it’s not clear whether the latter is actually true or not. See the introduction of Pollett for an overview of some known partial results.
|
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|
71,201 |
I'm wondering if there are examples of statements that have been proven whose consistency proofs came before the proofs of the statements themselves. More informally, I'm wondering how promising in general is the approach of attempting a consistency proof for a statement when faced with a statement that seems true but difficult to prove. Background: If a statement is provable from a set of axioms, then that statement is obviously consistent (assuming the set of axioms is consistent). So provability is stronger than consistency. This might lead one to think that constructing a consistency proof for a statement should be strictly easier than constructing a proof. Yet consistency proofs (at least oft-cited ones, for example those by Godel and Cohen about the Continuum Hypothesis) seem to require a high level of sophistication (though this might be a byproduct of the fact that consistency proofs like these are for the special class of statements that cannot be proven). For statements that can be proven then, are there cases where their consistency proofs are easier or came before the proofs themselves? Update: Thanks a lot, everyone, for the great answers so far. The number and existence of these examples is interesting to me, as well as the fact that they all rely on the same technique of first proving something using an additional axiom (an approach first suggested by Michael Greinecker). That hadn't occurred to me. I wonder if there are other approaches.
|
Here are my favorite examples of statements whose consistency was established and cherished before their proof. 1. The Keisler-Shelah isomorphism theorem stating that two elementarily equivalent structures have isomorphic ultrapowers (proved using in $ZFC+GCH$ by Keisler in 1964, and in $ZFC$ by Shelah in 1971). 2. Several algebraic results (normal forms, divison algorithms, etc) concerning the "left-distributive algebras of one generator" were first established by Richard Laver by assuming (very) large cardinals (known as (I3) in the literature). Later Patrick Dehornoy eliminated the large cardinal assumption in these results by giving a representation in the braid group. By the way, a left distributive algebra is a set with one binary operation * satisfying the left distributive law
$a*(b*c)=(a*b)*(a*c)$; the operation of conjugation in a group is an example. Here is an old paper of Laver about this topic; there is by now a large literature on the subject. 3. In some cases, the consistency proof of a statement $S$ can be combined with some absoluteness argument to yield a proof of $S$ (this was hinted at in the second example cited by Steprans). There are all sorts of absoluteness theorems in logic; the standard tool of the trade is the Shoenfield absoluteness theorem that shows that all sorts of consistency results can be translated into $ZFC$-proofs.
|
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|
71,339 |
This may not be a research level math question, but I believe it is still relevant to Math Overflow. What general resources exist for students in highschool who are very interested in Mathematics? What advice would you give to a young student to encourage them, and nurture their interest in mathematics? If a young high school student came to you and said they were very interested in math, and wanted to know what to do to keep learning, what would you tell them? Thank you for your help,
|
This is an answer to your question, What general resources exist for students in highschool who are very interested in Mathematics? A patient teacher is the best resource for an interested high school student. Fundamentally what students lack is not access to mathematical content (cf. Wikipedia and countless books) but of mathematical thinkers. The high school curriculum is rigid and this steers high school math teachers into a pretty doctrinal way of thinking (bless those teachers who break out of the mold); consequently, the students follow along because they are not exposed to anything else. The best thing you can do for a high school student is to show him or her how to think mathematically. Walk through problems with the student, explore until you find something you don't know and are curious about, ask the natural questions, and most importantly, try to set the student to up to ask the natural questions. Try to present things in a way so that you are not the lecturer who knows everything, but rather a fellow intellectual explorer. I definitely agree with Steve Huntsman: any project should feature a computational component. Students today aren't afraid of computers, and strong programming skills are invaluable in both the science and business worlds. As Steve says, implementing an algorithm forces students to really think, and isn't entirely unlike mathematical reasoning. Plus, serious proofs tend to be out of reach of even most undergraduates, whereas numerical simulations can let any student get a hands-on interaction with a mathematical problem. Let me demonstrate all of this with a timely anecdote. I am teaching this summer at the annual Stony Brook summer math camp . After my introductory probability class today, a student came up to me saying she wanted to do research in mathematics. She had done some computer modeling of molecular dynamics in biochemistry, but I have no training in that area and even if I did, I imagine the mathematical models would be quite complicated. Since any such model must contain some random dynamics (namely, diffusion), I decided to introduce random walks to her. I set up some definitions, went through some examples, and we played around with a simple, lattice-based model of molecular dynamics. After chatting for an hour or so, she asked the natural question: if we have a random walk on the 2-dimensional lattice, is it guaranteed to return? The answer is given by Pólya's theorem: a simple random walk is recurrent in 1 and 2 dimensions, but is transient in 3 dimensions and above. But a theorem statement is not a satisfactory answer, and I knew that I couldn't explain the proof of the theorem at the level of an advanced high school student. Thus her research project was born: understand the ideas behind an elementary proof of Pólya's theorem, and use a computer to estimate $p_d(n)$, the probability that a $d$-dimensional random walk returns to the origin in exactly $n$ steps. Voilà, a tailor-made project motivated by the student's interests, featuring both a mathematical and a computational component.
|
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|
71,630 |
My question originates from the book of Silverman "The Aritmetic of Elliptic Curves", 2nd edition (call it [S]). On p. 273 of [S] the author is considering an elliptic curve $E/K$ defined over a number field $K$ and he introduces the notion of a $v$-adic distance from $P$ to $Q$. This is done as follows: Firstly, let's fix an absolute value (archimedean or not) $v$ of $K$ and a point $Q\in E(K_v)$ (here $K_v$ is the completion of $K$ at $v$). Next let's pick a function $t_Q \in K_v(E)$ defined over $K_v$ which vanishes at $Q$ to the order $e$ but has no other zeroes. Now the $v$-adic distance from $P \in E(K_v)$ to $Q$ is defined to be $d_v(P, Q) := \min (|t_Q(P)|_v^{1/e}, 1)$. We will say that $P$ goes to $Q$, written $P~\xrightarrow{v}~ Q$, if $d_v(P, Q) \rightarrow 0$. Later in the text (among other places in the proof of IX.2.2) the author considers a function $\phi\in K_v(E)$ which is regular at $Q$ and claims that this means that $|\phi(P)|_v$ is bounded away from $0$ and $\infty$ if $P~\xrightarrow{v}~ Q$. I have a couple of questions about this: How does one choose a $t_Q$ that works? In the footnote in [S] it is demonstrated how one could use Riemann-Roch to pick a $t_Q$ that has a zero only at $Q$. It seems to me however that such a procedure will not make sure that $t_Q$ is defined over $K_v$ since $K_v$ is not algebraically closed. For $\phi$ as above which does not vanish nor has a pole at $Q$, how does one see that $|\phi(P)|_v$ is bounded away from $0$ and $\infty$ as $P~\xrightarrow{v}~ Q$? Do these $d_v$ have anything to do with defining a topology on $E(K_v)$? I assume not, since I don't see how to make sense of it; but then on the other hand they are called "distance functions"...
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You can choose $t_Q$ to be defined over $K_v$, since the divisor $n(Q_v)$ is defined over $K_v$, and for large enough $n$ there will be a global section. Note that Riemann-Roch works over non-algebraically closed fields this way. Or you can choose a basis defined over some finite Galois extension of $K_v$, and then taking appropriate linear combinations of the Galois conjugates, get a function defined over $K_v$. See Proposition II.5.8 in [S]. The function defined in the text is only a reasonable "distance function" in the sense that it measures the distance from $P$ to the fixed point $Q$. For the purposes of this proof, that's fine. If you want to define the $v$-adic topology, you need to be a little more careful. Locally around $Q$ you could use $$d_v(P_1,P_2)=min(|t_Q(P_1)-t_Q(P_2)|^{1/e},1)$$, but that still only works in a neighborhood of $Q$, i.e., in a set $$\{P : d_v(P,Q)<\epsilon\}$$ for a sufficiently small $\epsilon$. Using local height functions, more precisely the local height relative to the diagonal in $E(K_v)\times E(K_v)$, one gets a "good" distance function that is defined everywhere. See for example Lang's Fundamentals of Diophantine Geometry or the book Diophantine Geometry that Hindry and I wrote for the general construction of local height functions.
|
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|
71,727 |
Tate's thesis showed how to profitably analyze $\zeta$ functions of number fields in terms of adelic points on the multiplicative group. In particular, combining Fourier analysis and topology, Tate gave new and cleaner proofs of the finiteness of the class group, Dirichlet's theorem on the rank of the unit group, and the functional equation of the $\zeta$ -function. Weil's textbook Basic Number Theory re-presented algebraic number theory from the adelic perspective, showing how adelic methods could provide simple and unified proofs of all the results proved in a first course in algebraic number theory (and perhaps in a second one as well.) I have heard rumors that one can similarly rewrite the theory of elliptic curves in adelic terms, and that doing so gives intuition for the BSD conjecture. Franz Lemmermeyer's paper Conics, a poor man's elliptic curves provides a brief sketch. Is there a survey paper or textbook which lays this picture out in full, as Weil did for the multiplicative group, pointing out the connections between the adelic and the classical language at each step, and ideally discussing the connections with BSD? Note: This question has a peculiar history. See this meta thread if you are interested, but feel free to ignore the past and just answer the question if you are not.
|
I don't think that such a survey paper or textbook exists, but the closest thing I know of is "A note on height pairings, Tamagawa numbers, and the Birch and Swinnerton-Dyer conjecture" by Spencer Bloch, Invent. Math. v.58, no.1, pp. 65-76, 1980. Here's an abbreviated history, picking up where you left off: Takashi Ono wrote a paper "On the Tamagawa number of algebraic tori", Annals of Math., v.78, no. 1, July 1963. In that paper, Ono computes the volume of $T^1(A) / T(F)$, where $T$ is an algebraic torus over a number field $F$, and $A$ is the adele ring, and $T^1(A)$ denotes the intersection of kernels of $\vert \chi \vert$ as $\chi$ ranges over $F$-rational characters of $T$. Ono's formula states that this volume (called a Tamagawa number, but not to be confused with the local Tamagawa numbers $c_v$) equals $ \vert Pic_{tor}(T) \vert / \vert Sha(T) \vert$. The numerator is the order of the torsion subgroup of the Picard group of $T$. The denominator is the order of the Tate-Shafarevich group of $T$. Most of the arithmetic is contained in the normalization of the measure on the quotient space $T^1(A) / T(F)$ -- this normalization of measure uses the L-function (an Artin L-function) of $T$, and the special case $T = G_m$ corresponds to the Dirichlet class number formula for $F$. From looking at Ono's paper (an earlier Annals paper from 1961), it appears that Weil and Tate were influential in his work. Fast forwarding to 1980 (skipping lots of great things for reductive groups), here's a brief summary of what Bloch does (in the Inventiones paper mentioned above). He begins with an abelian variety $E$ over a global field $F$ (I already used $A$ for the adeles). Using the fact that the dual abelian variety $\hat E$ can also be viewed as $Ext(E, G_m)$, Bloch uses the Mordell-Weil lattice $L$ of $F$-rational points on $\hat E$ to construct an extension of algebraic groups over $F$:
$$1 \rightarrow T \rightarrow X \rightarrow E \rightarrow 1$$
in which $T$ is an $F$-split torus with character lattice $L$. Remarkably, Bloch proves that $X(F)$ is discrete and cocompact in $X(A)$. Moreover, most suggestively, Bloch proves that the BSD conjecture for $E$ is equivalent to the conjecture that the volume of $X(A) / X(F)$, with respect to a suitably normalized measure, equals $\vert Pic_{tor}(X) \vert / \vert Sha(X) \vert$. Of course, the meat of Bloch's approach is in the normalization of measure, which uses the L-function of $E$. I once gave a truly disastrous talk as a graduate student about Bloch's paper, in which all this normalization of measure stuff completely escaped me. I still find Bloch's paper very difficult and mysterious. It seems that it is mostly cited for its novel construction of height pairings, but not much has been done (publicly) with its interpretation of BSD.
|
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|
71,736 |
Is there a known formula for the number of closed walks of length (exactly) $r$ on the $n$-cube? If not, what are the best known upper and lower bounds? [Edit] Note: the walk can repeat vertices.
|
Yes (assuming a closed walk can repeat vertices). For any finite graph $G$ with adjacency matrix $A$, the total number of closed walks of length $r$ is given by $$\text{tr } A^r = \sum_i \lambda_i^r$$ where $\lambda_i$ runs over all the eigenvalues of $A$. So it suffices to compute the eigenvalues of the adjacency matrix of the $n$-cube. But the $n$-cube is just the Cayley graph of $(\mathbb{Z}/2\mathbb{Z})^n$ with the standard generators, and the eigenvalues of a Cayley graph of any finite abelian group can be computed using the discrete Fourier transform (since the characters of the group automatically given eigenvectors of the adjacency matrix). We find that the eigenvalue $n - 2j$ occurs with multiplicity ${n \choose j}$, hence $$\text{tr } A^r = \sum_{j=0}^n {n \choose j} (n - 2j)^r.$$ For fixed $n$ as $r \to \infty$ the dominant term is given by $n^r + (-n)^r$.
|
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|
71,765 |
Many of us presume that mathematics studies objects. In agreement with this, set theorists often say that they study the well founded hereditarily extensional objects generated ex nihilo by the "process" of repeatedly forming the powerset of what has already been generated and, when appropriate, forming the union of what preceded. But the practice of set theorists belies this, since they tend—for instance, in the theories of inner models and large cardinal embeddings—to study classes that, on pain of contradicting the standard axioms, are never "generated" at any stage of this process. In particular, faced with the independence results, many set theorists suggest that each statement about sets—regardless of whether it be independent of the standard axioms, or indeed of whether it be formalizable in the first order language of set theory—is either true or false about the class $V$ of all objects formed by the above mentioned process. For them, set theory is the attempt to uncover the truth about $V$. This tendency is at odds with what I said set theorists study, because proper classes, though well founded and hereditarily extensional, are not objects. I do not mean just that proper classes are not sets. Rather, I suggest that no tenable distinction has been, nor can be, made between well founded hereditarily extensional objects that are sets, and those that aren't. Of course, this philosophical claim cannot be proved. Instead, I offer a persuasion that I hope will provoke you to enlighten me with your thoughts. Suppose the distinction were made. Then in particular, $V$ is an object but not a set. Prima facie, it makes sense to speak of the powerclass of $V$—that is, the collection of all hereditarily well founded objects that can be formed as "combinations" of objects in $V$. This specification should raise no more suspicions than the standard description of the powerset operation; the burden is on him who wishes to say otherwise. With the powerclass of $V$ in hand, we may consider the collection of all hereditarily well founded objects included in it, and so on, imitating the process that formed $V$ itself. Let $W$ be the "hyperclass" of all well founded hereditarily extensional objects formed by this new process. Since we can distinguish between well founded hereditarily extensional objects that are sets and those that aren't, we should be able to mirror the distinction here, putting on the one hand the proper hyperclasses and on the other the sets and classes. Continuing in this fashion, distinguish between sets, classes, hyperclasses, $n$-hyperclasses, $\alpha$-hyperclasses, $\Omega$-hyperclasses, and so on for as long as you can draw indices from the ordinals, hyperordinals, and other transitive hereditarily extensional objects well ordered by membership, hypermembership, or whatever. It seems that this process will continue without end: we will never reach a stage where it does not make sense to form the collection of all well founded hereditarily extensional objects whose extensions have already been generated. We will never obtain an object consisting in everything that can be formed in this fashion. For me, this undermines the supposed distinction between well founded hereditarily extensional objects that are sets, and those that aren't. Having assumed the distinction made, we were led to the conclusion of the preceding paragraph. But that is no better than the conclusion that proper classes, including $V$ itself, are not objects. Indeed, it is worse, for in arriving at it we relegated set theory to the study of just the first two strata of a much richer universe. Would it not have been better to admit at the outset that proper classes are not objects? If we did that, would set theory suffer? In particular, how would it affect the idea that each statement about sets is either true or false?
|
Proper classes are not objects. They do not exist. Talking about them is a convenient abbreviation for certain statements about sets. (For example, $V=L$ abbreviates "all sets are constructible.") If proper classes were objects, they should be included among the sets, and the cumulative hierarchy should, as was pointed out in the question, continue much farther, but in fact, it already continues arbitrarily far. In particular, when I talk about statements being true in $V$, I mean simply that the quantified variables are to be interpreted as ranging over arbitrary sets. It is an unfortunate by-product of the set-theoretic formalization of semantics that many people believe that, in order to talk about variables ranging over arbitrary sets (or arbitrary widgets or whatever), we need an object, a set, that contains all the sets (or all the widgets or whatever). In fact, there is no such need unless we want to formalize this notion of truth within set theory. Anyone who wants to formalize within set theory the notion of "truth in $V$" is out of luck anyway, by Tarski's theorem on undefinability of truth. Considerations like these are what prompt me to view ZFC together with additional axioms (such as the universe axiom of Grothendieck and Tarski) as a reasonable foundational system, in contrast to Morse-Kelley set theory. A detailed explanation of how to use proper classes as abbreviations and how to unabbreviate statements involving them is given in an early chapter of Jensen's "Modelle der Mengenlehre". (The idea goes back at least to Quine, who used it not only for proper classes but even for sets, developing a way to understand talk about sets as being about "virtual sets" and avoiding any ontological commitment to sets.) Finally, I should emphasize, just in case it's not obvious, that what I have written here is my (current) philosophical opinion, not by any stretch of the imagination mathematical fact.
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|
71,909 |
I've been looking high and low for a mathematical book on String Theory. The only book I could find was "A Mathematical Introduction to String Theory" by Albeverio, Jost, Paycha and Scarlatti. I only stumbled upon this because I really like Jost's other books. After reading it, I found myself craving more. However, the above book is extremely short, and sadly doesn't cover a lot. I've been having trouble reading the current textbooks on String Theory. To me, it often seems that certain mathematical concepts are simply applied without checking or reasoning, something that has been bugging me ever since studying QFT. As I'm not a physicist, it's rather likely that I'm still lacking the intuition to see these things. My question is, are there any other introductory books/review-articles on String Theory written in a more mathematically rigorous way? By this I mean, books that are written in the style of a common math book? ("Definition-Theorem-Proof-Style")
|
There is the two volume set Quantum Fields and Strings: A Course for Mathematicians that attempts to bridge the gap. Here's an Amazon link: https://www.amazon.com/Quantum-Fields-Strings-Course-Mathematicians/dp/0821820141 . (If it is gauche to give an Amazon link, please change my post, o moderators!)
|
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|
71,950 |
In Hartshorne IV.2, notions related to ramification and branching are introduced, but only for curves. The main result is the Hurwitz formula. Now if you have a finite surjective morphism between nonsingular, quasi-projective varieties, then the notion of ramification (divisor) would still make sense and we can also still talk about the degree of a canonical divisor. It also seemed to me like no result in IV.2 really uses the fact that $X$ and $Y$ are of dimension $1$. So I ask, can I replace $f$ by a finite, dominant, separable morphism $X\to Y$ of nonsingular, quasi-projective varieties of arbitrary dimension? That is, of course, up to and including Proposition 2.3. If this is so, can we say anything about the degree of a canonical divisor in dimension greater than one? Maybe in special cases?
|
degree of the canonical divisor doesn't make any sense as already pointed out by Mohammed. On the other hand, by "purity of the branch locus", the branch locus, as well as the ramification locus of $f$ is a sum of irreducible divisors. Denote by $R_i$ the irreducible components of the ramification locus. Then, the local rings of the generic points of the $R_i$ are DVR's, and one can associate ramification indices $e_i$ to them (as explained in Hartshorne's book). Local computations show that
$$
\omega_X \cong f^*\omega_Y\otimes{\cal O}_Y(\sum_i (e_i-1)R_i)
$$
In fact, one checks this outside the intersections of the $R_i$, where these local computations are easy. This gives the desired isomorphism outside codimension $2$, and by reflexivity, the desired isomorphism holds everywhere.
|
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|
72,013 |
Sorry for asking a linear algebra question on a research forum, but this seems to be either a case of extreme blindness on my side, or a case of a result lying much deeper than it seems. The following theorem is "easily seen" according to a text I have been reading (more precisely, it is part of Proposition I.1.2 in that text): Theorem 1. Let $A$, $B$, $C$, $D$ be four vector spaces over a field $k$. Then, the canonical map $\mathrm{Hom}\left(A,C\right)\otimes\mathrm{Hom}\left(B,D\right) \to \mathrm{Hom}\left(A\otimes B,C\otimes D\right)$, $f\otimes g\mapsto \left(a\otimes b\mapsto f\left(a\right)\otimes g\left(b\right)\right)$ is injective. I see how this is trivial if $A$ and $B$ are finite-dimensional. I also see that it is indeed easy if $C$ and $D$ are finite-dimensional. But without finite-dimensionality conditions I have nowhere to start. The $\mathrm{Hom}$ functor does not commute with direct sums, while $\otimes$ does not commute with direct products (or does it over a field?), so there seems to be no easy way to reduce it to finite-dimensional cases. How can we proceed then? Also, is there any application of the above theorem outside of the two cases I mentioned? To make this more interesting, how much is saved if we let $k$ be a commutative ring with $1$, and require (say) flatness instead of freeness?
|
Suppose $\sum f_i\otimes g_i$ is in the kernel and assume that the $g_i$ are linearly independent. For every $a\in A$ and $\lambda\in C^*$, we have $\sum\lambda(f_i(a))g_i=0$, so by assumption $\lambda(f_i(a))=0$ for all $i$. Since $a$ and $\lambda$ were arbitrary, this implies $f_i=0$ for all $i$.
|
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|
72,020 |
Let $f:X\to Y$ be a generically finite proper morphism of varieties. There is some locus in $Y$ over which the fiber of $f$ is positive dimensional, so we blow it up, along with the preimage of it in $X$ to get a map $\tilde{f}:\tilde{X}\to\tilde{Y}$ which has finite fibers. Are there any nice conditions that will guarantee that the map $\tilde{f}$ is flat?
|
Suppose $\sum f_i\otimes g_i$ is in the kernel and assume that the $g_i$ are linearly independent. For every $a\in A$ and $\lambda\in C^*$, we have $\sum\lambda(f_i(a))g_i=0$, so by assumption $\lambda(f_i(a))=0$ for all $i$. Since $a$ and $\lambda$ were arbitrary, this implies $f_i=0$ for all $i$.
|
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|
72,052 |
I am having a problem which should not exist. I am reading what I believe to be an important paper by a person - let me call him/her $A$ - whom I believe to be a serious and talented mathematician. A lemma in this paper is proven by means of an argument which, if correct, is a highly elegant piece of mental acrobatics in the spirit of Grothendieck, where a complicated situation is reduced to a simple one by embedding the objects of study in much larger (but ultimately better) object. Unfortunately, the beauty of this argument is - for me - marred by a doubt about its correctness. In my eyes, the argument rests upon a confusion of two objects which are not equal and should not be, but have the same name by force of an abuse of notation going awry. A dozen of emails exchanged with $A$ did not clear up the situation, and I start feeling that this is unlikely to improve; what is likely is that after a few more mails the correspondence will degenerate into a flamewar (as any prolonged arguments with my participation seem to do, for some reasons unknown). The fact that $A$ is not a native English speaker adds to the difficulty. At this point, I can think of several ways to proceed: Let go. There is a number of reasons for me not to choose this option; first of all, I really want to know whether the proof of the lemma is correct or not (even though there seems to be a different proof in literature, although not of that beauty), but this has also become, for me, a matter of idealism and an exercise in tenacity (in its cheapest manifestation - it's not like writing emails is hard work...). Construct a counterexample. This is complicated by the fact that I am attacking the proof, not the theorem (which seems to be correct). Yet I think I have done so, and $A$ failed to really address the counterexample. But given the frequent misunderstandings between us (not least because of the language barrier) I am not sure whether $A$ has realized that I am talking counterexamples at all - and whether there is a way to tell this without switching to what will be probably understood as an aggressive tone. Request $A$ to break down the argument into simple steps, eschewing abuse of notations. This means, in the particular case I am talking about, requesting $A$ to write two pages in his/her free time and respond to some irritating criticism of these pages with the prospect of seeing them destroyed by a counterexample. I am not sure this counts as courteous. Besides, the paper is about 10 years old - most authors do not even bother answering questions on their work of such age. Go public (by asking on MO or similarly). This is something I really want to avoid as long as there is no other way. Neither criticizing $A$ as a person/scientist, nor devaluing the paper (which consists of far more than the lemma in question...) is among my goals; besides I cannot rule out as improbable that the error is on my side (and my experience shows that even in cases when I could rule this out, it still often was on my side). Have a break and return to the question in a month or so. I am expecting to hear this (seems to be a popular answer to lots of questions...) yet I am not sure how this can be of any use. These ideas are all I could come up with and none of them sounds like a good plan. What am I missing? Is my problem a common one, and if yes, does it have a time-tested solution? Can it be answered on this general scale? Is it a real problem or an artefact of my perception? PS. This is being posted anonymously in order to preserve genericity (of the author and, more importantly, of $A$).
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There are three separate issues here. 1) How to clarify whether the proof is correct? You should start with making a serious good will effort to understand what is written (which amounts to redoing all the bad notation, splitting things into small steps, etc. to the best of your abilities). If this fails, you should state as clearly as you can what exactly the problem with the argument is and hope that some expert will figure out who is right. Of course, first you should send the full account of your effort to the author reproducing all the parts of the proof you understand and showing clearly where you are stuck and why. Just to say "your notation is bad here so..." won't accomplish anything: at best, he'll make a local correction that will move the real issue somewhere else and you can play this shifting game forever. If he still fails to address the issue after that, request the help of some third party
sending the same account of your effort together with the paper. Again, it is important that you demonstrate your good will and decent understanding of what is written in the paper before you raise your objection. Without this, you just won't be taken seriously. Make sure that you understand everything that precedes the unclear/incorrect step and that you make it clear to everyone whom you want to ask that you understand it. Nobody pays attention to people coming out of nowhere with zero credentials and doubtful qualifications. If your first words are "I don't understand ... and I think it is wrong", the most likely answer will be "Go learn ... ". However if your first words are "This argument starts with using ... to establish ...", your general credibility goes up immediately (provided that what you are saying makes sense). The more times the person agrees with you on the issue before you raise the question, the more likely he is to take you and your objection seriously. It is your moral duty to make a real effort trying to understand the proof before making
any public comment on its correctness but it is also your moral duty to report a problem
with a proof when you are convinced that you see one. Note that it is completely normal
in mathematics to make bad mistakes occasionally and it is completely normal to fail
to understand correct arguments now and then. The priority/reputation chase has distorted the general attitudes beyond recognition, of course, but the heart of the matter is still the search for the knowledge, not building/preserving/destroying reputations and relationships. Even if you are wrong on the account that the proof has a gap, you may be right
on the account that it is unclear (assuming that you have a decent education in the subject, the fact that you fail to understand the argument is a clear indication that the paper is written not in an ideal way). So, the clarification may help innumerable poor souls (like graduate students) who may have the same difficulty but just do not dare to ask questions. You risk to look like a fool, of course, but the only way to avoid looking like a fool occasionally is to always be one. 2) How to avoid the confrontation? At some point there may be no way and all that you'll be able to write to the author something like "It seems very difficult for us to understand each other. Since the issue is principal, the best we can do is to seek an opinion of a third party. I'd appreciate your suggestions of whom we should ask. I'm thinking of (put the list of experts you know)". This may not save your good relationships but, at least, will clear you from any "doing things behind the curtains" charges. After that, send your doubts to both the people on your list and the people on his list, if he provides one. If he doesn't, it is his problem. There are three possibilities: 1) you'll be backed by some expert, which will make the author harder to ignore you; 2) someone will explain to you why the proof is correct, and 3) everyone will ignore your request. In that last case you may have to seek the opinion of general public but not before you double check your argumentation. 3) Is MO an appropriate place for this discussion? It is not what it was intended for but if you finally decide to seek the general public opinion and post your objection on arXiv or somewhere else (in the way I outlined above; let me emphasize once more that unless you are Terry Tao or Tim Gowers you should demonstrate both good understanding of the matter and your good will before anyone will bother to take a look at your objection in honest) I see nothing wrong with making a short post containing the corresponding link in this thread. In brief, if you really want to figure things out, I would advise that: a) you make a good effort putting all your thoughts together in written . Create a clear "case" starting from the beginning where you agree with A on everything and talk in the same terms and stopping where you have an objection. b) present this full writeup to A and wait for his comment. c) if it doesn't result in anything meaningful, present this writeup to a few experts or,
at least, people whom you feel to be more knowledgeable in the subject than yourself. d) if you are totally ignored, ask yourself why that may be the case and, if you see nothing wrong with your written argumentation (if you see nothing wrong with what you keep in your mind, there is a good chance that you are just blind), present it to the general public. I don't think asking a graduate student of A is a good idea. First, many graduate students are totally incompetent in anything beyond their thesis project and in such cases, you can just as well use a parrot for communication. Second, if the student is actually good and you are right, you'll put the student in the position where he will have to tell his adviser that the paper is wrong. This doesn't go well with many people.
|
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|
72,210 |
I do understand that my question might seem a little bit ignorant, but I thought about it a lot and still can't wrap my head around it. Analycity imposes very strong conditions on a map, from elementary ones like "locally zero implies globally zero", to a little bit more deep like the Hurwitz formula (in the complex case). Neither of above are true if we just assume smoothness. On one hand, it is quite easy to prove that smooth functions are dense in any "reasonable" function space (I guess it depends on what one considers reasonable, though…) - just convolve with smooth approximations of identity. Also, (although I do take it on faith), any map of two manifolds is homotopic to a smooth one and two homotopic smooth maps are actually smooth-homotopic. Because of above facts, it seems to me that smooth functions are "abundant" and are actually very close to topology, ie. mere continuity. On the other hand, when I recall basic calculus course, it always seemed like being differentiable even once is a "miracle", and being differentiable infinitely many times is a very, very strong condition, even more so in several variables. Why are objects so constrained, ie. smooth functions, so useful and also, so malleable?
|
Given a paracompact smooth manifold, you have smooth partitions of unity ( nLab ), but on a real analytic manifold (e.g. a complex manifold viewed as a real manifold) one doesn't have analytic partitions of unity (much less holomorphic, if you are in the complex case). That is, given any open cover on a smooth manifold, one can find a partition of unity subordinate to that cover - this is a very topological property. Using partitions of unity you can paste together local functions as desired. The existence of smooth partitions of unity comes down to the existence of a smooth (but not analytic!) bump function on $[-1,1]$. Edit: you can find details and formulas on Wikipedia . A related fact is that on a paracompact smooth manifold, the sheaf of real-valued functions is fine ( nLab , wikipedia ).
|
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|
72,259 |
I just ran into this deceptively simple looking question. Is it always possible to partition $\mathbb{R}$ (or any other standard Borel space) into precisely $\aleph_1$ Borel sets? On the one hand, this is trivial if the Continuum Hypothesis holds. Less trivially, this also follows from $\mathrm{cov}(\mathcal{M}) = \aleph_1$, $\mathrm{cov}(\mathcal{N}) = \aleph_1$, $\mathfrak{d} = \aleph_1$, and similar hypotheses. However, I can't think of a general argument that allows one to split $\mathbb{R}$ into precisely $\aleph_1$ pairwise disjoint nonempty Borel pieces. On the other hand, PFA or MM might give a negative answer but I don't see a good handle from that end either.
|
It suffices to express $\mathbb R$ as the union of $\aleph_1$ (not necessarily disjoint) Borel sets such that no countably many of them cover $\mathbb R$, because then you can list them in an $\omega_1$-sequence and subtract from each one the union of the previous ones. Partition $\mathbb R$ into a non-Borel $\Pi^1_1$ set $A$ (say the set of codes of well-orderings of $\omega$) and its complement. A classical theorem says that any $\Pi^1_1$ set is a union of $\aleph_1$ Borel sets, and so is every $\Sigma^1_1$ set. Apply that to $A$ and to $\mathbb R-A$ to get $\mathbb R$ as a union of $\aleph_1$ Borel sets. No countably many of them cover $\mathbb R$ because $A$ is not Borel and thus not a countable union of Borel sets.
|
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|
72,419 |
I'm a student (I've been studying mathematics 4 years at the university) and I like functional analysis and topology, but I only studied 6 credits of functional analysis and 7 in topology (the basics). What I am looking for is good books that I could understand to go deeper in this areas, what do you recommend? (I can read in Spanish, English, French and German)
|
I am an algebraist and not an analyst, however my favourite book on this area is "Walter Rudin: Functional Analysis".
|
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|
72,490 |
The question is not about where operads are used, I know that. It is about what makes them useful. For example, van Kampen diagrams are useful in combinatorial group theory because these are planar graphs and so one can use planar geometry (say, the Jordan lemma) to investigate the word problem in complicated groups. Similarly, asymptotic cones are useful in geometric group theory because they allow to study large scale properties of a discrete object (a group) by looking at small scale properties of a continuous object. I would like to know a similar answer for operads. Update Many thanks to everybody for your answers. Unfortunately I can accept only one. So I just accept the first answer.
|
Here are a couple 2-3 line answers to your question: 1) They allow you to treat various algebraic problems uniformly. For example, Commutative, Associative, and Lie algebras all have their own cohomology theories (Harrison, Hochschild, and Chevalley-Eilenberg respectively). These can all be seen as instances of a single operad cohomology. 2) One can use operads to construct cohomology classes for the Mapping Class group and Out(Fn) (and others). The idea is to use graphs "colored" by operads, and construct a chain complex out of these colored graphs that computes the desired cohomology. and perhaps the most classic answer: 3) They allow you to classify loop spaces and infinite loop spaces. For connected spaces, these are exactly classified as algebras over various operads.
|
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|
72,810 |
I was recently trying to learn a little bit about group cohomology, but one point has been confusing me. According to wikipedia ( https://en.wikipedia.org/wiki/Group_cohomology and some other sources on the internet), given a (topological) group $G$ , we have that the group cohomology $H^n(G)$ is the same as the singular cohomology $H^n(BG)$ (with coefficients in a trivial $G$ -module $M$ ). Moreover, it says that given any group $G$ , if we don't care about its topology, we can always give it the discrete topology and look at the cohomology of $K(G,1)$ . This seems to suggest that when $G$ has topology that we do care about, we can just look at $BG$ with whatever topology $G$ is supposed to have. The relevant citation in this section is a reference to a book called Cohomology of Finite Groups, but I was wondering if this result would work with groups such as $U(1)$ which are not finite? Moreover, it would seem then that there is some sort of natural way to define group cohomology to detect the topology of the group; for example maybe look at continuous $G$ -modules and continuous cochains. However, I heard that when doing this, one has to be careful because in general, the category of continuous $G$ -modules might not have enough injectives. Also, I found this article by Stasheff ( Link ) which seems to suggest that for continuous cohomology, the we might not have the equality $H^n(BG)=H^n(G)$ between singular and group cohomology. I was wondering if someone could explain these connections to me (including what "continuous cohomology" is) and/or clarify what is happening? It would also be great if someone could tell me how one might compute something like $H^n(U(1);M)$ where $U(1)$ carries the discrete topology. Thanks.
|
Short answer: you don't want to consider group cohomology as defined for finite groups for Lie groups like $U(1)$ , or indeed topological groups in general. There are other cohomology theories (not Stasheff's) that are the 'right' cohomology groups, in that there are the right isomorphisms in low dimensions with various other things. Long answer: Group cohology, as one comes across it in e.g. Ken Brown's book (or see these notes ), is all about discrete groups. The definition of group cocycles in $H^n(G,A)$ , for $A$ and abelian group, can be seen to be the same as maps of simplicial sets $N\mathbf{B}G \to \mathbf{K}(A,n)$ , where $\mathbf{B}G$ is the groupoid with one object and arrow set $G$ , $N$ denotes its nerve and $\mathbf{K}(A,n)$ is the simplicial set corresponding (under the Dold-Kan correspondence ) to the chain complex $\ldots \to 1 \to A \to 1 \to \ldots$ , where $A$ is in position $n$ , and all other groups are trivial. Coboundaries between cocycles are just homotopies between maps of simplicial spaces. The relation between $N\mathbf{B}G$ and $K(G,1)$ is that the latter is the geometric realisation of the former, and the geometric realisation of $\mathbf{K}(A,n)$ is an Eilenberg-MacLane space $K(A,n)$ , which represents ordinary cohomology ( $H^n(X,A) \simeq [X,K(A,n)]$ , where $[-,-]$ denotes homotopy classes of maps). This boils down to the fact that simplicial sets and topological spaces encode the same homotopical information. It helps that $N\mathbf{B}G$ is a Kan complex, and so the naive homotopy classes are the right homotopy classes, and so we have $$sSet(N\mathbf{B}G,\mathbf{K}(A,n))/homotopy \simeq Top(BG,K(A,n))/homotopy = [BG,K(A,n)]$$ In fact this isomorphism is a homotopy equivalence of the full hom-spaces, not just up to homotopy. If we write down the same definition of cocycles with a topological group $G$ , then this gives the 'wrong' cohomology. In particular, we should have the interpretation of $H^2(G,A)$ as isomorphic to the set of (equivalence classes of) extensions of $G$ by $A$ , as for discrete groups. However, we only get semi-direct products of topological groups this way, whereas there are extensions of topological groups which are not semi-direct products - they are non-trivial principal bundles as well as being group extensions. Consider for example $\mathbb{Z}/2 \to SU(2) \to SO(3)$ . The reason for this is that when dealing with maps between simplicial spaces , as $N\mathbf{B}G$ and $\mathbf{K}(A,n)$ become when dealing with topological groups, it is not enough to just consider maps of simplicial spaces; one must localise the category of simplicial spaces, that is add formal inverses of certain maps. This is because ordinary maps of simplicial spaces are not enough to calculate the space of maps as before. We still have $BG$ as the geometric realisation of the nerve of $\mathbf{B}G$ , and so one definition of the cohomology of the topological group $G$ with values in the discrete group $A$ is to consider the ordinary cohomology $H^n(BG,A) = [BG,K(A,n)]$ . However, if $A$ is also a non-discrete topological group, this is not really enough, because to define cohomology of a space with values in a non-discrete group, you should be looking at sheaf cohomology , where the values are taken in the sheaf of groups associated to $A$ . For discrete groups $A$ this gives the same result as cohomology defined in the 'usual way' (say by using Eilenberg-MacLane spaces). So the story is a little more complicated than you supposed. The 'proper ' way to define cohomology for topological groups, with values in an abelian topological group (at least with some mild niceness assumptions on our groups) was given by Segal in G Segal, Cohomology of topological groups, in: "Symposia Mathematica, Vol. IV (INDAM, Rome, 1968/69)", Academic Press (1970) 377–387 and later rediscovered by Brylinski (it is difficult to find a copy of Segal's article) in the context of Lie groups in this article .
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|
72,886 |
Allow me to quote a definition from Gelbart in " Modular Forms and Fermat's Last Theorem ": Definition. Let $E/\mathbb{Q}$ be an elliptic curve. We say that $E$ is modular if there is some normalised eigenform $$ f(z) = \sum_{i=1}^{\infty} \ a_ne^{2\pi inz} \in S_2(\Gamma_0(N),\epsilon), $$ for some level $N$ and Nebentypus $\epsilon$ , such that $$ a_q = q + 1 - \#(E(\mathbb{F}_q)) $$ for almost all primes $q$ . This is the basic question of the post: Why is the weight of $f$ taken to be 2? Can I instead take 3, or 4, or 5, or even 19/2, without disturbing the peace? I am aware of other definitions of modularity, some of which don't mention modular forms at all, but nonetheless I feel that weight 2 lurks beneath all of these. I think one approach would involve differentials, and the construction of Eichler-Shimura, but I'm not so sure. Further, perhaps there are several reasons which fit together to tell a nice story. Is it a corollary of this question that it doesn't matter what the weight is? Finally, can I replace $E$ above with any abelian variety, and ask the same question?
|
$\newcommand\Q{\mathbf{Q}}$
$\newcommand\Qbar{\overline{\Q}}$
$\newcommand\Gal{\mathrm{Gal}}$
$\newcommand\C{\mathbf{C}}$
$\newcommand\Sym{\mathrm{Sym}}$
$\newcommand\E{\mathcal{E}}$
$\newcommand\Betti{\mathrm{Betti}}$
$\newcommand\Z{\mathbf{Z}}$
$\newcommand\Hom{\mathrm{Hom}}$
$\newcommand\T{\mathbf{T}}$
To answer this question, it might be best to start with
the following: Q. What do the Galois representations attached to a variety know about the variety? In order make this more precise, let us introduce some notation. Fix a prime $p$ and a non-negative
integer $n$.
Let $X$ be a proper smooth scheme over $\Q$, and
let $V = H^n_{et}(X/\Qbar,\Q_p)$ denote the $n$th etale cohomology group
of $X$. The basic and fundamental properties of etale cohomology tell us that: $V$ is a vector space of dimension $H^n_{\Betti}(X(\C))$, where $H_{\Betti}$
denotes Betti (or singular) cohomology, and $X(\C)$ denotes the complex points of $X$ thought
of as a topological manifold. $V$ (with the $p$-adic topology) has a continuous action of $G_{\Q}:=\Gal(\Qbar/\Q)$. Grothendieck and Serre further conjecture that the $G_{\Q}$-representation $V$
is semi-simple.
The strongest possible conjecture one might make is to ask whether the functor from smooth projective varieties over $\Q$ to
semi-simple $G_{\Q}$-representations (or the collection of all such representations for $n \le 2 \cdot \mathrm{dim}(X)$)
is fully faithful. However, this is too much to ask, for the following reasons. (i). The target category is semi-simple, but the category of varieties is far from
semi-simple. (In particular, the existence of a map
$X \rightarrow Y$ does not imply the existence of a non-trivial map $Y \rightarrow X$.) (ii). Varieties built in a combinatorial way from projective spaces (think toric
varieties) tend to have etale cohomology groups indistinguishable from
products of projective spaces. This is because their cohomology groups are generated by geometric cycles, on which Galois
acts in a well understood way (essentially by some power of the cyclotomic character). These are - in some sense - manifestations of the same reason: A correspondence
in $X \times Y$ gives rise to a cohomology class in $H^*(X \times Y)$; then by
the Künneth formula, this leads to a relation between the cohomology
of $X$ and $Y$ even when there is not necessarily any non-trivial
map from $X$ to $Y$ (or vice versa).
In order to account for this, one can try to take the quotient category of the
category of algebraic varieties in which one is allowed to "break up" smooth
proper varieties into pieces given the existence of certain correspondences on $X$. There are a variety of ways in which one might do this. Conjecturally, these constructions are all essentially the same, and the corresponding category is the category of pure motives. The Tate conjecture now says that etale cohomology is a fully faithful functor from pure motives to semi-simple $G_{\Q}$-representations. Example If $E$ is an elliptic curve over $\Q$, and $n = 1$, then
the etale cohomology group $V$ is the (dual) of the usual representation
attached to the $p$-adic Tate module of $E$. Suppose that $E'$ is another
elliptic curve over $\Q$ with first etale cohomology group $V'$.
For curves, the theory of "motives" is essentially the theory of abelian varieties. (More generally, the theory of $H^1$ is essentially the theory of abelian
varieties, since, for any proper variety $X$, there is an isomorphism $H^1(X) \simeq H^1(A(X))$, where $A(X)$ is the Albanese of $X$.)
Tate's conjecture in this case says that
$$\Hom(E,E') \otimes \Q_p \rightarrow \Hom_{G_{\Q}}(V,V')$$
is an isomorphism. This is how you will see the Tate conjecture stated for elliptic
curves, for example, in AOEC. The Tate conjecture for abelian varieties is a theorem of Faltings.
(Suggestion: to understand what the Tate conjecture really is about, and why it is hard, you should really think about the special case of Elliptic curves.) If we now return to our question, we can (tautologically) say the following: assuming the Tate conjecture, the etale cohomology knows about the motive corresponding to the original variety. What does that really mean? One way of thinking about motives is as a ``universal cohomology theory''. In particular, we can recover from the motive not only the etale cohomology groups, but also the algebraic
de Rham cohomology groups. Recall that de Rham cohomology is another cohomology theory
that gives vector spaces of the "correct" dimension for a smooth proper variety $X/\Q$. The de Rham cohomology groups do not have associated Galois representations, but
they do have a Hodge filtration. Over $\C$, if one takes the associated graded of the Hodge filtration, one recovers the Hodge decomposition:
$$H^n_{dR}(X,\C) = \bigoplus_{p+q=n} H^{p}(\Omega^q_X).$$
The dimensions of the latter space are called the Hodge numbers $h^{pq}$.
So, assuming the Tate conjecture, from $V$ we can recover the underlying
motive, from which we may reconstruct the de Rham cohomology, and then the Hodge numbers. The Tate conjecture seems to be very hard.
However, Grothendieck asked the following: given
$V$, can we directly recover the (algebraic $p$-adic) de Rham cohomology along with its filtration without first constructing the motive? This was a great question, and the answer (yes!) constitutes one of the major achievements of $p$-adic Hodge Theory. I can do no more than give a cartoon description here. In order to do so, first recall the much more classical story connecting de Rham cohomology to Betti (singular) cohomology. These groups can both naturally be defined as vector spaces over $\Q$ (one has to define de Rham cohomology in the correct way), but the isomorphism relating these spaces comes from integrating forms over cycles. Yet these integrals are typically
transcendental numbers, so to pass from Betti to de Rham cohomology one first has to tensor with a field
bigger than $\Q$ which contains all these periods (usually, one simply tensors with $\C$).
In order to pass from etale cohomology to algebraic de Rham cohomology, one might ask for a period ring in which we can compare both groups. In this refined setting, the period ring should both have a Galois action and a filtration. The most basic verion of a period ring is $B_{HT}$, specifically,
$$B_{HT} := \bigoplus_{\Z} \C_p(n),$$
where $\C_p$ is the completion of $\Qbar_p$, and $\C_p(n)$ is $\C_p$ twisted
(as a local Galois module) by the $n$th power of the cyclotomic character.
The ring
$B_{HT}$ has a natural filtration (indeed, it is even graded). Now we can consider
$$D_{HT}(V) = (V \otimes B_{HT})^{\Gal(\Qbar_p/\Q_p)}.$$
The Galois group acts on both $V$ and $B_{HT}$. The result is a graded
(and so filtered) module. On the other hand, one can also consider the ring
$B_{HT} \otimes H^n_{dR}(X/\Q_p)$, where there is a natural way to make sense of the
corresponding filtration. An important theorem of Faltings then says that
$$H^{n}(X/\Qbar_p,\Q_p) \otimes B_{HT} = H^n_{dR}(X/\Q_p) \otimes B_{HT},$$
and $D_{HT}(V) = H^n_{dR}(X/\Q_p)$. In particular, from a geometric Galois
representation, we can recover the Hodge filtration and the Hodge numbers. Modular Forms. The Eichler-Shimura isomophism relates modular forms of weight $k \ge 2$
to $H^1(X_0(N),\Sym^{k-2}\Q)$. If $k = 2$, this is just $H^1(X_0(N),\Q)$. The Hecke algebra
$\T$ acts on $H^1_{\Betti}(X_0(N),\Q)$, and (since it is constructed functorially) also on the etale cohomology
$H^1(X_0(N),\Q_p)$. Now the Hodge decomposition of $H^1$ is $H^1 = H^{0,1} \oplus H^{1,0}$, where
$h^{0,1} = h^{1,0}$ is the genus of $X_0(N)$.
The Hecke algebra breaks up the cohomology into two dimensional pieces corresponding to the Galois representations
associated to eigenforms; it turns out that each two dimensional piece contains one dimension from
$H^{0,1}$ and one dimension from $H^{1,0}$. The result of Faltings above tells us that we can read off
that $h^{0,1} = h^{1,0} = 1$ directly from the Galois representation. For $k > 2$, recall that (technical issues aside) there is a universal elliptic curve $\E \rightarrow X_0(N)$.
The Kuga-Sato variety is (again, roughly) The $k-1$ dimensional variety $K = \E \times_X \E \ldots \times_X \E$ where
$X = X_0(N)$. There is a natural map $\pi: K \rightarrow X$.
The local system $\Sym^{k-2}(\Q^2_p)$ is trivialized over $K$, and so, using the proper base change theorem,
Deligne shows that $H^1(X_0(N),\Sym^{k-2}\Q_p)$ is a sub-quotient of the cohomology group $H^{k-1}(K,\Q_p)$.
(Warning: this requires more than simply a formal cohomological argument, it also requires some trickiness with weights to show
that terms on different diagonals the Leray spectral sequence don't "mix", and hence the sequence degenerates.) The Galois representation associated to a modular form
is now a two-dimensional piece of $H^{k-1}(K,\Q_p)$. Faltings proves that the corresponding "piece" of de Rham cohomology
seen by this representation is $H^{0,k-1} \oplus H^{k-1,0}$. In particular, the representation has Hodge numbers
$h^{0,k-1} = 1$ and $h^{k-1,0} = 1$. Given a Galois representation $V$, one can twist $V$ by the cyclotomic character. How does this effect the Hodge decomposition?
One can compute this on the Hodge side by seeing what happens to the cohomology of $X \times \mathbf{G}^1_m$ and comparing with
the Künneth formula. It turns out that $h^{p,q}(V(n)) = h^{p-n,q-n}$. Thus, if only know $V$ up to twist, we still recover some information
about the Hodge numbers. Returning to modular forms. The coefficients $a_p$ determine the Galois representation, by Cebotarev.
A modular form of weight $k$ has Hodge numbers $h^{0,k-1} = h^{k-1,0} = 1$. The determinant of the representation is the $k-1$th power
of the cyclotomic character (up to a finite character) which can be read off from the "degree". By twisting, we can easily change the determinant,
and change the Hodge numbers to $h^{-d,k-d-1} = h^{k-d-1,-d} = 1$. Yet,
it is clear that we cannot twist so that $h^{1,0} = h^{0,1} = 1$ unless $k = 2$. Thus, given a modular form of weight $k > 2$, it cannot be associated to
an elliptic curve even after twisting. This is Kevin's answer. Secondly, any motive has (conjecturally) an $L$-function. The recipe of building this $L$-function breaks up into two parts. The first involves the factors
at finite primes, which give rise to the Euler product. The second involves the infinite primes, which give rise to Gamma factors. The information
at $\infty$, however, (by Tate's conjecture) can be read off from the Galois representation, and the recipe of Deligne shows that it will exactly depend
on the Hodge numbers of the motive, and visa versa. Moreover, twisting by $\epsilon^k$ some power of the cyclotomic factor has the effect of replacing $L(s)$ by $L(s+k)$ (and shifting the corresponding central value) In particular, given an elliptic curve, one knows the Gamma factors (because one knows the Hodge decomposition of $E$), and one sees that even after twisting one cannot get Gamma factors that
"look like" the Gamma factors associated to a modular form of weight different from $2$. This is GH's answer. More generally, arithmetic conjectures of Langlands type imply that all motives should be "automorphic", and that the Hodge structure of the motive determines the infinity type of the automorphic form,
which in turn determines the Gamma factors. So, at least morally, given a pure irreducible motive $V$, we know that if it is automorphic, it must be automorphic of a particular weight determined by the underlying geometry of $V$. Of course, even before the result of Faltings, one had enough faith in terms of how these things were connected to be very confident that Elliptic curves over $\Q$ should
correspond exactly to weight two forms - GH's remark that "It was an experimental fact that the gamma factors are always the same, hence the precise form of the modularity conjecture was formulated, which then turned out to be right, namely it was proved by great efforts of great mathematicians" seems spot on. Related problems. Given a modular eigenform $f = \sum a_n q^n$ of weight four (in the arithmetic normalization), one can ask: is it possible to show that there does not exist a weight
two modular form
$g = \sum b_n q^n$ where $a_p = p b_p$ for all primes $p$ without using $p$-adic Hodge theory? I think this is not so easy. For example: (i) The arithmetic approach: The weight $4$ form $f$ would have the property that it is not ordinary at every prime, since clearly
$a_p = p b_p \equiv 0 \mod p$. One conjectures that a set of primes of density one are modular (or $1/2$ if $f$ has CM). Yet it is still unknown whether
any form of weight $\ge 4$ has even a single ordinary prime. (ii) The analytic approach: What do the distributions of coefficients of weight $2$ and $4$ forms look like? Sato-Tate says that the normalized coefficients satisfy a precise
distribution (now a theorem!). Yet the "normalized" coefficients of $f$ and $g$ are by construction exactly the same, so Sato-Tate says nothing. In particular,
it is hard to see any analytic estimates of functions involving the $a_p$ being able to distinguish two classes of numbers with the same
underlying distribution. A related argument: The Hasse
bound in weight four is satisfied by $p b_p$ if and only if the weight two Hasse bound is satisfied by $b_p$. Summary. Conjectures arise organically from heuristics and computation. I was "known" that Elliptic curves should be associated to weight two forms long before
one could actually formally prove that they weren't associated to twists of weight $4$ forms. To prove the latter fact, one has to use $p$-adic Hodge theory. (* I am not sure about a lot here, so I'm putting this community wiki. Also, there is no mention of weight 1 or half integers. Change whatever needs changing, or in the extreme cases, peacefully leave a comment to delete...)
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|
73,054 |
The classic reference of this topic is Serre's Algebraic Groups and Class Fields. However, many parts of this book use Weil's language, which I find quite hard to follow. Is there another reference to the topic, using a more modern language (schemes etc.)?
|
Have you looked at Katz-Lang Conrad's write-up Lang's BAMS article Ben-Zvi's notes (i seem to remember a video, but i could not find it.) also see his other lectures and videos Thesis of Peter Toth (follows Deligne's approach) Kerz's articles (amazing innovations in class field theory by Kerz and Wiesend); there is also an excellent Seminaire Bourbaki expose on this by Szamuely (in French). There are many other good references, but hope this can help.
|
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|
73,098 |
From parametric plots of $\zeta \left( \frac{1}{2} + it \right)$ it seems to be the case that: (1) except for $\zeta(\frac{1}{2})$ the Riemann zeta function does not attain any negative real value on the critical line. (2) the curve $(t, \zeta(1/2+it))$ is dense in the complex plane. Are these statements known to be false, if not, is there any proof affirming them?
|
A numerical counterexample to the first conjecture is $$
t = 282.4547208234621746108397940690599354\ldots
$$ where both gp and Wolfram Alpha agree that $\zeta(\frac12 + it)$ has negative real part $\simeq -0.02763$ and negligible imaginary part, so the actual zero of ${\rm Im}(\zeta(\frac12+it))$ near $t=295.5839\ldots$ yields a negative value of $\zeta(\frac12+it)$ . This was found by approximating $\zeta'(\frac12+it)$ at each of the first "few" zeros of $\zeta$ tabulated by Odlyzko in http://www.dtc.umn.edu/~odlyzko/zeta_tables/zeros1 and looking near the first zero (the 127th overall) at which $\zeta'$ has negative imaginary part. There are $22$ such zeros of the $649$ zeros whose imaginary part lies in $[0,1000]$ ; there's probably a counterexample near each of those, e.g. looking around the second such zero (#136) yields $$
t = 295.583906974228176092587915204356841\ldots
$$ with $\zeta(\frac12+it) \simeq -0.0169004$ . EDIT 1) Henry Cohn (in a comment below) provides gp code that looks for solutions in an interval by dividing it into segments $(t_0, t_0 + 0.01)$ , testing whether ${\rm Im}(\zeta(\frac12+it))$ changes sign between the endpoints, and if so whether the real part is negative at the crossing. Extending his computation to $0 \leq t \leq 1000$ finds the expected $22$ solutions; in particular $282.45472+$ seems to be the first. Once one has calculated an answer one can ask Google for its previous appearances. Google recognizes $282.45472$ from J.Arias-de-Reyna's paper "X-Ray of Riemann zeta function" ( http://arxiv.org/abs/math/0309433 ) where it appears (to within $10^{-5}$ ) as the first counterexample to "Gram's law" — see the plot on page 26 (thick and thin curves show where $\zeta(s)$ is real and imaginary respectively).
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|
73,104 |
Does anyone know an example of a curve $X$ over a perfect field $k$ such that if $\tilde{X}$ is its noramlisation, there exists a point $x \in X$ and a point $y \in \tilde{X}$ over $x$ such that $k(y) / k(x)$ is not trivial? (If we remove the hypothesis that it is over a field, there is such an example: $\mathbb{Z}[11\sqrt{3}] \to \mathbb{Z}[\sqrt{3}]$ and the points $(11, 11\sqrt{3}), (11)$.
|
A numerical counterexample to the first conjecture is $$
t = 282.4547208234621746108397940690599354\ldots
$$ where both gp and Wolfram Alpha agree that $\zeta(\frac12 + it)$ has negative real part $\simeq -0.02763$ and negligible imaginary part, so the actual zero of ${\rm Im}(\zeta(\frac12+it))$ near $t=295.5839\ldots$ yields a negative value of $\zeta(\frac12+it)$ . This was found by approximating $\zeta'(\frac12+it)$ at each of the first "few" zeros of $\zeta$ tabulated by Odlyzko in http://www.dtc.umn.edu/~odlyzko/zeta_tables/zeros1 and looking near the first zero (the 127th overall) at which $\zeta'$ has negative imaginary part. There are $22$ such zeros of the $649$ zeros whose imaginary part lies in $[0,1000]$ ; there's probably a counterexample near each of those, e.g. looking around the second such zero (#136) yields $$
t = 295.583906974228176092587915204356841\ldots
$$ with $\zeta(\frac12+it) \simeq -0.0169004$ . EDIT 1) Henry Cohn (in a comment below) provides gp code that looks for solutions in an interval by dividing it into segments $(t_0, t_0 + 0.01)$ , testing whether ${\rm Im}(\zeta(\frac12+it))$ changes sign between the endpoints, and if so whether the real part is negative at the crossing. Extending his computation to $0 \leq t \leq 1000$ finds the expected $22$ solutions; in particular $282.45472+$ seems to be the first. Once one has calculated an answer one can ask Google for its previous appearances. Google recognizes $282.45472$ from J.Arias-de-Reyna's paper "X-Ray of Riemann zeta function" ( http://arxiv.org/abs/math/0309433 ) where it appears (to within $10^{-5}$ ) as the first counterexample to "Gram's law" — see the plot on page 26 (thick and thin curves show where $\zeta(s)$ is real and imaginary respectively).
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73,121 |
Here it is mentioned that someone claims to have proven that there are no weakly inaccessibles in ZF. Question 1: What reasons are there to believe that weakly inaccessibles exist? Question(s) 2: Since all large cardinals are weakly inaccessible, this would have a profound effect on set theory. What are some of the most significant results whose only known proof assumes the existence of weakly inaccessibles? Might any of the arguments go through without their existence? For example, I've heard that the original proof of Fermat's Last Theorem (FLT) assumed (something equivalen to) a large cardinal, but it was then shown that the argument went through without such an assumption. Edit. I just added the phrase "whose only known proof" to Question 2 above, which is what I intended originally. The point of course, is that I want to know which results, if any, would be "lost" if weakly inaccessibles were lost. FLT is not an example of that, but would have been before it was known that weakly inaccessibles are not necessary in its proof.
|
As I pointed out in the meta thread, this question overlaps with a bunch of older MO questions. Arguments against large cardinals Nonessential use of large cardinals Inaccessible cardinals and Andrew Wiles's proof Reasons to believe Vopenka's principle/huge cardinals are consistent However, none of these questions directly address the particular case of the existence of inaccessible cardinals, which is of special interest as this is the weakest of all large cardinal hypotheses. This answer focuses on that case. Penelope Maddy gives several answers to Question 1 in §III of Believing the Axioms, I [JSL 53 (1988), 481-511, MR0947855 ].
In this wonderful paper, Maddy justifies many set theoretic axioms and hypotheses using five widely believed "rules of thumb": maximize , inexhaustibility , uniformity , whimsical identity , and reflection . Here is a brief summary of these five arguments as it pertains to the existence of inaccessible cardinals. The maximization argument. The maximize rule of thumb is perhaps best understood as the opposite of Occam's Razor .
However, blind application of this easily leads to contradictions.
Thus, the rule is generally understood as a pair of statements: thikness — powersets are very large; and tallness — there are lots and lots of ordinals.
The second easily leads to the existence of inaccessibles. The inexhaustibility argument. Maddy describes this one very well: "The universe of sets is too complex to be exhausted by any handful of operations, in particular by power set and replacement, the two given by the axioms of Zermelo and Fraenkel.
Thus there must be an ordinal number after all the ordinals generated by replacement and power set.
This is an inaccessible." (p. 502) The uniformity argument. Uniformity basically states that the richness of the universe should not concentrate in a small region, that if a certain property is found at a certain level of the cumulative hierarchy then analogue properties should also be found higher up.
Thus, there should be many cardinals that share the same properties as $\aleph_0$, such as the fact that $2^k < \aleph_0$ for every $k < \aleph_0$.
Combined with regularity, this leads to the existence of inaccessibles. The whimsical identity argument. This rule of thumb states that there should be no accidental identities, "like the identity between 'human' and 'featherless biped'." (p. 499)
It seems unlikely that $\aleph_0$ should be characterized as the unique regular cardinal $\kappa$ such that $2^\mu < \kappa$ for every $\mu < \kappa$ .
Therefore, there must be inaccessible cardinals. The reflection argument. This powerful rule of thumb is a generalization of Montague's Reflection Theorem, which states that for every first-order formula $\phi(\bar{x})$ of $V \vDash \phi(\bar{x})$ then there are arbitrarily large ordinals $\alpha$ such that $V_\alpha \vDash \phi(\bar{x})$.
The Reflection Principle generalizes this from first-order properties to arbitrary properties.
Thus, since $V$ is closed under replacement and powerset, there must be arbitrarily large ordinals $\alpha$ such that $V_\alpha$ is also closed under replacement and powerset.
These ordinals are inaccessibles. These five arguments have a lot in common, but the basic principles behind them are quite different.
I would contend that these are five distinct justifications for the existence of inaccessibles. Note that Maddy's paper has a sequel Believing the Axioms, II [JSL 53 (1988), 736-764, MR0960996 ]).
Let me also point out nother highly relevant paper: Kanamori and Magidor, The evolution of large cardinal axioms in set theory [LNM 669, 99-275, MR0520190 ]. Of course, detailed information can be found in Kanamori's The Higher Infinite [Perspectives in Mathematical Logic. Springer-Verlag, Berlin, 1994].
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|
73,205 |
The usual disclaimer applies: I'm new to all this stuff, so be gentle. It seems like the spectrum, as defined by Balmer, of the stable homotopy category of finite complexes is something like $M_{FG}$, the stack of formal groups (that is, $Spec L/ G$ where $L$ is the Lazard ring and $G$ acts by coordinate changes). I'm not actually sure if that's true, I don't think I've seen it written quite like that, but the picture of the spectrum in Balmer's paper looks an awful lot like how I'd imagine $M_{FG}$ looking. If the above is right, then there's another tensor triangulated category with the same spectrum, namely the derived category of perfect complexes on $M_{FG}$ (whatever that means for stacks...). So my question is: Just how far away is the stable homotopy category from actually being equivalent to this derived category? Is there a theorem to the effect that it can't be equivalent to such a thing? Do we even know that it's not equivalent? I've heard that chromatic homotopy theory is about setting up a rough dictionary between algebro-geometric terminology regarding $M_{FG}$ and the stable homotopy category, so I guess the question is about whether or not we can make the dictionary into a proper functor.
|
One useful thing to keep in mind is that the cohomological functor from the stable homotopy category to the category of quasi-coherent sheaves on the moduli stack $\mathcal{M}$ is not essentially surjective. For example, if you fix a prime $p$ and a height $n \geq 1$, then there is a closed substack $\mathcal{M}^{\geq n}$ consisting of formal groups over $\mathbb{F}_p$ having height $\geq n$. A standard problem in stable homotopy theory is to try to cook up finite spectra which map to the structure sheaf of $\mathcal{M}^{\geq n}$.
You can generally only do this when $p$ is large compared with $n$. For small values of $p$ you generally have to make do with finite spectra whose image is the structure sheaf of some nilpotent thickening of $\mathcal{M}^{\geq n}$. These can always be found (a deep result of Devinatz-Hopkins-Smith) and this is what gives you such a strong connection between the topology of $\mathcal{M}$ and the "spectrum" of the stable homotopy category. But you have to work hard for it, and the connection is much weaker (closed subsets of $\mathcal{M}$ have an interpretation in the stable homotopy category, rather than closed substacks) than what you would expect if Adams-Novikov spectral sequences were to degenerate.
|
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|
73,293 |
Let $M$ be a complex manifold, and $\omega$ a closed $2$-form. When is $\omega$ a Kähler form? I mean, when does there exist a Kähler metric for which $\omega$ is the corresponding form. I would (wildly) guess that necessary and sufficient conditions might be got from the Kähler identities.
|
I decided to make my comment into a more detailed answer. When $M$ has an almost complex structure $J$, then one can talk about smooth complex-valued differential forms of type $(p,q)$ in the usual way. A complex valued $2$-form $\omega$ is type $(1,1)$ if and only if it satisfies $\omega(JX,JY) = \omega(X,Y)$ for all smooth vector fields $X$ and $Y$ on $M$. If $\omega$ is a real $2$-form of type $(1,1)$, which means that $\overline \omega = \omega$, and if we define $g(X,Y) = \omega(X, JY)$, then it is easy to show that $g$ is a smooth, symmetric bilinear form on $M$. So it is a Riemannian metric if and only if it is positive definite. This is the definition of a positive $(1,1)$-form (that the associated $g$ is positive definite.) The triple of data $(J, \omega, g)$, where $J$ is an almost complex structure, $\omega$ is a real positive $(1,1)$-form, and $g$ is the associated Riemannian metric as defined above together define an almost Hermitian manifold . Now the condition for $M$ to be Kaehler is that $M$ be complex ($J$ is integrable) and that $d\omega = 0$. (These two conditions can be packaged together as $\nabla \omega = 0$ or $\nabla J = 0$, where $\nabla$ is the Levi-Civita connection of $g$.) Hence, if one is starting out with a complex manifold $M$, together with a closed real $2$-form, the only additional condition required to ensure that it defines a Kaehler metric is that it be a positive $(1,1)$-form.
|
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|
73,297 |
Well, the title clearly follows the title of this question. Why the objects so successfully defined by Grothendieck have been called "schemes"? In my opinion the original French word ( schéma ) doesn't help, by itself, to understand the motivations behind such a choice of nomenclature...
|
This is what Grothendieck says -- no difference with Andreas Blass, I just thought it might interest some as additional information -- in Récoltes et semailles (p. 31/32) [my emaphasize]: La notion de schéma est la plus naturelle, la plus "évidente" imaginable, pour englober en une notion unique la série infinie de notions de "variété" (algébrique) qu’on maniait précédemment (une telle notion pour chaque nombre premier (39)...). De plus, un seul et même "schéma" (ou "variété" nouveau style) donne naissance, pour chaque nombre premier p, à une "variété (algébrique) de caractéristique p" bien déterminée . La collection de
ces différentes variétés des différentes caractéristiques peut alors être visualisée comme une sorte d’ "éventail (infini) de variétés" (une pour chaque caractéristique). Le "schéma" est cet éventail magique, qui relie entre eux, comme autant de "branches" différentes, ses "avatars" ou "incarnations" de toutes les caractéristiques possibles. My (poor) translation: The notion of scheme is the most natural, the most "obvious" imaginable, to encompass in one unique notion the infinite series of notions of (algebraic) "variety", which one used before (such a notion for each prime number(39)...) Moreover, one and the same "scheme" (or "variety" of a new form) gives rise, for each prime number p, to a well-determined "(algebraic) variety of characteristic p." The collection of these different varieties of different characteristics can thus be seen as a sort of "(infinite) fan of varieties" (one for each characteristic). The "scheme" is this magic fan, which ties together, as many different "branches", its "avatars" or "incarnations" of all the possible characteristics. End of translation. [In particular the end might be a bit messed up, as éventail also has a botanic meaning and this might be the better one with the branches, but not sure.] Footnote 39 merely mentions that this is to include primes at infinity. P.S. In case somebody has suggestions for improvements of the translation, I'd appreciate them.
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|
73,321 |
Hi, the following statement appeared implicitly in a text I read and maybe you could just
give me a hint how to see this resp. give a reference: If you have two k-varieties $X$ and $Y$ (sufficiently nice) and you have a morphism $f:\ X \rightarrow Y$ between them, which is surjective and injective, then it is an isomorphism if $k$ is of characteristic zero. Thank you!
|
This is false. Consider a characteristic zero field $k$ and the cusp $C\subset \mathbb A^2_k$ with equation $y^2=x^3$ . Its normalization $n: \mathbb A^1_k \to C: t\mapsto (t^2, t^3)$ is bijective but not an isomorphism. "Ah, but Georges", you will say, "be attentive! The OP said nice varieties. Yours is ugly!" In that case Zariski's main theorem will come to your rescue. One version says that a birational morphism $f:Y\to X$ of $k$-varieties (in any characteristic) with finite fibers and $X$ normal is an isomorphism of $Y$ onto an open subset of $X$, hence an isomorphism if $f$ is bijective. "Aw, come on Georges, admit that you just dug up this birational stuff to make yourself look important!" Well, the theorem no longer holds without some such hypothesis, even in dimension zero. Just consider the bijective $\mathbb Q$-morphism $Spec(\mathbb Q(\sqrt 2)) \to Spec(\mathbb Q)$ of (singleton!) smooth schemes, which is not an isomorphism because it is not birational.(Of course you can inflate this to counterexamples in all dimensions) Some other counter-examples of bijective morphisms which are not isomorphisms, even over $\mathbb C$, are $Spec \mathbb C[\epsilon] \to Spec(\mathbb C)$ and $\mathbb G_m \bigsqcup Spec(\mathbb C)\to \mathbb A^1_\mathbb C$ (the evident morphism from the disjoint sum of a punctured affine line and a point onto the affine line).However the sources of those morphisms are respectively non reduced and reducible. Edit : Our friend Akhil gives a great argument (see his answer) showing that in ernest's case birationality is automatic. So, to sum up, we have the precise statement answering ernest's question: Proposition Let $k$ be an an algebraically closed field of characteristic zero and $f:X\to Y$ a morphism between integral $k$-schemes of finite type over $k$. Then if $f$ is bijective and $Y$ normal the morphism $f$ is an isomorphism. The case of characteristic $p$ As Akhil remarks, the Proposition is false in characteristic $p$. Consider an algebraically closed field $k$ of characteristic $p$ and the Frobenius morphism $f:\mathbb A^1_k \to \mathbb A^1_k:x\mapsto x^p$ with associated ring morphism $\phi:k[T] \to k[T]:P(T)\mapsto P(T^p)$ The morphism $f$ is bijective, but all fibers at closed points are non reduced of degree $p$ over $k$ .Indeed, let me denote for clarity by $A$ the $k$-algebra $\phi:k[T] \to A=k[T]$ above. Then the fibre of $f$ at the closed point $(T-a)$ in $\mathbb A^1_k$ is the affine $k$-scheme with algebra $A\otimes _{k[T]} \frac{k[T]}{(T-a)}=\frac{k[T]}{(T^p-a)}= \frac{k[T]}{(T-\sqrt[p] a)^p}$, so that the fiber is a single point but with non-reduced structure. This is an example where Grothendieck's introduction of non-reduced schemes helps dissipate a mystery: how can a bijective morphism have degree $p\gt 1$ ?
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|
73,347 |
Let $X_n$ be the "random Fibonacci sequence," defined as follows: $X_0 = 0, X_1 = 1$ ; $X_n = \pm X_{n-1} \pm X_{n-2}$ , where the signs are chosen by independent 50/50 coinflips. It is known that $|X_n|$ almost surely grows exponentially by a (much more general) theorem of Furstenberg and Kesten about random matrix products: the base of the exponent was determined explicitly by Viswanath to be $1.13\ldots$ I am not too proud to say that I learned all this from Wikipedia: http://en.wikipedia.org/wiki/Random_Fibonacci_sequence What I did not learn from Wikipedia, or any of the references I gathered therefrom, was: Question : what, if anything, do we know about the probability that $X_n = 0$ , as a function of $n$ ?
|
In response to Mark's comment, it is possible to determine that the probability of $X_n=0$ decays exponentially directly, and this is in fact easier than the theorems about the growth of these random sequences. Since we only care about testing if the sequence becomes zero, symmetry implies that we might as well reduce to the case $X_n=|X_{n-1}\pm X_{n-2}|$. If you see the paper "The random Fibonacci recurrence and the visible points of the plane" by T. Kalmar-Nagy, these sequences are in bijection with random walks on an infinite directed graph which looks like a trivalent tree with added parallel edges at every level. The probability you ask for can be translated into the question of the probability that a random walk in our graph returns to the origin after $n$ steps. There is a lot of machinery available for studying random walks on highly symmetric graphs (in this case we don't have a Cayley graph, but it's really close). I did the following simple estimate: The number of walks of length $3n$ starting from the origin is $2^{3n}$. For such a path to return to the origin exactly a third of its edges are directed south-west (I'm referring to the fig. 4 on page 4 of the paper I linked to). So the numbers of paths of length $3n$ with both endpoints at the origin is at most $\binom{3n}{n}\sim (27/4)^n$. So in particular the probability that $X_{3n}=0$ is bounded above by $(27/32)^n$. Added: Here's how one can get a lower bound as well. It is not hard to see that walks of length $3n$ in our graph are in bijection with sequences $\lbrace a_1,a_2,\dots,a_{3n}\rbrace$ which satisfy $a_i\in \lbrace 1,-\frac{1}{2}\rbrace$ and $$a_1+\cdots+a_{3n}= 0$$ $$a_1+\cdots+a_m\geq 0$$ for all $1\le m\le 3n$, and if $a_1+\cdots+a_m=\frac{1}{2}$ then $a_{m+1}=-\frac{1}{2}$. Enumerating these sequences exactly can get a bit tricky but there is a particularly nice subset if we restrict to the stronger condition $$a_1+\cdots+a_m > 1$$ for all $2\le m\le 3n-3$. Then we can enumerate these as Dyck paths on rectangles . The formula from that thread gives us $\frac{1}{3n-3}\binom{3n-3}{n-1}$. So to summarize we have
$$\frac{1}{(3n-3)2^{3n}}\binom{3n-3}{n-1}\le P(X_{3n}=0)\le \frac{1}{2^{3n}}\binom{3n}{n}$$
and in particular
$$\lim_{n\to \infty} \sqrt[n]{P(X_{3n}=0)}=\frac{27}{32}$$
which matches the experimental data in jc's answer.
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|
73,385 |
I'm teaching an introductory graph theory course in the Fall, which I'm excited about because it gives me the chance to improve my understanding of graphs (my work is in topology). A highlight for me will be to teach the Matrix-Tree Theorem, which I think is the only place that linear algebra is used in the course. Let κ(G) denote the number of spanning trees of G (the complexity of G), and let L(G) denote the Laplacian matrix of G. Finally, for a vertex v of G, let L(v|v) denote the Laplacian matrix with the row and column corresponding to v deleted. Matrix-Tree Theorem: κ(G)= det L(v|v). It seems a shame for Linear Algebra to be a prerequisite for my course. Anyway, I don't expect most of my students to be great Linear Algebra experts. Because my students might not remember things like Cauchy-Binet, but mainly so that I myself feel that I can really understand what I'm teaching, I wonder how the Matrix-Tree Theorem could be proven without ever mentioning matrices. On a planet with strong combinatorics and where linear algebra had not been discovered, how would they prove the Matrix-Tree Theorem? The RHS of the Matrix-Tree Theorem makes sense without ever mentioning matrices, via the Lindström-Gessel-Viennot (-Karlin-MacGregor) Lemma. Construct a graph H, with a source and a sink corresponding to each vertex of G, so that the signed sum of edge weights gives the entries of the Lagrangian matrix for G (surely there's a clever standard way to do this!) and define the determinant of H to be the signed sum of all n-tuples of non-intersecting paths from the sources to the sinks. This interpretation of the determinant seems a nice topic to teach. Maybe there's even an easier way. Cauchy-Binet becomes an elementary property of sets of non-intersecting paths in H, but I can't see how to free the rest of the proof of the Matrix-Tree Theorem from linear algebra. Question: Is there a proof of the Matrix-Tree Theorem which makes no mention of matrices? Is it simple enough that one could teach it in an introductory graph theory course? Note 1 : If the purpose of the Matrix-Tree Theorem is to efficiently calculate the complexity of a graph, then dodging linear algebra is surely counterproductive, because a determinant can efficiently be calculated in polynomial time. But, quite beside from my interest in "correctly-tooled" proofs and my teaching goals, I have vague dreams which are almost certainly nonsense about better understanding the Alexander polynomial as a quantum invariant along the lines of this question which might conceivably factor through the Matrix-Tree Theorem (or rather some version of Kirchhoff's formula), and I think I clearly want to stay in the diagrammatic world. Note 2 : This excellent post by Qiaochu Yuan suggests that the answer to my question is in Aigner. I'm travelling right now, and the relevant section isn't on Google Books, so I don't have access to Aigner for the next few weeks. If memory serves, Aigner still uses linear algebra, but maybe I'm wrong... anyway, if the answer turns out to be "look in Aigner" then please downvote this question, or if you could summarize how it's done, I would be happier. The answer will surely turn out to be "look in [somewhere]" anyway, because surely this is easy and well-known (just I don't know it)...
|
A combinatorial proof of the matrix-tree theorem can be found in the paper by D. Zeilberger A combinatorial approach to matrix algebra, Discrete Math. 56 (1985), 61–72.
The proof uses only the interpretation of the determinant as an alternating sum over permutations. Each term corresponds to a digraph, and the digraphs that do not correspond to trees cancel.
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73,450 |
Let $X$ be a (quasi-)projective, nonsingular, complex variety. Denote by $\mathcal{T}_X$ its tangent sheaf and by $X^{\mathrm{an}}$ its analytification. I am looking for a proof for the equality $\displaystyle \int_X c_n(\mathcal{T}_X) = \chi(X^{\mathrm{an}})$, i.e. the degree of the top chern class is equal to the topological Euler characteristic of $X$. There's Example 3.2.13 in Fulton's book on intersection theory which briefly mentions this, but it does not give a reference. Can someone help me out with one? Thanks in advance.
|
As an alternative to R. Budney's answer, one might also notice that the Gauss-Bonnet formula (the one you mention - mind that you must assume that $X$ is projective, otherwise the integral might not even make sense) is a consequence of the Hirzebruch-Riemann-Roch theorem. Indeed, the HRR theorem says
$$
\chi(V)=\int_{X}{\rm Td}({\rm T}X){\rm ch}(V)
$$
where $$\chi(V):=\sum_{l}{(-1)}^l{\rm rk}(H^l(X,V))$$ is the Euler characteristic of coherent sheaves. Now there is an universal identity of Chern classes
$$
{\rm ch}(\sum_{r}(-1)^r\Omega_X^r){\rm Td}(\Omega^\vee_X)=c^{\rm top}(\Omega^\vee_X)
$$
(called the Borel-Serre identity). Here $\Omega_X$ is the sheaf of differential of $X$ and thus $\Omega^\vee_X={\rm T}X$. Plugging the element $\sum_{r}(-1)^r\Omega_{X}^r$ into the HRR theorem, one gets
$$
\sum_{k,l}(-1)^{l+k}{\rm rk}(H^k(X,\Omega^l))=\int_{X}c^{\rm top}(TX)
$$
and by the Hodge decomposition theorem
$$
\sum_{k,l}(-1)^{l+k}{\rm rk}(H^k(X,\Omega^l))=\sum_{r}{(-1)}^r{\rm rk}(H^r(X({\bf C}),{\bf C}))
$$
where $H^r(X({\bf C}),{\bf C})$ is the $r$-th singular cohomology group.
The quantity $\sum_{r}{(-1)}^r{\rm rk}(H^r(X({\bf C}),{\bf C}))$ is the topological Euler characteristic, so this proves what you want.
The HRR theorem is proved in chap. 15 of Fulton's book (or in Hirzebruch's book "Topological methods...") and the Borel-Serre identity is Ex. 3.2.5, p. 57 of the same book.
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73,492 |
I am teaching Calc I, for the first time, and I haven't seriously revisited the subject in quite some time. An interesting pedagogy question came up: How misleading is it to regard $\frac{dy}{dx}$ as a fraction? There is one strong argument against this: We tell students that $dy$ and $dx$ mean "a really small change in $y$" and "a really small change in $x$", respectively, but these notions aren't at all rigorous, and until you start talking about nonstandard analysis or cotangent bundles, the symbols $dy$ and $dx$ don't actually mean anything. But it gives the right intuition! For example, the Chain Rule says $\frac{dy}{du} \cdot \frac{du}{dx}$ (under appropriate conditions), and it looks like you just "cancel the $du$". You can't literally do this, but it is this intuition that one turns into a proof, and indeed if one assumes that $\frac{du}{dx} \neq 0$ this intuition gets you pretty close. The debate about how rigorous to be when teaching calculus is old, and I want to steer clear of it. But this leaves an honest mathematical question: Is treating $\frac{dy}{dx}$ as a fraction the road to perdition, for reasons beyond the above, and which have not occurred to me?For example, what (if any) false statements and wrong formulas will it lead to? (Note: Please don't worry, I have no intention of telling students that $\frac{dy}{dx}$ is a fraction; only, perhaps, that it can usually be treated as one.)
|
You can think of $x$ and $y$ as smooth functions on a one-dimensional manifold of states of some system that you are thinking about, then $dx$ and $dy$ are differential forms. In any open region where $dx$ does not vanish we can say that $dy/dx$ is the unique smooth function such that $(dy/dx)dx=dy$; in other words, $dy/dx$ is $dy$ divided by $dx$. Of course you don't want to tell the students that, but it does clear up the logical question as asked. [Added later:] this approach also gives a clear picture of what goes wrong with partial derivatives: if your state space has dimension $n>1$, then $dy$ and $dx$ lie in a vector space of dimension $n$, and you cannot divide them to get a number. I think it's a bit fussy to worry too much about notation for derivatives in one variable, but traditional notation for partial derivatives is horrendous, especially in any context where you might want to hold different variables constant in different places, such as Maxwell's relations in thermodynamics ( http://en.wikipedia.org/wiki/Maxwell_relations )
|
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|
73,526 |
This is a soft question. How do people usually use arxiv to put their papers? At which stage does one usually put his/her paper/report there? Someone suggests me to submit a paper while putting it on arxiv. Is that the convention that people follow? Thank you! Anand
|
My comments above formulated as an answer: People typically post a preprint on the arxiv at the same time that they post it on their own homepage, with the goal of disseminating their work to their colleagues. (These days, posting on the web is more important than journal publication for sharing your work, and the arxiv is the central repository for math on the web.) This is often at the same time that they submit to a journal, although sometimes they wait for feedback (as Joe Silverman suggests in comments above), and sometimes they spend more time polishing their preprint before submitting it (as Darij Grinberg suggests, again in the comments above). The conclusion seems to be that it is standard to post on the arxiv as soon as one is ready to share one's work with colleagues, and that this is often at, or close to, the same time that one is ready to submit one's paper. In particular, it is quite common to post on the arxiv at the same time as submitting, or not long prior to submitting. (But there is nothing wrong with posting on the arxiv and then spending some more time polishing your preprint before submitting it to a journal.)
|
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|
73,651 |
The graph reconstruction conjecture claims that (barring trivial examples) a graph on n vertices is determined (up to isomorphism) by its collection of (n-1)-vertex induced subgraphs (again up to isomorphism). The way it is phrased ("reconstruction") suggests that a proof of the conjecture would be a procedure, indeed an algorithm, that takes the collection of subgraphs and then ingeniously "builds" the original graph from these. But based on some experience with a related conjecture (the vertex-switching reconstruction conjecture), I am led to wonder whether this is something that is simply true "by accident". By this I mean that it is something that is just overwhelmingly unlikely to be false ... there would need to be a massive coincidence for two non-isomorphic graphs to have the same "deck" (as the collection of (n-1)-vertex induced subgraphs is usually called). In other words, the only reason for the statement to be true is that it "just happens" to not be false. Of course, this means that it could never actually be proved.. and therefore it would be a very poor choice of problem to work on! My question (at last) is whether anyone has either formalized this concept - results that can't be proved or disproved, not because they are formally undecidable, but just because they are "true by accident" - or at least discussed it with more sophistication than I can muster. EDIT: Apologies for the delay in responding and thanks to everyone who contributed thoughtfully to the rather vague question. I have accepted Gil Kalai's answer because he most accurately guessed my intention in asking the question. I should probably not have used the words "formally unprovable" mostly because I don't really have a deep understanding of formal logic and while some of the "logical foundations" answers contained interesting ideas, that was not really what I was trying to get at. What I was really trying to get at is that some assertions / conjectures seem to me to be making a highly non-obvious statement about combinatorial objects, the truth of which depends on some fundamental structural understanding that we currently lack. Other assertions / conjectures seem, again, to me , to just be saying something that we would simply expect to be true "by chance" and that we would really be astonished if it were false. Here are a few unproved statements all of which I believe to be true: some of them I think should reflect structure and others just seem to be "by chance" (which is which I will answer later, if anyone is still interested in this topic). (1) Every projective plane has prime power order (2) Every non-desarguesian projective plane contains a Fano subplane (3) The graph reconstruction conjecture (4) Every vertex-transitive cubic graph has a Hamilton cycle (except Petersen, Coxeter and two related graphs) (5) Every 4-regular graph with a Hamilton cycle has a second one Certainly there is a significant chance that I am wrong, and that something that appears accidental will eventually be revealed to be a deep structural theorem when viewed in exactly the right way. However I have to choose what to work on (as do we all) and one of the things I use to decide what NOT to work on is whether I believe the statement says something real or accidental. Another aspect of Gil's answer that I liked was the idea of considering a "finite version" of each statement: let S(n) be the statement that "all non-desarguesian projective planes of order at most n have a Fano subplane". Then suppose that all the S(n) are true, and that for any particular n, we can find a proof - in the worst case, "simply" enumerate all the projective planes of order n and check each for a Fano subplane. But suppose that the length of the shortest possible proof of S(n) tends to infinity as n tends to infinity - essentially there is NO OTHER proof than checking all the examples. Then we could never make a finite length proof covering all n. This is roughly what I would mean by "true by accident". More comments welcome and thanks for letting me ramble!
|
This is a very interesting (yet rather vague) question. Most answers were in the direction of mathematical logic but I am not sure this is the only (or even the most appropriate) way to think about it. The notion of coincidence is by itself very complicated. (See https://en.wikipedia.org/wiki/Coincidence ). One way to put it on rigorous grounds is using probabilistic/statistical framework. Indeed, as Timothy mentioned it is sometimes possible to give a probabilistic heuristic in support of some mathematical statement. But its is notorious statistical problem to try to determine aposteriori if some events represent a coincidence. I am not sure that (as the OP assumes) if a statement is "true by accident" it implies that it can never be proved. Also I am not sure (as implied by most answers) that "can never be proved" should be interpreted as "does not follow from the axioms". It can also refers to situations where the statement admits a proof, but the proof is also "accidental" as the original statement is, so it is unlikely to be found in the systematic way mathematics is developed. In a sense (as mentioned in quid's answer), the notion of "true by accident" is related to mathematics psychology. It is more related to the way we perceive mathematical truths than to some objective facts about them. Regarding the reconstruction conjecture. Note that we can ask if the conjecture is true for graphs with at most million vertices. Here, if true it is certainly provable. So the logic issues disappear but the main issue of the question remains. (We can replace the logic distinctions by computational complexity distinctions. But still I am not sure this will capture the essence of the question.) There is a weaker form of the conjecture called the edge reconstruction conjecture (same problem but you delete edges rather than vertices) where much is known. There is a very conceptual proof that every graph with $n$ vertices and more than $n \log n$ edges is edge-reconstructible. So this gives some support to the feeling that maybe vertex reconstruction can also be dealt with. Finally I am not aware of a heuristic argument that "there would need to be a massive coincidence for two non-isomorphic graphs to have the same 'deck'" as the OP suggested. (Coming up with a convincing such heuristic would be interesting.) It is known that various graph invariants must have the same value on such two graphs.
|
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|
73,818 |
I know classification of 2 manifolds and geometrization for 3 manifolds.
Why for dimension great or equal to 4, this task become impossible? edit: Or should I ask "why geometrization won't be possible for 4 or higher dimension?"
|
I'm guessing that you heard this from someone whose reasoning goes "Every finite presentation of a group can be made to give the $\pi_1$ of a smooth 4-manifold. If we could put any 4-manifold into the Magic List of All, then we could recognize presentations of the trivial group. But no algorithm can do that." Often people worry about classifications of simply connected manifolds, and don't have to deal with this. (Of course in three dimensions this becomes Perelman's theorem.)
|
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|
73,945 |
Given a finitely generated $\def\CC{\mathbb C}\CC$-algebra $R$ and a $\CC$-point (maximal ideal) $p\in Spec(R)$, I define the singularity type of $p\in Spec(R)$ to be the isomorphism class of the completed local ring $\hat R_p$, as a $\CC$-algebra. Do there exist non-algebraic singularity types? That is, does there exist a complete local ring with residue field $\CC$ which is formally finitely generated (i.e. has a surjection from some $\CC[[x_1,\dots, x_n]]$), but is not the complete local ring of a finitely generated $\CC$-algebra at a maximal ideal? Googling for "non-algebraic singularity" suggests that the answer is yes, but I can't find a specific example. I would expect that it should be possible to write down a power series in two variables $f(x,y)$ so that $\CC[[x,y]]/f(x,y)$ is non-algebraic. What is a specific formally finitely generated non-algebraic singularity?
|
I got this example from Frank Loray. I'll explain the analytic version, but the formal variant works just as well. Let $U\subset \mathbb{C}$ be open. Choose two holomorphic functions $f$, $g$ which are algebraically independent over $\mathbb{C}$ (e.g. $f(z)=z$, $g(z)=e^z$). For simplicity, assume that $f$, $g$, $0$, $1$ never coincide (pairwise) on $U$. Now define $X\subset U\times\mathbb{C}^2$ (with coordinates $z$, $x$, $y$) as the union of $x=0$, $y=0$, $x=y$, $y=f(z)\,x$, and $y=g(z)\,x$. Thus, if we freeze $z$, we get five lines in the plane, with slopes $\infty$, $0$, $1$, $f(z)$ and $g(z)$. Globally, $X$ is the union of five copies of $U\times\mathbb{C}$ meeting along $Z:=U\times\mathbb{0}$. Now, the point is that the cross-ratio of four (ordered) lines through the origin in the plane is an intrinsically defined invariant. In particular, independently of the coordinates, we can recover $f$ and $g$ as holomorphic functions on the singular locus $Z$. If $X$ were isomorphic to a complex open subset of an algebraic variety, $f$ and $g$ would have to be algebraically dependent because $\dim Z=1$: contradiction.
|
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|
74,014 |
Some theorems are magical: their hypotheses are easy to meet, and when invoked (as lemmas) in the midst of an otherwise routine proof, they deliver the desired conclusion more or less straightaway—like pulling a rabbit from a hat. Here are five examples. Some are from outside of logic, but each is often useful within logic. Baire Category Theorem. In any completely metrizable topological space, each nonempty open set is nonmeager. Gödel's Diagonal Lemma. If a theory $T$ relatively interprets Robinson's arithmetic, then for each first order formula $\varphi(x,v_1,\dots,v_n)$ in the language of $T$, there is a $\psi(v_1,\dots,v_n)$ such that $T$ proves the sentence $\forall v_1 \dots \forall v_n[\psi(v_1,\dots,v_n) \leftrightarrow \varphi(\overline\psi,v_1,\dots,v_n)]$, where $\overline\psi$ is the code of $\psi$. König's Tree Lemma. Every finitely splitting tree of infinite height has an infinite branch. Knaster–Tarski Theorem. Every monotone nondecreasing operator on the powerset of a set has a fixed point. Mostowski Collapsing Lemma. If $E$ is a well founded, set like, and extensional binary class relation on a class $M$, then there is a unique transitive class $N$ such that $(M, E)$ is isomorphic to $(N, \in)$. Let's list other magical theorems that every logician can wield. Students among us will thereby learn of useful results that might otherwise escape their attention until much later. (There is related question here . But it and most of its answers don't focus on theorems useful in logic.) Please treat only one theorem per answer, and write as many answers as you like. Don't just link to Wikipedia or whatever; give a pithy statement. If possible, keep it informal. Bonus if the theorem isn't well known, or if you show it in action.
|
The Compactness Theorem Funny that no one mentioned it so far. I find the Compactness Theorem magical, actually: A first order theory $T$ has a model if and only if every finite subset of $T$ has a model. This let's you derive the finite form of Ramsey's theorem from the infinite form. That is magic.
For most applications of this kind, the Compactness Theorem for propositional calculus is actually enough. Compactness for first-order logic and propositional logic are actually equivalent (over ZF). In fact, there is a rather large collection of equivalent results: Boolean Prime Ideal Theorem The Ultrafilter Lemma The Stone Representation Theorem The Tychonoff Theorem for compact Hausdorff spaces The Compactness Theorem is strictly weaker than the Axiom of Choice, but it is not provable in plain ZF. This is often used to show that certain results are weaker than the Axiom of Choice. For example, it can be used to show that the existence of non-measurable sets is weaker than the Axiom of Choice (by an old result of Sierpiński).
|
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|
74,095 |
Is there a set convention for which name (maiden name or married name) a female married mathematician should use? While this question addresses women's maiden name it applies equally to men's maiden name when it differs from their married name. The question seeks for an advice for the dilemma: whether to use the maiden name or the new married name. For example, Fan Chung is married to Ron Graham, but she publishes under "Fan Chung." Vera T. Sós is another married woman who continued to use her maiden name, but the T. stands for Turán. Yet, I'm pretty sure that Emma Lehmer (née Trotskaia) published under her married name. Does it have something to do with the name under which the woman first publishes or the name under which name she receives her Ph.D.?
|
Like all questions involving names and marriage, there is no set convention (at least in the US). I know a male mathematician who publishes under his wife's last name which he took at marriage and I know people who have started publishing under a new name before they took it legally. As Ben says, there's also no rules for names that don't involve marriage: not only do people pick whether they use their full first name or a nickname, some people use initials, and some people use nicknames which are not related to their legal first name. Although there are no set rules or conventions, most people seem to agree that early on in your career it's unwise to change the name that you are publishing under. Your name is your brand and diluting it is likely to hurt you professionally. Thus there's a strong tendency for people to publish under a fixed name. Nonetheless this is not a fixed rule, a particularly striking example is a theorem that's changed names: Nichols-Richmond nee Nichols-Zoeller.
|
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|
74,212 |
Suppose we have integer matrices $A_1,\ldots,A_n\in\operatorname{GL}(n,\mathbb Z)$. Define $\varphi:F_n\to\operatorname{GL}(n,\mathbb Z)$ by $x_i\mapsto A_i$. Is there an algorithm to decide whether or not $\varphi$ is injective?
|
For $n=1, 2$ the answer is "yes" since the group is virtually free, for $n\ge 3$ the answer is not known (an open problem). Edit. In fact even for two $n\times n$-matrices the problem is open. Moreover the solution of the following ``easier" problem is not known: for which algebraic integers $\lambda$ the matrices $\left(\begin{array}{ll} 1 & 2\\\ 0 & 1 \end{array}\right)$ and $\left(\begin{array}{ll} 1 & 0\\\ \lambda & 1 \end{array}\right)$ generate a free group (see this paper, for example). The fact that this problem is easier follows from the trivial observation that the group generated by these two matrices is isomorphic to some effectively computable group of $n\times n$-integer matrices for some $n\ge 2$ (depending on the degree of the algebraic number $\lambda$).
|
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|
74,214 |
For an expository piece I'm writing, it would be useful to have good examples of the following phenomenon: (1) ${\cal X}$ is a parameterized family of somethings. (Varieties, schemes, manifolds, rings, groups, whatever.) (2) ${\cal X}_0$ is an member of this family. (3) ${\cal X}_0$ has some interesting property $P$ that, on the surface, appears to have nothing to do with the inclusion of ${\cal X}_0$ in the family ${\cal X}$. (4) Nevertheless, the only (or ``best'', or simplest, or most natural) way to prove that ${\cal X}_0$ has property $P$ is to invoke the existence of the family ${\cal X}$. So, to put this in the form of a question, what are some good examples of this phenomenon? I'm especially (but not exclusively) interested in examples that could be explained to an undergraduate. (It was hard to find the right tags for this question; feel free to change them.)
|
In a way, you can think of calculus as using exactly this sort of idea. Say you are interested in determining the area of the region consisting of points (x,y) with $a \le x \le b$ and $0 \le y \le f(x)$ . It turns out the best thing to do is to consider the family of regions $ S(c) $ consisting of points (x,y) with $ a \le x \le c, 0 \le y \le f(x) $, because then you can associate to this family the function A(c) = area of S(c). Then we know A'(c)=f(c), and this is a much easier problem to solve (and then just evaluate at c=b) than, say, trying to calculate the area of a single region directly, say by exhaustion.
|
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|
74,362 |
About a year ago, a colleague asked me the following question: Suppose $(R,+,\cdot)$ and $(S,\oplus,\odot)$ are two rings such that $(R,+)$ is isomorphic, as an abelian group, to $(S,\oplus)$, and $(R,\cdot)$ is isomorphic (as a semigroup/monoid) to $(S,\odot)$. Does it follow that $R$ and $S$ are isomorphic as rings? I gave him the following counterexample: take your favorite field $F$, and let $R=F[x]$ and $S=F[x,y]$, the rings of polynomials in one and two (commuting) variables. They are not isomorphic as rings, yet $(R,+)$ and $(S,+)$ are both isomorphic to the direct sum of countably many copies of $F$, and $(R-\{0\},\cdot)$ and $(S-\{0\},\cdot)$ are both isomorphic to the direct product of $F-\{0\}$ and a direct sum of $\aleph_0|F|$ copies of the free monoid in one letter (and we can add a zero to both and maintain the isomorphism). He mentioned this example in a colloquium yesterday, which got me to thinking: Question. Is there a counterexample with $R$ and $S$ finite?
|
There do exist pairs of finite unital rings whose additive structures
are isomorphic and whose multiplicative structures are isomorphic,
yet the rings themselves are not isomorphic. To see this, let $\mathbb F$ be a field and let $X = \{x_1,\ldots, x_n\}$
be a set of variables. The polynomial ring $\mathbb F[X]$
is graded by degree
$$
\mathbb F[X] = H_0\oplus H_1\oplus H_2\oplus\cdots.
$$
Let $Q(x_1,\ldots,x_n)$ be a quadratic form over $\mathbb F$.
Let
$$I = \mathbb F\cdot Q(X)\oplus H_3\oplus H_4\oplus\cdots$$
be the
ideal generated by $Q(X)$ and the homogeneous components
of degree at least $3$.
Let $S_{\mathbb F,Q}$ denote the $\mathbb F$-algebra
$\mathbb F[X]/I$. It is a commutative, local ring, which encodes
properties of the quadratic form $Q$. Two quadratic forms $Q_1$ and $Q_2$ are equivalent
if they differ by an invertible linear change of variables. Claim. Let $\mathbb F$ be a finite field of odd characteristic $p$.
Let $Q_1(x_1,\ldots,x_n)$ and $Q_2(x_1,\ldots,x_n)$ be
nonzero quadratic forms over $\mathbb F$. $S_{\mathbb F,Q_1}$ and $S_{\mathbb F,Q_2}$ have isomorphic
$\mathbb F$-space structures. If $n>4$ and $Q_1$ and $Q_2$ are nondegenerate, then
$S_{\mathbb F,Q_1}$ and $S_{\mathbb F,Q_2}$ have
isomorphic multiplicative monoids. $S_{\mathbb F,Q_1}\not\cong S_{\mathbb F,Q_2}$ as $\mathbb F$-algebras,
unless $Q_1$ is equivalent to a nonzero scalar multiple of $Q_2$. Proof. Exercise! \\ So let $\mathbb F = \mathbb F_3$ be the $3$-element field.
It is known that over a finite field of odd characteristic
the quadratic forms are classified by the dimension
and by the determinant of the form modulo squares.
The determinant of
$$
Q(x_1,\ldots,x_n)=a_1x_1^2+a_2x_2^2+\cdots+a_nx_n^2
$$
is $a_1\cdots a_n$. If
$\alpha\in \mathbb F_3^{\times}=\{\pm 1\}$,
then $\alpha\cdot Q$ has determinant
$\alpha^n a_1\cdots a_n=(\pm 1)^n a_1\cdots a_n$.
If $n$ is even, then the determinants of $Q$ and $\alpha\cdot Q$ will be equal,
so $Q$ will be equivalent
to $\alpha\cdot Q$ for every $\alpha\in \mathbb F_3^{\times}$.
This implies that, when working over $\mathbb F_3$ in an even dimension,
if $Q_1$ is not equivalent to $Q_2$, $Q_1$ will also
not be equivalent to any nonzero scalar
multiple of $Q_2$.
In particular, no scalar multiple of
$$
Q_1 = x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2,
$$
is equivalent to
$$
Q_2 = x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 - x_6^2
$$
over $\mathbb F_3$. For these forms we have
that $S_{\mathbb F,Q_1}$ and $S_{\mathbb F,Q_2}$ are nonisomorphic
finite unital rings with isomorphic additive and multiplicative structures.
(These rings have size $3^{27}$.) Minor side comment 1: If you allow nonunital rings, there is a pair of nonisomorphic $8$-element rings whose additive and multiplicative structures are isomorphic. Minor side comment 2: The solution to the exercise above (that is, the proof of the Claim)
can be found here .
|
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|
74,538 |
Countable models of PA fall into two categories: the standard one $(\omega, S)$ and the nonstandard ones (all the rest). The only way I've seen to construct a nonstandard model is through taking an ultraproduct or, equivalently, using the compactness theorem. My question is wether or not these are all the models there are? There are continuum many ultrafilters and continuum many nonstandard, countable models, but I don't know if there's a surjective correspondence.
|
An ultrapower will never yield a countable nonstandard model of PA --- either you will recover the standard model or the result will be uncountable. As far as the construction of a model is concerned, due to Tennenbaum's theorem (see http://en.wikipedia.org/wiki/Tennenbaum's_theorem ) you will never see a recursive nonstandard model of PA. Hence, in some sense, you will never construct a countable model of PA other than the standard model. On the other hand, if you consider the Henkin construction to be constructive enough for you, then by running his construction relative to the theory in ${\mathcal L}(+,\times,0,1,c,<)$ consisting of PA together with all the assertions $c > n$ for each $n \in {\mathbb N}$, then you would obtain a nonstandard model of PA.
|
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|
74,689 |
Weyl's theorem states that any finite-dimensional representation of a finite-dimensional semisimple Lie algebra is completely reducible. In my mind, the "natural" way to prove this result is by way of Lie groups. However, as a student, I first encountered Weyl's theorem in the textbook by Humphreys, in which he gives a purely algebraic proof. I remember finding this proof very mysterious, and in particular it seemed that Humphreys pulled the Casimir element out of a hat. Looking at the proof again now, I still find it somewhat mysterious, and if I had to present the proof in a graduate class, I don't think I would be able to motivate it very well. How would you motivate this purely algebraic proof of Weyl's theorem?
|
I think that there are two things to be motivated: one is the Casimir, and the other is the proof of semi-simplicity. First, for the Casimir, it might help to note that there is a ``formula-free" construction. A symmetric bilinear form $\kappa$ defines $\mathfrak{g}\simeq \mathfrak{g}^{\vee}$, and the Casimir is the image in $U(\mathfrak{g})$ of the element corresponding to the identity under the isomorphism $\mathfrak{g}\otimes\mathfrak{g}\simeq \mathfrak{g}^{\vee}\otimes\mathfrak{g}$. The idea of the proof is actually very simple: you construct an element of the center of the algebra which ``detects" the trivial representation. I.e., it acts by zero on the trivial representation and by a non-zero scalar on all simple (finite-dimensional) representations. (This is in exact analogy to what happens for finite groups in characteristic zero: then $1-\frac{1}{|G|}\sum_{g\in{G}}\delta_g$ has the same property). Once you have such a central element, let's call it $C$, the proof that finite-dimensional representations are semi-simple is easy. For all $V,W$, note that:
$$\operatorname{Ext}^i(V,W)=
\operatorname{Ext}^i(\mathbb{C},\underline{\operatorname{Hom}}(V,W))$$
(where $\underline{\operatorname{Hom}}$ is internal $\operatorname{Hom}$ relative to the usual tensor product of $\mathfrak{g}$-modules, and both $\operatorname{Ext}$s are of $\mathfrak{g}$-modules) because formation of internal $Hom$ is exact in both variables. Therefore, it's enough to show that $\operatorname{Ext}^1(\mathbb{C},V)=0$ for all finite-dimensional $\mathfrak{g}$-modules $V$ (substitute $\underline{\operatorname{Hom}}(V,W)$ for $V$). Clearly any such $V$ has finite length, so by devissage, it's enough to prove for simple modules. Either $V$ is trivial or it is not. If $V$ is non-trivial, then $C$ acts on $\operatorname{Ext}^i(\mathbb{C},V)$ by two different scalars: the one by which it acts on $V$ and by $0$ (by which it acts on $\mathbb{C}$). Therefore, this vector space must be zero. If $V=\mathbb{C}$, then $\operatorname{Ext}^1(\mathbb{C},\mathbb{C})=0$ since for any extension $E$, the homomorphism $\mathfrak{g}\to\operatorname{End}(E)$ maps to the 1-dimensional subspace sending $E$ to $\mathbb{C}$ and sending $\mathbb{C}$ to $0$, but since $\mathfrak{g}$ has no codimension 1 ideals this must be the trivial homomorphism. Of course, basically the same proof goes through for finite groups and is essentially the same as the usual proof.
|
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|
74,707 |
Once again I come to MO for help with something I'm writing for the public. Which habits of mathematicians -- aspects of the way we approach problems, the way we argue, the way we function as a community, the way we decide on our goals, whatever -- would you recommend that non-mathematicians adopt, at least in certain contexts? In other words: if you can imagine a situation in which someone came to you for advice, and you said, "Look, I think you should be a little more like a mathematician about this and...." what would be the end of the sentence?
|
In my experience, mathematicians will frequently argue (in general, not just in mathematics) by passing to an extreme case at the beginning. Non-mathematicians (again in my experience) sometimes object to such a mode of argument as invalid or irrelevant because such extreme hypotheticals are clearly unrealistic. I think that the mathematical idea of first setting all the parameters to their maximal, or minimal, values, and understanding that case, before trying to tune them to a more realistic choice of values (and seeing how the solution/context changes with the parameters) can sometimes be valuable (even though it involves as a first step considering a situation that may be very unrealistic).
|
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|
74,806 |
To put this question in precise language, let $X$ be an affine scheme, and $Y$ be an arbitrary scheme, and $f : X \rightarrow Y$ a morphism from $X$ to $Y$. Does it follow that $f$ is an affine morphism of schemes? While all cases are interesting, a counterexample that has both $X$ and $Y$ noetherian would be nice.
|
No, here is an example of a morphism $f:X\to Y$ which is not affine although $X$ is affine. Take $X=\mathbb A^2_k$, the affine plane over the field $k$ and for $Y$ the notorious plane with origin doubled: $Y=Y_1\cup Y_2$ with $Y_i\simeq \mathbb A^2_k$ open in $Y$ and $Y\setminus Y_i= \lbrace O_i\rbrace$, a closed rational point of $Y$. We take for $f:X\to Y$ the map sending $X$ isomorphically to $Y_1$ in the obvious way. Then, although the scheme $X$ is affine, the morphism $f$ is not affine because the inverse image $f^{-1}(Y_2)$of the affine open subscheme $Y_2\subset Y$ is $X \setminus \lbrace 0 \rbrace=\mathbb A^2_k \setminus \lbrace 0 \rbrace$, the affine plane with origin deleted, well known not to be affine.
|
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|
74,841 |
The background of my question comes from an observation that what we teach in schools does not always reflect what we practice. Beauty is part of what drives mathematicians, but we rarely talk about beauty in the teaching of school mathematics. I'm trying to collect examples of good, accessible proofs that could be used in middle school or high school. Here are two that I have come across thus far: (1) Pick's Theorem: The area, A, of a lattice polygon, with boundary points B and interior points I is A = I + B/2 - 1. I'm actually not so interested in verifying the theorem (sometimes given as a middle school task) but in actually proving it. There are a few nice proofs floating around, like one given in "Proofs from the Book" which uses a clever application of Euler's formula. A very different, but also clever proof, which Bjorn Poonen was kind enough to show to me, uses a double counting of angle measures, around each vertex and also around the boundary. Both of these proofs involve math that doesn't go much beyond the high school level, and they feel like real mathematics. (2) Menelaus Theorem: If a line meets the sides BC, CA, and AB of a triangle in the points D, E, and F then (AE/EC) (CD/DB) (BF/FA) = 1. (converse also true) See: http://www.cut-the-knot.org/Generalization/Menelaus.shtml , also for the related Ceva's Theorem. Again, I'm not interested in the proof for verification purposes, but for a beautiful, enlightening proof. I came across such a proof by Grunbaum and Shepard in Mathematics Magazine. They use what they call the Area Principle, which compares the areas of triangles that share the same base (I would like to insert a figure here, but I do not know how. -- given triangles ABC and DBC and the point P that lies at the intersection of AD and BC, AP/PD = Area (ABC)/Area(DBC).) This principle is great-- with it, you can knock out Menelaus, Ceva's, and a similar theorem involving pentagons. And it is not hard-- I think that an average high school student could follow it; and a clever student might be able to discover this principle themselves. Anyway, I'd be grateful for any more examples like these. I'd also be interested in people's judgements about what makes these proofs beautiful (if indeed they are-- is there a difference between a beautiful proof and a clever one?) but I don't know if that kind of discussion is appropriate for this forum. Edit : I just want to be clear that in my question I'm really asking about proofs you'd consider to be beautiful, not just ones that are neat or accessible at the high school level. (not that the distinction is always so easy to make...)
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Extending on Ralph's answer, there is a similar very neat proof for the formula for $Q_n:=1^2+2^2+\dots+n^2$. Write down numbers in an equilateral triangle as follows: 1
2 2
3 3 3
4 4 4 4 Now, clearly the sum of the numbers in the triangle is $Q_n$. On the other hand, if you superimpose three such triangles rotated by $120^\circ$ each, then the sum of the numbers in each position equals $2n+1$. Therefore, you can double-count $3Q_n=\frac{n(n+1)}{2}(2n+1)$. $\square$ (I first heard this proof from János Pataki). How to prove formally that all positions sum to $2n+1$? Easy induction ("moving down-left or down-right from the topmost number does not alter the sum, since one of the three summand increases and one decreases"). This is a discrete analogue of the Euclidean geometry theorem "given a point $P$ in an equilateral triangle $ABC$, the sum of its three distances from the sides is constant" (proof: sum the areas of $APB,BPC,CPA$), which you can mention as well. How to generalize to sum of cubes? Same trick on a tetrahedron. EDIT: there's some way to generalize it to higher dimensions, but unfortunately it's more complicated than this. See the comments below. If you wish to tell them something about "what is the fourth dimension (for a mathematician)", this is an excellent start.
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|
74,863 |
From time to time, I pretend to be an algebraic topologist. But I'm not really hard-core and some of the deeper mysteries of the subject are still ... mysterious. One that came up recently is the exact role of CW-complexes. I'm very happy with the mantra "CW-complexes Good, really horrible pathological spaces Bad." but there's a range in the middle there where I'm not sure if the classification is "Good" or just "Pretty Good". These are the spaces with the homotopy type of a CW-complex. In the algebraic topology that I tend to do then I treat CW-complexes in the same way that I treat Riemannian metrics when doing differential topology. I know that there's always a CW-complex close to hand if I really need it, but what I'm actually interested in doesn't seem to depend on the space actually being a CW-complex. But, as I said, I'm only a part-time algebraic topologist and so there may be whole swathes of this subject that I'm completely unaware of where actually having a CW-complex is of extreme importance. Thus, my question: In algebraic topology, if I have a space that actually is a CW-complex, what can I do with it that I couldn't do with a space that merely had the homotopy type of a CW-complex?
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Those of us who do algebraic topology too much should remember occasionally that topological spaces are, in general, terrible to work with. CW-complexes have a lot of properties that make them nice to work with in homotopy theory, such as being amenable to study by homotopy groups and such as being able to define maps out inductively. These properties are still possessed by objects with the homotopy type of a CW-complex. However, CW-complexes are also nice on the point-set level. They're compactly generated, locally contractible, and every compact subset is contained in a finite CW-subcomplex - and finite CW-complexes have almost every space-level regularity property that one can name. If one has the goal of doing homotopy theory, the fact that we have this large class of nice spaces means that (to a certain extent) we can set many point-set considerations aside. And sometimes these point-set considerations can be irritating (such as the smash product being nonassociative). Objects just in the homotopy type can be terrible. Take the cone on your favorite pathological space and you find something with the homotopy type of a nice CW-complex but terrible point-set behavior.
|
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|
75,038 |
Let $\omega$ be a closed non-exact differential $k$-form ($k \geq 1$) on a closed orientable manifold $M$. Question : Is there always a Riemannian metric $g$ on $M$ such that $\omega$ is $g$-harmonic, i.e., $\Delta_g \omega = 0$? Here $\Delta_g$ is the Laplace-deRham operator, defined as usual by
$\Delta_g = d \delta + \delta d$, where $\delta$ is
the $g$-codifferential. Note that non-exactness is important, since if $\omega$ were to be exact and harmonic, then by the Hodge decomposition theorem $\omega = 0$. For instance, if $\omega$ is a 1-form on the unit circle, then it is not hard to see that $\omega$ is harmonic with respect to some metric $g$ if and only if it is a volume form (i.e., it doesn't vanish). This observation generalizes to forms of top degree on any $M$. What can be said in general for forms which are not of top degree?
|
A closed $k$ -form is called intrinsically harmonic if there is some Riemannian metric with respect to which it is harmonic. E. Calabi ( Calabi, Eugenio , An intrinsic characterization of harmonic one-forms, Global Analysis, Papers in Honor of K. Kodaira 101-117 (1969). ZBL0194.24701 .)
showed that a one-form having non-degenerate zeros on a compact manifold without boundary is intrinsically harmonic if and only if it satisfies a property called transitivity . The precise statement and proof can be found in chapter 9 of M. Farber's book "Topology of closed one-forms". In what comes I am following Farber. That a closed one-form $\omega$ have non-degenerate zeros means that near each zero it can be written in the form $\omega = df$ with $f$ a Morse function. For such a one-form, the additional assumption of harmonicity means that the Morse index of a zero cannot be $0$ or $n$ (write $\omega = df$ near the zero; because $\omega$ is co-closed, $f$ is harmonic, so by the maximum principle cannot have a max or min at the zero). That $\omega$ be transitive means that for any point $p$ of $M$ which is not a zero of $\omega$ there is a smooth $\omega$ -positive loop $\gamma: [0, 1] \to M$ ; that is, $\gamma(0) = p = \gamma(1)$ , and $\omega(\dot{\gamma}(t)) > 0$ for $t \in [0, 1]$ . Then Calabi's theorem states that a closed one-form with non-degenerate zeros is intrinsically harmonic if and only if it is transitive.
Near a non-degenerate index $0$ zero of a closed one-form the one-form can be written in the form $\delta_{ij}x^{i}dx^{j}$ , for which it can be checked there are no positive loops beginning sufficiently near the origin. (If one can handle $k$ -forms then by Hodge duality one expects to be able to get somewhere with $(n-k)$ -forms. The intrinsic harmonicity of $(n-1)$ -forms was characterized in terms of transitivity in the thesis of Ko Honda, available on his web page). Evgeny Volkov ( Volkov, Evgeny , Characterization of intrinsically harmonic forms , J. Topol. 1, No. 3, 643-650 (2008). ZBL1148.57036 .)
weakens the non-degeneracy condition, replacing it with the condition that the closed one-form be locally intrinsically harmonic - that is, the restriction of the form to a suitable open neighborhood of its zero set is intrinsically harmonic. As far as I know, for higher degree forms nothing much is known at all, though for some special cases, like $2$ -forms on $4$ -manifolds, something more has been said. One imagines that with further assumptions on the form, perhaps more can be said - for example a symplectic form is always intrinsically harmonic (use the metric determined by a compatible almost complex structure). On the other hand, Volkov's paper exhibits a closed $2$ -form of rank $2$ on a $4$ -manifold which is transitive but not intrinsically harmonic.
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|
75,049 |
(This question was posed to me by a colleague; I was unable to answer it, so am posing it here instead.) Let $f: {\bf R}^n \to {\bf R}^n$ be an everywhere differentiable map, and suppose that at each point $x_0 \in {\bf R}^n$, the derivative $Df(x_0)$ is nonsingular (i.e. has non-zero determinant). Does it follow that $f$ is locally injective, i.e. for every $x_0 \in {\bf R}^n$ is there a neighbourhood $U$ of $x_0$ on which $f$ is injective? If $f$ is continuously differentiable, then the claim is immediate from the inverse function theorem . But if one relaxes continuous differentiability to everywhere differentiability, the situation seems to be much more subtle: In one dimension, the answer is "Yes"; this is the contrapositive of Rolle's theorem, which works in the everywhere differentiable category. (The claim is of course false in weaker categories such as the Lipschitz (and hence almost everywhere differentiable) category, as one can see from a sawtooth function.) The Brouwer fixed point theorem gives local surjectivity, and degree theory gives local injectivity if $\det Df(x_0)$ never changes sign. (This gives another proof in the case when $f$ is continuously differentiable, since $\det Df$ is then continuous.) On the other hand, if one could find an everywhere differentiable map $f: B \to B$ on a ball $B$ that was equal to the identity near the boundary of $B$, whose derivative was always non-singular, but for which $f$ was not injective, then one could paste infinitely many rescaled copies of this function $f$ together to produce a counterexample. The degree theory argument shows that such a map does not exist in the orientation-preserving case, but maybe there is some exotic way to avoid the degree obstruction in the everywhere differentiable category? It seems to me that a counterexample, if one exists, should look something like a Weierstrass function (i.e. a lacunary trigonometric series), as one needs rather dramatic failure of continuity of the derivative to eliminate the degree obstruction. To try to prove the answer is yes, one thought I had was to try to use Henstock-Kurzweil integration (which is well suited to the everywhere differentiable category) and combine it somehow with degree theory, but this integral seems rather unpleasant to use in higher dimensions.
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The usual reference to the proof is A. V. Cernavskii in "Finite-to-one open mappings of manifolds", Mat. Sb. (N.S.), 65(107) (1964), 357–369 and "Addendum to the paper "Finite-to-one open mappings of manifolds"", Mat. Sb. (N.S.), 66(108) (1965), 471–472. If I remember it correctly, he does not state it explicitly, but it follows from what is there.
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|
75,317 |
It is always a pain to move back and forth between definitions in algebraic geometry and complex analytic geometry. Dictionary is much easier when are working with (family of) smooth varieties but the pain grows exponentially when we include singular varieties. Here is some of that: Suppose $f: X\rightarrow Y$ is a map of possibly singular complex analytic varieties, then is there a simpler definition of flatness for $f$ ? this one should be hard to answer but how about following, ---Let $f:X\rightarrow Y$ be a family of curves. there might be multiple fibers, non-reduced fibers, nodal curves, cusp curves,... in the family, but fibers are connected. ---When this fibration is flat? What kind of bad fibers mentioned above are allowed in a flat family? ---Same question for higher dimensional family of analytic varieties? For simplicity you may assume that the base of fibration is smooth.
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Instead of trying to say what flatness in analytic geometry means I'll give you some street-fighting tricks for recognizing whether a morphism of analytic spaces ( not necessarily reduced) $f:X\to Y $ is , or has a chance to be, flat. a) A flat map is always open. . Hence, contraspositely, the embedding
$\lbrace 0\rbrace \hookrightarrow \mathbb C$ is not flat. More generally, the embedding of a closed (and not open !) subspace $X \hookrightarrow Y$ is never flat. b) An open map need not be flat: think of the open map $ Spec(\mathbb C) \to Spec(\mathbb C[\epsilon ]) $ from the reduced point to the double point, which is open but not flat. [It is not flat because the $\mathbb C[\epsilon]$- algebra $\mathbb C=\mathbb C[\epsilon]/(\epsilon)$ is not flat : recall that a quotient ring $A/I$ can only be flat over $A$ if $I=I^2$ and here
$I=(\epsilon) \neq I^2=(\epsilon)^2=(0)$] An example with both spaces reduced is the normalization [see g) below] $f:X^{nor} \to X $ of the cusp $X\subset \mathbb C^2$ given by the equation $y^2=x^3$ .That normalization is a homeomorphism and so certainly open, but it is not flat : this results either from g) or from h) below. c) Given the morphism $f:X\to Y $ , consider the following property: $\forall x \in X, \quad dim_x(X)=dim_{f(x)} (Y) +dim_x(f^{-1}(f(x)))$ $\quad (DIM) $ We then have: $ f \; \text {flat} \Rightarrow f \;\text { satisfies } (DIM)$ For example a (non-trivial) blowup is not flat. d) For a morphism $f:X\to Y $ between connected holomorphic manifolds we have: $$f \text { is flat} \iff f \text { is open} \quad \iff (DIM) \;\text {holds}$$
For example a submersion is flat, since it is open. e) Given two complex spaces $X,Y$ the projection $X\times Y\to X$ is flat ( Not trivial: recall that open doesn't imply flat!). f) flatness is preserved by base change. g) The normalization $f:X^{nor} \to X $ of a non-normal space is never flat. For example if $X\subset \mathbb C^2$ is the cusp $y^2=x^3$, the normalization morphism $\mathbb C\to X: t\mapsto (t^2,t^3)$ is not flat. h) Given a finite morphism $f:X\to Y $, each $y\in Y$ has a fiber $X(y) \subset X$ and for $x\in X$ we can define $\mu (x)= dim_{\mathbb C} (\mathbb C \otimes_{\mathcal O_{Y,y}} \mathcal O_{X,x})$ . Now for $ y\in Y$ we put $\; \nu (y)= \Sigma_{x\in X(y)} \mu(x)$ and we obtain : $$f \; \text{ flat} \iff \nu :Y\to \mathbb N \text { locally constant}$$ [Of course for connected $Y$, locally constant = constant] Bibliography: A. Douady, Flatness and Privilege , L'Enseignement Mathématique, Vol.14 (1968) G.Fischer, Complex Analytic Geometry , Springer LNM 538, 1976
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75,698 |
I'm looking for a list of problems such that a) any undergraduate student who took multivariable calculus and linear algebra can understand the statements, (Edit: the definition of understanding here is that they can verify a few small cases by themselves ) b) but are still open or very hard (say took at least 5 years to solve), Edit: c) and first proposed in the 20th century or later. Edit : my motivation is to encourage students to addict to solving mathematical problems. I know that there are many such problems in number theory and combinatorics, for trivial example, Fermat's last theorem. I'll be more interested in other fields, but less famous problems in number theory or combinatorics would be also welcome. For example, 1) The $n!$ conjecture can be stated in an elementary language, but had been notoriously hard. See section 2.2 in Haiman's paper http://arxiv.org/abs/math.AG/0010246 . 2) Let $r$ be any positive integer, and let $x_1, x_2$ be indeterminates. Consider the sequence $\{x_n\}$ defined by the recursive relation $$x_{n+1} =(x_n^r +1)/x_{n-1}$$
for any integer n. Prove that $x_n$ is of the form $P/Q$, where $P$ is a polynomial of $x_1$ and $x_2$ with non-negative coefficients, and Q is a monomial. This problem appeared as a special case of a conjecture by Fomin and Zelevinsky in the context of cluster algebras around 2001. They proved that $x_n$ can be written as $\frac{(\text{polynomial})}{(\text{monomial})}$. Proofs of positivity are recently obtained by Nakajima ( http://arxiv.org/abs/0905.0002 ) and Qin ( http://arxiv.org/abs/1004.4171 ). 3) Nagata conjecture : Let $r$ be a positive integer $\geq 10$, but not a square. Consider $r$ random points on the plane $R^2$. Let $m$ be any positive number. Prove that the degrees of plane curves passing through each of the $r$ points at least $m$ times are greater than $m\sqrt{r}$. See "Masayoshi Nagata, On the 14-th problem of Hilbert. Amer. J. Math. 81 (1959) 766–772". This is still wide open. I'll be grateful for any more examples.
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Here are five problems which you might like: (1) Are there 44 unit vectors in $\mathbb{R}^5$ where the dot product between each pair is less than $\frac{1}{2}$? I originally saw this formulation on this MO post . It is open, and related to the kissing number in 5 dimensions. (2) A Hadamard Matrix is a square matrix all of whose entries are $\pm1$, and whose rows are mutually orthogonal. Prove that there exists a Hadamard Matrix of size $4k$ for every $k\geq 1$. This is known as the Hadamard Conjecture . While it was considered by Hadamard in the 19th century, it is still open today. (3) Given $n$ distinct points in the plane, what is the minimum number of distinct distances between those points? This is a famous problem of Erdos , and while it has not completely been resolved, a near optimal bound belongs to Guth and Katz. (See Terence Tao's blog post ) (4) Prove that $$\sigma(n)\leq H_n +e^{H_n}\log H_n$$ where $\sigma(n)$ is the sum of divisors function and $H_n=1+\frac{1}{2}+\cdots+\frac{1}{n}$ is the $n^{th}$ Harmonic number. This problem is equivalent to the Riemann Hypothesis. This reformulation was done by Jeffrey Lagarias. (Since the reformulation was more recent, I think it qualifies for this list) (5) If $\alpha\neq 0,1$ is algebraic, and $\beta$ is irrational algebraic, will $\alpha^\beta$ be transcendental? This is the well known Hilbert's 7th Problem. This was resolved by Gelfond and Schneider in 1934. Related to this, there is the famous story of Hilbert giving a lecture in 1919 where he said that he might see the proof of the Riemann hypothesis in his life time, that the youngest members of the audience might live to see Fermat’s Last Theorem proved, but no one present in the hall would live to see a proof of transcendence of $2^{\sqrt{2}}$. Of course the exact opposite happened.
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75,795 |
On page 55 of Weibel's Introduction to homological algebra the following passage appears: Here are two consequences that use the fact that homology commutes with arbitrary direct sums of chain complexes I understand why homology commutes with arbitrary direct sums when the direct sum of a collection of monics is a monic (i.e the direct sum functor is exact) but I was under the impression that there were abelian categories where the direct sum functor is not exact. After a bit of thought, I realised that I don't know an example of an abelian category in which the coproduct functor is not exact. Sheaves of abelian groups on a fixed topological space give an example of an abelian category in which the product functor is not exact. Question 1: Is the passage from Weibel's book correct? If so, then why? Question 2: Is there an example of an abelian category where the direct sum functor is not exact?
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I couldn't think of a natural example of an abelian category in which direct sums are not exact (I think this is called axiom AB4). For example, sheaves of abelian groups and R-modules both have this property. However there are natural examples of abelian categories where direct products are not exact (i.e. not satisfying AB4*), for example, the category of abelian sheaves on a space. Taking the opposite category of such a category will then give an example of a category not satisfying AB4 (albeit, not a very nice one). Once you have such an example, homology of chain complexes in this category will not commute with direct sum: if $A_i \to B_i$ is a sequence of monos such that $\bigoplus (f_i :A_i \to B_i)$ is not a mono, then consider the sequence of two-term complexes $A_i \to B_i$. $H^0$ of each of these complexes is zero, but $H^0$ of the direct sum is the kernel of $\bigoplus f_i$. Here is one way to see that Sh(X) does not satisfy AB4* (probably not the easiest). Assume for simplicity X = [0,1]. Take a finite open cover, $\mathcal U_i$ of X by balls of radius $1/i$. Let $A_i$ be the sheaf $\prod _{U \in \mathcal U_i} j_{U!} \mathbb Z_U$. This has an epimorphism to $\mathbb Z_X$, but the direct product of all of them together is not epimorphic: taking sections over any open set $V$ will kill off any $A_i$ when no $1/i$-ball contains $V$. I hope this is correct!
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76,037 |
I wanted to put this originally on math.stackexchange, since I considered it to be a straightforward question and probably a fairly known fact. After I failed to solve the problem, I browsed through literature and what a surprise - two books claim it is primitive recursive, one resource claims it isn't, and neither one gives proof or reference. One paper also claims that inverse Ackermann function is slower than any primitive recursive function. If it were primitive recursive, I don't see why would that hold. Now, my questions would be: which one
is right - $Ack^{-1}$ is/isn't
primitive recursive, and is/isn't
slower than any primitive recursive
function. If it's a bad MO question, I'll migrate it to M.SE, no problem.
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The inverse Ackermann function is primitive recursive. One way to see this is to use the fact that a function $f$ is primitive recursive when and only when the graph of $f$ is primitive recursive, and $f$ is bounded above by some primitive recursive function. The graph of the Ackermann function is primitive recursive, i.e. the characteristic function of the set $\lbrace \langle x, y, z \rangle : z = A(x,y)\rbrace$ is primitive recursive. This is because checking that $A(x,y) = z$ is easy once $x, y, z$ are given. One can always construct a table of all previous values of $A$ used to justify that $A(x,y) = z$. If $z$ is indeed the correct answer, then the code for this table is not much bigger than $\langle x, y, z\rangle$ (smaller than $17^{17^{x+y+z}}$, for example). So, given a proposed triple $\langle x, y, z \rangle$, we can search for the relevant table and determine whether or not $A(x,y) = z$ is true in a primitive recursive fashion. Of course, the Ackermann function is not bounded above by a primitive recursive function, but that is the only thing that goes wrong. Since the graph of the Ackermann function is primitive recursive, then so is the graph of the inverse Ackermann function $Ack^{-1}(z) = \max\lbrace x : A(x,x) \leq z\rbrace$. Moreover, the growth rate of $Ack^{-1}$ is bounded by some primitive recursive function (e.g. the identity function). It follows that $Ack^{-1}$ is indeed primitive recursive.
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76,134 |
What is a purely topological characterisation of the real line( standard topology)?
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Here are a few examples that came up in a first search. Ward in "The topological characterization of an open linear interval", Proc. London Math. Soc.(2) 41 (1936), 191-198 proved the characterization of the real line as a connected, locally connected separable metric space, such that every point is a strong cut point (removing it leaves precisely two connected components). Franklin and Krishnarao proved that in this characterization "metric space" can be relaxed to "regular space", "On the topological characterization of the real line", Department of Mathematics, Carnegie-Mellon University, Report #69-36, 1969. On a different note, Thron and Zimmerman prove in "A characterization of order topologies by means of minimal T0-topologies", Proc. Amer. Math. Soc. 27, (1971), 161-167, that order topologies $\tau$ on a set $X$ can be characterized as the topologies for which $(X,\tau)$ is $T_1$ and $\tau$ is the least upper bound of two minimal $T_0$ topologies. (Minimal here means that the open sets form a nested family of sets and that the complements of the point closures form a base for the topology.) Similarly the reals can be characterized as a connected, separable, $T_1$ space, and $\tau$ is the least upper bound of two noncompact minimal $T_0$ topologies.
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76,295 |
What is an intuitive way to understand Haar measure as defined for random matrices, say, $N\times N$ orthogonal or unitary matrices? My understanding for what Haar measure means for $U(1)$ is that it can be thought of as a measure over a uniform distribution of phases on a circle, i.e. a matrix representing $M \in U(1)$ can be parameterized with an angle $\theta$ so that $$d\mu(M) = \frac {d\theta}{2\pi} $$ What is a correct generalization of this intuition to $N>1$? In particular, are there any explicit parameterizations of Haar measure that resemble writing down angles that are uniformly distributed? One possibility that came to mind is that eigenvectors, rows and/or columns have (generalized) phases that can be thought of as direction angles that are in some sense uniform over $SO(N)$ or $U(N)$? (Edit: seems like this would not be the case for rows and columns.) Another possibility I have thought of is using Givens rotations to parameterize an orthogonal matrix using the resulting rotation angles that bring it to the identity matrix. Are there any known results about the distribution of the Givens angles? It would seem plausible that they could be uniformly distributed, but given that the Givens rotations are usually applied in a particular fashion to achieve diagonalization, that could introduce correlations that would result in nonuniformity. (Caveat: I'm new to all this so I could very well be wrong about even trying to conceptualize such a question.)
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You want to think of the Haar measure $d\mu(U)$ as a way of measuring uniformity in the group $U(N)$ of unitary $N\times N$ matrices. To form your intuition, consider $N=1$. You then have $U=e^{i\phi}$, with $0<\phi\leq 2\pi$ and $d\mu(U)=d\phi$ measures the perimeter of the unit circle. This is a uniform measure, because $d(\phi+\phi_0)=d\phi$ for any fixed phase shift $\phi_0$. You could write the requirement of uniformity in the form $d\mu(UU_0)=d\mu(U)$, with $U_0=e^{i\phi_0}$ the unitary matrix corresponding to the phase shift $\phi_0$. Once your intuition is formed for $N=1$, you simply generalize to $N>1$ using the same definition of uniformity, $d\mu(UU_0)=d\mu(U)$ for any fixed $U_0\in U(N)$. For orthogonal (or symplectic) matrices you use the same definition of uniformity, with $U_0$ now restricted to the orthogonal or symplectic subgroup of $U(N).$ To explicitly write down the Haar measure $d\mu(U)$ in terms of the matrix elements of $U$ is only easily done for a few small values of $N$. (In particular, there is no relationship to random directions of rows or columns, as Yemon Choi pointed out.) You typically do not need such explicit expressions, since integrals with the Haar measure can be evaluated by using only the definition of uniformity. In response to the follow-up question: If you wish to evaluate Haar-measure integrals of polynomials of matrix elements of $U$, you can use the socalled Weingarten functions. http://en.wikipedia.org/wiki/Weingarten_function Here is a Mathematica program to generate these, http://arxiv.org/abs/1109.4244 If you need an explicit expression for the Haar measure, the steps to take are the following: 1) parameterize your matrix $U$ in terms of a set of real parameters $\{x_i\}$. 2) calculate the metric tensor $m_{ij}$, defined by $\sum_{ij}|dU_{ij}|^2 = \sum_{ij}m_{ij}dx_i dx_j$ 3) obtain the Haar measure by equating $d\mu(U) = ($Det $m)^{1/2}\prod_i dx_i$ This is the general recipe. In practice, for many parameterizations the answer is in the literature. In particular, for the Haar measure in Euler angle parameterizations see: http://arxiv.org/abs/math-ph/0205016 http://www.cft.edu.pl/~karol/pdf/ZK94.pdf
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|
76,349 |
I was woolgathering about the notion of a scheme, and it occurred to me that I know of no non-affine scheme $S$ that is the union of $Spec(O_K)$ 's of some number field $K$ (I allow $K$ to vary - so that $S$ might be $Spec(O_K)\cup Spec(O_L)$ for example). It would an interesting notion if one could patch rings of integers together to form some non-affine $1$ -dimensional normal scheme $S$ . The fact that I've never seen an example makes me think it's impossible. Question Is there a connected non-affine scheme $S$ such that it is the union of open subschemes of it that are $Spec$ 's of rings of integers of number fields? More pointedly, if $Spec(O_K)$ (the ring of integers of some number field $K$ ) is an open subscheme of a normal scheme $S$ then is it equal to it?
|
If $i: \mathrm{Spec}(O_K)\to S$ is an open immersion into a connected separate scheme $S$, then $i$ is an isomorphism. Indeed, the canonical morphism $\pi : \mathrm{Spec}(O_K)\to \mathrm{Spec}(\mathbb Z)$ is finite (hence proper) and can be decomposed into $i$ followed by the canonical morphism $S\to \mathrm{Spec}(\mathbb Z)$. As the latter is separated, this implies that $i$ is also proper, hence closed. The connectedness of $S$ implies that $i$ is onto.
|
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|
76,386 |
It is well known that given two Hilbert-Schmidt operators $a$ and $b$ on a Hilbert space $H$, their product is trace class and $tr(ab)=tr(ba)$. A similar result holds for $a$ bounded and $b$ trace class. The following attractive statement, however, is false: Non-theorem : Let $a$ and $b$ be bounded operators on $H$. If $ab$ is trace class , then $ba$ is trace class and $tr(ab)=tr(ba)$. The counterexample is $a=\pmatrix{0&0&0\\0&0&1\\0&0&0}\otimes 1_{\ell^2(\mathbb N)}$, $b=\pmatrix{0&1&0\\0&0&0\\0&0&0}\otimes 1_{\ell^2(\mathbb N)}$. I'm guessing that the following is also false, but I can't find a counterexample: Non-theorem? : Let $a$ and $b$ be two bounded operators on $H$. If $ab$ and $ba$ are trace class, then $tr(ab)=tr(ba)$.
|
EDIT: Bill Johnson has pointed out a gap in my initial answer. It seems that bridging this gap is just as difficult as proving that $tr(AB)=tr(BA)$. Below I give two other proofs of this equality. The flawed proof is also reproduced at the end. Proof 1 (alluded to by Bill in his comment to Gjergji's answer): It's well-known that $AB$ and $BA$ have the same nonzero eigenvalues (with the same multiplicities). Lidskii's trace formula then implies that $tr(AB)=tr(BA)$. QED. Proof 2 : This proof doesn't use Lidskii's formula. It's taken from Laurie, Nordgren, Radjavi, Rosenthal, On triangularization of algebras of operators. Reine Angew. Math. 327 (1981), 143–155. We'll need to rely on the fact that $tr(ST)=tr(TS)$ if one of $S$ and $T$ is trace class. Let $A=UP$ be the polar decomposition of $A$. Then $PB=U^\ast AB$ is trace class. But then one knows that $tr(UPB)=tr(PBU)$. So it suffices to prove that $tr(PBU)=tr(BUP)$. Therefore, we might as well assume that $A$ is positive. In this case we can let $P_n$ denote the spectral projection of $A$ onto $[1/n,\|A\|]$. Then $\lim_{n\to\infty} tr(P_nAB) = tr(AB)$. Now, for $T$ trace class $Q$ a projection we have $tr(QT)=tr(QTQ)=tr(TQ)$, whence $tr(P_n AB) = tr(P_n A P_n B P_n)$ since $P_n$ and $A$ commute. But the contraction $P_n B P_n$ of $B$ is trace class, for $P_n AB$ is trace class and the restriction of $P_n A$ to the range of $P_n$ is invertible. Thus $tr(P_n A P_n B P_n) = tr(P_n B P_n A) = tr(BA P_n)$ and consequently
$$ tr(AB) = \lim_{n\to\infty} tr(P_n AB) = \lim_{n\to\infty} tr(BAP_n) = tr(BA), $$
as desired. Finally, here is the flawed proof. Let $\{e_i\}$ be an orthonormal basis for $H$. Then
$$ tr(AB) = \sum_i \langle ABe_i, e_i \rangle = \sum_i \langle Be_i, A^\ast e_i \rangle. $$
But $\langle Be_i,A^\ast e_i \rangle = \sum_j \langle Be_i, e_j\rangle \overline{\langle A^\ast e_i, e_j\rangle}$, and therefore
$$\begin{align}
tr(AB) &= \sum_i \sum_j \langle Be_i, e_j\rangle \overline{\langle A^\ast e_i, e_j\rangle} \\
&= \sum_i \sum_j \overline{\langle B^\ast e_j, e_i\rangle} \langle A e_j, e_i \rangle \\
&= \sum_j \sum_i \overline{\langle B^\ast e_j, e_i\rangle} \langle A e_j, e_i \rangle \qquad (???) \\
&= \sum_j \langle Ae_j, B^\ast e_j \rangle \\
&= \sum_j \langle BAe_j, e_j \rangle \\
&= tr(BA).
\end{align}$$
|
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|
76,509 |
How to think about coalgebras? Are there geometric interpretations of coalgebras?
If I think of algebras and modules as spaces and vectorbundles, what are coalgebras and comodules? What basic examples of coalgebras should one keep in mind? Anything that helps to think about coalgebras without headache is welcome ;)
|
It seems like there are two basic sources of examples: 1. The basic structure you have on a space (set, scheme...) is the diagonal morphism $\Delta :X \to X\times X$. Functions on spaces are contravariant, which is why functions on a space form an algebra: $f.g = \Delta ^\ast (f \boxtimes g)$. We also have the morphism $\pi : X \to pt$, and the unit in this algebra is $\pi ^\ast 1$. If we chose some covariant linearization of our space (like measures, topological chains...) then this space would be a coalgebra, with comultiplication given by $\Delta_\ast$, and counit $\pi_\ast$. So, for example we have the coalgebra $C_\ast (X)$ of (say, singular) chains on a topological space. Such coalgebras are naturally cocommutative. In nice cases, we have a pushforward and a pullback (e.g. functions on a finite set, equipped with a measure), and the algebra and coalgebra structures together form a Frobenius algebra (the inner product is $\pi _\ast \Delta ^\ast$. 2. If our space $X$ is a group (or maybe just a monoid), then we have a multiplication map $m: X\times X \to X$, and then $m^\ast$ equips the space of functions (or, e.g. cochains) with the structure of a colagebra. If the multiplication on $X$ has a unit $e: pt \to X$, then $e^\ast$ is the counit of this algebra. This coalgebra will not be cocommutative in general (unless $(X,m)$ is commutative). If we also remember the ordinary multiplication structure (coming from $\Delta ^\ast$), then these two structures toegether form a Hopf algebra (also using the inversion $i:X \to X$ in the group). Julian's example is of this second kind. All the examples I know, are morally either of type 1 or type 2, but sometimes you need to have a very broad definition of "function"or "measure"!
|
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"https://mathoverflow.net/questions/76509",
"https://mathoverflow.net",
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|
76,585 |
Does any body know any reference in which the geometry of compactified moduli space of genus two curves ( Which is a three dimensional variety/stack/...) has been studied?
|
Genus 2 curves are hyperelliptic and so their coarse moduli space is just the Riemann-Hurwitz space $(\mathbb{P}^1)^6/(SL_2 \cdot S_6)$. So the description of $M_2$ is closedly linked with the invariants of binary sextic forms. The classic reference is the paper J. Igusa, Arithmetic Variety of Moduli for Genus Two , Annals of Mathematics, Vol. 72, No. 3 (1960), pp. 612-649. Brendan Hassett's paper Classical and minimal models of the moduli space of curves of genus two is also a nice paper studying explicit compactifications for $M_2$ and their birational geometry properties.
|
{
"source": [
"https://mathoverflow.net/questions/76585",
"https://mathoverflow.net",
"https://mathoverflow.net/users/5259/"
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|
76,591 |
This question is related to the question: Is there a $k$-structure for Hodge modules over a $k$-variety? . Suppose $K$ is a subfield of $\mathbb{C}$ and $M$ is a holonomic $D$-module "of geometric origin" on a smooth scheme $X/K$. By Saito's theory, the base-changed $D$-module $M_{\mathbb{C}}$ on $X\underset{\operatorname{Spec}(K)}\times \operatorname{Spec}(\mathbb{C})$ carries a natural mixed Hodge module structure. An ill-formed question: how does the $K$-rational structure on $M_{\mathbb{C}}$ interact with the Hodge structures? Here are some precise incarnations of that question. Saito's theory endows $M_{\mathbb{C}}$ with the weight filtration $W$ (by $D$-submodules) and a Hodge filtration (compatible with the filtration on differential operators). Are $W$ and $F$ defined over $K$? Or, a seemingly weaker question: does the action of $\operatorname{Gal}(\mathbb{C}/F)$ on (the mixed Hodge module of) de Rham cohomology of $M_{\mathbb{C}}$ preserve the induced filtrations on de Rham cohomology? (This is readily verified when $M=\mathcal{O}_X$).
|
Genus 2 curves are hyperelliptic and so their coarse moduli space is just the Riemann-Hurwitz space $(\mathbb{P}^1)^6/(SL_2 \cdot S_6)$. So the description of $M_2$ is closedly linked with the invariants of binary sextic forms. The classic reference is the paper J. Igusa, Arithmetic Variety of Moduli for Genus Two , Annals of Mathematics, Vol. 72, No. 3 (1960), pp. 612-649. Brendan Hassett's paper Classical and minimal models of the moduli space of curves of genus two is also a nice paper studying explicit compactifications for $M_2$ and their birational geometry properties.
|
{
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"https://mathoverflow.net/questions/76591",
"https://mathoverflow.net",
"https://mathoverflow.net/users/15630/"
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|
76,772 |
Let $P$ be a prime ideal in the polynomial ring $K\left[x_1,...,x_m\right]$ and $Q$ be a prime ideal in the polynomial ring $K\left[y_1,...,y_n\right]$. Is $P+Q$ a prime ideal in $K\left[x_1,...,x_m,y_1,...,y_n\right]$ ? For example for $Q=\left(y_1,...,y_n\right)$, it is easy to prove that. But how about general cases?
|
We have $K\left[x_1,...,x_m,y_1,...,y_n\right] \cong K\left[x_1,...,x_m\right] \otimes K\left[y_1,...,y_n\right]$ (where all tensor products are over $K$), and under this isomorphism, the ideal of $K\left[x_1,...,x_m,y_1,...,y_n\right]$ generated by $P+Q$ corresponds to $P\otimes K\left[y_1,...,y_n\right] + K\left[x_1,...,x_m\right]\otimes Q$. So your question is a particular case of the following one: Let $K$ be a field, and $A$ and $B$ be two $K$-algebras, and $P$ and $Q$ be two prime ideals of $A$ and $B$, respectively. When can we conclude that $P\otimes B+A\otimes Q$ is a prime ideal of $A\otimes B$ ? Since $P\otimes B+A\otimes Q$ is the kernel of the canonical map $A\otimes B\to \left(A/P\right)\otimes \left(B/Q\right)$ (this is a fact of linear algebra, and I can give a proof if needed), it is clear that $P\otimes B+A\otimes Q$ is a prime ideal of $A\otimes B$ if and only if $\left(A/P\right)\otimes \left(B/Q\right)$ is an integral domain. On the other hand, we know that $A/P$ and $B/Q$ are integral domains (since $P$ and $Q$ are prime ideals). So our question becomes: When is the tensor product of two integral domains (which both are $K$-algebras) again an integral domain? This is always true when $K$ is algebraically closed and one of the two integral domains is a finitely-generated $K$-algebra (see Proposition 4.15 (b) in J. S. Milne, Algebraic Geometry , Version 5.21 ). For $K$ not algebraically closed, the answer is "no" (not generally), even in your case as we can easily see (in fact, every finitely generated $K$-algebra which is an integral domain can be considered as a quotient of a polynomial ring $K\left[z_1,z_2,...,z_k\right]$ for some $k$ modulo some prime ideal, so the problems are equivalent). For a particularly simple counterexample to your claim, take $K=\mathbb R$, $n=1$, $m=1$, $P=\left(x_1^2+1\right)$ and $Q=\left(y_1^2+1\right)$ (in this case, $P\otimes Q\cong \mathbb C\otimes_{\mathbb R} \mathbb C$, and this is a commutative $\mathbb R$-algebra of dimension $4$, so it is clearly not an integral domain).
|
{
"source": [
"https://mathoverflow.net/questions/76772",
"https://mathoverflow.net",
"https://mathoverflow.net/users/18161/"
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|
76,955 |
A curve in the plane is determined, up to orientation-preserving
Euclidean
motions, by its curvature function, $\kappa(s)$ .
Here is one of my favorite examples, from
Alfred Gray's book, Modern Differential Geometry of Curves and Surfaces with Mathematica ,
p.116: Q1 .
Is there an analogous theorem stating that a surface in $\mathbb{R}^3$ is determined (in some sense) by its Gaussian curvature? I know such a reconstruction path (curvature $\rightarrow$ surface) is needed in
computer vision, and so there are approximation algorithms,
but I don't know what is the precise theorem underlying this work. Q2 .
Are there higher-dimensional generalizations, determining a
Riemannian manifold by its curvature tensor? I have no doubt this is all well known to the cognoscenti,
in which case a reference would suffice. Thanks! Addendum ( 4Oct11 ). Permit me to augment this question with a relevant reference
which loosens the notion of "determines" and answers my Q1 with that notion replaced
by "find some."
The paper by Gluck, Krigelman, and Singer, entitled "The converse to the Gauss-Bonnet Theorem in PL," J. Diff. Geom , 9(4): 601-616, 1974 , poses this question: Suppose that a closed smooth two-manifold $M$ and a smooth real-valued function $K \colon M \rightarrow \mathbb{R}$ are given, and that one is asked to find a Riemannian metric for $M$ having $K$ as its Gaussian curvature. [...] With these restrictions on $K$ [just elided], the problem has been completely solved for all closed smooth two-manifolds by: Melvyn Berger [...], Gluck [...],
Moser [...], Kazdan and Warner [...]. Recently Kazdan and Warner have obtained a uniform solution.
The problem for compact two-manifolds with boundary, however, seems not to have been addressed in the smooth category. The MathSciNet review of this paper was written by Gromov.
|
I'm not sure what you mean by "determining". One natural notion of equivalence is for two surfaces to be related by an ambient isometry (a euclidean motion). A basic result is that two surfaces in $\mathbb{R}^3$ are related by an isometry of $\mathbb{R}^3$ if and only if their first and second fundamental forms agree. A weaker condition is that of isometry. Two surfaces are said to be isometric if their first fundamental forms agree. Gauss's Theorema Egregium says that isometric surfaces have the same Gaussian curvature, but the converse is not true: there are examples of surfaces with the same Gaussian curvature, but which are not isometric. In dimension $\geq 4$ Kulkarni in his paper Curvature and Metric showed that a diffeomorphism which preserves the sectional curvature is an isometry, except possibly in the case of constant sectional curvature. In dimension $\leq 3$ there are counterexamples which are mentioned in his paper.
|
{
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|
77,029 |
Hello, Here is an interesting problem. It looks elementary, but it has taken me some efforts without solving it. Let $$
h(x) = e^{x^2/2} \Phi(x),\qquad \text{with}\quad \Phi(x):=\int_{-\infty}^x \frac{e^{-y^2/2}}{\sqrt{2\pi}} dy.
$$ The question is whether the function $h(x)$ is monotone increasing over $R$? Are there some work dealing with such function? It seems a quite easy problem. By taking the first derivative, we need to prove that $$
h(x)' = h(x) x + \frac{1}{\sqrt{2\pi}} \ge 0.
$$
which again, not obvious (for $x<0$). Some facts, that might be useful, are: $$
\lim_{x\rightarrow -\infty} h(x) =0, \quad \lim_{x\rightarrow -\infty} h(x)' =0.
$$ Thank you very much for any hints! Anand
|
We can write $h(x)=\frac 1{\sqrt{2\pi}}\int_{-\infty}^x \exp\left(\frac{x^2-y^2}2\right)dy$. Now put $t=x-y$. We get
\begin{align}
h(x)&=\frac 1{\sqrt{2\pi}}\int_0^{+\infty}\exp\left(\frac{x^2-(x-t)^2}2\right)dt\\\
&=\frac 1{\sqrt{2\pi}}\int_0^{+\infty}\exp\left(xt-\frac{t^2}2\right)dt.
\end{align}
We can differentiate under the integral thanks to the dominated convergence theorem. We get
$$h'(x)=\frac 1{\sqrt{2\pi}}\int_0^{+\infty}t\exp\left(xt-\frac{t^2}2\right)dt\geq 0.$$
Added later: we don't need to diffentiate. If $x_1\leq x_2$ then for $t\geq 0$ we have $e^{tx_1}\leq e^{tx_2}$ therefore $ h(x_1)\leq h(x_2) $.
|
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"https://mathoverflow.net/questions/77029",
"https://mathoverflow.net",
"https://mathoverflow.net/users/36814/"
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