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87,347 |
The group of Higman: $
\langle
\
a_0, a_1, a_2, a_3 \ | \
a_0 a_1 a_0^{-1}=a_1^2,
\ a_1 a_2 a_1^{-1}=a_2^2,
\ a_2 a_3 a_2^{-1}=a_3^2,
\ a_3 a_0 a_3^{-1}=a_0^2
\ \rangle .
$ Is it simple? What is actually known about it except the fact that it does not have non-trivial homomorphims into a finite group?
|
Higman's group is not simple. Indeed, if you look at Higman's paper, you will see that his group is an amalgamated product of two groups $K_{1,2}=\langle a_1, a_2, b_2\mid a_1^{-1}a_2a_1=a_2^2, a_2^{-1}b_2a_2=b_2^2\rangle$ and $K_{3,4}=\langle a_3, a_4, b_4\mid a_3^{-1}a_4a_3=a_4^2, a_4^{-1}b_4a_4=b_4^2\rangle$ with free amalgamated subgroups $\langle a_1, b_2\rangle$ and $\langle a_3, b_4\rangle$ with $a_1=b_4, b_2=a_3$. Now take $K_{1,2}$ and add the relations $w(a_1,b_2)=1, w(b_2,a_1)=1$ for some "complicated" word $w$. You get a non-trivial group $K$. Similarly imposing $w(a_3,b_4)=1=w(b_4,a_3)$ on $K_{3,4}$ you get a non-trivial group $K'$. Then there is a homomorphism from the Higman group to the amalgamated product of $K$ and $K'$ with a non-trivial kernel. The fact that $K$ (and, equally, $K'$) is not trivial is not obvious but it is not very difficult to prove. A simplification. Instead of adding two relations, in fact it is enough to add a relation $w(a_1,b_2)=w(b_2,a_1)$ and similarly for $a_3,b_4$. For example, $(a_1b_2)^3=(b_2a_1)^3$. Update 1. Probably the easiest way to be convinced that $K$ ($=K'$) is non-trivial is to input its presentation into GAP or Magma (you do not need 3 in the simplified relation above, 2 is enough). Update 2. An even easier way is to impose further relations $a_2=b_2=1$ on $K$ (as in the simplification above). The factor-group is the infinite cyclic group. Hence $K$ is infinite (and $K'$ is infinite too). Update 3. I forgot to mention that one also needs that after we impose the relation as in the simplification, we should make sure that the subgroup of $K$ generated by $a_1,b_2$ has an automorphism switching $a_1$ and $b_2$ (otherwise we cannot form the proper amalgamated product of $K$ and $K'$). That is why we cannot replace 3 in the Simplification above by $1$. Update 4. Gilbert Baumslag told me that Higman did not know whether his group is simple, but Paul Schupp proved it. And indeed, Schupp, Paul E. Small cancellation theory over free products with amalgamation. Math. Ann. 193 (1971), 255–264 proved much more: Higman's group is SQ-universal, so every countable group embeds into one of its quotients.
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87,480 |
Hartshorne at the end of page 76 of his Algebraic Geometry book gives an example of a scheme which is not an affine scheme. The scheme is constructed by gluing two affine lines together along their maximal ideals obtained by removing a point P. There's also a figure accompanying the example: __ _ __ _ __ _ __ : __ _ __ _ ___ Can someone please explain how to show that this is not an affine scheme?
|
Call $X $ your scheme over the field $k$, $P_1$ and $P_2$ the two special closed points, $A_1$and $A_2$ their respective open complements and $A_{12}=A_1\cap A_2$, so that $A_i\simeq \mathbb A^1_k$ and $A_{12}\simeq\mathbb G_m$, all affine schemes. Here are some (not independent) proofs that $X$ is not affine. Proof 1 The point $(P_1,P_2)\in X \times X $ is in the closure of the diagonal $\Delta_X\subset X \times X $, but $(P_1,P_2)\notin \Delta_X$ . So $\Delta_X$ is not closed, hence $X$ is not separated and a fortiori not affine Proof 2 The images of the restriction map $\Gamma(A_i,\mathcal O_X)=k[T] \to \Gamma(A_{12},\mathcal O_X)=k[T,T^{-1}]$ are both $k[T]$, and together do not generate $ k[T,T^{-1}]$. However, in an affine scheme (or more generally in a separated scheme) the ring of regular sections on the intersection of two open affines is generated by the images of the regular sections on the two opens. Proof 3 The two open immersions $\iota_j:\mathbb A^1_k \to X$ with respective image $A_j\subset X$ coincide on the open subscheme $\mathbb G_m\subset \mathbb A^1_k$ but are nevertheless distinct. This couldn't happen if $X$ were affine (or just separated). Proof 4 The cohomology vector space $H^1(X,\mathcal O_X)$ is infinite dimensional, whereas the cohomology of a coherent sheaf on an affine scheme vanishes in positive degree. In detail, consider the covering $\mathcal U=\lbrace A_1,A_2\rbrace$ of $X$. It is a Leray covering because $A_1,A_2,A_{12}$ are affine hence acyclic, for the coherent sheaf $\mathcal O_X$ (cf. Proof 2) . Thus Čech cohomology computes genuine cohomology. The Čech complex is the linear map $$C^0=\Gamma(A_1,\mathcal O_X)\times \Gamma(A_2,\mathcal O_X)=k[T]\times k[T]\stackrel {d^0}{\to} C^1=\Gamma(A_{12},\mathcal O^*_X)=k[T,T^{-1}]\to 0$$
given by $$d^0(P(T),Q(T)) =Q(T)-P(T) $$. Hence we get $H^1(X,\mathcal O_X)=k[T,T^{-1}]/k[T]$ Proof 5 The Čech complex above proves that $\Gamma(X,\mathcal O_X)=k[T]$ so that the restriction to the strictly smaller open affine subscheme $A_1\subsetneq X$ is bijective: $res: \Gamma(X,\mathcal O_X)\stackrel {\simeq}{\to} \Gamma(A_1,\mathcal O_X)$. This cannot happen for an affine scheme $X$. [In categorical language: $\Gamma$ is an anti-equivalence from the category of affine schemes to that of rings] Proof 6 Every global function $P(T)\in \Gamma(X,\mathcal O_X)=k[T]$ (see Proof 5) takes the exact same value at $P_1$ and $P_2$, namely $P(0)\in \kappa(P_1)=\kappa (P_2)=k$. In contrast given two closed points in an affine scheme , there exists a global regular function vanishing at the first one but not at the second.
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87,501 |
Has someone constructed a programming language that can construct all the algorithms in P, and no others? I'm interested in this restriction coming from the syntax naturally, as opposed to just being a normal Turing machine with a step-timer attached.
|
Yes, there is a whole research area devoted to this problem -- it's called "implicit complexity theory". The general idea is to use a lambda calculus based on linear logic. The linearity constraint on lambda-terms lets you control the complexity of cut-elimination (and hence of evaluation), giving natural programming languages that are complete for various complexity classes (such as PTIME, PSPACE, or LOGSPACE).
|
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|
87,627 |
Does anyone know when and why the Fraktur script was introduced for Lie and other algebras—$\mathfrak{g}$, $\mathfrak{gl}_n$, $X/\mathfrak{g}$,
$\mathfrak{g}\oplus\mathfrak{g}$, $\mathfrak{su}$, $\mathfrak{M}_g$, etc.?
And introduced by whom?
Is its use pretty much restricted to algebra, or was it in the past employed in, say, geometry as well, but has only survived to the current time within algebra?
(Or maybe it is currently used outside of algebra and I am just ignorant of those areas.) The typeface itself goes back to the 15th century.
The generally illuminating website " Earliest Uses of Various Mathematical Symbols " seems not to shed light on this issue. I find the Fraktur font adds a certain elegance and mystery to the subjects that utilize it!
I'm a bit envious, not working in those fields... —$\mathfrak{Joseph}$ :-)
|
Some of what's been said so far about the history makes good sense, but by no means all. Let me add my own perspective, for what it's worth. The font called Fraktur by LaTeX (also known as "gothic") was widely used historically in German printing (though I don't own a Gutenberg Bible). It naturally crept into mathematical usage and notation. For instance, the upper case Fraktur letter $\mathfrak{G}$ was commonly used to denote a group, while the ordinary italic or roman $G$ denoted an element of the group. This convention persisted among emigres like Walter Feit who grew up in Vienna (and escaped on the last children's train though his parents didn't). In his course at Yale which I took as a graduate student he filled the blackboard elegantly with ornate symbols, which I sort of learned to copy down (see his Benjamin lecture notes on character theory from that era). But I had actually encountered Fraktur when I first learned some German grammar in high school. It was a mediocre working class public school but located among various ethnic enclaves (including Italian and German), so those languages got taught for a while in two year sequences. The principal wouldn't let me and a classmate of German descent start with the second course, so we sat in the back of the classroom in the first year course and worked ahead on our own. The old German textbooks available in that postwar era were all in Fraktur, which had been promoted during the early Third Reich as the "correct" way to print the language of the master race. So I did learn to distinguish upper case B and V (in Fraktur $\mathfrak{B}$ and $\mathfrak{V}$, etc. The point is that group theory and Lie groups in particular were actively developed by German mathematicians in the nineteenth century; they were not inventing exotic notation when they used these particular letters as symbols. In number theory there is still a widespread tendency to use even lower case letters like $m, p, q$ ($\mathfrak{m}, \mathfrak{p}, \mathfrak{q}$ in Fraktur), which most people find impossible to imitate by hand. But for Hilbert and others this was quite natural notation, as was lower case $\mathfrak{f}$ for the German word 'Fuhrer' (printed with Umlaut over 'u'), now usually called the "conductor". By the way, in Lie algebra theory the lower case letter $g$ (or $\mathfrak{g}$) was naturally used because the Lie algebra was first regarded as an infinitesimal group.
|
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|
87,633 |
Hi all, As we know, every elliptic k3 surface admits an elliptic fibration over $P^1$, but generally how do we construct this fibration? For example, how to get such a fibration for Fermat quartic? Moreover, as we know all (elliptic) k3 surfaces are differential equivalent to each other, does this mean: topologically the elliptic fibration we get for each elliptic fibraion is the same, which is just the torus fibration over $S^2$ with 24 node singularities? Or, the totally space is the same, but different complex data(structure) provides different way or "direction" of projection onto $S^2$, thus induces different type of fibrations? Thanks!
|
Let $S$ be a smooth projective $K3$ surface, say over the complex numbers, and suppose that $S$ admits a (non-constant) fibration $\pi\colon S\to C$ over a curve $C$. By the universal property of the normalization, we can suppose that $C$ is normal, hence smooth. Now, this curve $C$ must be $\mathbb P^1$, since otherwise you would have (by pulling-back) some non-trivial global holomorphic $1$-form on $S$, contradicting $h^{1,0}(S)=0$. Next, this fibration is clearly given by the linear system $|\pi^*H^0(\mathbb P^1,\mathcal O(1))|\subset |L|$ of the pull-back $L:=\pi^*\mathcal O(1)$, which is spanned by two independent sections, say $\sigma$ and $\tau$. Now, take a general fiber: it is a smooth curve $F\subset S$ which is a divisor in the above-mentioned linear system, of the form $\{\lambda s+\mu t=0\}$. In particular $\mathcal O_S(F)\simeq L$. Moreover, by definition, $\mathcal O_F(F)\simeq L|_F=\pi^*\mathcal O(1)|_F$ which is trivial. By adjunction, $K_F\simeq (K_S\otimes\mathcal O_S(F))|_F\simeq\mathcal O_F$ is trivial, so that $F$ is an elliptic curve. Thus, any fibration of a smooth projective $K3$ surface is an elliptic fibration over $\mathbb P^1$, obtained as above. Now, let's consider the more specific case of the Fermat's quartic $S:\{x^4-y^4-z^4+t^4=0\}$ in $\mathbb P^3$ (it is the standard Fermat's quartic up to multiplying $y$ and $z$ by a $4$th root of $-1$). Then, we can factorize it in the following way
$$
(x^2+y^2)(x^2-y^2)-(z^2+t^2)(z^2-t^2)=0.
$$
This shows that, for $[\lambda:\mu]\in\mathbb P^1$, the complete intersection given by
$$
C_{[\lambda:\mu]}:=\begin{cases}
\lambda(x^2-y^2)=\mu(z^2+t^2) \\
\mu(x^2+y^2)=\lambda(z^2-t^2)
\end{cases}
$$
is contained in $S$. For generic $[\lambda:\mu]\in\mathbb P^1$, this is a smooth elliptic curve, since its tangent bundle fits in the following short exact sequence
$$
0\to T_{C_{[\lambda:\mu]}}\to T_{\mathbb P^3}|_{C_{[\lambda:\mu]}}\to\mathcal O_{C_{[\lambda:\mu]}}(2)\oplus\mathcal O_{C_{[\lambda:\mu]}}(2)\to 0.
$$
The function $[\lambda:\mu]$ defines a map from $S$ onto $\mathbb P^1$, which is the elliptic pencil on $S$ you were looking for. Note that, for $\lambda/\mu=0,\pm 1,\pm i,\infty$, $C_{[\lambda:\mu]}$ degenerates into a cycle of four lines. This gives you the 24 singularities.
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87,674 |
Is there any evidence for the classification of topological 4-manifolds, aside from Freedman's 1982 paper "The topology of four-dimensional manifolds", Journal of Differential Geometry 17(3) 357–453? The argument there is extraordinarily complicated and a simpler proof would be desirable. Is there evidence from any other source that would suggest that topological 4-manifolds are so much simpler than smooth 4-manifolds, or does it all hinge on Freedman's proof that Casson handles are homeomorphic to standard handles? My question is motivated from a number of points of view: The classification of topological 4-manifolds is now 30 years old and an easier version of the proof has not emerged. In contrast, Donaldson's invariants have been superseded by more easily computed invariants. This is a very unsatisfactory state of affairs for such a far-reaching topological result, particularly as it is so regularly used in proof-by-contradiction arguments against results in smooth 4-manifold theory. As the Bing topologists familiar with these arguments retire, the hopes of reproducing the details of the proof are fading, and with it, the insight that such a spectacular proof affords. I am delighted to see that the MPIM, Bonn is running a special semester on this topic next year. Hopefully this will introduce these techniques to a new generation of mathematicians (and save them from having to reinvent them!) It may be possible to refine the proof to gain more control over the resulting infinite towers - and perhaps get Hoelder maps rather than homeomorphisms, for example. This would require either a better exposition of the fundamental result or some new independent insight, which was the basis of my question.
|
After 7 months, over 800 MO views and (as suggested) emails to experts, the answer to the question is "No": other than Freedman's 1982 paper, there is no evidence what-so-ever that topological 4-manifolds are so much simpler (i.e. determined up to homeomorphism by their intersection form) than smooth 4-manifolds. Subsequent research (capped gropes etc) hinge on the key step in the paper - the removal of "gaps" in the "design". While this is a highly unsatisfactory state of affairs for the reasons mentioned in the original question, it is what it is.
|
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87,714 |
Suppose $F$ is a field, and $F_1, F_2$ are two extension fields of $F$. Is it always the case that there is a field $L$, containing three subfields $F, K_1, K_2$ and two ring isomorphisms $\varphi_{i}:F_i\rightarrow K_1$ fixing $F$? Note 1: We lose no generality assuming $F$, rather than an isomorphic copy of $F$, is a subfield of $L$. I ask this because I was wondering if there is a way to combine the reals and the $p$-adic numbers into a single extension of $\mathbb{Q}$. Note 2: I seem to recall someone telling me this couldn't be done (perhaps with additional topological data preserved). But I cannot seem to remember the reason why. In any case, I want to know if there is something other than topology which prevents it.
|
The tensor product $F_1 \otimes_F F_2$ is not 0, hence it has a quotient which is a field. This contains the images of both $F_i$.
|
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|
87,734 |
Sorry for the vague title but I couldn't find a better one. I want to compute the sum $S = \frac{1}{N}\sum_{i=1}^N c_i x_i$ where $c_i$s are known positive constants. The problem is that computing the value of each $x_i$ takes a huge amount of time. That's why I want to estimate the sum $S$ by using only a small subset of $x_i$s. How can I do this? I'm not sure if I can apply some sort of importance sampling here.
|
The tensor product $F_1 \otimes_F F_2$ is not 0, hence it has a quotient which is a field. This contains the images of both $F_i$.
|
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|
87,794 |
The question was asked by a student, and I did not have a ready answer. I can think of the German word ``Einheit'', but since in German that is not how the identity element of a group is called, I doubt that is the origin. Any ideas?
|
Heinrich Weber uses Einheit and e in his Lehrbuch der Algebra (1896).
|
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87,838 |
This question arises from the excellent question posed on math.SE
by Salvo Tringali, namely, Correspondence
between Borel algebras and topology .
Since the question was not answered there after some time, I am
bringing it up here on mathoverflow in the hopes that it may find an answer here. For any topological space, one may consider the Borel sets of the
space, the $\sigma$-algebra generated by the open sets of that
topology. The question is whether every $\sigma$-algebra arises in
this way. Question. Is every $\sigma$-algebra the Borel algebra of a
topology? In other words, does every $\sigma$-algebra $\Sigma$ on a set $X$
contain a topology $\tau$ on $X$ such that $\Sigma$ is the $\sigma$
algebra generated by the sets in $\tau$? Some candidate counterexamples were proposed on the math.SE
question, but ultimately shown not to be counterexamples. For
example, my
answer there shows that the collection of Lebesgue
measurable sets, which seemed at first as though it might be a
counterexample, is nevertheless the Borel algebra of the topology
consisting of sets of the form $O-N$, where $O$ is open in the
usual topology and $N$ is measure zero. A proposed counterexample
of Gerald Edgar's there, however, remains unresolved. And I'm not clear on the status of a related proposed counterexample of George Lowther's. Meanwhile, allow me to propose here a few additional candidate
counterexamples: Consider the collection $\Sigma_0$ of eventually periodic subsets of
$\omega_1$. A set $S\subset\omega_1$ is eventually periodic if
above some countable ordinal $\gamma$, there is a countable length pattern which is simply repeated up to
$\omega_1$ to form $S$. This is a $\sigma$-algebra, since it is closed under
complements and countable intersections (one may find a common
period among countably many eventually periodic sets by intersecting the club sets consisting of starting points of the
repeated pattern). Consider the collection $\Sigma_1$ of eventually-agreeing
subsets of the disjoint union $\omega_1\sqcup\omega_1$ of two copies of $\omega_1$. That is,
sets $S\subset \omega_1\sqcup\omega_1$, such that except for
countably many exceptions, $S$ looks the same on the first copy as it does on the
second. Another way to say it is that the symmetric difference of
$S$ on the first copy with $S$ on the second copy is bounded. This is a $\sigma$-algebra, since it is closed under complement and also under countable intersection, as the countably many exceptional sets will union up to a countable set. Please enlighten me by showing either that these are not actually
counterexamples or that they are, or by giving another
counterexample or a proof that there is no counterexample. If the answer to the question should prove to be affirmative, but only via strange or unattractive topologies, then consider it to go without saying that we also want to know how good a topology can be found (Hausdorff, compact and so on) to generate the given $\sigma$-algebra.
|
Unfortunately, I can only provide a reference but no ideas since I don't have the paper.
In "On the problem of generating sigma-algebras by topologies", Statist. Decisions 2 (1984), 377-388, Albert Ascherl shows (at least according to the summary to be found on MathSciNet)
that there are $\sigma$-algebras which can't be generated by a topology. Robert Lang (same journal 4 (1986), 97-98) claims to give a shorter proof. As suggested by Joel, I add the ideas of Lang's example. The underlying space is
$\Omega= 2^{\mathbb R}$, that is the space of all indicator functions, and the $\sigma$-algebra is $\mathcal A = \bigotimes_{\mathbb R} \mathcal P$ where $\mathcal P$
is the power set of the two element set. It is generated by the system $\mathcal E$ of the "basic open sets" of the product topology (prescribed values in a finite number of points).
This generator has the cardinality $c$ of the continuum and since the generated $\sigma$-algebra can be obtained in $\omega_1$ (transfinite) induction steps the cardinality
of $\mathcal A$ is also $c$. On the other hand, if $\mathcal T$ is a topology with $\mathcal A=\sigma(\mathcal T)$ then $\mathcal T$ separates points (this should follow from the "good sets principle"), in particular, for two distinct points of $\Omega$ the closures of the
corresponding singletons are distinct. Hence $\mathcal T$ has at least $|\Omega|=2^c$
elements.
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87,877 |
What's a quick way to prove the following fact about minors of an invertible matrix $A$ and its inverse? Let $A[I,J]$ denote the submatrix of an $n \times n$ matrix $A$ obtained by keeping only the rows indexed by $I$ and columns indexed by $J$. Then $$ |\det A[I,J]| = | (\det A) \det A^{-1}[J^c,I^c]|,$$
where $I^c$ stands for $[n] \setminus I$, for $|I| = |J|$. It is trivial when $|I| = |J| = 1$ or $n-1$. This is apparently proved by Jacobi, but I couldn't find a proof anywhere in books or online. Horn and Johnson listed this as one of the advanced formulas in their preliminary chapter, but didn't give a proof. In general what's a reliable source to find proofs of all these little facts? I ran into this question while reading Macdonald's book on symmetric functions and Hall polynomials, in particular page 22 where he is explaining the determinantal relation between the elementary symmetric functions $e_\lambda$ and the complete symmetric functions $h_\lambda$. I also spent 3 hours trying to crack this nut, but can only show it for diagonal matrices :( Edit: It looks like Ferrar's book on Algebra subtitled determinant, matrices and algebraic forms, might carry a proof of this in chapter 5. Though the book seems to have a sexist bias.
|
The key word under which you will find this result in modern books is "Schur complement". Here is a self-contained proof. Assume $I$ and $J$ are $(1,2,\dots,k)$ for some $k$ without loss of generality (you may reorder rows/columns). Let the matrix be
$$
M=\begin{bmatrix}A & B\\\\ C & D\end{bmatrix},
$$
where the blocks $A$ and $D$ are square. Assume for now that $A$ is invertible --- you may treat the general case with a continuity argument. Let $S=D-CA^{-1}B$ be the so-called Schur complement of $A$ in $M$. You may verify the following identity ("magic wand Schur complement formula")
$$
\begin{bmatrix}A & B\\\\ C & D\end{bmatrix} =
\begin{bmatrix}I & 0\\\\ CA^{-1} & I\end{bmatrix}
\begin{bmatrix}A & 0\\\\ 0 & S\end{bmatrix}
\begin{bmatrix}I & A^{-1}B\\\\ 0 & I\end{bmatrix}. \tag{1}
$$
By taking determinants, $$\det M=\det A \det S. \tag{2}$$ Moreover, if you invert term-by-term the above formula you can see that the (2,2) block of $M^{-1}$ is $S^{-1}$. So your thesis is now (2). Note that the "magic formula" (1) can be derived via block Gaussian elimination and is much less magic than it looks at first sight.
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87,919 |
I have been informed that there is a reference out there which specifically details what goes wrong with the mod 2 Moore spectrum, i.e. why it is not $A_\infty$ or something? I do not know the details, but I am interested in this from the point of view of trying to understand how to come up with the correct notion of ideals of spectra (in the sense of Smith and others), especially the ideal generated by multiplication by 2 on the sphere spectrum. I think it is a paper by Neeman. Does anyone know of this paper, or of other papers which might detail this situation carefully? Thanks! PS Is this question appropriate, as it is only a reference request, in the strongest sense of the phrase?
|
The actual statement is much stronger than you suggest, namely: the mod 2 Moore spectrum does not admit a unital multiplication (even if it is non-associative). I don't know a reference so I'll sketch the proof: Let $R$ be a spectrum with unital product, with unit map $\eta\colon S^0\to R$ and product map $\mu: R\wedge R\to R$. Then it is straightforward to show that if $n\eta=0$ in $\pi_0R$ for some integer $n$, then $n\cdot\mathrm{id}_R: R\to R$ is homotopic to the null map as well. (The key point is that proving this uses the existence of $\mu:R\wedge R\to R$ such that $\mu\circ (\eta\wedge \mathrm{id}_R) = \mathrm{id}_R$, but nothing about associativity of such $\mu$.) If $R$ is the mod $2$ Moore spectrum, with $\eta: S^0\to R$ the generator of $\pi_0R$, then you calculate that: $\pi_0R = Z/2$, but $\pi_2R = Z/4$, from which it follows that $2\eta=0$, $2\mathrm{id}_R\neq 0$. Therefore no such unital multiplication $\mu$ on $R$ can exist.
|
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|
88,056 |
I'm going through the crisis of being unhappy with the textbook definition of a differentiable manifold. I'm wondering whether there is a sheaf-theoretic approach which will make me happier. In a nutshell, is there a natural condition to impose on a structure sheaf $\mathcal{C}^k_M$ of a topological space $M$ that can stand-in for the requirement that $M$ be second-countable Hausdorff? Background Most textbooks introduce differentiable manifolds via atlases and charts . This has the advantage of being concrete, and the disadvantage of involving an arbitrary choice of atlas, which obscures the basic property that a differential manifold "looks the same at all points" (for $M$ connected, without boundary: diffeomorphism group acts transitively). And isn't introducing local coordinates "an act of violence"? I saw a much nicer definition of differentiable manifolds on Wikipedia, which I don't know a good textbook reference for. This definition proceeds via sheaves of local rings . The Wikipedia definition stated: A differentiable manifold (of class $C_k$) consists of a pair $(M, \mathcal{O}_M)$ where $M$ is a topological space, and $\mathcal{O}_M$ is a sheaf of local $\mathbb{R}$-algebras defined on $M$, such that the locally ringed space $(M,\mathcal{O}_M)$ is locally isomorphic to $(\mathbb{R}^n, \mathcal{O})$. [$\mathcal{O}(U)=C^k(U,\mathbb{R})$ is the structure sheaf on $\mathbb{R}^n$.] Beautiful, really! Entirely coordinate free. But isn't there a General Topology condition missing? I confirmed on math.SE (to make sure that I wasn't hallucinating) that this definition is indeed missing the condition that $M$ be second-countable Hausdorff. That indeed turned out to be the case, so I edited the Wikipedia definition to require $M$ to be second-countable Hausdorff. Why am I still not happy? The deep reason that we require a differentiable manifold to be paracompact, as per Georges Elencwajg's extremely informative answer , is that paracompactness makes sheaves of $C_M^k$-modules (maybe $k=\infty$) acyclic . This is a purely sheaf-theoretic property (a condition on the structure sheaf of $M$ rather than on $M$ itself), which quickly implies good things like that every subbundle of a vector bundle on $M$ be a direct summand. Is this in fact enough? If it were enough to require that $\mathcal{O}_M$ be acyclic , or maybe fine , then the nicest, most flexible (and, in a strange sense, most enlightening) definition of differentiable manifold, would be: Definition : A differentiable manifold (of class $C_k$) consists of a pair $(M, \mathcal{O}_M)$ where $M$ is a topological space, and $\mathcal{O}_M$ is an acyclic sheaf of local $\mathbb{R}$-algebras defined on $M$, such that the locally ringed space $(M,\mathcal{O}_M)$ is locally isomorphic to $(\mathbb{R}^n, \mathcal{O})$. Maybe the word acyclic should be fine . Maybe soft and acyclic. Maybe a bit more, but still something that can be stated in terms of the structure sheaf. Question : Can I put a natural sheaf-theoretic condition on $\mathcal{O}_M$ (acyclic? fine?) which ensures that $M$ (a topological space) must be a second-countable Hausdorff space? If not, would such a condition at least ensure that $M$ be a generalized differentiable manifold in some kind of useful sense? Update : This question really bothers me, so I've started a bounty. I'd like to narrow it down a little in order to make it easier to answer: Does acyclicity (or a slightly stronger condition) on $\mathcal{O}_M$ of a topological (Hausdorff?) space imply paracompactness (or a slightly weaker but still useful condition)? Hausdorff bothers me as well, of course; but a sheafy characterization of paracompactness somehow seems like it has the potential to be lovely and really enlightening.
|
Definition: ''A smooth manifold is a locally ringed space $(M;C^{\infty})$ which satisfies the conditions: Each $x \in M$ admits a neighborhood $U$, such that $(U,C^{\infty})$ is isomorphic to $(\mathbb{R}^n,C^{\infty})$ as a locally ringed space. The global sections of $C^{\infty}(M)$ separate points. The structure sheaf $C^{\infty}$ is fine as a sheaf of modules over itself. $C^{\infty}(M)$ has at most countably many indecomposable idempotents.'' Explanations: 1) is evident. 2) means that for $x \neq y \in M$, there exists $f \in C^{\infty}(M)$ with $f(x)=0$, $f(y) \neq 0$. This ensures the Hausdorff condition once we know that elements of $C^{\infty}(M)$ give rise to continuous maps $M \to \mathbb{R}$. This is as follows: $f \in C^{\infty}(M)$ given, $x \in M$. Pick a chart $h:U \to \mathbb{R}^n$; under this chart, $f|_U$ corresponds to a smooth function on $\mathbb{R}^n$, whose value at $h(x)$ does not depend on the choice of the chart. Call this value $f(x)$. Checking the continuity of $x \mapsto f(x)$ can be done in charts. 3.) By this I mean that for each open cover $(U_i)$, there is a partition of unity $\lambda_i$ with the usual properties and that $\lambda_i$ is a map of $C^{\infty} (M)$-modules. A standard argument shows that $\lambda_i$ is given by multiplication with a smooth function. Therefore, the underlying space $M$ is paracompact. 4.) An idempotent $p\neq 0 $ in a commutative ring $A$ is called indecomposable if $$
p=q +r; r^2 =r; q^2 =q , q \neq 0 \Rightarrow p = q
$$ holds. Indecomposable idempotents in $C^{\infty}(M)$ correspond to connected components. Therefore condition 4 means that $M$ has only countably many connected components. These conditions together imply that $M$ is Hausdorff and second countable, because a locally euclidean, connected and paracompact Hausdorff space is second countable, see Gauld, "Topological properties of manifolds", Theorem 7 (see http://www.jstor.org/stable/2319220 ). Paracompactness alone does not guarantee second countability, see $\mathbb{R}$ with the discrete topology. However, I think that sheaf theory and locally ringed spaces are the wrong software for differential geometry and differential topology.
|
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|
88,184 |
This is a question that has been winding around my head for a long time and I have not found a convincing answer. The title says everything, but I am going to enrich my question by little more explanations. As a layman, I have started searching for expositories/more informal, rather intuitive, also original account of non-commutative geometry to get more sense of it, namely, I have looked through The English translation of Review of non-commutative algebra by Alain Connes , Surveys in non-commutative geometry, Clay mathematics proceedings, Volume 6 , Non-commutative geometry by Alain Connes , Nevertheless, I am not satisfied with them at all. It seems to me, that even understanding a simple example, requires much more knowledge that is gained in grad school. Now for me, this field merely contains a lot of highly developed machineries which are more technical (somehow artificial) than that of other fields. The following are my questions revolving around the significance of this field in Mathematics . Of course, they are absolutely related to my main question. How can a grad student be motivated to specialize in this field? and What is (are) the well-known result (s), found solely by non-commutative geometric techniques that could not be proven without them?
|
$\DeclareMathOperator\coker{coker}$ I think I'm in a pretty good position to answer this question because I am a graduate student working in noncommutative geometry who entered the subject a little bit skeptical about its relevance to the rest of mathematics. To this day I sometimes find it hard to get excited about purely "noncommutative" results, but the subject has its tentacles in so many other areas that I never get bored. Before saying anything further, I need to say a few words about the Atiyah–Singer index theorem. This theorem asserts that if $D$ is an elliptic differential operator on a manifold $M$ then its Fredholm index $\dim(\ker(D)) - \dim(\coker(D))$ can be computed by integrating certain characteristic classes of $M$ . Non-trivial corollaries (obtained by "plugging in" well chosen differential operators) include the generalized Gauss–Bonnet formula, the Hirzebruch signature theorem, and the Hirzebruch–Riemann–Roch formula. It was quickly realized (first by Atiyah, I think) that the proof of the theorem can be viewed as a statement about the Poincaré duality pairing between topological K-theory and its associated homology theory (these days called K-homology). I wasn't around, but I'm told that people were very excited about Atiyah and Singer's achievement (understandably so!). People quickly began trying to generalize and strengthen the theorem, and my claim is that noncommutative geometry is the area of mathematics that emerged from these attempts. Saying that marginalizes the other important reasons for developing the subject, but I think it was Connes' main motivation and in any event it is a convenient oversimplification for a MO answer. It also helps me answer your first question by playing to my personal biases: when I was choosing an area of research I told my adviser that I was interested in learning more about that Atiyah–Singer index theorem and I was led inexorably toward the tools of noncommutative geometry. The origin of the relationship between NCG and Atiyah–Singer lies in equivariant index theory. Atiyah and Singer realized from the start that if $M$ admits an action by a compact Lie group $G$ and $D$ is invariant under the group action then it is better to think of the index of $D$ as a virtual representation of $G$ (i.e. an element of the $G$ -equivariant K-theory of a point) rather than as an integer. If $G$ is not compact then this doesn't really work, but the noncommutative geometers realized that $D$ does have an index in the K-theory of the reduced group C $^\ast$ -algebra $C_r^\ast(G)$ . Indeed, to a noncommutative geometer equivariant index theory is all about a map $K_\ast(M) \to K_\ast(C_r^\ast(M)$ where $K_\ast(M)$ is the K-homology of $M$ ; in the case where $M$ is the universal classifying space of $G$ , Baum and Connes conjectured that this map is an isomorphism. Proving this conjecture for more and more groups and understanding its consequences motivates a great deal of the development of NCG to this day. The conjecture is interesting in its own right if you already care about index theory, but even if you don't injectivity of the Baum–Connes map implies the Novikov conjecture (see Alain Valette's answer ) and surjectivity is related to the Borel conjecture. It has numerous other applications, for example to the theory of positive scalar curvature obstructions in Riemannian geometry or to the Kadison–Kaplansky conjecture in functional analysis (which would follow from surjectivity). Recently there has been a lot of interest in connections between the Baum–Connes conjecture and representation theory; the Baum–Aubert–Plyman conjecture in $p$ -adic representation theory has its origins in these sorts of considerations. Much of the rest of NCG can also be traced back to index theory. Kasparov's KK-theory arose as a way to understand maps and pairings between K-theory and K-homology, motivated in part by index theory. Connes' work on noncommutative measure theory arose from his work on index theory for measurable foliations (with applications to dynamical systems). Cyclic (co)homology was invented in part to gain access in a noncommutative setting to the Chern character map from K-theory to cohomology which translates the K-theoretic formulation of the index theorem into a cohomological formula. Connes' theory of spectral triples and noncommutative Riemannian geometry is based on the theory of Dirac operators which was invented by Atiyah and Singer to prove the index theorem. I guess my point with all of this is that all the esoteric machinery of NCG seems less artificial when viewed through the lens of index theory.
|
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88,252 |
Hi. I'll confess from the start to not being a logician. In fact this question came up not from research but during a discussion with a friend about whether the classical proof that $\sqrt{2}$ is irrational can be made acceptable to an intuitionist. (It can be.) The question is: Are there any "natural" statements which can be proven in Peano Arithmetic , but not in Heyting Arithmetic (Peano Arithmetic but with a logic that does not admit the law of the excluded middle)? In fact, any statements -- even pathological ones -- that can be proven in one but not the other would be interesting to me, since I wasn't able to come up with any. (Even after doing a few web searches!) But of course, the closer to the surface the better.
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The first example that occurs to me is (a formalization in the language of arithmetic, via coding, of) "For every Turing machine M and every input x, the computation of M on input x either terminates or doesn't terminate." With classical logic, this is trivially provable, as an instance of the law of the excluded middle. But it's not intuitionistically provable because the halting problem is undecidable. (In a bit more detail, if it were provable, then it would be recursively realizable, and the realizer would be an index for an algorithm that solves the halting problem.)
|
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|
88,368 |
This question was inspired by this blog post of Jordan Ellenberg. Define a "computable group" to be an at most countable group $G$ whose elements can be represented by finite binary strings, with the properties that There is an algorithm (by which I mean "Turing machine") which, when given a binary string, can determine whether that string lives in the group $G$ in finite time (i.e. $G$ is a computable set); There is an algorithm which, when given binary strings $x,y$ in $G$, can compute $x^{-1}$ and $xy$ in finite time (i.e. the group laws on $G$ are computable functions). In the blog post mentioned above, the group $SL_3({\bf Z})$ was discussed, which is certainly an example of a computable group, but one can of course devise many other examples of computable groups. But note that this condition is stronger than being recursively presented, since I am requiring the word problem to be decidable. Observe that if $G$ is a computable group, then there is an obvious algorithm which, when given any two elements $x,y$ of $G$ as input, which will halt in finite time if and only if $x,y$ fail to generate a free group, simply by evaluating all the nontrivial words of $x,y$ in turn and checking if any of them are the identity. My first question is if there is any computable group $G$ which is "Turing complete" in the sense that the halting problem for any Turing machine can be converted into a question of the above form. In other words, there would be an algorithm which, when given a Turing machine $T$, would return (in finite time) a pair $x_T, y_T$ of elements of $G$ with the property that $x_T, y_T$ generate a free group in $G$ if and only if $T$ does not halt in finite time. Or more informally: can a group be a universal Turing machine? I suspect the answer to this question is "yes", by doing something like taking G to be something like the group of reversibly computable operations on an infinite string of bits, though I wasn't able to quite push through the details. One might also modify the question by taking G to be a computable semigroup rather than a computable group, though I think this should not make too substantial of a difference. My second (and more vague) question is whether one can take $G$ to be a "non-artificial" group, which can be defined without reference to computability or Turing machines. (For instance, a group which is somehow constructed using Diophantine equations would qualify.)
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Update. Here is a more direct construction. (See edit history for previous version.) There is such a universal computable group as you request. Let $F$
be the free group on infinitely many generators $\langle
a_p\rangle_p$, indexed by the Turing machine programs $p$. Let $G$
be the quotient of this group by all the $k^{th}$ powers $a_p^k$, whenever the program $p$
halts (on trivial input) in exactly $k$ steps. Let us represent the group $G$ by reduced words in the generators
$a_p$ and their inverses, but in the case that we took the quotient
by $a_p^k$, then in these words we use exponents on $a_p$ in the interval
$(-k/2,k/2]$. (The reason for using this exponent format is that if we were to use only the positive powers of the finite-order generators, then we wouldn't be able to compute inverses in $G$, since we cannot compute whether $a_p$ has finite order or not.) First of all, we can computably recognize whether a
word in the generators fits this description, simply by checking
whether it is reduced and whether any of the exponents is too
large. The point of this last issue is that we can tell if the
exponent $a_p^r$ is too large by checking if program $p$ halts in
$2r$ steps or not. Similarly, we can easily compute the inverse of
a word from the word, and we can computably multiply words. Again,
whenever we have a word with some new exponents on the generators,
we need to check whether they reduce because of our quotient, and
this is possible by running the relevant computation for sufficient
number to steps to determine it. Thus, we have a computable representation of the group $G$. Finally, I claim that it is universal in the sense you requested.
Given any Turing machine program $p$, let $x_p=a_p$ and let
$y_p=a_q$ for some other program $q$ known not to halt. Thus, by
design, the group generated by $x_p,y_p$ will be the free group on
these generators if and only if $p$ does not halt. An essentially equivalent presentation of the group can be made without reference to Turing machines or computations, but only to Diophantine equations, simply by using the Diophantine representation of the universal Turing machine. That is, since every c.e. set is the solution set of a Diophantine equation, there is a fixed Diophantine equation $d(y,\vec x)=0$, such that Turing machine program $p$ halts on trivial input if and only if $d(p,\vec x)=0$ has a solution in the integers, viewing the program as its Gödel code. So we may define the group $G$ as above, with infinitely many generators $a_n$, but taking the quotient by $a_n^k$, if $k$ is the size of the smallest integer solution of $d(n,\vec x)=0$. I'm not sure this makes the group "natural," (and my opinion is that this word has no robust, coherent mathematical meaning), but it does omit any mention of Turing machines, using instead a fixed Diophantine equation. Lastly, let me observe that my group is not finitely generated, and
it may be interesting to have a finitely generated example, or even
a finitely presented example. I suspect that one can apply one of the embedding theorems to place this example into a finitely generated or even finitely presented group.
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|
88,420 |
I've looked on the web and haven't found a simple example.
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The one-point compactification of $\mathbb{Q}$ has the property that every compact subset is closed. So it is certainly a weak Hausdorff space. But it isn't Hausdorff, as $\mathbb{Q}$ isn't locally compact. Addendum Another example is the cocountable topology on an uncountable set. No two points have disjoint neighbourhoods, and the only compact subsets are the finite subsets.
|
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|
88,513 |
According to Wikipedia: "In mathematics, the homotopy principle (or h-principle) is a very general way to solve partial differential equations (PDEs), and more generally partial differential relations (PDRs)". I'd like to know if h-principle and theory from M. Gromov's "Partial Differential Relations" is a useful tool in the field of nonlinear PDE's. What type of problems can be attacked using h-principle? What type of results can be obtained?
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I'd like to know if h-principle and theory from M. Gromov's "Partial Differential Relations" is a useful tool in the field of nonlinear PDE's. Useful is a relative word. What type of problems can be attacked using h-principle? What type of results can be obtained? This is easier to answer. Here's my favourite example of an h-principle due to Gromov and Lees, independently. (I can recommend Lees's paper for brevity (when weighed against Gromov's book). Sadly the only place I know where it's available is in Duke Maths Journal and hence not free). Definition: An immersion $f: L\to\mathbf{C}^n$ of a closed $n$-manifold is called Lagrangian if $f^*\omega=0$ where $\omega$ is the standard symplectic 2-form $\sum_{i=1}^ndx_i\wedge dy_i$. The condition that an immersion be Lagrangian is a (very flexible) nonlinear PDE. Flexibility here means roughly that there are many solutions (for example, if you take any compactly supported function $H$ on $\mathbf{C}^n$ you can use it to construct a Hamiltonian vector field $X_H$ on $\mathbf{C}^n$ satisfying $\omega(X_H,V)=dH(V)$ for any $V$ and pushing $f$ around using the flow of such a vector field will give you more Lagrangians). More pertinently, flexibility means that Lagrangian immersions satisfy an h-principle... One might ask "How does the space of Lagrangian immersions sit inside the space of all smooth immersions?", but that would be the wrong question because a Lagrangian immersion has slightly more data than just the underlying smooth immersion. Namely, a Lagrangian immersion gives you a "Lagrangian Gauss map" which sends a point to the Lagrangian tangent space considered as a linear subspace in $\mathbf{C}^n$. One might then ask "How does the space of Lagrangian immersions sit inside the space of all smooth immersions equipped with an abstract Lagrangian Gauss map?". By abstract Lagrangian Gauss map, I mean that there is a map of bundles $F:TL\to T\mathbf{C}^n$ which lives over $f$ and which sends tangent spaces to Lagrangian tangent spaces. Note that $F$ doesn't have to be $df$! Now the h-principle tells you the answer to the second question: the space of Lagrangian immersions is a deformation retract of the space of smooth immersions with an abstract Lagrangian Gauss map! Moreover, any given immersion can be approximated by a Lagrangian immersion which is arbitrarily close to it (in the sense that it lives in an arbitrarily small neighbourhood, however the tangent spaces will vary wildly). Moremoreover, the Lagrangian Gauss map of the resulting immersion will be homotopic to the given abstract Gauss map. Not every immersion can be given such a Gauss map, but it's a topological condition to check: it's equivalent to triviality of the complexified tangent bundle. The same immersion can have different (non-homotopic) Lagrangian Gauss maps. Here is a simple example. Note that you don't have to start with an immersion because you can always approximate something by an immersion. So start with the map sending $S^1$ to the origin into $\mathbf{C}$. (Oriented) Lagrangian subspaces of $\mathbf{C}$ are just (oriented) lines through the origin.The oriented Lagrangian Grassmannian is therefore $S^1$ and any map $S^1\to S^1$ will do as the Lagrangian Gauss map. Suppose you take the trivial map $S^1\to S^1$ sending everything to a point. A nearby Lagrangian immersion whose Gauss map is homotopic to this is the figure 8 immersion. If instead you take the degree 1 map $S^1\to S^1$ then a nearby Lagrangian immersion would be the inclusion of a small circle centred at 0. If you've met it before, the degree of this Gauss map is (half) the minimal Maslov number. In higher dimensions there are various cohomological/homotopic Maslov invariants because the space of Lagrangian subspaces is more complicated (it's the homogeneous space $U(n)/O(n)$), but the most important is the analogue of this one. So the h-principle gives you a huge pool of solutions to your nonlinear PDE and tells you something about what they can look like. This is made all the more useful by the fact that a generic Lagrangian immersion has at worst double points and one can surger these double points to obtain embedded Lagrangian submanifolds, which are and have long been a beautiful and mysterious class of objects. What is most mysterious about them is that they don't satisfy an h-principle, so we don't know how to construct/classify them. Indeed there are examples due to Luttinger of smoothly embedded tori in $\mathbf{C}^2$ which are not isotopic to Lagrangian tori.
|
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|
88,996 |
Not every complex manifold is a Kähler manifold (i.e. a manifold which can be equipped with a Kähler metric). All Riemann surfaces are Kähler, but in dimension two and above, at least for compact manifolds, there is a necessary topological condition (i.e. the odd Betti numbers are even). This condition is also sufficient in dimension two, but not in higher dimensions. Therefore the task of finding examples of compact complex manifolds which are not Kähler is reduced to topological considerations. In the non-compact setting, we can also find such manifolds. For example, let $H$ be a Hopf surface, which is a compact complex surface which is not Kähler. Then for $k > 0$, $M_{k+2} = H\times\mathbb{C}^k$ is a non-compact complex manifold which is not Kähler - any submanifold of a Kähler manifold is Kähler, and $H$ is a submanifold of $M_{k+2}$. This generates examples in dimensions three and above. So I ask the following question: Does anyone know of some (easy) explicit examples of non-compact complex surfaces which are not Kähler?
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Following David Speyer's suggestion, let $X=\mathbb{C}^2-\{0\}/\lbrace(x,y)\mapsto (2x,2y)\rbrace$
be the standard Hopf surface. The image, $E$, of the $x$-axis is an elliptic curve.
Remove a point of $X-E$ to get $Y$. The second Betti number $b_2(Y)=0$ because it is homeomorphic to $S^3\times S^1-pt$. If $Y$ were Kähler then $\int_E\omega\not=0$,
where $\omega$ is the Kähler form, and this would imply that $b_2(Y)\not=0$.
|
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|
89,069 |
As every MO user knows, and can easily prove, the inverse of the matrix $\begin{pmatrix} a & b \\\ c & d \end{pmatrix}$ is $\dfrac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ . This can be proved, for example, by writing the inverse as $ \begin{pmatrix} r & s \\ t & u \end{pmatrix}$ and solving the resulting system of four equations in four variables. As a grad student, when studying the theory of modular forms, I repeatedly forgot this formula (do you switch the $a$ and $d$ and invert the sign of $b$ and $c$ … or was it the other way around?) and continually had to rederive it. Much later, it occurred to me that it was better to remember the formula was obvious in a couple of special cases such as $\begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix}$ , and diagonal matrices, for which the geometric intuition is simple. One can also remember this as a special case of the adjugate matrix. Is there some way to just write down $\dfrac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ , even in the case where $ad - bc = 1$ , by pure thought—without having to compute? In particular, is there some geometric intuition, in terms of a linear transformation on a two-dimensional vector space, that renders this fact crystal clear? Or may as well I be asking how to remember why $43 \times 87$ is equal to $3741$ and not $3731$ ?
|
Think about $\left({\phantom-d\phantom--b\atop-c\phantom{--}a}\right)$
as $tI - A$ where $t=a+d$ is the trace of $A$.
Since $A$ satisfies its own characteristic equation
( Cayley-Hamilton ),
we have $A^2 - t A + \Delta \cdot I = 0$
where $\Delta = ad-bc$ is the determinant. Thus $\Delta \cdot I = t A - A^2$.
Now divide both sides by $\Delta \cdot A$ to get $A^{-1} = \Delta^{-1}(tI-A)$, QED .
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89,274 |
It is well known that when $X$ is a compact space (or locally compact space), the dual space of $C(X)=\{f |f:
X\rightarrow \mathbb{C} \text{ is continuous and bounded} \}$ is $M(X)$, the space of Radon measures with bounded variation. However, according to my knowledge, there are few books which discuss the case when X is noncompact, for example a complete separable metric space. Even for the simplest example, when taking $X=\mathbb{R}$, what does $(C(X))^{*}$ mean? Any advice and reference will be much appreciated.
|
What you state in the first paragraph is the Riesz Representation Theorem (see wikipedia ). This is valid for all locally compact Hausdorff spaces; so in particular for $\mathbb R$ (ah, I guess, if you look at $C_0(\mathbb R)^*$ ). If $X$ is any topological space, then of course we can talk of $C^b(X)$ (the bounded continuous functions on $X$ ). This is still a commutative C $^*$ -algebra, and so is isomorphic to $C(K)$ , where $K$ is some compact Hausdorff space. The process of moving from $X$ to $K$ is functorial; purely at the topological level it corresponds to constructing the Stone-Cech compactification (see wikipedia ). Point evaluation at $x\in X$ induces a character on $C^b(X) = C(K)$ and hence a point $k$ of $K$ ; we thus get a (continuous) map $X\rightarrow K$ . This is injective if $X$ is completely regular; but it can fail to be injective (basically, we might lack enough continuous functions to separate points of $X$ ). Back to your question: $C^b(X)^* = C(K)^* = M(K)$ . For $\mathbb R$ , we find that $K$ is nothing but $\beta\mathbb R$ the Stone-Cech compactification (quite a large space!)
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|
89,324 |
The function $\Gamma(s)$ does not have zeros, but $\Gamma(s)\pm \Gamma(1-s)$ does. Ignoring the real solutions for now and assuming $s \in \mathbb{C}$ then: $\Gamma(s)-\Gamma(1-s)$ yields zeros at: $\frac12 \pm 2.70269111740240387016556585336 i$
$\frac12 \pm 5.05334476784919736779735104686 i$
$\frac12 \pm 6.82188969510663531320292827393 i$
$\frac12 \pm 8.37303293891455628139008877004 i$
$\frac12 \pm 9.79770751746885191388078483695 i$
$\frac12 \pm 11.1361746342106720656243966380 i$
$\frac12 \pm 12.4106273718343980402685363665 i$ $\dots$ and $\Gamma(s)+\Gamma(1-s)$ gives zeros at: $\frac12 \pm 4.01094805906156869492043027819 i$
$\frac12 \pm 5.97476992595365858561703252235 i$
$\frac12 \pm 7.61704024553573658642606787126 i$
$\frac12 \pm 9.09805003388841581320246381948 i$
$\frac12 \pm 10.4760650707765536619292369200 i$
$\frac12 \pm 11.7804020877663106830617193188 i$
$\frac12 \pm 13.0283749883477570386353012761 i$ $\dots$ By multiplication, both functions can be combined into: $\Gamma(s)^2 - \Gamma(1-s)^2$ After playing with the domain of $s$ and inspecting the associated 3D output charts, I now dare to conjecture that all 'complex' zeros of this function must have a real part of $\frac12$. Has this been proven? If not, appreciate any thoughts on possible approaches. Thanks!
|
In the first part, we show that there are no zeros for $z = s + i t$
with $|t| \ge 4$ . Let $\psi(z):= \Gamma'(z)/\Gamma(z)$ be the digamma function.
If $z = s + i t$, then
$$\frac{d}{ds} |\Gamma(z)|^2 = \frac{d}{ds} \Gamma(z) \Gamma(\overline{z})
= |\Gamma(z)|^2 \left(\psi(z) + \psi(\overline{z})\right).$$
(Both $\Gamma(z)$ and $\psi(z)$ are real for real $z$, and so satisfy
the Schwartz reflection principle.)
The product formula for the Gamma function implies that there is an identity
$$\psi(z) = - \ \gamma + \sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{z + n} \right)
= 1 - \gamma + \sum_{n=1}^{\infty} \left(\frac{1}{n + 1} - \frac{1}{z + n} \right),$$
and hence
$$\psi(z) + \psi(\overline{z}) = 2(1 - \gamma) +
\sum_{n=1}^{\infty} \left(\frac{2}{n + 1} - \frac{1}{z + n} - \frac{1}{\overline{z} + n} \right).$$
Suppose that $z = s + i t$, and that $s \in [0,1]$.
Then
$$ \frac{2}{n + 1} - \frac{1}{s + i t + n} - \frac{1}{s - i t + n}
= \frac{2(s^2 + t^2 + n s - s - n)}{(1+n)(n^2 + 2 n s + s^2 + t^2)} \ge
\frac{-2}{(n^2 + t^2)}.$$
(The last inequality comes from ignoring
all the positive terms in the numerator, and then
setting $s = 0$ in the denominator.)
It follows that
$$\psi(z) + \psi(\overline{z}) \ge 2(1 - \gamma) -
\sum_{n=1}^{\infty} \frac{2}{n^2 + t^2},$$
which is positive for $t$ big enough, e.g. $|t| \ge 4$.
On the other hand,
$$\psi(z + 1) + \psi(\overline{z} + 1) = \psi(z) + \psi(\overline{z}) + \frac{1}{z} + \frac{1}{\overline{z}} =
\psi(z) + \psi(\overline{z}) + \frac{2s}{|z|^2}.$$
In particular, if $\psi(z) + \psi(\overline{z})$ is positive for $s \in [0,1]$ for some particular $t$, it is positive for
all $s$ and that particular $t$.
It follows that, if $|t| > 4$, that $|\Gamma(s + it)|^2$ is increasing as a function of $s$.
In particular, if $|t| > 4$, then any equality
$$|\Gamma(s + i t)| = |\Gamma(1 - (s + i t))| = |\Gamma(1 - s + i t)|$$
implies that $s = 1/2$. The second part is a continuation of the argument above, which completes the argument. (merged from a different answer.) Let $C_n$ denote the square with vertices $[n \pm 1/2, \pm 4 I]$ for a positive integer $n$.
We have the following inequalities for $z \in C_n$ and $n \ge 15$:
$$|\sin(\pi z)| \ge 1, \quad z \in C_n.$$
$$|\Gamma(z)| \ge \frac{1}{2} \Gamma(n - 1/2),$$
$$|\Gamma(1-z)| \le \frac{\pi}{\Gamma(n - 1/2)} \le 1,$$
$$|\psi(1-z)|, |\psi(z)| \le 2 \log(n), $$ The first is easy, the second follows from Stirling's formula (this requires $n$ to be big enough, and also
requires $z$ to have imaginary part at most $4$), the third follows from the previous
two by the reflection formula for $\Gamma(z)$, the last follows by induction and by the formula
$\psi(z+1) = \psi(z) + 1/z$.
It follows that
$$\left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\Gamma'(z)}{\Gamma(z)} - \frac{d/dz (\Gamma(z) + \theta
\cdot \Gamma(1-z))}{\Gamma(z) + \theta\cdot \Gamma(1-z)} \right|$$
$$= \left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\theta \Gamma(1-z) (\psi(1-z) + \psi(z))}
{\Gamma(z) + \theta \cdot \Gamma(1-z)} \right|$$
$$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)}
\oint_{C_n} \frac{1}
{|\Gamma(z) + \theta \cdot \Gamma(1-z)|}$$
$$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \cdot \frac{1}{1/2 \Gamma(n - 1/2) + 1} \ll 1,$$
where $\theta = \pm 1$ (or anything small) and $n \ge 15$, where the final inequality holds by a huuuge margin.
It follows that
$\Gamma(z) + \theta \cdot\Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the
number of poles in $C_n$. Since $\Gamma(z)$ has no zeros and poles in $C_n$, it follows that $\Gamma(z) + \theta\cdot\Gamma(1-z)$
has the same number of zeros and poles. It has exactly one pole, and thus exactly one zero.
If $\theta = \pm 1$ (and so in particular is real), by the Schwarz
reflection principle, this zero is forced to be real.
By symmetry, the same argument applies in the region $z = s + i t$ with $|t| \le 4$ and
$s \le -15$.
Combined with the above argument, this reduces the claim to $z = s + i t$ with $|s| \le 15$ and $|t| \le 4$ where the
claim can be checked directly. Hence all the zeros outside the box $z = s + it$ with $|t| \le 4$ and $|s| \le 15$ are either in $\mathbf{R}$, or lie on the line $1/2 + i \mathbf{R}$. EDIT To clarify, I didn't actually check that there were no ``exceptional'' zeros in the box $\pm 15 \pm 4 I$, since I presumed that the original poster had done so.
If $F(z) = \Gamma(z) - \Gamma(1-z)$, then computing the integral
$$\frac{1}{2 \pi i} \oint \frac{F'(z)}{F(z)} dz$$
around that box, one obtains (numerically, and thus exactly) $1$. There are (assuming
the OP at least computed the critical line zeros correctly) $2$ zeros in that range on the critical line. Along the real line in that range, there are $30$ poles and $25$ zeros. This means that there must be $1 + 30 - 25 = 6$ unaccounted for zeros. For such a zero
$\rho$ off the line, by symmetry one also has $\overline{\rho}$, $1 - \rho$ and
$1 - \overline{\rho}$ as zeros. Hence there must be either $1$ or $3$ pairs of zeros on the critical line, and either $1$ or $0$ quadruples of roots off the line. Varying the parameters of the integral, one can confirm there is a zero with $\rho \sim 2.7 + 0.3 i$, which is one of the four
conjugates of the root found by joro. A similar argument applies
for $\Gamma(z)+\Gamma(1-z)$. Hence: Any zero of $\Gamma(z) - \Gamma(1-z)$ is either in $\mathbf{R}$, on the line $1/2 + i \mathbf{R}$, or is one of the four exceptional zeros $\{\rho,1-\rho,\overline{\rho},1-\overline{\rho}\}$. A similar calculation implies the same
for $\Gamma(z) + \Gamma(1-z)$, except now with an exceptional set
$\{\mu,1-\mu,\overline{\mu},1-\overline{\mu}\}$.
|
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|
89,345 |
Every manifold that I ever met in a differential geometry class was a homogeneous space: spheres, tori, Grassmannians, flag manifolds, Stiefel manifolds, etc. What is an example of a connected smooth manifold which is not a homogeneous space of any Lie group? The only candidates for examples I can come up with are two-dimensional compact surfaces of genus at least two. They don't seem to be obviously homogeneous, but I don't know how to prove that they are not. And if there are two-dimensional examples then there should be tons of higher-dimensional ones. The question can be trivially rephrased by asking for a manifold which does not carry a transitive action of a Lie group. Of course, the diffeomorphism group of a connected manifold acts transitively, but this is an infinite-dimensional group and so doesn't count as a Lie group for my purposes. Orientable examples would be nice, but nonorientable would be ok too.
|
$\pi_2$ of a Lie group is trivial, so $\pi_2(G/H)$ is isomorphic to a subgroup of $\pi_1(H)$, which is finitely generated (isomorphic to $\pi_1$ of a maximal compact subgroup of the identity component of $H$). But $\pi_2$ of a closed manifold is often not finitely generated. For example, the connected sum of two copies of $S^1\times S^2$ has as a retract a punctured $S^1\times S^2$, which is homotopy equivalent to $S^1\vee S^2$ and so has universal cover homotopy equivalent to an infinite wedge of copies of $S^2$. EDIT This ad hoc answer can be extended as follows: All I really used was that $\pi_2(G)$ and $\pi_1(G)$ are finitely generated. But $\pi_n(G)$ is finitely generated for all $n\ge 1$ (reduce to simply connected case and use homology), so $\pi_n(G/H)$ is finitely generated for $n\ge 2$. That leads to a lot more higher-dimensional non simply connected examples.
|
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|
89,844 |
I stumbled upon the fact that the Bolza surface can be obtained as the locus of the equation, $y^2 = x^5-x$ Its automorphism group has the highest order for genus $2$, namely $48$. I recognized $x^5-x$ as a polynomial invariant of the octahedron. (In fact, the Bolza surface is connected to the octahedron.) If we use the analogous polynomial invariant of the icosahedron, then does the genus 5 surface, $y^2 = x(x^{10}+11x^5-1)$ have special properties? How close does the order of its automorphism group get to the bound $84(g-1)$? (For $g = 5$, this would be $336$.) POSTSCRIPT : My thanks to Noam Elkies for the highly detailed answer below. The background to this question is an identity I found involving $x^{10}+11x^5-1$. Define, $a = \frac{r^5(r^{10}+11r^5-1)^5}{(r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1)^2}$ and, $w = \frac{r^2(r^{10}+11r^5-1)^2(r^6+2r^5-5r^4-5r^2-2r+1)}{r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1}$ then, $w^5-10aw^3+45a^2w-a^2 = 0$ for arbitrary r . This in fact is the Brioschi quintic form which the general quintic can be reduced into. Two of the polynomials are easily recognizable as icosahedral invariants, while $r^6+2r^5-5r^4-5r^2-2r+1$ is a polynomial invariant for the octahedron. Other than in formulas using Ramanujan's continued fractions , I wondered where else those polynomials appear. Since the Bolza surface involved an invariant of the octahedron, it was reasonable to consider if using the corresponding one for the icosahedron would also be special. As Elkies wonderfully showed, it turns out that it is.
|
Yes, this Riemann surface, call it $C: y^2 = x^{11}-11x^6-x$,
is quite special: not only does it have the maximal number of automorphisms
for a hyperelliptic surface of genus $5$, but it is a modular curve
in at least two ways, both of which exhibit its full automorphism group. One is a classical (elliptic) modular curve of level $10$, intermediate between
$X(5)$ and $X(10)$, with $[C:X(5)] = 2$ (the hyperelliptic map) and
$[X(10):C] = 3$ (a cyclic cover); this modular curve parametrizes
elliptic curves $E$ with full level-$5$ structure and odd ${\rm Gal}(E[2])$,
or equivalently full level-$5$ structure and square $j(E)-12^3$.
Explicitly, $E$ has Weierstrass equation $Y^2 = X^3 - A(x)X/48 + B(x)/864$
where $A(x) = x^{20} + 228x^{15} + 494x^{10} - 228x^5 + 1$ and
$$
x^{30} - 522x^{25} - 10005x^{20} - 10005x^{10} + 522x^5 + 1
$$
are polynomials with roots at the $20$- and $30$-point orbits of $A_5$.
We have $A^3 - B^2 = 12^3 (x^{11}-11x^6-x)^5$,
so $j - 12^3 = B^2/(x^{11}-11x^6-x)^5$. The corresponding congruence
subgroup $\Gamma$ of ${\rm SL}_2({\bf Z})$ is the index-$2$ subgroup of
$\Gamma(5)$ consisting of matrices that reduce mod $2$ to the index-$2$
subgroup of ${\rm SL}_2({\bf Z}/2{\bf Z})$, with $[\Gamma : \Gamma(10)] = 3$.
This $\Gamma$ is normal in ${\rm SL}_2({\bf Z})$, and the quotient group
is ${\rm Aut}(C)$. Another modular approach to $C$ is via the $(2,3,10)$ triangle group,
call it $G^*$, which appears in class VIII of the nineteen commensurability
classes tabulated in Takeuchi, K.: Commensurability classes of arithmetic triangle groups, J. Fac. Sci. Univ. Tokyo 24 (1977), 201-212. According to Takeuchi's table, $G^*$ is the normalizer of the unit-norm
group $G_1$ of a maximal order in a quaternion algebra over
${\bf Q}(\sqrt 5)$ ramified over one real place and the prime $(\sqrt 5)$. Moreover $G_1$ is the $(3,3,5)$ triangle group, contained in $G^*$ with
index $2$. Let $G_5$ be the normal subgroup of $G_1$ consisting of
units congruent to $1 \bmod (\sqrt 5)$. Then
$G^*/G_5 \cong \lbrace \pm 1 \rbrace \times A_5$, and
the quotient of the upper half plane $\cal H$ by $G_5$ has genus $5$,
so must be our $C$. Moreover, ${\cal H} / G_5$ has no elliptic points,
so this identifies the image of the fundamental group $\pi_1(C)$ in
${\rm Aut}{\cal H} = {\rm SL}_2({\bf R})$ with an arithmetic congruence group. P.S. Roy Smith already noted that if we allow also non-hyperelliptic
Riemann surfaces then the maximal number of automorphisms for genus $5$
is not $120$ but $192$. An explicit model for a Riemann surface $S$
with $192$ automorphisms is the intersection of three quadrics
$$
y^2 = x_0 x_1,
\phantom{and}
{y'}^2 = x_0^2 - x_1^2,
\phantom{and}
{y''}^2 = x_0^2 + x_1^2
$$
in ${\bf P}^4$. Then $(x_0:x_1:y:y':y'') \mapsto (x_0:x_1)$
gives a normal cover $S \rightarrow {\bf P}^1$ with Galois group
$N = ({\bf Z}/2{\bf Z})^3$ acting by arbitrary sign changes on
$y,y',y''$, ramified above the vertices of a regular octahedron,
with each of $x_0 x_1, x_0^2 - x_1^2, x_0^2 + x_1^2$ vanishing
on an opposite pair of vertices. I claim that there is an exact sequence
$1 \rightarrow N \rightarrow {\rm Aut}(S) \rightarrow S_4 \rightarrow 1$,
so in particular $\#({\rm Aut}(S)) = 2^3 4! = 192$. Indeed let
$G$ be the subgroup of ${\rm Aut}(S)$ that stabilizes the span of
$\lbrace x_0, x_1 \rbrace$. Then $G$ contains $N$ as the kernel of
a homomorphism $G \rightarrow {\rm Aut}({\bf P}^1)$ given by the
action on $(x_0:x_1)$. The image is contained in the
group $S_4$ of rotations of the octahedron, and indeed equals $S_4$
because any rotation permutes the three opposite pairs of vertices
and thus lifts to ${\rm Aut}(S)$. Therefore ${\rm Aut}(S)$ contains
a group $G$ of order $2^3 4! = 192$, and by the Hurwitz bound
this must be the full group of automorphisms, QED
|
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|
90,009 |
Is it possible to get widely available math software (Maple/Matlab/Mathematica, etc) to symbolically differentiate vector and scalar functions of matrices, returning the result in terms of the original matrices and vectors involved? I have in mind the simple sort of rules collected here for example. On a few separate occasions I've scoured around the internet for such a thing and only turned up a bunch of incomplete threads of various vintage (like this or this or this ). So my main question is if I am missing the right keywords to find what is obvious to people who use this sort of functionality all the time, and what platform it is available on if so. If in fact this sort of functionality is not available in any of the commonly used software my question is if this is because of some sort of practical obstruction I am not seeing or simply because the problems for which it would be useful are simple enough to be done by hand (which is what I've ended up doing after I spend 4 hours searching for the "easy" way).
|
Indeed, I was having the same problem. Hence, I implemented a matrix calculus toolbox myself. You can find it at www.matrixcalculus.org . It can compute vector and matrix derivatives and will return the result in terms of the original vectors and matrices involved.
|
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|
90,012 |
Or can you give me a good place to read about things related to assembly map, besides wikipedia? I am specially interested in the case of algebraic K-theory. Would appreciated if you could provide examples.
|
If you are given a homotopy functor $L$ from spaces to spectra, the assembly map is a natural map of spectra
$$
H_\bullet(X;L) \to L(X) ,
$$
where the domain is a homology theory. This homology theory is the "homology of $X$ with coefficients in $L$," that is, $X_+ \wedge L(\text{pt})$. This map is a universal approximation to $L$ on the left by a homology theory in the homotopy category of functors. The map can most easily be described as follows: let $\Delta_X$ be the category whose objects are maps $\sigma: \Delta^n \to X$ (where $n$ can vary) and whose morphisms are given by restricting to a face of $\Delta^n$. Then there is an evident map
$$
\underset{\sigma\in \Delta_X}{\text{hocolim}} L(\Delta^n) \to L(X)
$$
where the map $L(\Delta^n) \to L(X)$ is given by applying $L$ to $\sigma$. The domain of this map is a homology theory, and in fact it's the homology of $X$ with coefficients in $L$. Notes Algebraic K-theory is the case of the functor $K: X \mapsto K(\Bbb Z[\pi(X)])$ where $\pi(X)$ denotes the fundamental groupoid. If we restrict to $X = B\pi$ for $\pi$ a discrete group, we get $H_\bullet(B\pi;K) \to K(\Bbb Z[\pi])$. An important case is that of surgery theory. In that case, the construction needs to be modified slightly: the category of spaces needs to be replaced with the category of spaces equipped with stable spherical fibration (this is the over-category
$TOP_{/BG}$, where $BG$ classifies stable spherical fibrations). In the surgery theory case, if $X = B\pi$, the Novikov conjecture can be stated: it says that the assembly map is a split surjection on rational homotopy. The Borel conjecture states that the assembly map is a weak equivalence. Of course, these statements are only known to be true under certain assumptions. When $L = A$ is Waldhausen's algebraic $K$-theory of spaces functor, the homotopy fiber of the assembly map at a manifold $X$ is the moduli space for the $h$-cobordisms relative to $X$ after a suitable stabilization with respect to dimension. This is a very deep result, the details of which have only just recently been written down by Jahren, Rognes and Waldhausen. The assembly map was first due to Quinn. The formulation I gave appears in a paper by Weiss and Williams: Assembly, in Novikov conjectures, index theorems and rigidity, Vol. 2 (Oberwolfach, 1993), 332--352, London Math. Soc. Lecture Notes 227, Cambridge Univ. Press, Cambridge, 1995. Here's the reason that the domain of the map constructed above is a homology theory: there is a natural transformation
$$
\underset{\sigma\in \Delta_X}{\text{hocolim}} L(\Delta^n) \to \underset{\sigma\in \Delta_X}{\text{hocolim}} L(\text{pt})\, .
$$
As $L(\Delta^n) \to
L(\text{pt})$ is a weak equivalence (since $L$ is a homotopy functor), the same is true after
taking hocolim. Since the second of these hocolims is that of a constant functor, it's given by $|\Delta_X|_+ \wedge L(\text{pt})$ (this is a standard exercise). Finally, the realization of the nerve, $|\Delta_X|$ is weak equivalent to $X$, also by a standard argument.
|
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|
90,021 |
Dear All, here is the question: Does there exist a finitely generated group $G$ with a proper subgroup $H$ of finite index, and an (onto) homomorphism $\phi:G\to G$ such that $\phi(H)=G$? My guess is "no", for the following reason (and this is basically where the question came from): in Semigroup Theory there is a notion of Rees index -- for a subsemigroup $T$ in a semigroup $S$, the Rees index is just $|S\setminus T|$. The thing is that group index and Rees index share the same features: say for almost all classical finiteness conditions $\mathcal{P}$, which make sense both for groups and semigroups, the passage of $\mathcal{P}$ to sub- or supergroups of finite index holds if and only if this passage holds for sub- or supersemigroups of finite Rees index. There are also some other cases of analogy between the indices. Now, the question from the post is "no" for Rees index in the semigroup case, so I wonder if the same is true for the groups. Also, I beleive the answer to the question may shed some light on self-similar groups.
|
Here is a proof that there is no such finitely generated group. It's similar to Mal'cev's proof that finitely generated residually finite groups are non-Hopfian. First, note that $\ker\phi$ is not contained in $H$---otherwise, $|\phi(G):\phi(H)|=|G:H|$. Let $k\in\ker\phi\smallsetminus H$. Because $\phi$ is surjective, there are elements $k_n$ for each $n\in\mathbb{N}$ such that $\phi^n(k_n)=k$. Let $\eta:G\to\mathrm{Sym}(G/H)$ be the natural action by left translation. Then the homomorphisms $\eta\circ\phi^n$ are all distinct. Indeed, $\eta\circ\phi^n(k_n)=\eta(k)\neq 1$ because $k\notin H$, whereas $\eta\circ\phi^{m}(k_n)=\eta(1)=1$ for $m>n$. But there can only be finitely many distinct homomorphisms from a finitely generated group to a finite group.
|
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|
90,070 |
Question: Let $A\in\mathbb{R}^{n \times n}$ be an orthogonal matrix and let $\varepsilon>0$ . Then does there exist a rational orthogonal matrix $B\in\mathbb{R}^{n \times n}$ such that $\|A-B\|<\varepsilon$ ? Definitions: A matrix $A\in\mathbb{R}^{n \times n}$ is an orthogonal matrix if $A^T=A^{-1}$ A matrix $A\in\mathbb{R}^{n \times n}$ is a rational matrix if every entry of it is rational.
|
Yes. It is a theorem of Cayley that the mapping $S \rightarrow (S-I)^{-1}(S+1)$ gives a correspondence between the set of $n\times n$ skew-symmetric matrices over $\mathbb{Q}$ and the set of $n\times n$ orthogonal matrices which do not have one as an eigenvalue. Since the mapping is nice, and rational skew-symmetric matrices are dense in the set of all skew-symmetric matrices, you have your result. For more, see the very nice paper by Liebeck and Osborne
|
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|
90,251 |
I have spoken to one professor so far about this, which of course was helpful, and so I am looking for additional opinions: To work with topological tools that were built via analysis, should I be a "master" at that analysis? By this I mean, for instance, to use Seiberg-Witten Theory and Floer Homologies. As an "entering" graduate student I am "purely" a pure topologist, as in I have no real training in analysis but Algebraic Topology under my belt for $\approx 6$ years. Now learning Seiberg-Witten Floer Homology and other Floer homologies, I tend to put all/most of the analysis (ex: compactness of moduli spaces) in a black box, and then continue to "learn". As a result, I am unsure if I am kind of wasting my time, i.e. if I can still utilize the theories effectively (and of course, I would like to extend theories). Is there a "good" balance between 1) simply accepting the analysis and 2) being able to do the analysis with both hands tied behind your back (as Kronheimer-Mrowka seem to do in their Monopoles and 3-Manifolds book)? I am unsure how to make this question less vague / more precise, but I feel that there is a good underlying question here that can have an informative response.
|
I am very sad. We wrote "Monopoles and Three Manifolds" with the idea
that a good graduate student who had read something like Warner's book
(through the chapter on Hodge theory) could reasonably read much of the book.
Oh well.
|
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|
90,252 |
I have a question about the possibility to apply/restate the Kleene fixed point theorem on recursive subsets of computable functions. I don't know if this is trivial and/or if related questions have been already solved. Any suggestion is more than welcome. Kleene fixed point theorem states that for every total computable function $T_i \simeq \varphi:\mathbb{N}\rightarrow \mathbb{N}$ there exists some $v\in\mathbb{N}$ such that
$\forall x, T_{\varphi(v)}(x)\simeq T_v(x)$. That is, the $v$-th ($T_v$) Turing Machine (TM) computes the same function as the $\varphi(v)$-th TM, and $v$ is thus a "fixed point" for the transformation $\varphi$. Consider now a total computable function $f:\mathbb{N}\rightarrow \mathbb{N}$. The recursive (infinite) set $E=f(\mathbb{N})$ represents the subset of TMs/computable functions, $T_{f(1)}, T_{f(2)}, ..$, we want to consider. Since $E$ is a recursive set, we can use the notation $T^E_i$ to denote $T_{f(i)}$, i.e. $T^E_i$ is the i-th TM/function in $E$. Let $T^E_{i}\simeq\phi$ be a total computable function (if any). Now, it is possible to restate the Kleene fixed point theorem on the set $E$? That is, is it true that there exists some $v\in \mathbb{N}$ such that: $\forall x: T^E_{\phi(v)}(x)\simeq T^E_{v}(x)$? Note that we cannot apply here directly the Kleene's theorem since the property above translates into $\exists v \forall x: T_{f(\phi(v))}(x)\simeq T_{f(v)}(x)$ and the original (standard?) proof seems hard (if not impossible) to recode in this new setting (unless $f$ is assumed to be surjective, in which case the problem becomes trivial). I don't expect this property to be true for all the possible recursive functions $f$ (it's just my non-expert opinion), so my question is: is it possible to characterize the class of mappings $f$ for which the above property is true? For example, is this property true if we assume that the range of $f$ is a set of total functions?
|
I am very sad. We wrote "Monopoles and Three Manifolds" with the idea
that a good graduate student who had read something like Warner's book
(through the chapter on Hodge theory) could reasonably read much of the book.
Oh well.
|
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|
90,455 |
Integration on an orientable differentiable n-manifold is defined using a partition of unity and a global nowhere vanishing n-form called volume form. If the manifold is not orientable, no such form exists and the concept of a density is introduced, with which we can integrate both on orientable and non-orientable manifolds.
My question is: On a non-orientable n-manifold, every n-form vanishes somewhere, but shouldn't I be able to chose an n-form with say a countable number of zeros, which would then constitute a set of measure zero and thus allow me to use n-forms (with zeros) for global integration also on non-orientable n-manifolds?
|
The problem is that there is no way to figure out signs - It would be like trying to integrate a function from $\mathbb{R}$ to $\mathbb{R}$ without knowing whether you were moving forward or backward. What you CAN actually integrate are pseudo-differential forms. The whole point of choosing an orientation is to turn a differential form into a psuedo-differential form. For those, I recommend the wonderful short story by John Baez found here: https://groups.google.com/group/sci.physics.research/msg/3c6a1a7237b66c8c?dmode=source&pli=1
|
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|
90,551 |
This principle claims that every true statement about a variety over the complex number field $\mathbb{C}$ is true for a variety over any algebraic closed field of characteristic 0. But what is it mean? Is there some "statement" not allowed in this principle? Is there an analog in char p>0? Is there reference about this topic? I tried to find some but in vain. Thanks:)
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The Lefschetz principle was formulated and illustrated the first time in: S. Lefschetz, Algebraic Geometry , Princeton University Press, 1953. The basic idea is that every equation over some algebraically closed field of characteristic $0$ only involves finitely many elements, which generate a subfield isomorphic to a subfield of $\mathbb{C}$. But as Seidenberg points out in A. Seidenberg, Comments on Lefschetz's principle , American American Monthly (65), No. 9, Nov. 1958, 685 - 690 Lefschetz has not given a rigorous proof and it is not clear at all if it holds when analytical methods over $\mathbb{C}$ are used. Tarski's classical result that the theory of algebraically closed fields of characteristic $0$ admits quantifier elimination and therefore all models are elementary equivalent is called the "Minor Lefschetz principle", because it does not apply to prominent examples such as Hilbert's Nullstellensatz. A precise formulation, with a short proof, which works in every characteristic, can be found here: Paul C. Eklof, Lefschetz's Principle and Local Functors , Proc. AMS (37), Nr. 2, Feb. 1973, online In the language of that paper, the principle states the following: Let $F$ be a functor from universal domains of characteristic $p$ ( = algebraically closed field of infinite transcendence degree over $\mathbb{F}_p$) to some category of many-sorted structures with embeddings, which satisfies the following finiteness condition: If $K \subseteq L$ is an extension, then every finite subset of $F(L)$ is already included in the image of a subextension of finite transcendence degree over $K$. Then, for all $K,L$, we have that $F(K)$ and $F(L)$ are elementary equivalent. For a specific statement one wants to prove using the Lefschetz princple, one can take $F(K)$ to be the collection of all "relevant algebraic geometry over $K$". A generalization is treated in: Gerhard Frey, Hans-Georg Rück, The strong Lefschetz Principle in Algebraic Geometry , manuscripta math. (55), 385 - 401 (1986)
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|
90,700 |
Where is number theory used in the rest of mathematics? To put it another way: what interesting questions are there that don't appear to be about number theory, but need number theory in order to answer them? To put it another way still: imagine a mathematician with no interest in number theory for its own sake. What are some plausible situations where they might, nevertheless, need to learn or use some number theory? Edit It was swiftly pointed out by Vladimir Dotsenko that the borders between number theory and algebraic geometry, and between number theory and algebra, are long and interesting. One could answer the question in many ways by naming features on that part of the mathematical landscape. But I'd be most interested in hearing about uses for number theory that aren't so obviously near the borders of the subject. Background In my own work, I often find myself needing to learn bits and pieces of other parts of mathematics. For instance, I've recently needed to learn new bits of analysis, algebra, topology, dynamical systems, geometry, and combinatorics. But I've never found myself needing to learn any number theory. This might very well just be a consequence of the work I do. I realized, though, that (independently of my own work) I knew of no good answer to the general question in the title. Number theory has such a long and glorious history, with so many spectacular achievements and famous results, that I thought answers should be easy to come by. So I was surprised that I couldn't think of much, and I look forward to other people's answers.
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Here are a few examples. In some, number theory provided an essential motivation. In the others, it plays a more direct role. 1) Are there nonisometric Riemannian manifolds that are isospectral (eigenvalues of the Laplacian match, including multiplicities)? An example was given by Milnor in the 1960s, which depended on prior work of Witt involving theta-functions (modular forms) of lattices. In the 1980s, Sunada created examples systematically by exploiting the analogy with the number theorist's construction of pairs of nonisomorphic number fields that have the same zeta-function. These number field pairs are found with Galois theory (find a finite group $G$ admitting a pair of nonconjugate subgroups having appropriate properties and then find a Galois extension of the rationals with Galois group isomorphic to $G$). There is a well-known analogy between Galois theory and covering spaces, and Sunada used this to translate the group-theoretic conditions for Galois groups into the setting of Riemannian manifolds. For more on this story, see the Wikipedia page here , where you'll see that the nonisometric isospectral pairs found between the work of Milnor and Sunada were closely related to other parts of number theory (quaternion algebras over the rationals). 2) Lens spaces are distinguished from each other using quadratic residues. 3) Knot theory uses continued fractions. See one of the answers to the MO question here . (Some of the other answers to that question could also be regarded as more applications of number theory, to the extent that you consider finite continued fractions to be number theory.) 4) The construction of Ramanujan graphs uses number theory. Also look here . 5) Frobenius proved that the only ${\mathbf R}$-central division algebras that are finite-dimensional are ${\mathbf R}$ and the quaternions. If you want to see infinitely many other examples of noncommutative division rings that are finite-dimensional over their centers, especially if you want examples that are more than 4-dimensional, you probably should learn number theory since the simplest examples come from cyclic Galois extensions of the rationals. Verifying the examples really work requires knowing a rational number is not a norm from a particular number field, and that amounts to showing a certain Diophantine equation has no rational solutions. 6) The classical induction theorems of Artin and Brauer about representations of finite groups were motivated by the desire to prove Artin's conjecture on Artin $L$-functions. Although number theory appears in the proof in the context of algebraic integers, the main point I want to make is that a conjecture from number theory provided an essential motivation to imagine the theorems might be true in the first place. 7) Several concepts of general importance in mathematics were originally developed within number theory. The most prominent example is ideals, which were first defined by Dedekind in his work on algebraic number theory. The first examples of finite abelian groups were unit groups mod $m$ and class groups of quadratic forms. The first finitely generated abelian groups to be studied as such were unit groups in number fields (Dirichlet's unit theorem). The first application of the pigeonhole principle was in Dirichlet's proof of the solvability of Pell's equation. The motivation for Steinitz's 1910 paper setting out a general theory of fields was Hensel's creation of $p$-adic numbers.
|
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|
90,707 |
General question : Given a vector bundle $E \rightarrow M$ on a complex manifold $M$, and a connection $\nabla$ on $E$, is it possible to find an Hermitian structure on $E$ such that $\nabla$ is the associated metric connection (i.e. the unique connection compatible with both the metric and the complex structure)? Specific question : Given a line bundle $L \rightarrow X$ on a compact Riemann surface $X$ equipped with a flat connection $\nabla$, is it possible to find an Hermitian structure on $L$ such that $\nabla$ is the associated metric connection? Motivation : I´m trying to prove that a degree zero line bundle on a compact Riemann surface always admits an harmonic Hermitian metric. By a classical result of Weil and Atiyah every degree zero vector bundle admits a flat connection, moreover I think (though I still did not prove it) that the flatness condition on the connection could be translated (by computation on the vector fields $\partial z$ and $\partial \overline{z}$) into the harmonicity condition on the "associated" (in the sense of the question) Hermitian metric. The question is clearly related to the well discussed question: When can a Connection Induce a Riemannian Metric for which it is the Levi-Civita Connection . But I´m not able to adjust the proof given in the mentioned question to an answer for my own. I´m asking also a general version of the question because I´m just curious about the answer. Thank you for your time! Edit : As pointed out by David Speyer the metric connection is constructed taking as input an Hermitian structure on the bundle and not an Hermitian metric on the manifold. I changed both the questions consequently.
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Here are a few examples. In some, number theory provided an essential motivation. In the others, it plays a more direct role. 1) Are there nonisometric Riemannian manifolds that are isospectral (eigenvalues of the Laplacian match, including multiplicities)? An example was given by Milnor in the 1960s, which depended on prior work of Witt involving theta-functions (modular forms) of lattices. In the 1980s, Sunada created examples systematically by exploiting the analogy with the number theorist's construction of pairs of nonisomorphic number fields that have the same zeta-function. These number field pairs are found with Galois theory (find a finite group $G$ admitting a pair of nonconjugate subgroups having appropriate properties and then find a Galois extension of the rationals with Galois group isomorphic to $G$). There is a well-known analogy between Galois theory and covering spaces, and Sunada used this to translate the group-theoretic conditions for Galois groups into the setting of Riemannian manifolds. For more on this story, see the Wikipedia page here , where you'll see that the nonisometric isospectral pairs found between the work of Milnor and Sunada were closely related to other parts of number theory (quaternion algebras over the rationals). 2) Lens spaces are distinguished from each other using quadratic residues. 3) Knot theory uses continued fractions. See one of the answers to the MO question here . (Some of the other answers to that question could also be regarded as more applications of number theory, to the extent that you consider finite continued fractions to be number theory.) 4) The construction of Ramanujan graphs uses number theory. Also look here . 5) Frobenius proved that the only ${\mathbf R}$-central division algebras that are finite-dimensional are ${\mathbf R}$ and the quaternions. If you want to see infinitely many other examples of noncommutative division rings that are finite-dimensional over their centers, especially if you want examples that are more than 4-dimensional, you probably should learn number theory since the simplest examples come from cyclic Galois extensions of the rationals. Verifying the examples really work requires knowing a rational number is not a norm from a particular number field, and that amounts to showing a certain Diophantine equation has no rational solutions. 6) The classical induction theorems of Artin and Brauer about representations of finite groups were motivated by the desire to prove Artin's conjecture on Artin $L$-functions. Although number theory appears in the proof in the context of algebraic integers, the main point I want to make is that a conjecture from number theory provided an essential motivation to imagine the theorems might be true in the first place. 7) Several concepts of general importance in mathematics were originally developed within number theory. The most prominent example is ideals, which were first defined by Dedekind in his work on algebraic number theory. The first examples of finite abelian groups were unit groups mod $m$ and class groups of quadratic forms. The first finitely generated abelian groups to be studied as such were unit groups in number fields (Dirichlet's unit theorem). The first application of the pigeonhole principle was in Dirichlet's proof of the solvability of Pell's equation. The motivation for Steinitz's 1910 paper setting out a general theory of fields was Hensel's creation of $p$-adic numbers.
|
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|
90,779 |
Are there applications of the Zariski topology in mathematics that are not within the scope of algebraic geometry (including schemes and algebraic groups) ? There is an older question with a similar title ( What is the Zariski topology good/bad for? ) but the answers given there are mainly concerned with the geometry stuff.
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Given two $n\times n$ matrices $A,B$ over a field $k$ let's prove that the characteristic polynomials of $AB$ and $BA$ are equal: $\chi(AB)=\chi(BA)$. Since the characteristic polynomial of a matrix obviously doesn't change under field extension , we may and do assume $k$ algebraically closed If $A$ is invertible, the result is clear because $\chi(BA)=\chi (A(BA)A^{-1})=\chi (AB) $. Now fix $B$ and consider the set $F\subset M_n(k)$ of all $A$ for which $\chi(AB)=\chi(BA)$. It is closed in the Zariski topology of $M_n(k)\cong \mathbb A^{n^2}(k)=k^{n^2}$ (because the characteristic polynomial of a matrix $M$ has as coefficients polynomials in the entries of $M$). Since, as we have just seen,it contains the open non-empty set of invertible matrices $A$, it is dense by irreducibility of $\mathbb A^{n^2}(k)$ (which requires that $k$ be algebraically closed). Since $F$ is closed and dense, we have $F=\mathbb A^{n^2}(k)$: all matrices $A$ satisfy $\chi(AB)=\chi(BA)$ Many theorems in elementary linear algebra can similarly be proved by using the Zariski topology on $M_n(k)$
|
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|
90,820 |
Clearly I first need to formally define what I mean by "junk" theorem. In the usual construction of natural numbers in set theory , a side-effect of that construction is that we get such theorems as $2\in 3$, $4\subset 33$, $5 \cap 17 = 5$ and $1\in (1,3)$ but $3\notin (1,3)$ (as ordered pairs, in the usual presentation). Formally: Given an axiomatic theory T, and a model of the theory M in set theory, a true sentence $S$ in the language of set theory is a junk theorem if it does not express a true sentence in T. Would it be correct to say that structural set theory is an attempt to get rid of such junk theorems? EDIT: as was pointed out $5 \cap 17 = 5$ could be correctly interpreted in lattice theory as not being a junk theorem. The issue I have is that (from a computer science perspective) this is not modular: one is confusing the concrete implementation (in terms of sets) with the abstract signature of the ADT (of lattices). Mathematics is otherwise highly modular (that's what Functors, for example, capture really well), why not set theory too?
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What you are describing is the idea of "breaking" an abstraction. That there is an abstraction to be broken is pretty much intrinsic to the very notion of "model theory", where we interpret the concepts in one theory in terms of objects and operations in another one (typically set theory). It may help to see a programming analogy of what you're doing: uint32_t x = 0x12345678;
unsigned char *ptr = (unsigned char *) &x;
assert( ptr[0] == 0x12 || ptr[0] == 0x78 ); // Junk!
const char text[] = "This is a string of text.";
assert( text[0] == 84 ); // Junk!
// Using the GMP library.
mpz_t big_number;
mpz_init_ui(big_number, 1234);
assert(big_number[0]._mp_d[0] == 1234); // Junk! All of these are examples of the very same thing you are complaining about in the mathematical setting: when you are presented with some sort of 'type', and operations for working on that type, but it is actually implemented in terms of some other underlying notions. In the above: I've broken the abstraction of a uint32_t representing a number modulo $2^{32}$, by peeking into its byte representation and extracting a byte. I've broken the abstraction of a string being made out of characters, by using knowledge that the character 'T' and the ASCII value 84 are the same thing In the third, I've broken the abstraction that big_number is an object of type integer, and peeked into the internals of how the GMP library stores such things. In order to avoid "junk", I think you are going to have to do one of two things: Abandon the notion of model entirely Realize that you were actually lying in your theorems: it's not that $2 \in 3$ for natural numbers $2$ and $3$, but $i(2) \in i(3)$ for a particular interpretation $i$ of Peano arithmetic. Maybe making the interpretation explicit would let you be more comfortable? (Or, depending on exactly what you mean by the notation, the symbols $2$ and $3$ aren't expressing constants in the theory of natural numbers, but are instead expressing constants in set theory.)
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|
90,862 |
Many famous problems in mathematics can be phrased as the quest for a specific construction. Often such constructions were sought after for centuries or even millennia and later proved impossible by taking a new, "higher" perspective. The most obvious example would be the three geometric problems of antiquity: squaring of circles, duplication of cubes and trisection of angles by ruler and compass alone. Closely related to these three is the construction of regular $n$-gons for general $n$. Later we have the solution of an arbitrary algebraic equation by means of radicals or the expression of the circumference of an ellipse by means of elementary functions. In the 20th century we have Hilbert's tenth problem: find an algorithm to determine whether a given diophantine equation has solutions. All these constructions turned out to be impossible, but the futile search produced some new and great mathematics: Galois theory, group theory, transcendental numbers, elliptic curves ... But I am looking for examples which are in some sense the opposite of the above: where somebody turned up with an ingenious construction in a problem where it had been generally believed that no such construction should exist. Ideally, this construction should have made new interesting questions and methods turn up, but I am also interested in isolated results that may just be counted as funny coincidences. To make my point clearer, let me present my own two favourite examples. (1) Belyi's Theorem : If $X$ is a smooth projective algebraic curve defined over a number field, there exists a rational function on $X$ whose only singular values are $0$, $1$ and $\infty$. --- According to his Esquisse d'un programme , Grothendieck had thought about this problem shortly before but found the statement so bold that he even felt awkward for asking Deligne about it. To put the theorem into context, the converse statement (that every curve which admits a rational function with only these three singular values can be defined over a number field) had been known before by abstract nonsense and is quite straightforward to deduce from deep results in Grothendieck-style algebraic geometry. Belyi's proof, however, was completely elementary, constructive, and tricky. Also it is more important than it might seem at first sight since it opens up a very strict, and equally unexpected, connection between the topology of surfaces and number theory. (2) Julia Robinson's theorem about the definability of integers : Suppose you want to single out $\mathbb{Z}$ as a subset of $\mathbb{Q}$, using as little structure as possible. The result in question is at least to me absolutely striking. I do not know if it was so unexpected to the experts at that time, but the construction is in any case really ingenious. It says that there exists a first-order formula $\varphi$ in the language of rings (i.e. only talking about elements, not subsets, and using only logical symbols and multiplication and addition, and the symbols $0$ and $1$) such that for a rational number $r$, $\varphi (r)$ is true if and only if $r$ is an integer. Robinson's original formula is
$$\varphi (r)\equiv\forall y\forall z(\psi (0,y,z)\wedge\forall x(\psi (x,y,z)\longrightarrow \psi (x+1,y,z))\longrightarrow\psi (r,y,z))$$
with
$$\psi (x,y,z) \equiv \exists a\exists b\exists c(2 + x^2yz = a^2 + yb^2-zc^2).$$
Since this is not my area of research I do not attempt to estimate the historical importance of this discovery, but it seems to me that it is of great weight in the intersection of number theory and logic. So I hope these two examples make it clear what I am after, and I am looking forward to reading your examples.
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I am not expert in the field, but it seems that Nash's embedding theorem is considered very unexpected for two reasons: first because at that time people thought that Riemannian manifolds were a so general object that nobody believed that they could be actually embedded (as smoothly as possible) in an Euclidean space. Second, because Nash's proof used new and unexpected techniques. If you read Gromov's interview for the Abel prize http://www.ams.org/notices/201003/rtx100300391p.pdf , at p.394, second column, third answer, he said: At first, I looked at one of Nash’s papers and thought it was just nonsense. But Professor Rokhlin said: “No, no. You must read it.” I still
thought it was nonsense; it could not be true. But
then I read it, and it was incredible. It could not
be true but it was true.
|
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90,872 |
This question is a follow-up to my previous question . The statement of the question is the title. Note that the $4$-dimensional real projective space is non-orientable and a characteristic class argument gives that it does not embed in $7$-space. Right now, I am more interested in orientable $4$-manifolds.
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This is true if and only if $X^4$ is spin and its signature vanishes. This is on p. 345 in Gompf/Stipsicz (4-manifolds and Kirby calculus), who cite Ruberman: Imbedding four-manifold and slicing links, 1982. EDIT Of course I mean that $X^4$ CAN be embedded in 6-dimensional space iff the conditions are met.
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90,876 |
Update (21st April, 2019). Removed the reference / initial trigger behind my question (please see comment thread below for the reasons). Am retaining, of course, the actual question, noted both in the title as well as in the post below. The key questions of this post are the following: 1. How "disastrous" would inconsistency of ZFC really be? 2. A slightly more refined question is: what would be the major consequences of different types of alleged inconsistencies in ZFC ? Old material (the "no-longer relevant" part of the question). I was happily surfing the arXiv, when I was jolted by the following paper: Inconsistency of the Zermelo-Fraenkel set theory with the axiom of choice and its effects on the computational complexity by M. Kim , Mar. 2012. Abstract. This paper exposes a contradiction in the Zermelo-Fraenkel set theory with the axiom of choice (ZFC). While Godel's incompleteness theorems state that a consistent system cannot prove its consistency, they do not eliminate proofs using a stronger system or methods that are outside the scope of the system. The paper shows that the cardinalities of infinite sets are uncontrollable and contradictory. The paper then states that Peano arithmetic, or first-order arithmetic, is inconsistent if all of the axioms and axiom schema assumed in the ZFC system are taken as being true, showing that ZFC is inconsistent. The paper then exposes some consequences that are in the scope of the computational complexity theory. Now this seems to be a very major claim, and I lack the background to be able to judge if the claim is true, or there is some subtle or even obvious defect in the paper's arguments. But picking on this paper itself is not the purpose of my question. If you feel that my questions might not admit "clearly right" answers, I will be happy to make this post CW.
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I'm confident that ZFC is consistent, but one can imagine an inconsistency. Like François said, it would probably be handled pretty well. I'd divide the possibilities into four cases: A technicality, like separation vs. comprehension in ZFC. This would be an important thing to get right, but it would have little impact on the theorems mathematicians prove. (For example, Frege's system was inconsistent, but his mistake didn't propagate.) A topic requiring serious clarification, like infinitesimals in the 1600's. The intution was right, but it took some genuine work to turn this intuition into actual theorems with rigorous proofs. A topic that fundamentally cannot be clarified, where some part of mathematics just turns out to be defective. For example, imagine if cardinals beyond $\aleph_0$ were inherently self-contradictory, and no clarification could save them. This would require huge modifications to set theory. It could turn out that we have no idea what any of mathematics really means. For example, if Peano Arithmetic were inconsistent, then it would call into question the whole axiomatic approach to mathematics. It would be tantamount to saying that the natural numbers as we understand them do not exist. (Some parts of the axiomatic approach could still survive, but I don't think it would be wise to trust anything if we couldn't even get the consistency of PA right.) My feeling is that 1 is very unlikely, 2 would be among the biggest shocks in the history of mathematics, 3 is difficult to imagine, and 4 is so extreme that if I read a proof of the inconsistency of PA, I'd be more likely to decide that I had gone crazy than that PA was actually inconsistent.
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|
90,972 |
I am investigating solutions to Fermat's equation
$$x^n+y^n=z^n$$
with $x,y,z$ in the Gaussian integers, excluding solutions in excluding $\mathbb{Z}$ or $i\mathbb{Z}$ . I have found out that there are only trivial solutions for the n=3 and n=4 cases, e.g. here . I would be grateful if you let me know of the current status or if it is already a theorem. P.S.: This same question was asked on Math.SE but it has now drowned under the fold and I thought I will have better chances of getting answers here.
|
This is still way open, I should think. "Elementary" methods won't even solve the analogous problem over $\mathbf{Z}$, so you need to use "modular form" methods. The problem is that even if the result were to follow from a Frey curve argument and a potential theorem of the form "all sufficiently nice Galois representations come from automorphic forms", we're a long way from establishing such a theorem. A few more details: a key problem is that there are two natural candidates for where the automorphic forms will come from, and neither is good enough. The first is the group $GL(2)/\mathbf{Q}(i)$. This is a very natural place to look, but the problem is that this group does not admit Shimura varieties, so it's very hard to even go the "easy" way and to attach a Galois representation to an algebraic automorphic representation (i.e. do what Deligne did), let alone to do what Wiles did. There are some theorems of this nature, but they all have a self-duality hypothesis built into them, which will not hold for the Tate module of the Frey curve in general. The second place to look is rank 2 unitary groups for the extension $\mathbf{Q}(i)/\mathbf{Q}$. These do have Shimura varieties and their geometry is understood fairly well nowadays, and there are are very strong theorems attaching Galois representations to these automorphic forms (there are even very strong theorems for rank $n$ unitary groups nowadays -- see for example the book by Harris and Taylor, but things have moved on even further since then, e.g. because of recent work of Sug-Woo Shin). However we then run into the same problem -- the unitary groups have some extra symmetry and this means that the associated Galois representations have some extra symmetry (they need to be essentially conjugate self-dual), and this extra symmetry will not in general be true for the Galois representations attached to the Tate module of the Frey curve. So one needs a good new idea before we can push forward what one might call "Wiles' strategy" in this situation. This is in marked contrast to the case of totally real fields, where a lot of the machinery works fine and it would not surprise me nowadays if FLT could be proved for several totally real fields. As has been implicitly mentioned in the comments, Jarvis and Meekin did this for $\mathbf{Q}(\sqrt{2})$, and this was a few years ago now, and modularity lifting theorems have moved on tremendously since then, so it would not surprise me if the experts could prove these results for other totally real number fields now. However somehow, after the Jarvis-Meekin work, which is a proof that the machine can be made to work in other cases, perhaps the interesting question now is not something like "is FLT true for $\mathbf{Q}(\sqrt{5})$?" (which one could perhaps hope to answer, perhaps with a lot of work, but with basically existing methods and a lot of hard graft to deal with the small $n$ cases) but more like "is some slightly weakened version of FLT true for all totally real fields?" or some such thing. You need to weaken it a bit because the case $n=3$ is an elliptic curve and it has positive rank for lots of totally real $F$, so now you need to deal with $n=4$ and $n=9$ and $n=6$ by hand, or just declare that you're only interested in $x^p+y^p=z^p$ with $p\geq5$ prime; and then you'll sometimes get reducible mod $p$ Galois representations -- so perhaps the correct weakening is something like "$x^p+y^p=z^p$ has no solutions for $p$ sufficiently large (depending on $F$)". Even then there may be problems in "case 2". My impression is that the machinery being developed now is not really being developed with generalisations of FLT in mind, but it's generalising this "Wiles machine" in different directions, for example to prove things like the Fontaine-Mazur conjecture and the Sato-Tate conjecture. This is the direction deemed "trendy" -- and in a sense I can see why, because is FLT over a specific number field other than the rationals really a "natural" question? And FLT over a general number field is obviously false, so now you have to weaken things etc etc. On the other hand hard conjectures like Sato-Tate do look to me like natural questions. IMPORTANT EDIT: I wrote this answer a long time ago -- what is it, 9 days now? -- but life moves on, and I hear from my spies in Toronto that Richard Taylor yesterday announced some results joint with Harris, Lan and Thorne, where they claim that they can attach Galois representations to (not necessarily self-dual) cohomological cuspidal automorphic representations on $GL(n)$ over totally real and CM fields. In particular apparently the theory for $GL(2)/\mathbf{Q}(i)$ is now up to about the state that the theory for $GL(2)/\mathbf{Q}$ was in the late 60s. So give it another 23 or so years and we should have FLT for $\mathbf{Q}(i)$!
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|
90,977 |
Today I entered the following expression in maple:
$$a_i = H_{10^i} - ln(10^i) - \gamma$$
Here $H_j$ equals $\sum_{k=1}^{j} 1/k$ and $\gamma$ is the Euler-Mascheroni constant. When I computed $a_n$ for $i = 0$ to $10$ I obtained the following results: $i = 0$; 4.227843350984671393934879099175975689578406640600764011942327651151323 * $10^{-1}$ $i=1$; 4.9167496072675423629464709201487329610707429399557393414873118115813 * $10^{-2}$ $i=2$; 4.991666749996032162622676207122311664609813510982102304110919767206 * $10^{-3}$ $i=3;$ 4.999166666749999960317501984051226762153678825611388678758121701133 * $10^{-4}$ $i=4$; 4.99991666666674999999960317460734126976551226762154503821179264423 * $10^{-5}$ $i=5$; 4.9999916666666667499999999960317460321626984126226551226762154523 * $10^{-6}$ $i=6$; 4.999999166666666666749999999999960317460317501984126984051226523 * $10^{-7}$ $i=7$; 4.99999991666666666666674999999999999960317460317460734126984123 * $10^{-8}$ $i=8$;
4.9999999916666666666666667499999999999999960317460317460321623 * $10^{-9}$ $i=9$;
4.999999999166666666666666666749999999999999999960317460317423 * $10^{-10}$ $i=10$;
4.99999999991666666666666666666674999999999999999999960317423 * $10^{-11}$ So we see that the periodic strips of ...99999..., of ...66666... and ...99999... an many other periods increase for even larger $i$. The question is now: Is there any rule behind it that the remainder term $a_i$ behaves that way?
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Yes, there is a rule. There are results that are finer than merely $H_k - \ln k - \gamma$ tends to $0$ and explain this pattern. More specifically, let us consider some more terms of the asymtotic expansion of $H_k$ . One has for example
$$H_k = \ln k + \gamma + \frac{1}{2k} - \frac{1}{12k^2} + O(k^{-4}) $$
and this is even true with a small implied constant, or more precisely this is true with $O(k^{-4})$
replaced by $x_k\frac{1}{120}k^{-4}$ with $0 \le x_k \le 1$.
Thus the error to be expected when doing the calculation in the question is
$$\frac{1}{2k} - \frac{1}{12k^2}$$ up to something still (much) smaller. This is precisely what one sees; if one chooses for $k$ a power of $10$ one sees a nice pattern (the $10$ being special due to the fact that one has the decimal representation; if one chooses a different base for the representation, powers of that base become special); it is the beginning of the decimal representation of $$\frac{1}{2} 10^{-j} - \frac{1}{12} 10^{-2j} ;$$
how long it is really just this can also be known from the estimate of the error mentioned above. One can continue on this, as it is known that
$$H_k = \ln k + \gamma + \frac{1}{2k} - \sum_{i=1}^{n-1} \frac{B_{2i}}{2i k^{2i}} + O(k^{-2n})$$
and more precisely the $O(n^{-2k})$ can be replaced by $x_{k,n}( -\frac{B_{2n}}{2n}) k^{-2n} $ with
$0\le x_{k,n} \le 1$ where the $B$'s are the Bernoulli numbers ; some care is needed if one would want to try to see more complex patterns as the Bernoulli numbers while small at first then grow very fast, so that then the implied constant is large and the $k$ needs to be sufficiently large (relative to the $n$) to see the pattern for all the terms. Besides the approximation I mentiond above there are various other approximations known. Also, questions like this are closely linked, essentially equivalent, to questions on the Digamma function .
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90,980 |
In an anwswer to a question on our sister site here I mentioned that a reduced commutative ring $R$ has zero Krull dimension if and only if it is von Neumann regular i.e. if and only if for any $r\in R $ the equation $r=r^2x$ has a solution $x\in R$. A user asked in a comment whether this implies that an arbitrary product of zero-dimensional rings is zero dimensional. I answered that indeed this is true and follows from von Neumann regularity if the rings are all reduced but I gave the following counterexample in the non reduced case: Let $R$ be the product ring $R=\prod_{n=1}^\infty \mathbb Z/2^n\mathbb Z$. Every $\mathbb Z/2^n\mathbb Z$ is zero dimensional but $R$ has $\gt 0$ dimension. My argument was that its Jacobson radical $Jac(R)=\prod_{n=1}^\infty Jac (\mathbb Z/2^n\mathbb Z)=\prod_{n=1}^\infty2\mathbb Z/2^n\mathbb Z$ contains the non-nilpotent element $(2,2,\cdots,2,\cdots)$. However in a zero dimensional ring the Jacobson radical and the nilpotent radical coincide and thus $R$ must have positive dimension. My question is then simply: we know that $dim(R)\gt 0$, but what is the exact Krull dimension of $R$ ? Edit Many thanks To Fred and Francesco who simultaneously (half an hour after I posted the question!) referred to an article by Gilmer and Heinzer answering my question . Here is a non-gated link to that paper. Interestingly the authors, who wrote their article in 1992, explain that already in 1983 Hochster and Wiegand had outlined (but not published) a proof that $R$ was infinite dimensional. Already after superficial browsing I can recommend this article, which contains many interesting results like for example infinite-dimensionality of $\mathbb Z^{\mathbb N}$. New Edit As I tried to read Hochster and Wieland's article, I realized that it refers to an article of Maroscia to which I have no access. Here is a more self-contained account of some of Hochster and Wieland's results.
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The ring $R$ is infinite-dimensional. More generally, the product of a family of zero-dimensional rings has dimension $0$ if and only if it has finite dimension. This is proven as Theorem 3.4 in R. Gilmer, W. Heinzer, Products of commutative rings and zero-dimensionality, Trans. Amer. Math. Soc. 331 (1992), 663--680.
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91,326 |
Are there any examples of non-Hilbert normed spaces which are isomorphic (in the norm sense) to their dual spaces? Or, is there any result in Functional Analysis which says that if a space is self-dual it has to be Hilbert space. Since, we want isomorphism in the norm sense, examples like $\mathbb{R}^{n}$ are ruled out. The norms of the space and its dual have to be equal and not just equivalent. Thank you.
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I have two, and perhaps infinitely many, examples in finite dimension $n$ . n=2 . Take $X={\mathbb R}^2$ with $\ell^1$ -norm $$\|x\|_1=|x_1|+|x_2|.$$ Then $X^*={\mathbb R}^2$ has the $\ell^\infty$ -norm $$\|y\|_\infty=\max(|y_1|,|y_2|).$$ I turns out that $$\|x\|_1=\max(|x_1+x_2|,|x_1-x_2|)$$ and thus $X'$ is isometric to $X$ , via $x\mapsto(x_1+x_2,x_1-x_2)$ . More generally, suppose that in $\mathbb R^n$ , we have a convex polytope $T$ that is self-dual and is symmetric under $x\leftrightarrow-x$ . Let $\|\cdot\|_T$ be the gauge associated with $T$ . Then $X=(\mathbb R^n, \|\cdot\|_T)$ is isometric to $X'$ because $T$ is the unit ball of $X$ and $T'=T$ is that of $X'$ . For instance, if n=4 , the polyoctahedron (= octaplex) has these properties, thus there is an $\mathbb R^4$ that is isometric to its dual, yet is not Hilbert. If $n\ge3$ , the simplex is self-dual but not centro-symmetric. This raises two questions: Does there exist other centro-symmetric self dual convex polytopes? Maybe there exist one in any even dimension ... Is it possible to deform the examples above so as to replace the polygone/-tope by a ball with a smooth boundary?
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91,385 |
Where can I find a proof of the de Rham-Weil theorem? Does anyone know?
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Of course Weil did it (although I'm not able to give you the ref right away), and I even lectured on it in Paris 40 years ago. His method is very simple :
(1) You first prove that your compact manifold X can be endowed with a Riemann structure (obvious locally, global result by using a smooth partition of unity).
(2) By the general theory of Riemann spaces, there are plenty of "convex" open sets U in X, i.e. such that any two points of U can be joined by one and only geodesic arc in U. Such an open set is clearly homotopic to a point, so that every closed diff form on it is exact; the intersection of any two such "convex" sets is also convex.
(3) That being said, choose a finite covering of X by convex open sets U(i) - no math available - and let omega (idem) be a form of degree p on X. There are forms omega(i) of degree p-1 in the U(i) such that
omega = d[omega(i)] in U(i).
Since U(i) inter U(j) = U(i,j) is convex, there are forms omega(i,j) of degree p-2 in the U(i,j) such that
omega(j) - omega(i) = d[omega(i,j)] in U(i,j).
By standard de Rham, there are forms omega(i,j,k) in the U(i,j,k) = U(i) inter U(j) inter U(k) such that
omega(j,k) - omega(i,k) + omega(i,j) = d[omega(i,j,k)] in U(i,j,k),
and so on. In this way, you eventually get forms of degree O whose alternate sums are closed, hence CONSTANTS, in the p to p intersections of the U(i). If you consider the abstract simplicial complex defined by the covering (U(i)), i.e. its so-called "nerve" (nerf in French), you thus associate to omega a cocyle of degree p (with real coefficients) of this complex. Etc, etc. You can do it all by your little self - matter of patience. Before Grothendieck's Tohoku, there was a Cartan Seminar on sheaves theory (ca. 1948-49) in which everybody, at any rate in France, learned the theory, including Groth and myself. I even wrote a book on the subject (Théorie des Faisceaux, Paris, Hermann, 1957), which was still on sale (and found customers) two or three years ago, and possibly still is. Don't infer from my answer I'm still doing mathematics. Roger Godement, Paris.
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91,423 |
Imagine a particle in the complex plane, starting at $c_0$ , a Gaussian integer ,
moving initially $\pm$ in the horizontal
or vertical directions. When it hits a Gaussian prime , it turns left $90^\circ$ .
For example, starting at $12 - 7 i$ , moving initially $+x$ , this closed circuit results: Instead, starting at $3+5 i$ , again $+x$ , this (pleasingly symmetric!) closed cycle results: Here's another (added later), starting at $5+23 i$ : (Gaussian primes $a+bi$ off-axes have $a^2+b^2$ prime; on axis, $\pm(4n+3)$ prime.) My question is, Q0 . What's going on? More specifically, Q1 . Does the spiral always form a cycle? Q2 . Have these spirals been investigated previously? (I am about to step on a plane; apologies for not acknowledging responses!)
... Later: Q3 . Under what conditions is the spiral (assuming it closes) symmetric w.r.t. reflection in a
horizontal (as is $12-7 i$ ), or reflection in a vertical (as is $5 + 23 i$ ),
or reflection in both (as is $3 + 5i$ )?
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Expanding slightly on Greg Martin's answer, the symmetry applies across the imaginary axis as well as the real axis, so the only way a path can avoid closing up is if it crosses at most one axis at most once (and if it does cross an axis, the path -- if extended in the backward as well as forward direction -- will be mirror symmetric with respect to the axis it crosses). Note that within a quadrant the horizontal (respectively vertical) steps are alternately toward and away from the imaginary (respectively real) axis. Intuitively the steps, on average, get larger the further you are from the axes. So if you're in the first quadrant and take a step to the right and then a step up, you expect your next step, to the left, to be somewhat larger than your previous step to the right (and similarly with the next step down, assuming you're still in the first quadrant). So in a sense the axes are exerting kind of a gravitational tug on sort of a random walk. Of course, nothing of the sort is literally going on. Still, it seems hard to imagine the stars (I mean Gaussian primes) aligning to keep a random walk going in perpetuity. It seems more likely one might encounter closed paths that don't have any nice symmetry, such as
$$(a,b) \rightarrow (a+4,b) \rightarrow (a+4,b+8) \rightarrow (a-2,b+8) \rightarrow$$ $$(a-2,b+4) \rightarrow (a+2,b+4) \rightarrow (a+2,b+6) \rightarrow (a,b+6) \rightarrow (a,b)$$ (Sorry, if someone could replace that with a picture, that would be helpful.) As "unknown" pointed out, there are certainly closed square paths that stay in the first quadrant. There are also rectangles, such as the $2\times4$ rectangle with $8+13i$ for its lower left hand corner and the $2\times6$ one starting at $14+19i$. (I spotted these in a picture of Gaussian primes of norm less than 1000 in the paper "A Stroll Through the Gaussian Primes" by Gethner, Wagon, and Wick.) One might expect a souped-up (supped-up?) $k$-tuple conjecture to predict the existence of any closed-path pattern that isn't forbidden by the usual suspects. (Part of the souping up, though, is that not only are there primes at the specified corners, but everything else along the edges is composite.) All in all, a nice problem -- and the spirals, reminiscient of Celtic knots, are really lovely!
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91,583 |
I would like to work a theorem on a article who deals with the rank one symmetric spaces. i looked up the definition of symmetric spaces of rank one, but I did not find a satisfactory definition then what is the meaning of rank, intuitively and mathematically? please if anybody already worked with rank one symmetric spaces..
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First the algebraic definition. A (non-compact) symmetric space is of the form $G/K$, where $G$ is a (non-compact) semisimple Lie Group defined over $\mathbb{R}$, and $K$ is a maximal compact subgroup of $G$. Then the rank of a symmetric space is the dimension of the "maximal $\mathbb{R}$-split torus", i.e. the maximal dimension of an abelian diagonalizable over $\mathbb{R}$ subgroup of $G$. The geometric meaning is that the rank is the dimension of the maximal flat submanifold of the symmetric space. If the rank is $1$, then the maximal flats are geodesics, and the symmetric space turns out to be negatively curved. If the rank is larger then one, then the symmetric space is only non-positively curved. However, higher-rank symmetric spaces have spectacular rigidity properties (e.g. Margulis superrigidity, arithmeticity and the normal subgroup property come to mind). There are only three families of rank 1 symmetric spaces, 1) hyperbolic $n$-space, corresponding to the Lie group $SO(n,1)$. 2) complex hyperbolic $n$-space, corresponding to the Lie group $SU(n,1)$. 3) quaternionic hyperbolic $n$-space, corresponding to the Lie group $Sp(n,1)$. There is also one exceptional example: 4) the Cayley upper half plane, corresponding to the Lie group $F_4^{-20}$. The spaces 3) and 4) have some but not all of the rigidity properties of higher rank (in particular, superrigidity and arithmeticity, but not the normal subgroup property).
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91,597 |
When I study group theory, I find that there are some mysterious things. For example, Daniel Gorenstein, in his paper On a Theorem of Philip Hall , mentioned the unpublished lecture notes of Philip Hall. Many other famous group theorists also confirmed that these notes are important to their work. Since the notes are not published, I can not find way to see them. But I am very curious about what's in these notes. In these notes, are there any theorems about groups that are not otherwise known or do not appear in published books or papers?
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It may help to have some explicit bibliographic references, though some items are by now out of print and may be difficult to locate even through libraries. First, the 1966 paper by Gorenstein is located online here : Gorenstein, Daniel.
On a theorem of Philip Hall.
Pacific J. Math. 19 (1966), 77–80. Presumably he is referring to the 1957 lectures by Hall, which circulated in small numbers and were later published at QMC in London (where I may even have seen them, since I was lecturing there in 1969): Hall, Philip. The Edmonton notes on nilpotent groups. Queen Mary College Mathematics Notes. Mathematics Department, Queen Mary College, London 1969 iii+76 pp. These typewritten notes have largely disappeared from view by now, but fortunately a version got included in the 1988 volume of collected works: Hall, Philip. The collected works of Philip Hall.
Compiled and with a preface by K. W. Gruenberg and J. E. Roseblade. With an obituary by Roseblade. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1988. xii+776 pp. ISBN 0-19-853254-7. Here is a brief excerpt from the review in Mathematical Reviews : "The works appear in chronological order and are reproduced in their original format apart from the Edmonton notes on nilpotent groups (based on Hall’s lectures given at the summer seminar of the Canadian Mathematical Congress held in 1957) which are a corrected reprint of the third edition (Queen Mary College, 1979)." The book is also hard to find now outside some libraries, but does exist. As reviews indicate, much but not all of the material in Hall's original lecture notes has appeared in various research papers. Good luck. ADDED: I'm sorry to have gone off on the wrong trail in response to the question raised here. Certainly Philip Hall's influence on other group theorists was not limited to his formally published work, though I suspect the informal stuff left out of his collected works might be covered indirectly through the papers he inspired other people to write. Though not many people directly involved in that era remain, it would for example be interesting to know what John Thompson's recollections are. Perhaps it may inspire some finite group theorists to suggest better answers if I quote more precisely the "theorem of Philip Hall" which Gorenstein revisited from the viewpoint of the 1963 Feit-Thompson paper: "If $P$ is a $p$-group with no noncyclic characteristic abelian subgroups, then $P$ is the central product of subgroups $P_1$ and $P_2$, where $P_1$ is extra-special and either $P_2$ is cyclic or $p=2$ and $P_2$ is dihedral, generalized quaternion, or semi-dihedral."
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91,646 |
What is the relationship between the surreal numbers and non-standard analysis? In particular, is there a transfer principle for surreal numbers they way there is for NSA? A specific situation in which such a transfer principle would be useful arose in the thread Uniformizing the surcomplex unit circle ; can the surjectivity of the map $t \mapsto e^{it}$ from the reals to the complex unit circle be transferred to the surreals? Presumably, one would need a definition of the map that was in some sense first-order; what sorts of definitions count as first-order? It is not clear to me how definitions involving the two-sided bracket operation can be fit into a first-order framework.
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In the final section of my paper “The Absolute Arithmetic Continuum and the Unification of All Numbers Great and Small” (The Bulletin of Symbolic Logic 18 (2012), no. 1, pp. 1-45, I not only point out that the real-closed ordered fields underlying the hyperreal number systems (i.e. the nonstandard models of analysis) are isomorphic to initial subfields of the system of surreal numbers, but that the system of surreal numbers itself is isomorphic to the real-closed ordered field underlying what may be naturally regarded as the maximal hyperreal number system in NBG (von-Neumann-Bernays-Gödel set theory with global choice)—i.e., the saturated hyperreal number system of power On, On being the power of a proper class in NBG. It follows immediately from the latter that the ordered field of surreal numbers admits a relational extension to a model of non-standard analysis and, hence, that in such a relational extension the transfer principle does indeed hold. By the way, by an initial subfield, I mean a subfield that is an initial subtree. Discussions of surreal numbers (including most of the early discussions) that downplay or overlook the marriage between algebra and set theory that is central to the theory overlook many of the most significant features of the theory. In addition to the paper listed above, this marriage of algebra and set theory is discussed in the following papers which are found on my website http://www.ohio.edu/people/ehrlich/ “Number Systems with Simplicity Hierarchies: A Generalization of Conway’s Theory of Surreal Numbers,” The Journal of Symbolic Logic 66 (2001), pp. 1231-1258. Corrigendum, 70 (2005), p. 1022. “Conway Names, the Simplicity Hierarchy and the Surreal Number Tree”, The Journal of Logic and Analysis 3 (2011) no. 1, pp. 1-26. “Fields of Surreal Numbers and Exponentiation” (co-authored with Lou van den Dries), Fundamenta Mathematicae 167 (2001), No. 2, pp. 173-188; erratum, ibid. 168, No. 2 (2001), pp. 295-297.
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91,649 |
I have been wondering about something for a while now, and the simplest incarnation of it is the following question: Find a finite group that is not a subgroup of any $GL_2(q)$. Here, $GL_2(q)$ is the group of nonsingular $2 \times 2$ matrices over $\mathbb{F}_q$. Maybe I am fooled by the context in which this arose, but it seems quite unlikely that "many" finite groups are subgroups of $GL_2(q)$. Still, I can't seem to exclude a single group, but I am likely being stupid. Motivation: This arose when I tried to do some explicit calculations related to Serre's modularity conjecture. I wanted to get my hands on some concrete Galois representations, and play around with the newforms associated to it. Call it recreational if you like (it certainly is!). In Serre's original paper [1], there is a wonderful treatment of some explicit examples. He uses the observation that the Galois group over $\mathbb{Q}$ of $$x^7 -7x +3$$ is isomorphic to $PSL_2(7)$. This gives him a surjection $G_{\mathbb{Q}} := \mbox{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\rightarrow PSL_2(7)$, which he combines with the character associated to $\mathbb{Q}(\sqrt{-3})$ to obtain a homomorphism $G_{\mathbb{Q}} \rightarrow PSL_2(7) \times C_2$. This homomorphism he can then lift to $GL_2(49)$ using a very clever calculation in the Brauer group. The way I see it (without claiming this is justified) is that it is unclear or even provably impossible to embed $PSL_2(7)$ directly into some $GL_2(q)$, and for that reason we need a nice lifting. This raises the natural question of when this little trick with the Brauer group is "necessary". What can be said in general about subgroups of $GL_2(q)$? Can we exclude any particular finite group or families of finite groups? Can we perhaps even classify the possible isomorphism types of such subgroups? Of course this is all becomes even more natural to ask in the light of the inverse Galois problem. How many Galois groups can we "see" in the two-dimensional representations? Remark. Of course, the question could be approached the other way. If one has the (perhaps not unrealistic?) hope of realising every $GL_2(q)$ as a Galois group over $\mathbb{Q}$, how many groups do you automatically realise over $\mathbb{Q}$ by taking quotients of $GL_2(q)$? Which groups $G$ admit a surjection $GL_2(q) \rightarrow G \rightarrow 1$ for some $q$? Reference: [1] Serre, Jean-Pierre , "Sur les représentations modulaires de degré 2 de Gal(Q/Q)" , Duke Mathematical Journal 54.1, (1987): 179–230
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Well, ${\rm GL}(2,q)$ has Abelian Sylow $p$-subgroups for every odd prime $p.$ The symmetric group $S_{n}$ has non-Abelian Sylow $p$-subgroups for each prime $p$ such that $p^2 \leq n.$ Hence thesymmetric group $S_{25}$ is not a subgroup of any ${\rm GL}(2,q)$ since it has non-Abelian Sylow $3$-subgroups and non-Abelian Sylow $5$-subgroups ( even the alternating group $A_{25}$ will do). To answer directly the question about ${\rm PSL}(2,7),$ if $q$ is a power of $2,$ then all odd order subgroups of ${\rm GL}(2,q)$ are Abelian, so ${\rm PSL}(2,7)$ can't be a subgroup of such a ${\rm GL}((2,q)$ since ${\rm PSL}(2,7)$ contains a non-Abelian subgroup of order $21.$ On the other hand, if $q$ is odd, then the alternating group $A_4$ is not a subgroup of ${\rm GL}(2,q)$ ( for example, any Klein 4-subgroup of such a ${\rm GL}(2,q)$ contains a cenral element of order $2,$ whereas $A_4$ has no central element of order $2$), but ${\rm PSL}(2,7)$ has a subgroup isomorphic to $A_4.$
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91,685 |
The odds of two random elements of a group commuting is the number of conjugacy classes of the group $$ \frac{ \{ (g,h): ghg^{-1}h^{-1} = 1 \} }{ |G|^2} = \frac{c(G)}{|G|}$$ If this number exceeds 5/8, the group is Abelian (I forget which groups realize this bound). Is there a character-theoretic proof of this fact? What is a generalization of this result... maybe it's a result about semisimple-algebras rather than groups?
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If $c(G)> 5|G|/8$, then the average character has a dimension-squared of less than $8/5$, so at least $4/5$ of the characters are dimension $1$ (since the next-smallest dimension-squared is $4$), so the abelianization, which has one element for each 1-dimensional character, is more than half the size of the group, so the commutator subgroup has size smaller than $2$ and so is trivial.
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91,852 |
Let $M$ be a smooth manifold, and let $TM$ denote its tangent bundle. Under what conditions does $TM$ admit a flat connection $\omega$? Edit: Formerly, I asked about a flat connection on the frame bundle, but Deane Yang points out that a connection on the frame bundle is the same thing as one on the tangent bundle. I am imposing no other assumptions on the manifold other than smoothness, and I am seeking what assumptions may obstruct the existence of a flat connection.
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The question of existence of flat connection on tangent bundles of manifolds was studied quite extensively. Milnor proved in one of his early papers that surfaces (compact without boundary) of non-zero Euler characteristic don't admit such a connection. A result of Smillie can be used to rule out existence of flat connection on tangent bundles of many even dimensional manifolds; a manifold $M^n$ that admit such a connection should satisfy the condition $|\chi(M^n)|\le \frac{||M^n||}{2^n}$, where $||M^n||$ denotes the simplicial norm of $M^n$. You can check http://www.ihes.fr/~gromov/PDF/4[35].pdf , page 229 for a short proof. Also, Smillie constructed examples of manifolds of non-zero Euler characteristics that admit flat connection on their tangent bundle: http://www.springerlink.com/content/g6804q4u77327887/ The following recent article of Goldman will be relevant Milnor's seminal work on flat manifolds and bundles http://arxiv.org/abs/1108.0216
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92,140 |
A friend of mine just asked me this very question. While I had some training in combinatorics, I have never heard of the "Seetapun Enigma", which, supposedly, is related to the Ramsey's theorem. A quick Google search provides nothing useful, nor was Math.StackExchange of any help. Does anyone know what it is, and what is the significance of its solution/proof/refutation.
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The question seems to be about the following special form of Ramsey's Theorem: $\mathsf{RT}^2_2$: for every $2$-coloring of the unordered pairs from $\mathbb{N}$ there is an infinite subset of $\mathbb{N}$ for which all unordered pairs receive the same color. which is a special case of $\mathsf{RT}^n_k$: for every $k$-coloring of the unordered $n$-tuples from $\mathbb{N}$ there is an infinite subset of $\mathbb{N}$ for which all unordered $n$-tuples receive the same color. The computability strength of infinite Ramsey's theorem was first studied by Jockusch (1972). When interpreted in modern terminology that didn't exist then, Jockusch's result is that $\mathsf{RT}^n_k$ is equivalent to $\mathsf{ACA}_0$ whenever $n \geq 3$ and $k \geq 2$. The equivalence is over the standard base system $\mathsf{RCA}_0$ which is assumed in the rest of this post. As a corollary, $\mathsf{ACA}_0$ proves $\mathsf{RT}^2_k$ for all $k \geq 2$. Later, Hirst (1987) characterized the strength of principles of the form $\mathsf{RT}^1_k$. The separate results of Jockusch and Hirst leave a gap for exponent $2$, and in particular for $\mathsf{RT}^2_2$. The exact reverse mathematics strength of $\mathsf{RT}^2_2$ is somewhat mysterious, although I don't know that anyone calls it an "enigma". It has proven to be a particularly difficult open problem. The first result was due to Seetapun (published as Seetapun and Slaman (1995)), who showed that $\mathsf{RT}^2_2$ does not imply $\mathsf{ACA}_0$. The fact that this seemingly weak result was all that could be obtained hints at the difficulty of finding the exact strength of $\mathsf{RT}^2_2$ with known methods. Seetapun's proof used an intricate forcing argument. The ideas behind this argument have been progressively clarified and extended, and are now well understood; the most recent paper on this is by Dzhafarov and Jockusch (2009). The principle $\mathsf{WKL}_0$ says that every infinite subtree of $2^{<\mathbb{N}}$ has an infinite path. $\mathsf{WKL}_0$ is one of the "big five" systems of reverse mathematics, and is the natural comparison point for principles weaker than $\mathsf{ACA}_0$ such as $\mathsf{RT}^2_2$. Cholak, Jockusch, and Slaman (2001) made the next significant progress on $\mathsf{RT}^2_2$. Among many other new results they showed that $\mathsf{RT}^2_2$ is not provable in $\mathsf{WKL}_0$, because $\mathsf{WKL}_0$ does not prove the principle $\mathsf{COH}$ which is provable from $\mathsf{RT}^2_2$. The principle $\mathsf{COH}$ is a formalized statement of a theorem from recursion theory about the existence of $r$-cohesive sets. The results I have mentioned left the question open whether $\mathsf{RT}^2_2$ implies $\mathsf{WKL}_0$. This was recently solved by Liu in 2011. Liu showed in a still-unpublished paper that $\mathsf{RT}^2_2$ does not imply $\mathsf{WWKL}_0$, which is the restriction of $\mathsf{WKL}_0$ to trees of positive measure, and which is strictly weaker than $\mathsf{WKL}_0$. Thus, combining results, $\mathsf{RT}^2_2$ and $\mathsf{WKL}_0$ are mutually independent. As I understand it, Liu proved this independently while a student at Central South University (China), without an advisor in logic or any graduate training in logic. Liu presented his result at the Reverse Mathematics workshop at University of Chicago in September 2011. The slides from that talk are available online, but they are quite technical. The proof uses another intricate forcing argument. As I understand it, Liu's paper was submitted to a journal some time before the workshop, the results have been verified by referees, and the paper will be published once it is in final form. Citations Cholak, Peter A.; Jockusch, Carl G.; Slaman, Theodore A. On the strength of Ramsey's theorem for pairs .
J. Symbolic Logic 66 (2001), no. 1, 1–55. MR1825173 (2002c:03094) Dzhafarov, Damir D.; Jockusch, Carl G., Jr. Ramsey's theorem and cone avoidance .
J. Symbolic Logic 74 (2009), no. 2, 557–578. MR2518811 (2010e:03052) Hirst, Jeffry Lynn. Combinatorics in subsystems of second-order arithmetic .
PhD Thesis, The Pennsylvania State University. 1987. 153 pp. Jockusch, Carl G., Jr. Ramsey's theorem and recursion theory .
J. Symbolic Logic 37 (1972), 268–280. MR0376319 (51 #12495) Seetapun, David; Slaman, Theodore A. On the strength of Ramsey's theorem .
Special Issue: Models of arithmetic.
Notre Dame J. Formal Logic 36 (1995), no. 4, 570–582. MR1368468 (96k:03136)
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|
92,206 |
The title essentially says it all. Consider the category $\mathfrak{Top}_2$ of triples $(J,e_0,e_1)$ where $J$ is a topological space, and $e_i \in J$. There is an obvious generalization of the definition of homotopic maps. Suppose we have selected $(J,e_0,e_1)\in \mathfrak{Top}_2$. We could say that two continuous maps $f,g:X\to Y$ are "$J$-homotopic" if there is a continuous map $h:X\times J\to Y$ such that $h(x,e_0) = f(x)$ and $h(x,e_1) = g(x)$. We could then define $\pi_1 (X,x)$ to be the set of continuous maps $f:J\to X$ satisfying $f(e_0)=f(e_1)=x$, with $J$-homotopic maps identified. Here in order to define composition of paths in the naive way, we need to have picked some homeomorphism continuous map from $J$ to $(\{0\}\times J\cup \{1\}\times J)/((\{0\},e_1) = (1,e_0))$, taking $e_0$ to $e_0\times 0$ and $e_1$ to $e_1\times 1$. I have two questions: Can $([0,1],0,1)$ be characterized as an object in $\mathfrak{Top}_2$ in a purely categorical manner? When is $\pi_1 (X,x)$ a group? For that matter, when is $\pi_1 (X,x)$ associative? Essentially, the question comes down to: what properties of $[0,1]$ are needed in order to do homotopy theory?
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The answer to 1 is yes. For the purpose of this answer, a bipointed space is a topological space $J$ equipped with distinct closed points $e_0$ and $e_1$. As you say, for any bipointed space $J = (J, e_0, e_1)$, we can form a new bipointed space $J \vee J$ by taking the disjoint union of two copies of $J$, identifying the first $e_1$ with the second $e_0$, and giving the resulting space the obvious pair of basepoints. Theorem: In the category of bipointed spaces $J$ equipped with a map $J \to J \vee J$, the terminal object is the bipointed space $([0, 1], 0, 1)$ equipped with the map "multiplication by 2" from $[0, 1]$ to $[0, 1] \vee [0, 1] \cong [0, 2]$. Or informally: $[0, 1]$ has the structure needed in order to be able to define and compose paths, and is universal as such. The theorem is proved here , and is a variant of a result of Peter Freyd's (which characterized the interval set-theoretically and order-theoretically, but not topologically). The idea that $[0, 1]$ is universal with the structure needed for homotopy theory is expanded on in these talk slides .
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|
92,337 |
Hello, I'm looking for an invariant to distinguish the homeomorphism types of homotopy equivalent spaces. Specifically, how does one show that the total spaces of the tangent bundle to $S^2$ and the trivial bundle $S^2 \times R^2$ are not homeomorphic? (I am not asking for a proof that $TS^2$ is not the trivial bundle.) Also, is there a way to reduce the question, "Are the total spaces of two vector bundles homeomorphic" to "Are the associated sphere bundles homeomorphic"? In the case of $TS^2$ and $S^2\times R^2$ it's not too difficult to show that the sphere bundles are not homeomorphic, and I'm wondering if there's a way to leverage that. Thanks, Zygund
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This is more or less equivalent to Ryan's comment but with more details and a slightly different point of view. Let $X$ be the total space of the tangent bundle, and put $Y=S^2\times\mathbb{R}^2$. If $X$ and $Y$ were homeomorphic, then their one-point compactifications would also be homeomorphic. We will show that this is impossible by considering their cohomology rings. Put $X'=\{(p,q)\in S^2\times S^2 : p+q\neq 0\}$. There is a homeomorphism $f:X\to X'$ given by $f(u,v)=((\|v\|^2-1)u+2v)/(\|v\|^2+1)$ (a variant of stereographic projection). It follows that $X_\infty$ can be obtained from $S^2\times S^2$ by collapsing out the antidiagonal. We have $H^*(S^2\times S^2)=\mathbb{Z}[a,b]/(a^2,b^2)$ and it follows that $H^*(X_\infty)$ is the subring generated by $1$, $a+b$ and $ab$. In particular, the squaring map from $H^2$ to $H^4$ is nonzero. However, $Y$ can be identified with $(S^2\times S^2)\setminus (S^2\times\{point\})$, so $H^*(Y_\infty)$ is isomorphic to the subring generated by $1$, $a$ and $ab$, so the squaring map $H^2\to H^4$ is zero. Note that the tangent bundle plus a rank-one trivial bundle is trivial, so the suspensions of $X_\infty$ and $Y_\infty$ are homeomorphic.
|
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|
92,422 |
Question. Suppose that $M$ is a closed connected topological manifold and $G$ is its group of homeomorphisms (with compact-open topology). Does $G$ (as a topological group) uniquely determine $M$ ? One can ask the same question where we regard $G$ as an abstract group (ignoring topology), replace topological category by smooth category (here one can equip $G=\mathrm{Diff}(M)$ with a finer structure of a Frechet manifold), varying degree of smoothness, dropping compactness assumption, recovering $M$ up to homotopy, etc. I do not know how to answer any of these questions. I do not even know if one can recover the dimension of $M$ from its group of homeomorphisms. In low dimensions, or assuming that $M$ has a locally-symmetric Riemannian metric, and if $\dim(M)$ is given, I know few things. For instance, among 2-dimensional manifolds one can recover $M$ from $G$ since $G/G_0$ is the mapping class group $Mod(M)$ of $M$ and one can tell the genus of $M$ from maximal rank of free abelian subgroups of $Mod(M)$ . Same for, say, closed hyperbolic manifolds with non-isomorphic isometry groups. However, given, for instance, two closed hyperbolic 3-manifolds $M_1, M_2$ with trivial isometry groups, I do not know how to distinguish $M_i$ 's by, say, $\mathrm{Homeo}(M_i)$ (the problem reduces to a question about homeomorphism groups of the unit ball commuting with $\pi_1(M_i)$ , $i=1,2$ , but I do not see how to solve it). Update: Results quoted by Igor and Martin give the complete answer in topological and smooth category in the strongest possible form (much more than I expected!). Positive answer is also known in the symplectic category, but, apparently, is open for contact manifolds and their groups of contactomorphisms. Another reference in the smooth case, sent to me by Beson Farb is Banyaga, Augustin , The structure of classical diffeomorphism groups, Mathematics and its Applications (Dordrecht). 400. Dordrecht: Kluwer Academic Publishers. xi, 197 p. (1997). ZBL0874.58005 .
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Answer is: Yes, one can recover $M$ if it is a compact manifold. See J. V. Whittaker: On Isomorphic groups and homeomorphic spaces, Annals of Math 1963. EDIT Actually, one knows a lot more, see, for example
Tomasz Rybicki
Journal: Proc. Amer. Math. Soc. 123 (1995), 303-310.
MSC: Primary 58D05; Secondary 17B66, 22E65, 57R50
MathSciNet review: 1233982 And references therein... ANOTHER EDIT A quite different proof of a stronger theorem (actually a large set of theorems) than Whittaker's (actually, Whittaker's paper seems to be rather badly written) is given by Matatyahu Rubin in
Rubin, Matatyahu(3-SFR)
On the reconstruction of topological spaces from their groups of homeomorphisms.
Trans. Amer. Math. Soc. 312 (1989), no. 2, 487–538.
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|
92,439 |
A recent article in the New York Times, http://www.nytimes.com/2012/03/27/science/emmy-noether-the-most-significant-mathematician-youve-never-heard-of.html?pagewanted=all says, among other things, "Noether was a highly prolific mathematician, publishing groundbreaking papers, sometimes under a man’s name, in rarefied fields of abstract algebra and ring theory." This is the first I have ever heard of Emmy Noether publishing under a male pseudonym, and I ask whether anyone can confirm, or refute, the assertion in the Times. I wonder if the author is confusing Emmy with her mathematician father Max; or if the author has in mind times when Noether gave lectures that were advertised as Hilbert's; or if the author has in mind Sophie Germain, who wrote under the name M. LeBlanc. EDIT: I have an answer from the writer, and it appears that Zsban hit the nail on the head in a comment. The writer says her point was badly phrased, and she was referring to Noether's letting (male) students and colleagues publish her ideas as if those ideas were their own. My thanks to all who have contributed here.
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I have a copy of her biography, Emmy Noether, 1882-1935 by Auguste Dick (translated to English by H.I. Blocher). Appendix A contains a list of 43 publications, apparently complete, and not one is indicated as being published pseudonymously. Of course a few had male co-authors, but that is not the same at all. Also, I skimmed the text of the book and could find no reference to such a thing. If Natalie Angier, the author of the New York Times article, is aware of a pseudonymous Noether paper, she would seem to be the only one. I agree with Allen Knutson that a letter to the paper's corrections department is in order.
|
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|
92,451 |
Simpliefied setup. Assume I am given some function f(t).
I know that it is constructed as $f(t) = \sum_{k=1...M} C_k exp(2 \pi~ i~ w_k t ) + noise(t)$.
where $noise(t)$ is some random set of numbers depending on $t$ (white noise if you like).
I need to estimate $C_k$ and $w_k$. Solution: I will make Fourier Transform $\hat f(\lambda)$ and will pick up those $\lambda= w_k$ such that $\hat f(\lambda) > threshold$.
How should I choose this threshold ? (The question is probably not 100% well-posed, but hope idea is clear). True setup. Now assume that $t$ is discrete e.g. $t= l/N$, for some $N$ and $l=1...N$. And the problem is that $w_k$ are do not of the form $l/N$. Difficulty So the functions $exp(2 \pi~ i~ w_k t )$ are NOT orthogonal on the
discrete set $t=l/N$. So there are some problems with making Fourier transform approach.
|
I have a copy of her biography, Emmy Noether, 1882-1935 by Auguste Dick (translated to English by H.I. Blocher). Appendix A contains a list of 43 publications, apparently complete, and not one is indicated as being published pseudonymously. Of course a few had male co-authors, but that is not the same at all. Also, I skimmed the text of the book and could find no reference to such a thing. If Natalie Angier, the author of the New York Times article, is aware of a pseudonymous Noether paper, she would seem to be the only one. I agree with Allen Knutson that a letter to the paper's corrections department is in order.
|
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|
92,454 |
I'm currently reading The book of Jacob Lurie, 'Higher Topos Theory', and I'm a little confused by the relation between classical topos and $\infty$-topos : to an $\infty$-topos I can attach the ordinary topos of its $0$-truncated objects. And to a classical topos I have several way to associate $\infty$-topos 'above' it. Jacob Lurie (in his book, section 6.4) present this relation as similar to the relation between a classical topos and its locale of sub-terminal objects. In this situation, I know I can have plenty of topoi (a proper class) that are associated to the same locale, even if this locale is just a point. But I have no idea of what happens in the case of $\infty$-Topos : I have seen that in some cases there might be several non equivalent $\infty$-topos above a same ordinary topos, but I see them more like "different ways of doing homotopy theory in the internal logic of $X$ because some classical result of homotopy theory (like Whitehead's theorem) may fail in the internal logic" rather than completely different objects that just share a small property" (like the class of topoi whose locale of subterminal objects is reduced to a point is just the class of topos whose internal logic is two-valued) So for example :
Are there several (non equivalent) $\infty$-topoi, whose topos of $0$-truncatued objects is the topos of set ? If it's the case, can I have an example ? Are we able to 'classify' them ?
|
For any space $X$, there's an $\infty$-topos of spaces fibered over $X$. The underlying
ordinary topos is the category of representations of the fundamental groupoid of $X$.
So if $X$ is simply connected, this is just the category of sets. But the $\infty$-topoi are different for different values of $X$ (two spaces $X$ and $Y$ yield equivalent $\infty$-topoi if and only if $X$ and $Y$ are homotopy equivalent).
|
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|
92,624 |
For $\mathbb{C}$-valued functions, why are $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar{z}}$ defined as
$$
\frac{\partial}{\partial z}=
\frac{1}{2}\left(
\frac{\partial}{\partial x} - i
\frac{\partial}{\partial y}
\right)
\quad \quad
\frac{\partial}{\partial \bar{z}}=
\frac{1}{2}\left(
\frac{\partial}{\partial x} + i
\frac{\partial}{\partial y}
\right)
$$
where $x$ and $y$ are the real and imaginary parts, respectively. It seems to be me that the inventor of this notation could have easily reverse the signs: Let $\frac{\partial}{\partial z}$ have a $+$ in the middle and let $\frac{\partial}{\partial \bar{z}}$ have a $-$ in the middle instead. Is there any reason that they are defined the way they are now? The same question extends to the $\partial$ and $\bar{\partial}$ operators. One possible guess is that in complex analysis, one usually works with holomorphic functions, so one operator is used much more often than the other. We probably want the more frequently used one to be the one that is easier to write. But I really doubt this is the reason. After all, it's just one extra stroke.
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If $f:\mathbb{C}\to\mathbb{C}$ is any smooth function and $z\in\mathbb{C}$, the derivative $df_z$ of $f$ at $z$ is a $\mathbb{R}$-linear operator from the tangent space $T_{z}\mathbb{C}$ to $\mathbb{C}$ (and $T_{z}\mathbb{C}$ can of course be canonically identified with $\mathbb{C}$ since $\mathbb{C}$ is a vector space). Said differently, $df$ is a complex-valued differential $1$-form on $\mathbb{C}$ (i.e. an element of the space $\Omega^1(\mathbb{C};\mathbb{C})=\Omega^1(\mathbb{C})\otimes_{\mathbb{R}} \mathbb{C}$). Any $\mathbb{R}$-linear operator $A$ from one complex vector space to another can be written uniquely as $A=B+C$, where $B$ is complex-linear (i.e. $B(iv)=iBv$ for all $v$) and $C$ is complex anti-linear (i.e. $B(iv)=-iBv$). Specifically, let $Bv=\frac{1}{2}(Av-iAiv)$ and $Cv=\frac{1}{2}(Av+iAiv)$. The operators $\partial$ and $\bar{\partial}$ can then be characterized as follows: for a function $f:\mathbb{C}\to\mathbb{C}$, one has the unique decomposition $$ df=\partial f+\bar{\partial} f $$ where the complex-valued $1$-forms $\partial f$ and $\bar{\partial}f$ are such that, at each $z\in\mathbb{C}$, $(\partial f)_z$ is complex-linear and $(\bar{\partial} f)_z$ is complex anti-linear. At least to me that justifies the notation--it makes more sense to have $\partial f$ be the linear part of $df$ and $\bar{\partial} f$ be the antilinear part than the other way around. As for $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar{z}}$, note first that as a special case of the above discussion one has functions $z:\mathbb{C}\to\mathbb{C}$ (which is just the identity) and $\bar{z}:\mathbb{C}\to\mathbb{C}$, and therefore complexified one-forms $dz,d\bar{z}\in\Omega^1(\mathbb{C};\mathbb{C})$. (More specifically, since $z=x+iy$ and $\bar{z}=x-iy$, one has $dz=dx+idy$ and $d\bar{z}=dx-idy$, so for $dz$ and $d\bar{z}$ the signs on $i$ are what you wanted them to be.) At each point of $\mathbb{C}$, $\{dz,d\bar{z}\}$ is a basis (over $\mathbb{C}$) for the complexified cotangent space. The operators $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial\bar{z}}$ are then naturally defined as the complexified vector fields such that at every point the basis $\{\frac{\partial}{\partial z},\frac{\partial}{\partial\bar{z}}\}$ for the complexified tangent space is the dual basis to the basis $\{dz,d\bar{z}\}$ for the complexified cotangent space. Imposing this dual basis requirement, the formulas $dz=dx+idy$ and $d\bar{z}=dx-idy$ then readily yield the standard formulas for $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial\bar{z}}$, and give rise to the pleasant identity, for any smooth $f:\mathbb{C}\to\mathbb{C}$, $$
df=\partial f+\bar{\partial} f=\frac{\partial f}{\partial z}dz+\frac{\partial f}{\partial{\bar{z}}}d\bar{z} $$ just like one would get if $z$ and $\bar{z}$ were real coordinates on a real two-manifold.
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|
92,696 |
Can you make an example of a great proof by induction or construction by recursion? Given that you already have your own idea of what "great" means, here it can also be taken to mean that the chosen technique : is vital to the argument; sheds new light on the result itself; yields an elegant way to fulfill the task; conveys a powerful and simple view of an intricate matter; is just the only natural way to deal with the problem. Here induction and recursion are meant in the broadest sense of the words, they can span from induction on natural numbers to well-founded recursion to transfinite induction, and so on... Elementary examples are especially appreciated, but non-elementary ones are welcome too!
|
A Classic: Fix a positive integer $n$. Show that it is possible to tile any $2^n \times 2^n$ grid with exactly one square removed using 'L'-shaped tiles of three squares. It serves as a wonderful introductory example to proof by induction. Indeed, the proof can almost be represented with two appropriate figures. Yet, for those just learning induction, it is a significant problem where the application of the inductive hypothesis is far from obvious.
|
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|
92,755 |
It is well known that most topological spaces can be studied via their algebra of continuous real-valued (or complex-valued) functions. For instance, in the setting of compact Hausdorff spaces, there is a complete dictionary between topological properties of the space $X$ and corresponding algebraic properties of the algebra $C(X)=C(X,\mathbb{R})$, and two such spaces $X$ and $Y$ are homeomorphic if and only if their algebras $C(X)$ and $C(Y)$ are isomorphic. A similar theory is available in the locally compact Hausdorff case, by replacing $C(X)$ with the algebra of real-valued functions on $X$ which vanish at infinity. Furthermore, these algebras are precisely all the commutative C*-algebras (at least if you take complex-valued functions). Thus if one wishes to study locally compact Hausdorff spaces, he might as well study commutative C*-algebras. Is there an analogous algebraic approach to the theory of "nice" metric spaces (e.g. compact, locally compact, connected, etc.)? More specifically, is there a natural function space one can attach to every nice metric space which essentially contains all data about the metric space, up to isometry? I'll welcome any references related to this question. EDIT: Many of the answers referred me to Nik Weaver's (and others') theory of "Lipschitz algebras". I took a look at his book and while very interesting (and well written), it doesn't seem to be quite what I am looking for. More specifically, it seems that his constructions of Lipschitz algebras only characterize the metric space up to bi-Lipschitz equivalence (which he calls a quasi-isometry), while I'm looking for natural function spaces that would characterize the space up to isometry. His results in section 1.8 of the book show that two (pointed) complete metric spaces have isomorphic Lipschitz function algebras if and only if they are bi-Lipschitz equivalent, rather than isometric. He does have a characterization up to isometry (using a different Lipschitz algebra) for the class of metric spaces of diameter less than 2 (section 1.7) , but his reduction of a general metric space (say complete) to a metric space of diameter $\le 2$ is of course far from preserving isometries (rather preserving Lipschitz mappings). So it seems that I am looking for a somewhat more "rigid" construction. Any ideas about that? Thanks.
|
A very complete reference is the book "Lipschitz Algebras" by Nik Weaver. In there you will find various types of spaces of Lipschitz functions that can be associated to a metric space, and several results of the kind you are asking about. From the book's introduction: Thus, there is a robust duality
between metric properties of $X$ and
algebraic properties of $Lip_0(X)$.
The set of weak* continuous
homomorphisms from $Lip_0(X)$ into the
scalars can be isometrically
identified with the completion of $X$
(Theorem 4.3.2); Lipschitz maps from
$X$ into $Y$ correspond to weak*
continuous homomorphisms from
$Lip_0(Y)$ into $Lip_0(X)$ (Corollary
4.2.9); closed subsets of $X$ correspond to weak* closed ideals of
$Lip_0(X)$ (Theorem 4.2.5); and
nonexpansive images of $X$ correspond
to weak* closed, self-adjoint
subalgebras of $Lip_0(X)$ (Theorem
4.1.10).
|
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92,771 |
Let $X$ be a smooth projective variety over $\mathbb{C}$. The following information is all equivalent (any of these numbers can be computed by a linear equation from any of the others): the arithmetic genus of $X$ the constant coefficient of the Hilbert polynomial of $X$ $\chi(X, \mathscr{O}_X)$ the "Todd genus" $\int_X \operatorname{td}(T_X)$, where $T_X$ is the tangent bundle of $X$ and $\operatorname{td}$ denotes the Todd class. Is there a geometric characterization for any of these numbers? If I understand correctly, characteristic classes (and in particular, Todd classes) can be defined entirely from the topology of $X$, or at least its structure as a smooth manifold. [ Edit : This is not true --see the answer of "anonymous." If I understand correctly, the Todd class of a complex vector bundle is a smooth invariant. However, different complex structures on the same real manifold $X$ can give rise to non-isomorphic complex vector bundle structures on $T_X$; in fact, a complex vector bundle structure on $T_X$ is, by definition, an almost complex structure on a real manifold $X$.] Thus, in some sense, item 4 provides a "geometric characterization" for the arithmetic genus of $X$ (and the other items on the list). However, I personally find this description so far abstracted from actual geometric properties of $X$ as to be hardly geometric at all. (If anyone disagrees with me and can articulate a geometric intuition for the Todd genus, that would be a reasonable answer.) By comparison, I do consider the following characterizations of various properties "geometric": The self-intersection number of the diagonal embedding of $X$ into $X \times X$. (the Euler characteristic) The number of points in which a general linear space of complementary dimension meets $X \subset \mathbb P^n$. (the degree of $X \hookrightarrow \mathbb P^n$) The genus of the curve $X \cap L$, where $L$ is a general linear space of dimension one greater than $\operatorname{codim} X$. (I don't know of a standard name for this, but in a particular sense, it is one of the coefficients of the Hilbert polynomial of $X$.) The maximum number of copies of $S^1$ that can be removed from $X$ without disconnecting it. (the genus of $X$ if $X$ is a smooth curve, i.e., Riemann surface) Note that either the first or the last point gives a geometric characterization for (information equivalent to) the genus of a curve. Without one of these, I would not consider the third bullet a "geometric characterization" of anything. In a way, this provides part of my motivation for asking this question. Let $L_k$ be a general linear space of codimension $k$ in $\mathbb P^n$. Unless I am mistaken, knowing the Hilbert polynomial for $X$ is equivalent to knowing the arithmetic genus of $X \cap L_k$, for every $k \leq n$ such that this intersection is nonempty, via the formula
$$ \chi(\mathscr O_X(n)) = \sum_{k \ge 0} \chi(\mathscr O_{X \cap L_k}) \binom{n+k-1}{k}\;\text.$$ Thus, a geometric characterization for arithmetic genus would automatically give a geometric characterization for the Hilbert polynomial. (Again, in some sense, this is already provided by the Hirzebruch-Riemann-Roch Theorem; but I find this formula so abstracted as to be hardly geometric at all.)
|
First let me note that there is an unfortunate clash in terminology: the arithmetic genus of a smooth complex projective variety $X$ of dimension $n$ can mean either a) The number $\chi (X, \mathcal O_X)$: the Hirzebruch arithmetic number in which you are interested . b) The number $p_a(X)=(-1)^n(\chi (X, \mathcal O_X)-1)$, the Severi arithmetic genus, which has historical precedence but of course was defined non-cohomologically. For example, for projective space we have $\chi (\mathbb P^n, \mathcal O_{\mathbb P^n})=1$ but $p_a (\mathbb P^n)=0$. Hirzebruch introduced his definition mainly because it has the powerful multiplicativity property $$\chi (X\times Y,\mathcal O_{X\times Y})= \chi (X,\mathcal O_{X})\cdot \chi ( Y,\mathcal O_{ Y})$$ which certainly is a step toward the geometric interpretation you are seeking. Another step in the right direction is that for a finite covering $X\to Y$ of degree $d$ we have the pleasant relation $\chi (X,\mathcal O_{X})=d\cdot \chi ( Y,\mathcal O_{ Y})$. But the most important geometric property is that $\chi (X,\mathcal O_{X})$ is a birational invariant, because each number $dim_\mathbb C H^i(X,\mathcal O_{X})$ is already a birational invariant. Arithmetic genus is reasonably easy to compute: for a hypersurface $H\subset \mathbb P^n$ of degree $d$ you have $p_a(H)=\binom {d-1}{n}$, which for $n=2$ gives the well-known elementary formula $p_a(C)=\frac {(d-1)(d-2)}{2}$ for the plane curve $C$. [This formula (and others) can be found in Hartshorne, Chapter I, Exercise 7.2, page 54] For a surface you have Max Noether's formula $\chi (S, \mathcal O_S)=\frac {c_1^2(S)+c_2(S)}{12}$, where $c_2(S)$ (=second Chern class of $S$) is also the purely topological Euler-Poincaré characteristic of $S$, equal to the alternating sum of the Betti numbers of the underlying toplogical space.$S_{top}$. Finally, Fulton has given an axiomatic characterization of the arithmetic genus in algebraic geometry over an arbitrary algebraically closed field here . In a sense it may be considered an explanation of the geometric significance of the arithmetic genus: if you want it to satisfy certain geometric properties, the definition is forced upon you. Edit (added by Charles with Georges's permission):
Fulton's axiomatic characterization may be described as follows: There is a unique assignment of a number $\mathcal{A}(X)$ to every [smooth irreducible projective variety over a fixed algebraically closed field] (hereafter simply "variety"), such that the following three axioms are satisfied: $\mathcal{A}$ respects isomorphism classes. If $X$ is a point, then $\mathcal{A}(X) = 1$. Let $X$, $Y$, and $Z$ be (smooth) varieties of the same dimension.
Suppose that $X$, $Y$, and $Z$ can be embedded as codimension-one subvarieties
of a common (smooth) variety $W$, such that $X$ and $Y+Z$ are linearly equivalent as divisors in $W$, and $Y$ and $Z$ intersect transversely in a disjoint union of (smooth) varieties
$V_1, \dotsc, V_{\ell}$. Then $$\mathcal{A}(X) = \mathcal{A}(Y) + \mathcal{A}(Z) - \sum_i \mathcal{A}(V_i).$$ This assignment takes $X$ to its "Hirzebruch arithmetic number" $\mathcal{A}(X) = \chi(X, \mathcal{O}_X)$.
|
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|
92,813 |
It is well known that a (real) vector bundle $\pi : E\to B$ over a topological space (or manifold) $B$ is a fibre bundle whose fibres $$F=\pi^{-1}(x), \ \ \ x\in B $$ over any $x\in B$, are diffeomorphic to a vector space $V$. On the other hand, a principal $G$-bundle is a fibre bundle $\pi : P\to B$ over $B$ with a right free action of a Lie group $G$ on $P$ such that for any open set $U\subset B$, the locally trivial fibrations defined by:
$$
\Phi_{U} : \pi^{-1}(U)\to U\times G, \ \ \Phi_{U}(p)=(\pi(p), \varphi_{U}(p)).
$$
Here
$$
\ \varphi_{U} :\pi^{-1}(U)\to G
$$
is a $G$-equivariant map, that is $\varphi_{U}(pg)=\varphi_{U}(p)g$, for all $p\in\pi^{-1}(U)$ and $g\in G$. In the last case the fibers are submanifolds of $P$ which are always diffeomorphic with the structure group $G$. Although for any vector bundle $\pi : E\to B$, its fibers $F\cong V$ can be considered as Lie group with operation the vector addition, in general we do not include the vector bundles
as examples of principal $G$-bundles (Although, to every vector bundle we can associate the frame bundle which is a ${\rm GL}_{n}\mathbb{R}$-principal bundle, but I don't speak here about associated bundles ). My question is about a good explanation about the fact that IN GENERAL vector bundles (themselves) do not give examples of principal $G$-bundles . For example, the tangent bundle $TM$ of a smooth manifold is a prototype example of a vector bundle, but itself it cannot be considered as a principal bundle for a Lie group $G$, is this true?
Thus I ask: Which is the basic difference between a vector bundle an a $G$-bundle, which does not allows us (almost always??), to consider the vector bundles themselves as examples of $G$-bundles? For example, a $G$-bundle is trivial (isomorphic to the product bundle), if and only if it admits a global section , but I think that this is not true for vector bundles . A second question is about examples of vector bundles which can be considered the same time as $G$-bundles for some Lie group (I think that such an example is a cylinder)
|
The difference is that, for a vector bundle, there is usually no natural Lie group action on the total space that acts transitively on the fibers. The fact that all of the fibers are, individually Lie groups, doesn't mean that there is a Lie group that acts on the whole space, restricting to each fiber to be a simply transitive action. The simplest example of this is the nontrivial line bundle over the circle. Another example is the tangent bundle of $S^2$.
|
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|
92,818 |
As usual I consider a semigroup to be a structure $(A, +)$ such that $+$ is an associative binary function over the set $A$. The notion of linearly-ordered semigroup corresponds to structures of the form $(A, + , \leq)$ such that $(A,+)$ is a semigroup, $\leq$ is a linear order on $A$, and this order $\leq$ is compatible with the binary operation $+$ (i.e., if $a \leq b$ and $a' \leq b'$, then $a + a' \leq b + b'$). I am interested on known answers to the following questions (they follow the same pattern): Is there some "useful" characterization of semigroups which can be linearly ordered? To be more precise, for which semigroups $(A, +)$ there is a linear order $\leq$ such that $(A, +, \leq)$ is a linearly ordered semigroup? Is there some "useful" characterization of commutative semigroups which can be linearly ordered? Perhaps it is worth pointing out that for the case of commutative groups it is well known that the criteria for admitting a linear order coincides with being torsion-free.
|
The difference is that, for a vector bundle, there is usually no natural Lie group action on the total space that acts transitively on the fibers. The fact that all of the fibers are, individually Lie groups, doesn't mean that there is a Lie group that acts on the whole space, restricting to each fiber to be a simply transitive action. The simplest example of this is the nontrivial line bundle over the circle. Another example is the tangent bundle of $S^2$.
|
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"https://mathoverflow.net/questions/92818",
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|
92,939 |
Just a new guy in optimization. Is it true that all convex optimization problems can be solved in polynomial time using interior-point algorithms?
|
No, this is not true (unless P=NP). There are examples of convex optimization problems which are NP-hard.
Several NP-hard combinatorial optimization problems can be encoded as convex optimization problems over cones of co-positive (or completely positive) matrices. See e.g. " Approximation of the stability number of a graph via copositive programming ", SIAM J. Opt. 12(2002) 875-892 (which I wrote jointly with Etienne de Klerk). Moreover, even for semidefinite programming problems (SDP) in its general setting (without extra assumptions like strict complementarity) no polynomial-time algorithms are known, and there are examples of SDPs for which every solution needs exponential space.
See Leonid Khachiyan, Lorant Porkolab. " Computing Integral Points in Convex Semi-algebraic Sets ". FOCS 1997: 162-171 and Leonid Khachiyan, Lorant Porkolab " Integer Optimization on Convex Semialgebraic Sets ". Discrete & Computational Geometry 23(2): 207-224 (2000). M.Ramana in " An Exact duality Theory for Semidefinite Programming and its Complexity Implications " Mathematical Programming, 77(1995) shows that SDP lies either in the intersection of NP and co-NP, or outside the union of NP and coNP, and nothing better than this is known. In " Semidefinite programming and arithmetic circuit evaluation " Discrete Applied Mathematics, 156(2008) Sergey P. Tarasov and Mikhail N. Vyalyi show that SDP can be used to compare numbers represented by arithmetic circuits. (The latter is regarded as one of hard problems).
|
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"https://mathoverflow.net/questions/92939",
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|
93,034 |
I am currently trying to understand the notion of criticality (as discussed, e.g., in Terence Tao's book on nonlinear dispersive equations) from a physical viewpoint. That's why i'm interested in the question which PDE from physics (apart from the Navier-Stokes equations) and geometry are supercritical with respect to some symmetry and all (known) controlled quantities.
|
Generally speaking, supercriticality occurs when the dimension and/or the nonlinearity exponent is sufficiently large. Sigma field models such as the harmonic map, wave map, or Schrodinger map equations become supercritical in three and higher spatial dimensions. (The critical two-dimensional case is probably the most interesting.) Einstein's equations of general relativity also becomes supercritical in three and higher spatial dimensions. The closely related Ricci flow used to be supercritical in three and higher dimensions, until Perelman discovered some new scale-invariant controlled quantities; now I would classify it as critical in three dimensions at least, and possibly in higher dimensions (though it is not as clear there whether Perelman's quantities are coercive enough to fully control the dynamics at small scales). (In general, elliptic and parabolic equations can defy to some extent the criticality classification arising from dimensional analysis, due to powerful monotonicity formulae such as those arising from the maximum principle.) Yang-Mills equations are supercritical in five and higher spatial dimensions, and similarly for related equations such as the Maxwell-Klein-Gordon equations. (Yang-Mills theory becomes particularly interesting in the critical four-dimensional case, what with its instantons, self-dual and anti-self-dual solutions, etc.) For pure power nonlinearity interactions (with a term of the form $|\phi|^p$ in the Hamiltonian), one typically has supercriticality once the exponent p becomes large enough, although the precise threshold of p depends on the dimension and on the precise model. For example, $|\phi|^4$ models generally become supercritical in five and higher spatial dimensions. Milder nonlinearities, such as Hartree-type nonlinearities, tend to be less supercritical than power nonlinearities. Navier-Stokes is supercritical in three and higher dimensions. One can certainly perform the relevant dimensional analysis on other fluid equations (e.g. quasi-geostrophic), but I don't recall the exact numerology off-hand. But in the absence of viscosity (e.g. for the Euler equations), there is now a two-parameter family of scaling invariances, and there is not really a well-defined notion of criticality, subcriticality, or supercriticality in this case. For systems of coupled equations (e.g. Zakharov type models) for which there is no natural scaling, it becomes more difficult (and perhaps even impossible) to cleanly make the division into subcritical, critical, and supercritical equations; the distinction is most useful for simplified model equations.
|
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|
93,282 |
This was sparked because I wanted to compute $\pi_2(Sym^2(\Sigma_2))$ via $Sym^2(\Sigma_2)\approx \mathbb{T}^4$# $\bar{\mathbb{C}P}^2$. We know how to compute $\pi_1$ of $M$ # $N$ via van-Kampen's theorem. But what about higher homotopy groups? I looked in the literature and google without luck, and so I am wondering if no such procedure exists. Are there any results for calculating $\pi_n$ of connected sums? There was mention of "higher van Kampen theorem"... has this actually been used to do such computations? I'd be interested in references if not just examples.
|
The 2nd homotopy group of a connect sum is fairly reasonable to compute. $\pi_i X$ is isomorphic to $\pi_i \tilde X$ provided $i \geq 2$ and $\tilde X$ indicates any covering space of $X$, so we might as well take the universal cover. By the Hurewicz theorem, $\pi_2 \tilde X$ is isomorphic to $H_2 \tilde X$. In the case of a connect sum, the universal cover has a very nice description (take disjoint unions of the universal covers of the punctured manifolds and glue them together appropriately). Since $\mathbb CP^2$ is simply connected this is a fairly easy thing to compute. The universal cover looks like $\mathbb R^4$ with a $\mathbb CP^2$ summand at every integer lattice point. So, $$\pi_2 ((S^1)^4 \# \mathbb CP^2) \simeq \bigoplus_{\pi_1 T^4} \pi_2 \mathbb CP^2$$ i.e. a direct sum over $\mathbb Z^4$ of copies of the integers, i.e. $\mathbb Z[t_1^\pm, t_2^\pm, t_3^\pm, t_4^\pm]$ a laurent polynomial ring in four variables. $\pi_1$ acts by multiplication by units in the Laurent polynomial ring. Higher homotopy groups in general can be fairly painful to compute but $\pi_2$ is usually quite reasonable, like this case.
|
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|
93,289 |
$\DeclareMathOperator{\Spec}{Spec}$
[Edit] Martin pointed out that $\dim A = 0$ does not imply that $\Spec A$ is discrete. Therefore I changed the wording of question 2.[/Edit] With dimension of a ring I mean the Krull-dimension. It is well-known that for a commutative ring $A$ the following are equivelent $A$ is noetherian and $\dim A = 0$; $A$ is artinian. It is easy to think of noetherian rings that are not artinian ($\mathbb{Z}$). However I cannot find an example of a $0$-dimensional ring that is not artinian. Questions What is an example of a commutative ring $A$ with $\dim A = 0$ that is not artinian (or equivalently, not noetherian)? A related question is: Give an example of an affine scheme $X$, such that $X$ is discrete as topological space, but $\mathcal{O}_X(X)$ is not noetherian/artinian. Yet another question: Why does the converse of proposition 8.3 in Atiyah-MacDonald fail for a ring $A$ with $\dim A = 0$? (The proposition says that artinian rings have finitely many maximal ideals.) I have tried various constructions, but they all fail somehow.
|
Take any compact totally disconnected Hausdorff space $X$ (for example the Cantor set, or the one-point compactification of $\mathbb{N}$). Then $\mathcal{C}(X,\mathbb{F}_2)$ is a ring whose spectrum is homeomorphic to $X$. In particular, this ring is zero-dimensional, but this ring is noetherian iff $X$ is finite. More generally, a commutative ring is called von Neumann regular when for every $x$ we have $x^2 | x$ (in particular, boolean rings qualify). Equivalently, every localization at a prime ideal is a field. In particular, they are zero-dimensional (in fact, they are precisely the reduced zero-dimensional rings). It is easy to check that these rings are closed under infinite products. In particular, an infinite product of fields is a zero-dimensional ring, which is not noetherian. If the index set is $I$, the spectrum is the space of ultrafilters on $I$. EDIT: It is even more trivial to give non-reduced examples. If $V$ is any $k$-module, then $A=k \oplus V$ is a $k$-algebra (with $V^2=0$). Then $A_{\mathrm{red}}=k$ is a field, in particular $\mathrm{Spec}(A)$ is just a single point. If $V$ is not noetherian as a module, it is clear that $A$ won't be noetherian as a ring.
|
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|
93,330 |
Consider a compact Riemann surface $X$ of genus $g$. It is well-known that its fundamental group $\pi_1(X)$ is the free group on the generators $a_1,b_1,...,a_g,b_g$ divided out by the normal subgroup generated by the single relator $[a_1,b_1]\cdot \ldots\cdot [a_g,b_g]$. (This has of course nothing to do with the complex structure of $X$, but may be computed by considering the underlying topological manifold as a cell complex.) This group is trivial for $g=0$ and free abelian on two generators for $g=1$. For $g\geq 2$, however, I had always taken for granted that it is not free but I have just realized that I cannot prove that. So, although I guess the answer is no, I'll ask my official question in an open way : Is $\pi_1(X)$ free for $g\geq 2$ ? Edit: Users have now brilliantly solved the problem in multiple ways. Non-freeness is definitely established, with 12 proofs plus sketches of proofs in the comments! It is clearly impossible to select in a reasonable way an answer for "acceptance" among all these great answers . Since the software forces me to make only one choice, I have chosen Daniel's answer because of its merit, but also because it acknowledges Vitali's contribution: Vitali was the first to sketch a solution (in the comments) .
|
As per Theo's request, I'm posting this as an answer, though it's largely an expansion on Vitali's comment. Let $F_n$ be the free group on $n$ letters; $K(F_n, 1)$ is a wedge of $n$ circles and so has vanishing cohomology in degrees $>1$. On the other hand, if $X$ is a compact Riemann surface of genus $g>1$, $X$ is a $K(\pi_1(X), 1)$ as its universal cover is the upper-half plane, which is contractible. But then $$H^2(\pi_1(X), \mathbb{Z})=H^2_{sing}(X, \mathbb{Z})=\mathbb{Z},$$ which is non-zero. In particular, $\pi_1(X)$ has non-vanishing cohomology in degree $2$ and so is not free, as Vitali says.
|
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|
93,828 |
Friedman, in _ Lectures notes on enormous integers shows that TREE(3) is much larger than n(4), itself bounded below by $A^{A(187195)}(3)$ (where $A$ is the Ackerman function and exponentiation denotes iteration). But actually, using the fast-growing hierarchy, $n(p)$ is smaller than $f_{\omega^{\omega^\omega}}(p)$ , shown by Friedman in Long finite sequences , Journal of Combinatorial Theory, Series A 95 Issue 1 (2001) pp. 102-144, doi: 10.1006/jcta.2000.3154 , author pdf while it seems that TREE grows faster than $f_{\Gamma_0}$ ( ${\Gamma_0}$ being the Feferman-Schütte ordinal). So it could well be that in fact TREE(3) is larger than, say, n(n(4)), or even any number expressible by iterations of n. What is known on this question? For reference, I should have added that TREE(3) is the incredibly (at first, or even second look) large answer to the question : "which is the length of the longest sequence $(T_2,T_3,T_4,\dots,T_n)$ of labeled trees such that $T_k$ has at most $k$ nodes labeled $a$ or $b$ , and $T_i$ is not a subtree of $T_j$ for $i < j$ ?". Here, trees are rooted trees, and are treated as poset on their sets of vertices. A tree $T$ is called a subtree of $T'$ if there is an inf-preserving embedding from $T$ into $T'$ , (that is, an injective map $h:Vertices(T) \to Vertices(T')$ such that $h(\inf(x,y)) = \inf((h(x), h(y))$ ) that respects the labeling by $a$ or $b$ .
|
I believe I can state with some confidence that TREE(3) is larger than $f_{\vartheta (\Omega^{\omega}, 0)} (n(4))$ , given a natural definition of $f$ up to $\vartheta (\Omega^{\omega}, 0)$ . I can state with certainty that TREE(3) is larger than $H_{\vartheta (\Omega^{\omega}, 0)} (n(4))$ , where H is a certain version of the Hardy hierarchy. To obtain this result, I will first define a version of TREE(n) for unlabeled trees: Let tree(n) be the length of the longest sequence of unlabeled rooted trees $T_1, T_2, \ldots, T_m$ such that $T_i$ has less than or equal to $n+i$ vertices and for no $i, j$ with $i < j$ do we have $T_i$ homeomorphically embeddable into $T_j$ . (Note the term "embeddable" rather than "subtree"; the terms are different, and I believe using "subtree" would lead to infinite sequences.) In order to obtain a long sequence of trees, we will define a well-order on unlabeled rooted trees. This definition will be by induction on the sum of the heights of the two trees being compared. Define an immediate subtree of a rooted tree $T$ to be a full subtree starting at one of its children. Given two rooted trees $S, T$ , we define $S = T$ if the two trees are identical. We define $S \leq T$ if $S = T$ or $S < T$ . Given two rooted trees $S, T$ , we define $ < $ as follows. Say $S < T$ if $S \leq T_i$ , where $T_i$ is an immediate subtree of $T$ . Similarly, say $T < S$ if $T \leq S_i$ , where $S_i$ is an immediate subtree of $S$ . Otherwise, compare the number of children of $S$ and $T$ . If $S$ has more children than $T$ , then $S > T$ , and vice versa. Otherwise, suppose $S$ and $T$ both have $n$ children. Let $S_1, S_2, \ldots, S_n$ and $T_1, T_2, \ldots T_n$ be the immediate subtrees of $S$ and $T$ respectively, ordered from smallest to largest. Compare $S_1$ to $T_1$ , then $S_2$ to $T_2$ , etc., until we get a pair of unequal trees $S_i$ and $T_i$ . If $S_i > T_i$ then $S > T$ , and vice versa. Of course, of all pairs of immediate subtrees are equal, then $S$ and $T$ will be equal. This gives a linear order on unlabeled rooted trees, and one can prove that this is a well-order. Further, this well-ordering has order type $\vartheta(\Omega^\omega,0)$ . This definition is a modification of a well-ordering of ordered rooted trees due to Levitz, and expounded on in papers by Jervell. From this well-ordering we can define fundamental sequences for ordinals up to $\vartheta (\Omega^{\omega}, 0)$ . Simply put, given an ordinal $\alpha$ , let $\alpha[n]$ be the largest ordinal less than $\alpha$ corresponding to a tree of $n$ vertices or less. From this, we can define our version of the Hardy hierarchy: $H_0(n) = n$ $H_{\alpha + 1}( n) = H_{\alpha}( n+1)$ For $\alpha$ a limit ordinal, $H_{\alpha}( n) = H_{\alpha[n+1]}( n+1)$ Note the $n+1$ 's in the last line - this differs from the usual values of $n$ . Of course, this will only make the functions larger. $H_{\alpha}( n)$ for $\alpha < \vartheta (\Omega^{\omega}, 0)$ is the final index $m$ in the sequence of trees $T_n, T_{n+1}, \ldots, T_m$ where $T_n$ corresponds to $\alpha$ and $T_i$ is the largest tree with at most $i$ vertices that is smaller than $T_{i-1}$ , and $T_m$ is the tree with one vertex. Thus $H_{\vartheta (\Omega^{\omega}, 0)}( n)$ will be the final index $m$ in the sequence of trees $T_{n+1}, T_{n+2}, \ldots, T_m$ where $T_{n+1}$ is arbitrary. Thus tree(n) $\geq H_{\vartheta (\Omega^{\omega}, 0)}( n) - n$ . So where does TREE(3) come in? Harvey Friedman himself explains in a post to the Foundations of Mathematics message boards: http://www.cs.nyu.edu/pipermail/fom/2006-March/010260.html In the post he explains why a proof of the theorem "TREE(3) exists" in the theory $ACA_0 + \Pi^1_2\text{-}BI$ must have more than $2\uparrow\uparrow 1000$ symbols. He does this by showing that TREE(3) must be very large - specifically, he constructs a sequence of more than $n(4)$ rooted trees labeled from {1,2,3} such that $T_i$ has at most $i$ vertices, for no $i, j$ with $i < j$ do we have $T_i$ homeomorphically embeddable into $T_j$ , and each tree contains either a 2 label or a 3 label. We can obviously continue this with tree( $n(4)$ ) trees with all labels 1. Thus we have TREE(3) $\geq$ tree $(n(4)) + n(4) \geq H_{\vartheta (\Omega^{\omega}, 0)}(n(4))$ In fact, we can do somewhat better than this; we can replace the $n(4)$ above by $F(4)$ , where $F(4)$ is defined as the length of the longest sequence of sequences $x_1, x_2, \ldots x_n$ from {1,2,3,4} such that $x_i$ has length $i+1$ and for no $i,j$ with $i < j$ do we have $x_i$ a subsequence of $x_j$ . I can prove much better bounds for $F(4)$ than Friedman's lower bound for $n(4)$ ; specifically, $F(4) > f_{\omega^2 + \omega + 1}f_{\omega^2 + \omega + 1}f_{\omega^2 + \omega}f_{\omega^2 + 1}f_{\omega^2 + 1}f_{\omega^2}f_{ \omega + 1}f_{ \omega + 1}f_{\omega}(30)$ But such specificity is perhaps unwarranted given how far from TREE(3) it may be.
|
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|
93,916 |
There is a literature dealing with
$$
\sum_{k\le x}d(f(k))
$$
where $f$ is an irreducible polynomial and $d(n)$ is the number of divisors of $n$. Erdos 1952 shows that the sum $\asymp x\log x,$ which was improved to $Ax\log x+O(x\log\log x)$ by Bellman-Shapiro (cited in Scourfield). But these results only apply to irreducible polynomials. What asymptotics are known for $\sum_{k\le x}d(k^2)$? Are there good methods for calculating this sum quickly? The literature includes: Dirichlet 1850, Voronoi 1903 and van der Corput 1922, Kolesnik 1969, Huxley 1993, Nowak 2001 (linear); Scourfield 1961, Hooley 1963, McKee 1995, McKee 1997, McKee 1999, Broughan 2002 (quadratic). The sequence is in the OEIS as A061503 but there is no real information there.
|
In the case you are interested in there is a simple generating (Dirichlet) series:
$$ \sum_{n=1}^\infty \frac{d(n^2)}{n^s} = \frac{\zeta^3(s)}{\zeta(2s)}.$$
From this you can either use a convolution argument or a Perron formula type argument to get an asymptotic formula. In particular, I believe it follows that
$$\sum_{n\leq x} d(n^2) = \frac{3}{\pi^2}x \log^2 x +O(x \log x). $$
With more work, you can get lower-order terms of size $\asymp x \log x$ and $\asymp x.$ Edit: There seems to some disagreement on whether the coefficient of the leading order term is $\frac{3}{\pi^2}$ or $\frac{3}{2\pi^2}$. I believe that $\frac{3}{\pi^2}$ is correct. Here are three bits of reasoning:
From the generating series, we have
$$ \sum_{n\leq x}d(n^2) = \sum_{n\leq x} \sum_{k\ell^2=n} d_3(k)\mu(\ell) = \sum_{\ell\leq \sqrt{x}} \mu(\ell) \sum_{k\leq x/\ell^2} d_3(k) $$
where $d_3(n)$ denotes the number of ways to write $n$ as a product of three positive divisors and $\mu(\ell)$ is the Moebius function. By a standard estimate
$$ \sum_{k\leq x/\ell^2} d_3(k) = \frac{x}{2\ell^2}\log^2(x/\ell^2) + O\left(\frac{x\log x}{\ell^2} \right)$$
from which it follows that
$$ \sum_{n\leq x}d(n^2) = \frac{x \log^2 x}{2} \sum_{\ell \leq \sqrt{x}} \frac{\mu(\ell)}{\ell^2} +O(x\log x) = \frac{3 x}{\pi^2}\log^2 x +O(x\log x).$$
Alternatively, a Perron formula (e.g. Prime Number Theorem) type argument can be used to show that
$$ \sum_{n\leq x}d(n^2) = \text{Res}_{s=1} \frac{\zeta^3(s)}{\zeta(2s)} \frac{x^s}{s} +o(x) = \frac{3x}{\pi^2}\log^2 x + O(x\log x).$$
Moreover, in Mathematica, you can use DivisorSigma[0, n^2] to calculate $d(n^2)$. For $x=1,000,000$ I get that
$$ \frac{\pi^2}{3x\log^2x}\sum_{n\leq x} d(n^2) \approx 1.27305392....$$
The slow convergence to 1 is from the influence of lower-order terms. Notice, however, that the value it is not anywhere near $1/2$. However, if I define $F(x)$ to be the residue of$\frac{\zeta^3(s)}{\zeta(2s)}\frac{x^s}{s}$ at $s=1$, I get that
$$ \frac{1}{F(x)}\sum_{n\leq x} d(n^2) \approx 1.0000073....$$
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|
94,226 |
The following problem has bothered me for a long time. Let us imagine a point on the real axis. At the beginning, it is located at point $O$. Then it will "walk" on the real axis randomly in the following way. For every step of the "walk", it will choose a real number $\Delta x$ uniformly from the interval $[-1,1]$, turn right, and move $\Delta x$ unit. Once it reaches the left side of the point $O$, it will "die" immediately. Our task is find out the probability of the point is alive after $n$ steps of "walk" $P_n$.
I guess that $P_n=\frac{(2n)!}{4^n (n!)^2}$, but I can't prove this or explain why it is true.
|
Here is an argument that proves the conjecture. More generally, it shows that if $X_1,\dots,X_n$ is a "generic" sequence of positive real numbers, and we form a sum by permuting the terms randomly and putting random signs on the terms (uniform distribution over all possibilities), then the probability that all partial sums are positive is given by $P_n$ as expressed in the OP. By "generic" I mean that no nonempty signed subsequence sums to zero. First consider a special case where the statement clearly holds: Assume that $X_1>>X_2>>X_3>>\dots >>X_n$ in the sense that the sign of any partial sum will depend only on the sign of the dominating term. A valid signed permutation (one where all partial sums are positive) of $n$ terms must then be formed by inserting the term $X_n$ into a valid signed permutation of $X_1,\dots,X_{n-1}$. Conversely if we insert $X_n$ into a valid signed permutation of $X_1,\dots,X_{n-1}$, then the only way we can turn this into a non-valid signed permutation is if we insert $X_n$ first and with a negative sign. Therefore the probability that a signed permutation of $n$ terms is valid is $$\frac12\cdot \frac34\cdot\frac56\cdots\frac{2n-1}{2n},$$ as required. Next, let us move the point $(X_1,\dots,X_n)$ in $\mathbb{R}^n$ and see how the probability of a valid signed permutation might change. We need only consider what happens when we pass through a hyperplane given by a signed subsequence being equal to zero. Instead of introducing double indices let me take a generic example: Suppose we pass through the hyperplane where the sum $X_3+X_5+X_6-X_2-X_9$ changes sign. At the moment where the sign change occurs, the only signed permutations that go from valid to invalid or the other way are those where this particular sum occurs as the five first terms (in some order). Now for every signed permutation that goes from valid to invalid, there will be a corresponding one that goes from invalid to valid, by reversing those five first terms and changing their signs. For instance, if $$(X_5+X_6-X_9+X_3-X_2) + \dots$$ goes from valid to invalid, then $$(X_2-X_3+X_9-X_6-X_5) + \dots$$ will go from invalid to valid. The conclusion is that the probability of a signed permutation being valid (having all partial sums positive) will never change, and will therefore always be equal to the product given in the OP. UPDATE:
Instead of looking at what happens when we cross one of these hyperplanes, here is another way of seeing that $P_n$ is independent of $X=(X_1,\dots,X_n)$, which has the additional feature of showing that $$P_0P_n + P_1P_{n-1}+\cdots + P_nP_0 = 1.$$ Suppose that $X_1,\dots,X_n$ are generic as described above. Let $P_n(X)$ denote the probability that a random signed permutation will have all partial sums positive, and suppose for induction on $n$ that we have shown that $P_k(X) = P_k$ is independent of $X$ for $k=1,\dots,n-1$. Now let $Y_1,\dots, Y_n$ be a random signed permutation of $X$, and consider the distribution of the $k$ for which the partial sum $Y_1+\dots+Y_k$ is minimized. I claim that the probability that this occurs for a particular $k$ is $P_kP_{n-k}$, since it means that all consecutive sums of the form $Y_{k+1}+Y_{k+2}+\dots+Y_m$ for $m>k$ must be positive, while similarly all sums of the form $Y_m + Y_{m+1}+\dots+Y_k$ for $m\leq k$ must be negative. Since generically the minimum is unique, we have $$P_0P_n(X) + P_1P_{n-1}+\cdots + P_n(X)P_0 = 1,$$ which shows that $P_n(X)$ is independent of $X$. I like this very much, since the equation $P_0P_n + P_1P_{n-1}+\cdots + P_nP_0 = 1$ is crucial for my proof in the Monthly December 2007 of the Wallis product formula for pi, http://www.math.chalmers.se/~wastlund/monthly.pdf . See also Don Knuth's 2010 christmas tree lecture "Why pi?" under "Computer musings" at http://scpd.stanford.edu/knuth/index.jsp .
I did a proof by induction and some algebraic manipulation in the Monthly paper, and Knuth used power series and the binomial theorem some 20-25 minutes into his talk, but the argument motivated by this question is much nicer!
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|
94,517 |
This question has been bugging me for some time. Take the hamiltonian for the hydrogen atom: $$\hat{H}=-\frac{1}{2}\nabla^2-\frac{1}{r},$$ acting on (a domain contained in) $L^2(\mathbb{R}^3)$. It is standard fact that this is an unbounded operator which has a countable infinity of eigenvalues, all of which are negative and which accumulate around 0, and has a continuous spectrum on the whole of $(0,\infty)$. Physically, the former are bound states which correspond to elliptic Keplerian orbits in the classical problem, and the latter are unbound states and correspond to hyperbolic orbits. I also know that the spectrum of all unbounded operators is a closed set, so that 0 is definitely in $\sigma\left(\hat{H}\right)$. My question is then: to what part of the spectrum does 0 belong to (i.e. point, continuous, residual)? What are the corresponding eigenfunctions? What kind of degeneracy does it have? (I would expect it to admit a common eigenvector with any $l,m$ angular momentum numbers, but I'm far from sure.) How do the eigenfunctions correspond to the nearby bound and unbound states?
|
First of all, the Hamiltonian in question is defined on $L^2(\mathbb R^3)$ , not on $L^2(\mathbb R)$ . This is important because in the one-dimensional case the potential would have a non-integrable singularity which complicates things seriously. On $L^2(\mathbb R^3)$ , the operator, defined as a closure from $C_0^\infty$ , is selfadjoint. This is proved, for example, in the book by T. Kato, Perturbation theory for linear operators , Springer, 1966. Thus the residual spectrum is impossible. A rigorous calculation of eigenvalues and eigenfunctions can be found in the books L. D. Faddeev and O. A. Yakubovskii, Lectures on quantum mechanics for mathematics students . American Mathematical Society, 2009; L. A. Takhtajan, Quantum mechanics for Mathematicians , American Mathematical Society, 2008. The point 0 is an accumulation point of negative eigenvalues and the limit point of continuous spectrum, thus it belongs to the essential spectrum.
|
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|
94,537 |
Consider a connected holomorphic manifold $X$ and its ring of holomorphic functions $\mathcal O(X).$ My general question is simply: in which cases is the Krull dimension $\dim \mathcal O(X)$ known? Of course if $X$ is compact $\mathcal O(X)=\mathbb C$ and that dimension is $0$. There are also quite a lot of non-compact manifolds with $\mathcal O(Z)=\mathbb C$: For example if $X$ is connected of dimension $\geq 2$ and $Y\subset X$ is an analytic subset of codimension at least $2$ ( or a small compact ball) , you will still have $\mathcal O(X\setminus Y)=\mathbb C$ . But apart from these trivial examples I can't compute a single Krull dimension $dim \mathcal O(X)$ for, say, Stein manifolds of positive dimension. Just in order to ask something definite, let me pose the ridiculous-sounding question: Does there exist a connected holomorphic manifold $X$ with $0\lt \dim \mathcal O(X)\lt \infty$ ?
|
It follows from the proof in Sasane's paper that Krull dimension of a (connected) complex manifold $M$ is infinite iff $M$ admits a nonconstant holomorphic function $F: M\to {\mathbb C}$. Namely, using Sard's theorem find a sequence of points $a_k \in F(M)$ which are regular values of $F$ and so that $(a_k)$ converges to a point in $({\mathbb C}\cup \infty) \setminus F(M)$. Then, pick regular points $b_k\in V_k:=F^{-1}(a_k)$ of $F$ and define multiplicity of zero for a holomorphic function $h: M\to {\mathbb C}$ with respect to the germ of $V_k$ at $b_k$. (I.e., multiplicity of $h$ is determined by the largest $m$ so that $h=(F-a_k)^m g$ on the level of germs at $b_k$.) Now, the same proof as in Sasane's paper goes through, where you will be using functions $f_n\circ F$ instead of Sasane's functions $f_n$. The point is that Sasane's argument is essentially local at zeroes of the functions $f_n$. Actually, what Sasane proves is a lemma about a commutative ring $R$ with a sequence of valuations $m_k$ for which there exists a sequence of elements $f_i\in R$ so that $m_k(f_i)$ grows slower than $m_k(f_{i+1})$ for every $i$ as $k\to \infty$ (more precisely, in his case, the growth rate of $m_k(f_i)$ is $k^{i+1}$). Under this assumption, Krull dimension of $R$ is infinite. Edit : I finally wrote a detailed proof here . Edit. I wrote a proof that the Krull dimension of $H(M)$ (when it is positive) has cardinality at least continuum. The new proof uses surreal numbers instead of ultralimits. For the sake of completeness I am keeping the older proof as well.
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|
94,742 |
According to Wikipedia False proof For example the reason validity fails may be a division by zero that is hidden by algebraic notation. There is a striking quality of the mathematical fallacy: as typically presented, it leads not only to an absurd result, but does so in a crafty or clever way. The Wikipedia page gives examples of proofs along the lines $2=1$ and the primary source appears the book Maxwell, E. A. (1959), Fallacies in mathematics. What are some examples of interesting false proofs?
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My favorite example is the following proof of the Cayley-Hamilton theorem, which caused me some disconcertion when I was a student. Let $A$ be a square matrix, and call $p(t) = \det(tI - A)$ its characteristic polynomial. Then $p(A) = \det(AI-A) = 0$.
|
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|
95,261 |
Dear all, The "Bernstein center" of a $p$-adic reductive group appears frequently in the literature of automorphic forms, often without a precise definition. For example, in page 233 of Moeglin-Waldspurger's classic "Spectral decomposition and Eisenstein series" , the couple tell us : "...in particular the centre of enveloping algebra acts on $\delta$ via a character at the infinite places and the Bernstein centre does so at the finite places... " So one may guess that it is some analogy of "the centre of enveloping algebra" at fintie places. My questions are: What is the definition of the Bernstein centre of a p-adic reductive group. What is the original motivation to introduce it ? What role does it play in the theory of automorphic forms ? Could you explain these in some concrete example ,say $GL_2$ ? Please feel free to choose part of the questions to reply.
Any comments and references (in English) will also be very welcome ! Thank you very much in advance!
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Abstract Definition .
Let $Rep(G)$ be the abelian category of smooth complex representations of our $p$-adic group $G$. The Bernstein center is the endomorphism ring $\mathfrak Z(G)$ of the identity functor of $Rep(G)$. So it acts on any smooth representation, and this action commutes with any $G$-morphism. As a projective limit . Let $H$ be a compact open subgroup. Letting $\mathfrak Z(G)$ act on the permutation representation $\mathbb C[G/H]$ gives a morphim to the center $Z(G,H)$ of the Hecke algebra $\mathcal H(G,H)$. This yields an isomorphism $\mathfrak Z(G) \simeq \lim\limits_{\leftarrow H} Z(G,H)$ where transition maps are given by applying idempotents. Geometric realization . (here "geometric" is in the sense of Trace formulas, i.e. on the side of harmonic analysis). $\mathfrak Z(G)$ acts on the regular representations $C^\infty_c(G)$. The pairing $(z,f)\mapsto z.f(1)$ embeds $\mathfrak Z(G)$ as a set of distributions on $G$. The image is the convolution algebra of "essentially compact invariant distributions". Spectral realization . By Schur's lemma (which holds in this context), $\mathfrak Z(G)$ acts on any irreducible representation $\pi$ via a character $\theta_\pi:\mathfrak Z(G)\longrightarrow \mathbb C$. This is sometimes called the "infinitesimal character" of $\pi$, by analogy with the Archimedean situation, although there is nothing "infinitesimal" here. We get in this way a realization of $\mathfrak Z(G)$ as an algebra of continuous functions on the smooth dual $\hat G$ of $G$ equipped with the Fell topology. So far, nothing deep. Now, two major achievements in the representation theory of p-adic groups are the Bernstein theorem which describes the spectral realization explicitly, the Harish Chandra Plancherel formula which provides a link between both realizations. Let me try to describe Bernstein's result. Bernstein first splits the category $Rep(G)$ as a (infinite) product of indecomposable abelian subcategories (called "blocks"). Accordingly, the smooth dual decomposes into infinitely many connected components, and the center decomposes as an infinite product of rings. The simplest example is that of a compact $G$ (e.g. the kernel of the norm map in a division algebra). In this case, $\hat G$ is discrete and the center is a product of copies of $\mathbb C$ indexed by the set of classes of irreps.
The next example is that of a compact-mod-center $G$ (e.g. the unit group of a division algebra). In this case there is an action of the group $\Psi(G)$ of unramified characters of $G$ on $\hat G$. Note that $\Psi(G)$ is naturally an algebraic torus over $\mathbb C$ because $G$ mod its maximal compact subgroup is a free abelian group of finite type. Now, connected components are the orbits of $\Psi(G)$, and the topology is the homogeneous space topology. In particular each such orbit carries a natural structure of an algebraic variety over $\mathbb C$. Finally, $\mathfrak Z(G)$ is the direct product of the ring of regular functions on these orbits. Let us go to the general case.
Assume first that $G$ is semisimple. Then each supercuspidal representation gives an isolated point in the smooth dual $\hat G$, because such representations are both projective and injective objects. Therefore, denoting by $Cusp(G)$ the set of (isom classes of) supercuspidal irreps, the ring $\mathbb C^{Cusp(G)}$ is a factor of $\mathfrak Z(G)$. If $G$ is reductive, then $Cusp(G)$ is still open and closed in $\hat G$, but is not discrete. As above, $\Psi(G)$ acts on $Cusp(G)$ and the latter is the disjoint union of orbits under $\Psi(G)$ with the natural quotient topology. The corresponding product of ring of regular functions on each orbits is then a factor of $\mathfrak Z(G)$, the "cuspidal" part $\mathfrak Z(G)_{cusp}$ of $\mathfrak Z(G)$. Now remains the crucial step of describing the non-cuspidal part of the center. Bernstein uses parabolic induction from Levi subgroups. One quick way of stating the final result is :
$$ \mathfrak Z(G)\simeq \left( \prod_M \mathfrak Z(M)_{cusp} \right)^G$$
Here the product is over all Levi subgroup and $G$ acts by conjugacy. In order to get something less frightening, one can fix a maximal split torus $T$, restrict the product to those $M$'s that contain $T$ (finitely many) and take $N_G(T)$-invariants. In particular, connected components are labeled by conjugacy classes of $\Psi(M)$-orbits
in $Cusp(M)$.
Suppose $G$ is split for simplicity. A particularly interesting component is that which corresponds to the $\Psi(T)$ orbit of the trivial character of $T$. Its contribution to
$\mathfrak Z(G)$ is isomorphic to $\mathbb C [X(T)]^W$.
This component contains the unramified representations and the action of $\mathbb C[X(T)]^W$ on each such representation is given by its Satake parameter. In fact the corresponding connected component of $\hat G$ is the set of irreps that have non-trivial invariant under an Iwahori-subgroup, and one recovers the fact that the center of the Hecke-Iwahori algebra identifies with $\mathbb C[X(T)]^W$.
Finally, note that this ring looks like a "group version" of the center of an enveloping algebra, so the analogy with the Archimedean context is even deeper than expected. Edit : answers to comments. On the link with Plancherel formula. Denote by $\hat G_u$ the unitary smooth dual inside $\hat G$. The abstract Plancherel theorem says that there is a measure $d\pi$ on $\hat G_u$ such that for any $f \in C^\infty_c(G)$ (even for $f$ in the Schwartz algebra) we have
$$ \int_{\hat G_u} Tr(\pi(f)) d\pi = f(1). $$
Note that $d\pi$ depends on the choice of a Haar measure on $G$ and the same choice is done to let $f$ act on $\pi$.
Let $z\in \mathfrak Z(G)$, applying the formula to $z.f$ yields the formula
$$ \int_{\hat G_u} Tr(\pi(f)) \theta_\pi(z) d\pi = (z.f)(1) $$
which expresses the distribution associated to $z$ (geometric realization) in terms of the
its action on $\hat G_u$.
Of course this is only useful if one has an explicit formula for the measure $d\pi$. Such a formula is provided by Harish Chandra's theorem. The first point is that the Plancherel measure is supported in the tempered spectrum.
The latter decomposes as a disjoint union of infinitely many connected components in a similar fashion to the full smooth dual, except that now one considers $\Psi_u(M)$-orbits of discrete series, where $\Psi_u(M)$ now denotes unitary unramified characters (a compact torus inside $\Psi(M)$). So each component is a quotient of some homogeneous space under some $\Psi_u(M)$. The second point is that on such a tempered component, the Plancherel measure is absolutely continuous w.r.t. to the natural Lebesgue measure, and in fact given by some rational function. The precise computation of this rational function is not given by Harish Chandra, and in fact I don't know if it is known in general (maybe OK for classical groups). Relation to Galois representations Assume here that $G=GL_n$ and consider the Langlands correspondence $\pi \mapsto \sigma(\pi)$ between irreducible representations of $G$ and Weil-Deligne representations. Then $\pi$ and $\pi'$ lie in the same Bernstein component if and only if the restriction to inertia of the Weil group action on $\sigma(\pi)$ and $\sigma(\pi')$ are isomorphic. I guess that they are tempered iff the monodromy filtration of $\sigma(\pi)$ is pure of weight $0$ and in this case $\pi$ and $\pi'$ are in the same tempered component iff restrictions of $\sigma(\pi)$ and $\sigma(\pi')$ to (Inertia $\times \langle N\rangle$) are isomorphic. Applications in global context In the Langlands-Kottwitz approach to counting points on Shimura varieties in hyperspecial level, a certain spherical function (depending on the Shimura datum) has to be plugged in the Trace formula. In Iwahori level, work of Haines, Rapoport, Kottwitz shows that one should plug an element of the centre of the Iwahori-Hecke algebra. They then conjectured that similar phenomena should happen in deeper level. More generally, recent work of Scholze shows (if I understood something, maybe he will contradict me soon!) that if one fixes an element $\tau$ in the Weil group (but outside the inertia group) and one considers the function
$\pi \mapsto Tr(\tau, \sigma(\pi))$, then one gets an element of $z_\tau\in \mathfrak Z(G)$ such that for $H$ a congruence subgroup, the trace of $\tau$ on the cohomology of a (suitable) Shimura variety of local level $H$ can be computed using a trace formula evaluated at a test function whose local factor is (associated to) the function $z_\tau.e_H$. I'm not able to provide more details on it, but the slogan is : central elements (should) provide good test functions in counting points on Shimura varieties.
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95,371 |
For a complex manifold $M$, one can consider (A) its de Rham cohomology, or (B) its Dolbeault cohomology. I'm looking for some motivation as to why one would bother introducing Dolbeault cohomology. To be more specific, here are some straight questions. What can Dolbeault tell us that de Rham can't? Does there exist some simple relationship between these two cohomologies? When are they equal? Do things become simpler for the Kahler case? What happens for the projective spaces? Why does nobody talk about the holomorphic cohomology?
|
Let $\Omega^{p,q}(M)$ be the $C^{\infty}$ $(p,q)$-forms. One always has a double complex with $\Omega^{p,q}(M)$ in position $(p,q)$. The cohomology in the $q$ direction is Dolbeault cohomology, the cohomology of the total complex is deRham cohomology. (In each case, essentially by definition.) Whenever you have a double complex, you get a spectral sequence. On the first page of the spectral sequence is Dolbeault cohomology; the spectral sequence converges to deRham cohomology. This is sometimes called the Hodge-de Rham spectral sequence, and sometimes called the Frolicher spectral sequence. If $M$ is compact Kahler (in particular projective) then the spectral sequence collapses at the first page; all maps between Dolbeault groups are zero. So $H^k(M) \cong \bigoplus_{p+q=k} H^{p,q}(M)$ in this case. If $M$ is Stein (in particular, affine) then the only nonzero Dolbeault groups are the $H^{p,0}(M)$, corresponding to holmorphic $p$-forms. The next page takes the cohomology of the complex of holomorphic $p$-forms; I'll term this ``holomorphic deRham". After that, there are no further maps, so holomorphic deRham equals deRham. In general, the spectral sequence can be arbitrarily nondegenerate .
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|
95,701 |
I am given to understand that the celebrated open problem (MLC) of the Mandelbrot set's local connectness has broader and deeper significance deeper than some mere curiosity of point-set topology. From http://en.wikipedia.org/wiki/Mandelbrot_set I see that conjecture has implications concerning the structure of the Mandelbrot set itself, but I don't think I grasp its broadest implications for complex dynamics and matters beyond. Request: Could someone explain the proper current context in which to view MLC and/or how the world would look it its full glory if MLC has a positive answer? Alternatively, please give a pointer to somewhere in the literature that does the same.
|
If a connected compact $K \subset C$ is locally connected then the Riemann map
$h\colon C \setminus \Delta \to C \setminus K$ extends continuously to $\partial \Delta$. For each $z \in \partial K$, the boundary of the convex hull of $h^{-1}(\{z\})$ is the union of a set $\Lambda_z$ of chords; the union of these $\Lambda_z$ over all $z \in \partial K$ is a closed set $\Lambda_K$ of disjoint chords; it is called a lamination of $\Delta$. We can reconstruct the convex hulls of each $h^{-1}(\{z\})$ from $\Lambda_K$, and when we collapse every convex hull to a point, we obtain a topological model for $K$. In the case where $K$ is the Mandelbrot set $M$, the lamination $\Lambda_M$ can be described combinatorially, so MLC would mean that we know the topology of $M$. There is a second answer which is more subtle and more important. For each $c \in M$, the filled Julia set $K_c$ of $z \mapsto z^2 + c$ is compact and connected; if it is locally connected the resulting lamination $\Lambda_c \equiv \Lambda_{K_c}$ is, in the right sense, invariant under $z \mapsto z^2$ on $\partial \Delta$. Even if $K_c$ is not locally connected, there is a way of defining what the lamination would be if $K_c$ were locally connected. Every invariant lamination appears as $\Lambda_c$ for some c, and MLC is equivalent to the statement there is a unique $c$ with a given lamination. We think of $\Lambda_c$ as describing the combinatorics of $K_c$, and we think of this uniqueness conjecture as "combinatorial rigidity"---two maps of the form $z \mapsto z^2 + c$ are conformally conjugate (and hence equal) if they are "combinatorially equivalent". (Actually, if $z \mapsto z^2 + c$ has an attracting periodic cycle, then the set of combinatorial equivalent parameters form an open subset of $C$, so the statement of combinatorial rigidity must be suitably modified in that case. It is known that structural stability is open and dense in the family of maps $z \mapsto z^2 + c$, so combinatorial rigidity implies that every $z \mapsto z^2 + c$ in this open and dense set must have an attracting periodic cycle; this is the implication that Eremenko alluded to in his answer. ) In this sense MLC is closely analogous to Thurston's Ending Lamination Conjecture (proven by Brock, Canary, and Minsky), which says, broadly speaking, that a finitely generated Kleinian group is determined by the topology of its quotient and the ending laminations of its ends, which are also, when viewed appropriately, invariant laminations of the disk. There is a third answer which is more historical and empirical. We can prove MLC and combinatorial rigidity "pointwise" (or "laminationwise") by proving that for a given invariant lamination $\Lambda$, it appears as the lamination $\Lambda_c$ for a single $c$. This has been done in great many cases, first by Jean-Christophe Yoccoz, and then by Mikhail Lyubich, the author of this post, Genadi Levin, and Mitsuhira Shishikura. To prove this combinatorial rigidity for a given $c$ seems to require a detailed understanding of the geometry of the associated dynamical system, and this almost always leads to further results. So proving MLC would most likely mean having a thorough understanding of the geometry and dynamics of every map $z \mapsto z^2 + c$.
|
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|
95,743 |
We know that
$$
\sum_{n \le x}\frac{1}{n\ln n} = \ln\ln x + c_1 + O(1/x)
$$ where $c_1$ is a constant. Again Mertens' theorem says that the primes $p$ satisfy
$$
\sum_{p \le x}\frac{1}{p} = \ln\ln x + M + O(1/\ln x).
$$ Thus both these divergent series grow at the same rate. Mertens' theorem was proved without using the prime number theorem, some 25 years before PNT was proved. However from these two examples, we cannot conclude that $$
\lim_{n \to \infty} \frac{p_n}{n\ln n} = 1
$$
otherwise Mertens' would have been the first to prove PNT. My question is - based on the above two series, what are the technical difficulties that prevent us from reaching the conclusion that $p_n/n\ln n = 1$. There may be counter examples with other series, so such conclusions may not be true in general. However I am not interested in the general case. Instead I am asking only in case of the sequence $1/n\ln n$ and $1/p_n$.
|
Here is a heuristic that I plan to make into a blog post some day. Suppose that there were only finitely many primes with first digit $9$. Is your estimate good enough to see that? To be more precise, suppose that there were no primes between $9 \times 10^k$ and $10^{k+1}$ for all sufficiently large $k$. And suppose that the number of primes between $a$ and $b$, for $10^k \leq a < b \leq 9 \times 10^k$ is $\approx \frac{\log 10}{\log 9} \int_{a}^b dt/\log t$ (when $a$ is not too close to $b$). We'll see later where the fraction $\log 10/\log 9$ comes from. The first thing to note is that this would violate the prime number theorem. In this scenario, we would have $\pi(9 \times 10^k) = \pi(10 \times 10^k)$ for $k$ large. But the prime number theorem says that
$$\pi(10\times 10^k) - \pi(9 \times 10^k) \sim \frac{10 \times 10^k}{k \log 10 + \log {10}} - \frac{9 \times 10^k}{k \log 10 + \log {9}} \sim \frac{10^k}{k \log 10} \to \infty.$$
So proving the prime number theorem involves disproving this ridiculous scenario. Now, let's see that the scenario is consistent with $\sum_{p \leq N} 1/p = \log \log N + M + O(1/\log N)$. The sum over the primes between $10^k$ and $10^{k+1}$ would be roughly
$$\frac{\log 10}{\log 9} \int_{10^k}^{9 \times 10^k} \frac{dt}{t \log t} = \frac{\log 10}{\log 9} \left( \log \log (9 \times 10^k) - \log \log 10^k \right)$$
$$=\frac{\log 10}{\log 9} \left( \log( k \log 10+\log 9) - \log (k \log 10) \right) = \frac{\log 10}{\log 9} \left( \log \left( 1+\frac{\log 9}{k \log 10} \right) \right) $$
$$ = \frac{\log 10}{\log 9} \frac{\log 9}{k \log 10} + O(1/k^2)= \frac{1}{k} + O(1/k^2)$$ So
$$\sum_{p \leq 10^{n+1}} \frac{1}{p} = \sum_{j=1}^n \left( \frac{1}{j} + O(1/j^2) \right)=$$
$$\log n + B + O(1/n) = \log \log 10^{n+1} + C + O(1/\log 10^{n+1}).$$ Very important exercise left for you: If you redo this computation for $\sum_{p \leq 9 \times 10^k} 1/p$, you get $\log \log (9 \times 10^k) + C + O(1/\log(9 \times 10^k))$ for the same constant $C$. The point is that $\log \log 10^{k+1} - \log \log (9 \times 10^k) = O(1/\log 10^k)$, so this estitmate is consistent with $\sum_{p \leq N} 1/p$ not growing at all between $9 \times 10^k$ and $10 \times 10^k$. This trick is useful for refuting other simple approaches to the PNT. For example, the "primes hate to start with $9$ scenario" is also consistent with $\sum \log p/p^s = 1/(s-1) + O(1)$ as $s \to 1^{+}$, so that is also not enough to prove PNT.
|
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95,837 |
I am looking for examples of theorems that may have originally had a clunky, or rather technical, or in some way non-illuminating proof, but that eventually came to have a proof that people consider to be particularly nice. In other words, I'm looking for examples of theorems for which have some early proof for which you'd say "ok that works but I'm sure this could be improved", and then some later proof for which you'd say "YES! That is exactly how you should do it!" Thanks in advance. A sister question: Examples of major theorems with very hard proofs that have NOT dramatically improved over time
|
[ Edit: This answer seems to fit the title of the question, though not the actual question in the body.] Resolution of singularities in algebraic geometry seems like a good example. Hironaka's original proof was over 200 pages and hard to understand: "Even A. Grothendieck [in Actes du Congrès International des Mathématiciens (Nice, 1970), Tome 1, 7--9, Gauthier-Villars, Paris, 1971; MR0414283 (54 #2386)] admitted openly that he did not completely understand Hironaka's proof." That quote is from Dan Abramovich's Math Review of the book Lectures on resolution of singularities by Kollár; the review goes on to say "One can [nowadays] devote a few weeks in a first course on algebraic geometry to give just a complete proof of resolution of singularities in characteristic 0 (Chapter 3 of the present book, which is largely self-contained)." I know almost nothing about this topic, but some names I know associated to the various approaches to simplification of Hironaka's proof are Bierstone, Milman, Encinas, Villamayor, Hauser, Cutkosky, Włodarczyk, Kollár, Cossart, Piltant... Please tell me any I missed!
|
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|
95,865 |
It seems to me that almost all conjectures (hypotheses) that were widely believed by mathematicians to be true were proved true later, if they ever got proved. Are there any notable exceptions?
|
In 1908 Steinitz and Tietze formulated the Hauptvermutung ("principal conjecture"), according to which, given two triangulations of a simplicial complex, there exists a triangulation which is a common refinement of both. This was important because it would imply that the homology groups of a complex could be defined intrinsically, independently of the triangulations which were used to calculate them. Homology is indeed intrinsic but this was proved in 1915 by Alexander, without using the Hauptvermutung, by simplicial methods. Finally, 53 years later, in 1961 John Milnor (some topology guy, apparently) proved that the Hauptvermutung is false for simplicial complexes of dimension $\geq 6$ .
|
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95,939 |
What is the difference between holonomy and monodromy? And what is the simplest example in which one is trivial and the other is not?
|
Holonomy= monodromy iff the bundle is flat. In general, monodromy group is the quotient of holonomy group by the normal subgroup formed by parallel transports along homotopically trivial loops. One of the simplest examples when two groups are different is the holonomy of the tangent bundle of the standard Riemannian metric on the 2-sphere. Then monodromy is trivial since sphere is simply-connected, while holonomy group is $SO(2)$. Two more remarks. First, the conundrum: What is the holonomy of a complete hyperbolic surface $S$? The answer: It depends who you ask. A differential geometer (like Robert Bryant) would think of the tangent bundle and his answer would be $SO(2)$ or $O(2)$ (depending on orientability). A hyperbolic geometer (like William Thurston) would think of the hyperbolic structure as a special $(X,G)$-structure and answer: $\pi_1(S)\subset PSL(2, {\mathbb R})$.
(An $(X,G)$ structure could be regarded as a flat $X$-bundle with a section transversal to the flat connection, so holonomy of the flat bundle is the holonomy of the structure.) If you were to ask me, I would say "It depends ..." Second remark: For cultural, historic, etc. reasons, given a flat bundle, differential geometers and topologists tend to use the word "holonomy," while people in algebraic geometry, complex analysis, singularity theory, tend to use the word "monodromy."
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95,974 |
It is a well-known and deep${}^\ast$ theorem that if a group $G$ has cohomological dimension one then it must be free. This was proved in the late 60's by Stallings (for finitely generated groups) and Swan. The proofs in the original articles are well-written and informative, bringing together a lot of ideas from topology, algebra and set theory. However there seems to be some scope for the proofs to be shortened, for instance by a clever choice of projective resolution, or the use of non-abelian cohomology. My questions then is Have any alternative (shorter) proofs of the Stallings-Swan theorem appeared since 1969? ${}^\ast$I am prepared to accept that the answer may be 'no', but in that case I wonder if someone could offer an explanation of why this theorem is so "deep", ie why there cannot exist some "quick trick" proof.
|
The heart of the matter is the Stallings' "ends of groups" theorem: A finitely-generated group with infinitely many ends splits as graph of groups with finite edge groups. In addition, one also has to show that the decomposition process terminates for your group (this property is called accessibility ). Neither one has a quick an dirty proof and for a good reason. a. Dunwoody has shown that accessibility fails for some finitely-generated groups with torsion, so something interesting (Grushko's theorem) is going on even in the easy part of the proof. b. The ends of groups deals with the key problem of geometric group theory: Relating geometric properties of a group and its algebraic structure. Each time one manages to recover algebraic structure of a group from geometric information about the group, some minor (or major) miracle has to happen and, to the best of my knowledge (with few trivial exceptions) there are no easy proofs of the results of this type. The "shortest" proof of the "Ends of groups theorem" is due to Gromov, see pages 228-230 of his essay on hyperbolic groups. The trouble with Gromov's proof is that it relies on a compactness property ("obvious" to Gromov) for a certain family of harmonic functions (in addition, a construction of the tree was missing in his proof, and this requires some trickery if one uses harmonic functions for finitely-generated groups). This compactness result (as far as I know) has no easy proof. I wrote a (somewhat long) proof in http://arxiv.org/pdf/0707.4231.pdf , Bruce Kleiner managed to shorten it to about 7 pages (this is not published), but his proof is still not quick. Dunwoody's proof (see John Klein's excellent comments) improved on the Stallings' proof, but his proof is still quite complicated. Niblo's proof in http://eprints.soton.ac.uk/29820/1/Stallingstheorem.pdf provides another geometric argument using Sageev's complex, but Niblo's paper is still 20 pages long. Just to indicate how nontrivial Stallings' theorem is, consider the question: Is it true that every finitely generated group $G$ of homological dimension 1 is free? This is false for infinitely generated groups (like $G={\mathbb Q}$) and is true for finitely-presented groups (simply since in this case cohomological dimension is also 1). Otherwise, this problem is open since Stallings' theorem (and not for the lack of trying!).
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|
96,078 |
Products, are very elementary forms of categorical limits. My question is whether in the category of groups, semi-direct products are categorical limits. As was pointed in: http://unapologetic.wordpress.com/2007/03/08/split-exact-sequences-and-semidirect-products/ Bourbaki (General Topology, Prop. 27) gives a universal property: Let $f \colon N \to G$, $g \colon H \to G$ be two homomorphisms into a group $G$, such that $f(\phi_h(n)) = g(h)f(n)g(h^{-1})$ for all $n \in N$, $h \in H$. Then there is a unique homomorphism $k \colon N \rtimes H \to G$ extending $f$ and $g$ in the usual sense. However, I remain unsatisfied. The condition $f(\phi_h(n)) = g(h)f(n)g(h^{-1})$ is a condition on elements of groups, rather than a condition that says that some diagram is commutative. So the question remains: are semi-direct products in the category of groups categorical limits?
|
This is a partial answer, summing up some of my comments. The semi-direct product is not a limit, but rather it is a colimit. The reason is that the universal property cited above describes maps on the semi-direct product. In the special case that $\phi$ is the trivial action, the semi-direct product becomes the direct product $N \times H$ and the universal property is not just the usual universal property as a product, but rather as a representing object of the pairs of morphisms on $N,H$ which commute pointwise. In a general semi-direct product, this commutation is twisted by an action of $H$ on $N$. So basically the idea is that we have the coproduct $N * H$ of the two groups (which is usually called the free product, which is quite unfortunate), and we impose the relation $h n h^{-1} = \phi_h(n)$. The universal property of $N \rtimes H$ is equivalent to the isomorphism $$N \rtimes H = (N * H) / \{h n h^{-1}= \phi_h(n)\}_{h \in H, n \in N},$$ which exhibits $N \rtimes H$ as a special colimit of some diagram associated to $N,H,\phi$. However, this still uses elements in the relations. I think we cannot get rid of these elements, unless we use $2$-colimits. See below. Actually this isomorphism is used very often in group theory in order to recoqnize groups given by some presentation as a semi-direct product. For example, the dihedral group $D_n = \langle r,s : r^n = s^2 = 1, srs=r^{-1} \rangle$ is $\mathbb{Z}/n \rtimes \mathbb{Z}/2$. On the other hand, there is a purely category-theoretic construction which is due to Grothendieck: Let $I$ be a small category and $F : I \to \mathsf{Cat}$ be a diagram of small categories. The Grothendieck construction $\int^I F$ is the category of pairs $(i,x)$, where $i$ is an object of $I$ and $x$ is an object of $F(i)$. A morphism $(i,x) \to (j,y)$ is a pair $(a,f)$, consisting of a morphism $f : i \to j$ and a morphism $a : F(f)(x) \to y$ in $F(j)$. The composition is defined by the rule $(a_2,f_2) \circ (a_1,f_1) = (a_2 \circ F(f_2)(a_1),f_2 \circ f_1)$. Now if $H$ is a monoid, considered as a category with just one object $*$, and $F : H \to \mathsf{Cat}$ is a diagram such that $F(*)=N$ is just a monoid, then $F$ corresponds to a homomorphism of monoids $H \to \mathrm{End}(N)$ and the Grothendieck construction $\int^H N$ has just one object, thus corresponds to a monoid, namely what is usually called the semi-direct product $N \rtimes H$. This is shown by the multiplication rule above. Back to the general case of a diagram $F : I \to \mathsf{Cat}$, the Grothendieck construction $\int^I F$ is the lax 2-colimit of $F$. I don't know the original reference right now, but a very comprehensive account on that is the Appendix A in "The stack of microlocal sheaves" by I. Waschkies. The choice of the morphism $a : F(f)(x) \to y$ in the definition above is precisely the reason for the "2" here. If it was the identity, we would get the usual colimit. Thus, the semi-direct product $N \rtimes H$ is the lax $2$-colimit of the diagram $N : H \to \mathrm{Cat}$.
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96,219 |
I suppose most of you are familiar with the Mathematics Subject Classification (MSC) . Particularly, when submitting an article for publication one has to
choose appropriate classification codes. But I am wondering if it does play a significant role in searching for
literature on a specific subject, e.g. through MathSciNet or ZBMATH.
Or do you prefer searching by subject terms? Thank you very much!
|
At first, I thought this was a silly question, and it seemed highly implausible that searching by MSC code could actually be good for anything. Certainly it's a terrible way to find specific information, but I just gave it a try and it's quite a bit more useful than I thought. I put 52C17 in MathSciNet and got a list of 731 papers, and I've just been browsing through them and learning about a number of interesting papers I was unaware of. I should do this more often. So my answer is that it's a waste of time if you're looking for anything in particular, but it can actually be a good way to get an overview of what's been happening recently in an area.
|
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96,378 |
Let $M$ be a compact complex connected [but not necessarily kähler ] $n$-manifold, and suppose we have a holomorphic map $$(\mathbb{C}^*)^n \to M$$ such that the image is open. Is the image necessarily dense in $M$? Motivation: My intuition (which comes from the algebraic world) says that the answer ought to be "yes." On the other hand, I know that many properties of smooth algebraic varieties do not hold for complex manifolds in general. Knowing whether this statement has a counterexample would improve my intuition about the complex world.
|
An example exists already for $M={\mathbb C}P^2$, furthermore, there exists an injective holomorphic map $f: {\mathbb C}^2\to {\mathbb C}^2\subset {\mathbb C}P^2$ whose image is open but not dense. Recall that a domain $\Omega$ in ${\mathbb C}^2$ is called a Fatou-Bieberbach (FB) domain if $\Omega\ne {\mathbb C}^2$ and there exists a biholomorphic map $f: {\mathbb C}^2\to \Omega$. First examples of FB domains were constructed by Fatou and Bieberbach and it is a bit of an industry to construct FB domains with interesting properties. For instance, B. Stensönes constructed in her paper ("Fatou-Bieberbach domains with smooth boundary", Annals of Math, vol. 145, 1997, 365-377) FB domains in ${\mathbb C}^2$ whose boundaries are smooth. If you look at Proposition 3.1 (part v) of her paper, it shows that the complement of her FB domain has nonempty interior. I am pretty sure that one can find earlier examples as well. Now, if you want the domain of $f$ to be $({\mathbb C}^\times)^2$, just restrict the above holomorphic map.
|
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96,809 |
Recently I encountered a new phenomenon when I tried to submit a paper to arXiv. The paper was an erratum to another, already published, paper and will be published separately. I got a message from arXiv saying that I need to join the erratum with the original file. I was a little surprised receiving a reply from, obviously, a human being. Although I thought the request was a bit silly, I did what was requested, submitted the joint paper (the original union the errata), and forgot about it. But today I got a call from another mathematician. She tried to submit a paper with a title "... II". The paper "... I" was already in the arXiv and submitted to a (very good) journal. Both papers solve similar but different problems. One of these problems is at least 40 years old. Her submission was denied: she got a request from the arXiv to submit a union of that new paper and the old paper instead. This is quite silly. Is there now a special person in the arXiv who is making these decisions? It looks like there has been a change in how arXiv is managed. I understand that this is not a research question, and I make it a community Wiki. I post it here because several frequent MO users are affiliated with arXiv.
|
I'm still the chair of the math arXiv advisory committee, which admittedly hasn't done a whole lot lately, and one of the global math moderators. No, there has not been any dramatic change in the management of the arXiv at Cornell. If anything, I wish that by now more might have changed. The arXiv has always had the bare minimum funding, sometimes less than the bare minimum. They have never had polished public relations to properly explain small changes in policy. (Actually even wealthy Internet companies sometimes stir up confusion when they make changes.) At some informal level, they/we have always worried about duplicate submissions, and near duplicates, and errata posted as new papers. And yes there is a new text overlap tool to detect both plagiarism and self-plagiarism. There is no good, rigorous way to draw the line for any of these issues. (Just as there isn't at MathOverflow --- what exactly is an "exact duplicate" of a previous question?) Regardless, if your submission is rejected, you do have the right to "file" an appeal with the Cornell staff. If it is a plausibly sane appeal, then they should show it to the math moderators and/or the math advisory committee, more likely the former these days. One perfectly valid consideration is to have the arXiv correspond to what is published in journals. Although there are cases where strict adherence to that rule is untenable. For instance, my mother and I have a joint paper in the Annals of Mathematics that appeared twice just because the first time, the paper had TeX symbol encoding errors. Also, I personally think that this posting is reasonable for MathOverflow. However, it would have been better with a less suspecting tone. The arXiv doesn't always make the best impression, but long-time users know that actually it has gotten better over the years. For a long time it had a reputation as a "user belligerent" web site. Even then, it was still a force for good, obviously.
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96,832 |
1) Given $p\in (1,\infty)$. 2) Let us fix two, non-isometric subspaces $X,Y\subseteq \ell_p$ isomorphic to $\ell_p$. 3) Are there an $\varepsilon\in (0,1)$ and an isomorphism $S\colon X\to Y$ such that $$(1-\varepsilon)\|x\|\leqslant \|Sx\|\leqslant (1+\varepsilon)\|x\|$$ holds for each $x\in X$?
|
I'm still the chair of the math arXiv advisory committee, which admittedly hasn't done a whole lot lately, and one of the global math moderators. No, there has not been any dramatic change in the management of the arXiv at Cornell. If anything, I wish that by now more might have changed. The arXiv has always had the bare minimum funding, sometimes less than the bare minimum. They have never had polished public relations to properly explain small changes in policy. (Actually even wealthy Internet companies sometimes stir up confusion when they make changes.) At some informal level, they/we have always worried about duplicate submissions, and near duplicates, and errata posted as new papers. And yes there is a new text overlap tool to detect both plagiarism and self-plagiarism. There is no good, rigorous way to draw the line for any of these issues. (Just as there isn't at MathOverflow --- what exactly is an "exact duplicate" of a previous question?) Regardless, if your submission is rejected, you do have the right to "file" an appeal with the Cornell staff. If it is a plausibly sane appeal, then they should show it to the math moderators and/or the math advisory committee, more likely the former these days. One perfectly valid consideration is to have the arXiv correspond to what is published in journals. Although there are cases where strict adherence to that rule is untenable. For instance, my mother and I have a joint paper in the Annals of Mathematics that appeared twice just because the first time, the paper had TeX symbol encoding errors. Also, I personally think that this posting is reasonable for MathOverflow. However, it would have been better with a less suspecting tone. The arXiv doesn't always make the best impression, but long-time users know that actually it has gotten better over the years. For a long time it had a reputation as a "user belligerent" web site. Even then, it was still a force for good, obviously.
|
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|
96,914 |
What are good/interesting examples of theorems than can be proven classically, but not constructively, and have applications in e.g. physics?
|
In general it is very difficult to be sure that a theorem cannot be constructivised in some form that preserves its applicability. As you will notice most of the answers offered have comments attesting this fact. One reason for this is that many non-constructive theorems in analysis become constructive when they are relaxed a little bit. A typical example is the mean value theorem. It does not hold constructively as usually stated, but it its $\epsilon$ version does: if $f$ is continuous and $f(0) < 0 < f(1)$ then for every $\epsilon > 0$ there is $x \in [0,1]$ such that $|f(x)| < \epsilon$. Many other theorems can be relaxed in this way: Hahn-Banach, Brouwer fixed-point, etc. Moreover, such relaxed versions often make more sense in applications than their exact versions, for example because we need to take into account noise, errors, or bounded numerical precision. Another reason is that for applications we typically do not need a theorem in its full generality because we have extra information, which allows for a specialized constructive version. For example, while the general mean value theorem fails constructively, it holds for locally non-constant maps. Special versions of Hahn-Banach holds constructively, and they will typically suffice in concrete situations. There is a third reason why it is difficult to find classical theorems with applications that cannot be constructivized. Most applications belong to the fields of physics and computer science, which are both very naturally constructive. Physics is constructive by its very nature because "everything is continuous" in the real world, while in computer science "everything is computable". These are two main motivations for intuitionistic mathematics (namely Brouwerian continuity principles or sheaf-theoretic models, and computable interpretations of intuitionistic mathematics). Lastly, there remains the simple fact that one has to perform an exhaustive literature search to be sure that a theorem has not been constructivized. A lot more has been constructivized than one would think, and the only obstacle seems to be lack of man power.
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97,041 |
The basic question that I have is in the title, but let us make it more rigorous below. Let $N=\{1, 2, ..., n\}$, and put the (normalized) counting measure, $\mu_n$, on $N\times N$. Let $\mathcal{S}_n= \{ (a, b)\in N\times N: gcd(a, b)=1\}$ and $x_n=\mu_n(\mathcal{S}_n).$ Then what is the assymptotic behavior of $x_n$ as $n\rightarrow\infty$.
|
The probability tends to $\frac{1}{\zeta(2)}=\frac{6}{\pi^2}$ as was mentioned by Qiaochu. This actually generalizes to arbitrary number fields, and is a less commonly known fact. In fact in any number field, the probability that two ideals are relatively prime is given by $1/\zeta_K(2)$ , where $\zeta_K$ is the Dedekind zeta function of the number field $K$ . And is proven in a similar way to the classical result. Here is a reference: "The probability of relative primality of Gaussian integers" . For example the analogous probability for Gaussian integers is $6/(\pi^2G)$ where $G=1-\frac{1}{3^2}+\frac{1}{5^2}+\cdots$ is the Catalan constant .
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97,092 |
I would like to get a better idea of how badly compactness fails in $\mathcal{L}_{\omega_1\omega}$. Let $\Gamma$ be an arbitrary set of sentences from $\mathcal{L}_{\omega_1\omega}$. Let the underlying signature $\tau$ also have arbitrary cardinality. Is there some cardinal $\kappa $ such that if every $\Delta\subseteq\Gamma$ where $|\Delta|\leq\kappa$ is satisfiable, then $\Gamma$ is satisfiable? It is relatively easy to show that any such $\kappa$ would need to be $\geq \beth_{\omega_1}$, but I am unsure of how to proceed beyond there. If there is no such $\kappa$, I am also interested in weakening the question.
|
I like the question very much. First, let me mention briefly that the question has a flaw in the
quantifier order, since you have first fixed the theory $\Gamma$
and then ask for a cardinal $\kappa$ such that if all subtheories
$\Delta\subset\Gamma$ of size at most $\kappa$ are consistent,
then $\Gamma$ is consistent. This is trivially affirmative, since
we may simply let $\kappa=|\Gamma|$,
in which case $\Delta=\Gamma$ is one of the allowed subtheories. The actual question here is the following (and note that I replace your $\leq\kappa$ with $\lt\kappa$, since this is how one usually frames it with weakly and strongly compact cardinals): Question. Is there are a cardinal $\kappa$ such that if
$\Gamma$ is any $L_{\omega_1,\omega}$ theory in any signature, and
every $\kappa$-small subtheory is consistent, then $\Gamma$ is
consistent? Let us call this property the $\kappa$-compactness property for
$L_{\omega_1,\omega}$. Just to be clear, $L_{\omega_1,\omega}$ is
the infinitary language in which one is allowed to form countable
conjunctions and disjunctions, but still only finitely many
quantifiers at a time. Meanwhile, $L_{\kappa,\lambda}$ allows
conjunctions and disjunctions of size less than $\kappa$ and
blocks of quantifiers of size less than $\lambda$. One
instinctively thinks of the following large cardinals: A cardinal $\kappa$ is weakly compact if and only if it is
uncountable and $L_{\kappa,\kappa}$ has the $\kappa$-compactness property for theories in a language of size at most $\kappa$. A cardinal $\kappa$ is strongly compact if and only if it is uncountable and
$L_{\kappa,\kappa}$ has the $\kappa$-compactness property for any
theory without any size restriction. One crucial difference between $L_{\omega_1,\omega}$ and
$L_{\kappa,\kappa}$ or even $L_{\omega_1,\omega_1}$ is that in
$L_{\omega_1,\omega_1}$, one can express the assertion that a
relation is well-founded, since you can say that it has no
infinite descending sequence. This does not seem possible to
express in $L_{\omega_1,\omega}$, because one can quantify only
finitely many variables at a time. Theorem. If $\kappa$ is strongly compact, then
$L_{\omega_1,\omega}$ has the $\kappa$-compactness property. Proof. This is immediate, since any $L_{\omega_1,\omega}$ theory
is also a $L_{\kappa,\kappa}$ theory. QED Theorem. If $L_{\omega_1,\omega}$ has the $\kappa$-compactness
property, then there is a measurable cardinal. Proof. Suppose that $L_{\omega_1,\omega}$ has the
$\kappa$-compactness property. Let $\Gamma$ be the theory
including the following assertions: the full $L_{\omega_1,\omega}$ diagram of the structure $\langle
\kappa,\in,\hat A\rangle_{A\subset \kappa}$, in the language with a
predicate $\hat A$ for each $A\subset\kappa$ and constants
$\hat\alpha$ for each $\alpha\in\kappa$. the assertions $c\neq \hat\alpha$ for each $\alpha\in\kappa$. Note that any $\kappa$-small subtheory of $\Gamma$ is consistent,
since we may interpret $c$ inside $\kappa$ if only fewer than
$\kappa$ many $\alpha$ are excluded. So by the
$\kappa$-compactness property, $\Gamma$ has a model $\langle
M,\hat\in,\hat A^M\rangle$. Let $U$ be the set of $A$ for which
$M\models c\in\hat A$. This $U$ is an ultrafilter and it is
countably complete, since the assertions $(\forall x. \bigwedge_n
x\in A_n)\to x\in A$, whenever $A=\cap A_n$, are part of $\Gamma$.
It is nonprincipal since $c\neq \hat\alpha$ for any $\alpha$. So
there is a countably complete nonprincipal ultrafilter, and hence
there is a measurable cardinal, since the degree of completeness
of such an ultrafilter is always measurable. QED In particular, the hypothesis is strictly stronger than a weakly
compact cardinal. I'm not yet sure of the exact strength, but I'm inclined to think it is equiconsistent with a strongly compact cardinal, in light of the following observation: Theorem. If $\kappa$ is the least measurable cardinal, then
$L_{\omega_1,\omega}$ has the $\kappa$-compactness property if and
only if $\kappa$ is strongly compact. Proof. Suppose $\kappa$ is the smallest measurable cardinal and
$L_{\omega_1,\omega}$ has the $\kappa$-compactness property. Fix
any regular cardinal $\theta\geq\kappa$, and let $\Gamma$ be the
$L_{\omega_1,\omega}$ theory of $\langle\theta,\in,\hat\alpha,\hat
A\rangle_{\alpha\in\theta,A\subset\theta}$, plus the assertion
$\hat\alpha\lt c$ for each $\alpha\in\theta$. This theory is
$\kappa$-satisfiable, and indeed, $\theta$-satisfiable. So it is
satisfiable, and there is therefore a model $\langle
M,\in^M,\hat\alpha^M,\hat A^M,c^M\rangle$. Let $U$ be the set of
$A\subset \theta$ such that $M\models c\in \hat A$. This is a
countably complete nonprincipal uniform ultrafilter on $\theta$.
Since $\kappa$ is the least measurable cardinal, it follows that
$U$ must be $\kappa$-complete. So we have a $\kappa$-complete
nonprincipal uniform ultrafilter on every regular
$\theta\geq\kappa$. By a theorem of Menas, this implies that
$\kappa$ is strongly compact. QED Note that results of Magidor show that it is indeed possible for the least measurable cardinal to be strongly compact.
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|
97,381 |
I need this result for something else. It seems fairly hard, but I may be missing something obvious. Just one non-trivial solution for any given $c$ would be fine (for my application).
|
[edited again mainly to add the Euler link, see last paragraph] Yes, and indeed there are infinitely many rational points:
the birationally equivalent Diophantine equation
given by J.Ramsden in his partial answer to his own question,
$$
X + Y = Z + T,
\phantom{and}
XYZT = c,
$$
was already studied by Euler (in the equivalent form $xyz(x+y+z)=a$),
who in 1749 obtained the rational curve of solutions
$$
X = 6\frac{cr(2r^4+c)^2}{(r^4-4c)(r^8+10cr^4-2c^2)},
\phantom{and}
Y = -2\frac{r^8+10cr^4-2c^2}{3r^3(r^4-4c)},
$$ $$
Z = -3\frac{r^5(r^4-4c)^2}{2(2r^4+c)(r^8+10cr^4-2c^2)},
\phantom{and}
T = \frac{r^8+10cr^4-2c^2}{r^3(2r^4+c)}.
$$
Inverting J.Ramsden's birational transformation
$$
(X,Y,Z,T) = \left(
\frac{x-1}{z},
\frac{y+1}{z},
\frac{x+1}{z},
\frac{y-1}{z}
\right)
$$
then yields
$$
x = \frac{(r^8+2c^2)(r^8-44cr^4-2c^2)}{(r^8+10cr^4-2c^2)^2},
\phantom{and}
y = \frac{7r^4+8c}{9r^4},
\phantom{and}
z = \frac{-4(r^4-4c)(2r^4+c)}{3r(r^8+10cr^4-2c^2)}.
$$ Euler's solution is unpublished and somewhat mysterious;
he gave the formulas in a letter to Goldbach but didn't explain
how he found them, writing only that he obtained the solution
"endlich nach vieler angewandter Mühe" [at last, after applying much effort].
The curve is singular, and finding such a curve requires
going somewhat beyond the usual manipulation of
elliptic fibrations on K3 surfaces, which for these surfaces will not work unless
$c$ is a square (the case in which R.Kloosterman already found a nonsingular
rational curve of solutions). I've lectured on Euler's surface and solution
a number of times over the past few years; here 's
the latest iteration. Thanks to Franz Lemmermeyer for bringing
Euler's letter to my attention. Added later: The same lecture notes show on page 43 a somewhat simpler curve that I found on Euler's surface,
which also yields a somewhat simpler curve of solutions of
$(x^2-1)(y^2-1)=cz^4$:
$$
x = \frac{u^{12}+48cu^8-48c^2u^4+128c^3}{u^4(u^4-12c)^2},
\phantom{and}
y = \frac{5u^4+4c}{3u^4-4c},
\phantom{and}
z = \frac{4(u^4+4c)}{u(u^4-12c)}.
$$
For $c=2$, we get at $u=2$ the solution $(26, 11/5, 6)$ which is almost
as simple as the solution $(5/3,17,4)$ that Mark Sapir noted. Added later yet: Here's a transcription of Euler's 1749 letter . See the last page.
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|
97,464 |
The last remaining problem in this whole "everything is a sphere" business, is the Smooth Poincare Conjecture in dimension 4: If $X\simeq_\text{homo.eq.} S^4$ then $X\approx_\text{diffeo} S^4$. Freedman showed that this holds if we replace "diffeomorphism" by "homeomorphism", so another viewpoint would be that $S^4$ has no exotic smooth structures. Restated: If $X$ is a connected, closed smooth 4-manifold with $\pi_1X=1$ and $H_*X\cong H_*S^4$, then $X\approx_\text{diffeo}S^4$. This was told to me by Michael Hutchings, who motivationally remarked a way to solve it (Edit: not an original suggestion of his): Find a symplectic structure on $X-\lbrace pt\rbrace$ which is standard near the puncture-point. Then we're done by "Recognition of $\mathbb{R}^4$": Let $(M,\omega)$ be a noncompact symplectic 4-manifold such that $H_\ast(M)\cong H_\ast (pt)$. Suppose there exist compact sets $K_0\subset M$ and $K_1\subset\mathbb{R}^4$ and a symplectomorphism $\phi:(M-K_0,\omega)\to(\mathbb{R}^4-K_1,\omega_\text{std})$. Then $\phi$ extends to a symplectomorphism $(M,\omega)\to(\mathbb{R}^4,\omega_\text{std})$, after removing slightly bigger compact sets. I am ignorant to the size of this wall (the conjecture) and the ability to make an indent in it. But the above idea is pretty cool, even if not anymore helpful than the original statement. The Poincare Conjecture (in dimension 3) was solved using Hamilton's idea of Ricci flow. This leads me to ask: Is there another idea proposed to tackle this conjecture? Or a failed attempt?
|
In principle the Ricci flow (with surgery) could also be used to prove the smooth Poincare' conjecture in dimension $4$. There are some major problems to be overcome in this approach (problems which did not arise in dimension $3$, such as the absence of Hamilton-Ivey pinching estimates, and the new "hole-punch" singularities, as opposed to the "neck-pinch") and I know that some people are indeed working on these issues. If you are interested, there are quite a few papers on Ricci flow on $4$-manifolds, starting with the work of Hamilton ( here and here ) and more recent work of Chen-Zhu ( here and here ) and many others.
|
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|
97,495 |
There are the following Nakayama style lemmata: (the classical Nakayama lemma) Let $R$ be a commutative ring with $1$ and $M$ a finitely generated $R$-module. If $m_1, \ldots, m_n$ generate $M$ modulo $I$, where $I \subset \mathrm{Jac}(R)$, then they generate $M$. (the graded Nakayama lemma) See How to memorise (understand) Nakayama's lemma and its corollaries? . (the filtered Nakayama lemma) See How to memorise (understand) Nakayama's lemma and its corollaries? . (the topological Nakayama lemma, see [Neukirch, Schmidt, Wingberg], Cohomology of Number Fields, (5.2.18)): Let $\mathcal{O}$ be a commutative local ring complete in the $\mathfrak{m}$-adic topology with finite residue field of characteristic $p$. Assume $G$ is a pro-$p$-group and $M$ is a compact $\mathcal{O}[[G]]$-module. If $M/\mathfrak{m}$ is a finitely generated $\mathcal{O}[[G]]$-module, so is $M$. (Burnside's basis theorem, see also http://groupprops.subwiki.org/wiki/Burnside%27s_basis_theorem ) Let $G$ be a group such that its Frattini subgroup $\Phi(G)$ is finitely generated. Then a subset of $G$ generates $G$ iff it generates it modulo $\Phi(G)$. [tbc] Now my question is: Is there a common categorical version, like there is a categorical generalisation of Baer's criterion (In a suitable abelian catgory, an object $I$ is injective iff for all subobjects $U$ of a generator $G$ and morphisms $U \to I$ there is a lift $G \to I$)?
|
Let me describe a common generalization of Nakayama's lemmas and Burnside's basis theorem which may shed some light here. Let $G$ be a group and $P$ a set of endomorphisms of $G$. A $P$-subgroup will be a subgroup of $G$ which is closed under acting by elements of $P$. We'll call $\mathbb{Fr}_P(G)$ the "$P$-Frattini subgroup of $G$", defined as the intersection of all maximal $P$-subgroups. Note the following special cases: When $P$ is empty then $\mathbb{Fr}_P(G)$ is the Frattini subgroup of $G$. When $R$ is a ring, $G$ its additive group and $P$ its multiplicative semigroup then $\mathbb{Fr}_P(G)$ is $J(R)$, the Jacobson radical of $R$. When $P$ is as above, and $G$ is an $R$-module then $\mathbb{Fr}_P(G)$ contains $J(R)G$. Let's denote the smallest $P$-subgroup containing a set $S$ by $\langle S\rangle_P$, and call an element $x\in G$, a non-generator if $G=\langle S,x \rangle_P$ always implies $G=\langle S\rangle_P$. We have the following theorem: The set of non-generators of $G$ is precisely $\mathbb{Fr}_P(G)$. By taking $P$ empty we obtain Burnside's basis theorem. By taking $G$ to be an $R$-module and $P$ to be the multiplicative semigroup of $R$ we recover Nakayama's lemma. If $R$ is a graded ring and we take $P$ to be the semigroup of elements of positive degree and $G$ to be a graded $R$-module, we recover the graded version of Nakayama's lemma, something similar should hold for the filtered version. Surely someone has taken up this point of view (Which I learned from Gruenberg's "Cohomological topics in group theory") to define a Frattini object for a large class of categories?
|
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|
97,512 |
While browsing the Net for some articles related to the history of the Whittaker-Shannon sampling theorem, so important to our digital world today, I came across this passage by H. D. Luke in The Origins of the Sampling Theorem : However, this history also reveals a process which is often apparent in theoretical problems in technology or physics: first the practicians put forward a rule of thumb, then the theoreticians develop the general solution, and finally someone discovers that the mathematicians have long since solved the mathematical problem which it contains, but in " splendid isolation ." Other interesting examples? (Matrices and Bohr's Quantum Mechanics of course. Someone could elaborate on the sampling theorem if they wish.)
|
Cormack and Hounsfield received the 1979 Nobel prize in medicine for their work on CT scans. Cormack, a physicist, published his mathematical work on this in 1963, to essentially no response. Hounsfield, an engineer, built the first CT scanner in 1971 unaware of Cormack's work. Cormark included the following in his Nobel prize speech: "If a fine beam of gamma-rays of intensity $I_0$ is incident on the body and the emerging intensity is $I$ , then the
measurable quantity is $g = \ln(I_0/I) = \int_L f ds$ , where $f$ is
the variable absorption coefficient along the line $L$ .
Hence if $f$ is a function in two dimensions, and $g$ is
known for all lines [...], the question is: Can $f$ be
determined if $g$ is known? This seemed like a problem
which would have been solved before, probably in the
19th century, but a literature search and enquiries of
mathematicians provided no information about it.
Fourteen years would elapse before I learned that Radon
had solved this problem in 1917." Fourteen years after Cormack's work means 1977, so Radon's work was rediscovered by the people involved with creating CT scan technology only after CT scan's had been around for several years. (Search on "Radon transform" for more information.) Radon's work was rediscovered multiple times: Cramer and Wold (1936) in probability theory, Ambartsumian (1936) in astronomy, Bracewell (1956) in astronomy, De Rosier and Klug (1968) in chemistry. In fact, Radon's basic idea was worked out before Radon , by Funk (1916) and Lorentz (1905). This work of Lorentz was unpublished, but a formula he found is mentioned in a paper by Bockwinkel in 1906. More on this history is in Cormack's survey paper Computed tomography: some history and recent developments , pp. 35--42 in "Computed tomography: Proceedings of Symposia in Applied Mathematics" 27, AMS, 1983. Shortly before the work of Cormack, Oldendorf (a medical doctor in LA) published a paper in 1961 describing a crude CT scanner he had built out of household parts, such as model railroad tracks (!) but it went unnoticed. Hounsfield acknowledged it, but Oldendorf was not included in the Nobel prize list with Cormack and Hounsfield. He once said in an interview "I think Professor Cormack was selected [for the Nobel prize] because he worked out all the line integrals mathematically. [...] I didn't provide any mathematical treatment of it, and that apparently carried a lot of weight with the Nobel committee. See https://en.wikipedia.org/wiki/William_H._Oldendorf for more on his story. The mathematical and engineering concepts in CT scan technology, with applications
to medical imaging, were worked out in an obscure journal in Kiev by S. T. Tetelbaum
in 1957-58, before Oldendorf!
|
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|
97,830 |
Homotopy groups of Lie groups I asked it also there, and I still don't know the answer, so I try again. I would like to know a closed manifold (possibly of low dimension) such that $\pi_2(\textrm{Diff}(M))\neq 0$ .
|
$\newcommand{\Diff}{\mathrm{Diff}}$ Probably the simplest such manifold is $S^1 \times S^2$ , whose diffeomorphism group has the homotopy type of $O(2) \times O(3) \times \Omega SO(3)$ . This has $\pi_2$ equal to $\pi_2\Omega SO(3)=\pi_3 SO(3) = {\mathbb Z}$ . The $\Omega SO(3)$ term is realized by rotating the $S^2$ slices of $S^1\times S^2$ by an element of $SO(3)$ that varies as one goes around the $S^1$ factor of $S^1 \times S^2$ . This calculation was originally done in a paper of mine: On the diffeomorphism group of $S^1\times S^2$ , Proc. Amer. Math. Soc. 83 (1981), 427-430, doi: 10.1090/S0002-9939-1981-0624946-2 . An updated version of this paper is posted on my webpage . Added later: In higher dimensions, spheres provide interesting examples. By an elementary argument there is a homotopy equivalence $\Diff(S^n) \simeq O(n+1) \times \Diff_\partial(D^n)$ for all $n$ , where the subscript $\partial$ denotes diffeomorphisms that restrict to the identity on the boundary of $D^n$ . The question is whether $\Diff_\partial(D^n)$ is contractible. The status of this is: true for $n\le3$ , unknown for $n=4$ , and false for each $n\ge 5$ . The noncontractibility can be deduced from the sequence $$\cdots\ \to \pi_2 \Diff_\partial(D^{n-2})\to \pi_1 \Diff_\partial(D^{n-1})\to \pi_0 \Diff_\partial(D^n) = \Theta_{n+1}$$ where $\Theta_{n+1}$ is the group of exotic (n+1)-spheres and the equality $\pi_0 \Diff_\partial(D^n) = \Theta_{n+1}$ is assuming $n\ge 5$ . Usually $\pi_0 \Diff_\partial(D^n)$ is nonzero since exotic spheres exist in most dimensions greater than $6$ , and in the rare dimensions in which they don't exist one can appeal to known results about how far some elements of $\Theta_{n+1}$ pull back in the sequence above. For example, Cerf's pseudoisotopy theorem says the map from $\pi_1 \Diff_\partial(D^{n-1})$ to $ \pi_0 \Diff_\partial(D^n)$ is surjective for all $n\ge 6$ , so in particular $\pi_1 \Diff_\partial(D^5)$ is nonzero since it maps onto $\Theta_7={\mathbb Z}/28$ . A paper of Crowley and Schick posted on the arXiv last month shows there is a nontrivial element of $\Theta_{8k+2}$ that pulls all the way back to $\pi_{8k-6} \Diff_\partial (D^7)$ , hence $\Diff_\partial(D^n)$ has infinitely many nontrivial homotopy groups for all $n\ge 7$ . Since the groups $\Theta_{n+1}$ are finite (apart perhaps from the unknown $\Theta_4$ ), these constructions don't give nontrivial rational homotopy groups of $\Diff_\partial(D^n)$ , but there is another construction that does, coming from algebraic K-theory. Using Waldhausen's big machine, Farrell and Hsiang in 1978 computed $\pi_i \Diff_\partial (D^n) \otimes{\mathbb Q}$ in a stable range $n\gg i$ to be ${\mathbb Q}$ for $i\equiv 3$ mod $4$ and $n$ odd, and $0$ otherwise. (This is the result mentioned in Vitali Kapovitch's answer.)
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|
98,007 |
This is a direct (and obvious) generalization of the recent MO question, " Covering disks with smaller disks ": How many balls of radius $\frac{1}{2}$ are needed to cover completely a ball of radius 1? The answer may be in the paper "Covering a Ball with Smaller Equal Balls in $\mathbb{R}^n$ ,"
by Jean-Louis Verger-Gaugry, which I cannot immediately access ( Springer link here ).
If anyone knows the answer for $\mathbb{R}^3$ , I'd be curious to learn of it.
And it would be especially interesting if there were a proof as satisfying as Noam Elkies's for seven half-disks covering one in $\mathbb{R}^2$ . Thanks! Update . [ 29May2012 ] A series of contributions by Will Jagy, Gerhard Paseman, Karl Fabian, and zy, have reduced the upper bound from $56$ to $33$ to $22$ , with a lower bound of $16$ . Update . [ 5Aug2012 ] Ed Wynn settled the question with a careful analysis: $21$ balls are needed, and suffice!
|
Here is (I'm fairly sure) an optimal solution, building on the ideas in other answers. We know that we can generate maximum 30° caps on the sphere, by placing the half-balls at radius sqrt(3)/2. We deduce from http://neilsloane.com/coverings/index.html that we will need 20 half-balls to cover the outer surface, plus a 21st to cover the centre. So, 21 is a lower bound. To get a 21-ball solution, scale the 20 coordinates of http://neilsloane.com/coverings/dim3/cover.3.20.txt to that radius, and add one at the origin. Note that the balls at sqrt(3)/2 give caps on the central half-ball with the same angle as the outer caps, so the cover works on the inside too. Here are the coordinates of 21 half-balls to cover the ball: {{+0.060111,-0.479945,-0.718359},{-0.559060,+0.562759,+0.347496},{-0.164091,-0.848680,-0.053063},{-0.509173,-0.375273,-0.591535},{-0.140963,+0.208121,+0.828743},{+0.690970,+0.512659,-0.098697},{+0.377263,-0.700146,+0.342736},{-0.250648,-0.504068,+0.658097},{-0.743210,+0.007958,+0.444494},{+0.550076,+0.074555,-0.664724},{+0.521328,-0.588831,-0.362624},{+0.088527,+0.791576,+0.339956},{-0.158465,+0.221391,-0.822116},{-0.718586,-0.473046,+0.099309},{-0.339530,+0.726217,-0.327609},{+0.439846,-0.237215,+0.707294},{+0.853710,-0.140538,+0.037800},{+0.536789,+0.335659,+0.590924},{+0.243253,+0.709202,-0.433429},{-0.778146,+0.197643,-0.324694},{+0.000000,+0.000000,+0.000000}} There is some "fat" in the surface covers (using 30° caps when 29.6..° is required), and each non-central sphere bulges convexly out of the cone that it's responsible for, so it's easy to believe that this is indeed a cover. Also I've checked it thoroughly, though I'll be grateful if someone checks it with less makeshift methods. At the positions stated, the radius of the balls can be as low as 0.49812. By moving the original coordinates further inwards, to radius 0.8595, the radius of the balls can be as low as 0.49439. Ed Wynn, 4 August 2012. [ Graphic added by J.O'Rourke :]
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|
98,229 |
In other words: What is $\mathrm{Ext}_{\mathcal{A}}^{4,t}(\mathbb{Z}/2,\mathbb{Z}/2)$? If the 4-line is not known, how much is known about it? Here, $\mathcal{A}$ is the 2-primary Steenrod algebra, $4$ is the homological degree corresponding to the Adams filtration, and $t$ is the internal grading degree. Those $\mathrm{Ext}$ groups make up the fourth row of the classical Adams spectral sequence $E_2 = \mathrm{Ext}_{\mathcal{A}}^{s,t}(\mathbb{Z}/2,\mathbb{Z}/2)$ converging to the 2-adic completion of the $(t-s)^{\mathrm{th}}$ stable homotopy group of the sphere. For context, the 1-line is generated by the classes $h_i$, $i \geq 0$, ($\mathrm{deg}\: h_i = (1,2^i)$), the 2-line is generated by the product classes $h_i h_j$, subject to the relations $h_i h_{i+1} = 0$ and $h_i h_j = h_j h_i$, the 3-line is generated by two sets of classes, the product classes $h_i h_j h_k$, subject to the relations implied by $h_i h_{i+2}^2 = 0$, $h_{i+1}^3 = h_i^2 h_{i+2}$, $h_i h_{i+1} = 0$, and $h_i h_j = h_j h_i$, the Massey products $\langle h_{i+1},h_i,h_{i+2}^2 \rangle$.
|
The 4-line is determined by Wen-Hsiung Lin in "$Ext_A^{4,*}({\bf Z}/2,{\bf Z}/2) $ and $Ext_A^{5,*}({\bf Z}/2,{\bf Z}/2) $", Topology and its Applications (2008) vol 155 no.5 pp 459-496. He gives a basis for the indecomposable elements in $Ext_A^{4,*}$ and generators and relations for the quotient of $Ext_A^{s,*} $ for $s \le 4$ by the indecomposables of $Ext_A^{4,*}
|
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|
98,366 |
If $X$ is a nonsingular algebraic (or analytic) variety over $\mathbb C$ or $\mathbb R$ then it is certainly $C^\infty$ over the reals. The converse is false for a silly reason : in the real or complex affine plane with coordinates $x,y$ the variety $x^2=0$ is singular since it is not reduced, but set-theoretically it is the $y$-axis, a $C^\infty$ submanifold of the plane. So let me concentrate on reduced varieties. Over the reals you have the disturbing phenomenon that the plane curve $C$ defined by $y^3+2x^2y-x^4=0$ is algebraically singular at the origin, but your favourite computer graphics system won't show you that, because the curve is actually $C^\infty$. Worse: it is the graph of a real analytic function! (I learned this example in Milnor's Singular Points of Complex Hypersurfaces ) I think this pathology is impossible over $\mathbb C$ but I cannot find a reference. So let me ask the official question: Can a reduced algebraic (or analytic) variety which has a singularity at a point be $C^\infty$ at that point? (Information on the more complicated real case welcome, of course) Edit Francesco's answer shows that indeed a singular complex variety cannot be smooth. This is quite interesting, because many books will give you the Jacobian criterion for a map $\mathbb C^n\to \mathbb C^k$ to define a submanifold but none (to my knowledge) adds that this Jacobian criterion also allows to prove nonsmoothness (modulo some technicalities), which is paradoxically a more delicate question. Second Edit (June 29th,2012)) Here is a proof more geometric than Milnor's that the curve $X$ defined by $y^3+2x^2y-x^4=0$ is an analytic submanifold of $\mathbb R^2$. The curve $X$ is actually rational: it is the image of $\mathbb R$ under $\phi (t)=(t(t^2+2),t^2(t^2+2))$. The analytic morphism $\phi$ is injective, proper and immersive, hence an embedding into $\mathbb R^2$ with closed image $X$. [It boils down to the fact that the polynomial $t^3+2t$ has positive derivative $3t^2+2$ and is thus strictly increasing!]
|
NB: This answer is directed to the questions about the real case, not the complex case, which was already treated by Francesco. Added 5 July 2021: Because of some questions I have received over the years since this was written, I have realized that there are a few places where the logic and flow are not completely transparent, so I have decided to make a few small changes to clarify those points. In some sense, the reason you are running into these 'problems' is that you are working with the ring of real polynomials rather than the ring of (germs of) real-valued analytic functions, which is also a UFD. For example, it is not hard to show that there is a (unique) real-valued, real-analytic function $f$ defined in a neighborhood of $0$ and satisfying $f(0)=\frac12$ such that $$
y^3 + 2x^2y-x^4 = \bigl(y - x^2f(x^2)\bigr)\bigl(y^2 + x^2f(x^2) y + x^2/f(x^2)\bigr),
$$ so $y^3 + 2x^2y-x^4$ is reducible in the ring of real-valued, real analytic functions defined on a neighborhood of the origin. The curve $y = x^2f(x^2)$ is smooth (in fact, real-analytic, of course), but the $y$ -discriminant of the quadratic factor is $x^4f(x^2)^2-4x^2/f(x^2) = -8x^2 + \cdots$ , so the only real point of $y^2 + x^2f(x^2) y + x^2/f(x^2)=0$ near the origin is the origin itself. (The quadratic factor is irreducible in the ring of real-valued analytic functions defined on a neighborhood of the origin.) Thus, a more easily approached question is: Suppose that the origin is a singular zero of an irreducible element $f$ in the ring of real-valued analytic functions defined on a neighborhood of the origin (i.e., real-analytic germs). Can the zero locus of $f$ be a nonsingular real-analytic hypersurface near the origin? The answer to this question is 'no' because near a nonsingular point of such a hypersurface, there will always be a real-analytic function $g$ that vanishes on it whose differential at that point is nonvanishing. However, this will imply that $g$ is a factor of $f$ , which is assumed to be irreducible. There remains the question of whether the zero locus of an irreducible real-analytic germ $f$ that is singular at the origin could contain a smooth hypersurface passing through the origin.
The answer to this is also 'no', but it takes a little work to see this. The case of a curve is not hard, using some standard facts about resolution of curve singularities: If $f(x,y)$ is a nonzero, real-valued analytic function defined on a neighborhood of the origin in $\mathbb{R}^2$ that is irreducible in the ring of analytic germs at the origin and satisfies $f_x(0,0)=f_y(0,0)=0$ , then the locus $f(x,y)=0$ cannot contain a smoothly embedded curve passing through the origin. A sketch of a proof is as follows: If the origin is not isolated, then $f(z,w)$ is a $\mathbb{C}$ -valued analytic function defined on a neighborhood of the origin in $\mathbb{C}^2$ that is also irreducible in this larger ring, and hence there is a neighborhood of the origin in $\mathbb{C}^2$ such that, in this neighborhood, the locus $f(z,w)=0$ can be parametrized by an embedded disc in $\mathbb{C}$ in the form $(z,w) = (a(\tau),b(\tau))$ where $a$ and $b$ are analytic functions of $\tau$ for $|\tau| < 1$ with $a(0)=b(0)=0$ . By a (real) rotation, we can assume that $a$ vanishes to a lower order, say $k>1$ , than $b$ does. Thus, we can reparametrize in $\tau$ so that $a(\tau) = \tau^k$ for some $k>1$ . In particular, the real locus will be parametrized by some curves of the form $\tau = \omega\,t$ where $t$ is real and $\omega^k = \pm 1$ . Choosing one such curve and replacing $t$ by $t/\omega$ , we can assume that $(a(t),b(t))$ is real for all small real $t$ , and that this parametrizes a 'branch' of the real locus that passes through the origin. In particular, the coefficients of $b$ are real, so our curve is parametrized in the form $$
(x,y) = \bigl(\ t^k,\ b_l t^l + b_{l+1} t^{l+1} + \cdots\ \bigr)
$$ where $l>k$ and, because of the embeddedness property of the disk, the greatest common divisor of $k$ and those $m$ for which $b_m\not=0$ must be $1$ . Thus, the curve is expressed in the form $$
y = b_l\ x^{l/k} + b_{l+1}\ x^{(l+1)/k} + \cdots
$$ where at least one of the exponents in this series is not an integer. It follows that the function on the right hand side of this equation cannot be smooth at $x=0$ , even though, since $l>k$ , it is $C^1$ . One conclusion of all this is that, if $g$ is a real-valued smooth function on a neighborhood of $0\in \mathbb{R}$ such that $g(0)=0$ and such that there exists a nontrivial real-analytic $f$ defined on a neighborhood of $(0,0)$ such that $f\bigl(x,g(x)\bigr)\equiv0$ for $x$ in the domain of $g$ , then $g$ must actually be real-analytic in a neighborhood of $0$ . (Note, however, that there do exist such 'real-analytically constrained' $g$ that, for some $m>0$ , are $C^m$ but not $C^{m+1}$ at $x=0$ .) Now, an easy argument shows that this $1$ -variable fact implies the corresponding $n$ -variable fact: If $g$ is a real-valued smooth function on a neighborhood of $0\in \mathbb{R}^n$ such that $g(0)=0$ and such that there exists a nontrivial real-analytic $f$ defined on a neighborhood of $(0,0)\in\mathbb{R}^n\times\mathbb{R}$ such that $f\bigl(x,g(x)\bigr)\equiv0$ for $x$ in the domain of $g$ , then $g$ must actually be real-analytic in a neighborhood of $0\in\mathbb{R}^n$ . (Basically, the hypotheses and the $1$ -variable result imply that $g\circ x$ is real-analytic for any real-analytic germ of a curve $x:(\mathbb{R},0)\to(\mathbb{R}^n,0)$ , and this easily implies that $g$ itself is real-analytic in a neighborhood of $0\in\mathbb{R}^n$ .) Thus, we have the answer to the question Suppose that the origin is a singular zero of an irreducible element $f$ in the ring of real-valued analytic functions defined on a neighborhood of the origin. Can the zero locus of $f$ be a nonsingular smooth hypersurface near the origin? The answer is 'no', because smooth would imply real-analytic , and we have already seen that this cannot happen.
|
{
"source": [
"https://mathoverflow.net/questions/98366",
"https://mathoverflow.net",
"https://mathoverflow.net/users/450/"
]
}
|
98,385 |
The example shown below (courtesy of David Eppstein) is a common example of a cubic graph that admits no perfect matching : (source: uci.edu ) Are there other examples of cubic graphs that do not admit a perfect matching and , unlike the above example, do not contain a vertex that lies at the intersection of three bridges (i.e. an edge whose removal increases the number of connected components in the graph)?
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Substitute your central vertex in your graph with a 3-cycle $abc$ so that the graph stays cubic. Now subdivide each edge in this 3-cycle. So we have new vertices $u$ connected to $a$ and $b$, $v$ connected to $b$ and $c$, $w$ connected to $c$ and $a$. Now add a final vertex $x$ and connect it to $u,v$ and $w$. This graph has exactly three bridges, none of which intersect the other at a vertex, and moreover has no perfect matching! One result which relates the existence of a perfect matching in a cubic graph and its bridges is the following theorem of Petersen from "Die theorie der regularen graphen", Acta Math. 15
(1891), 163-220: Theorem : Every cubic graph with at most two bridges contains a perfect matching. As well as this strengthening by Errera, "Du colorage des cartes", Mathesis 36 (1922), 56-60: Theorem : If all the bridges of a connected cubic graph $G$ lie on a single path of $G$, then $G$ has a perfect matching. So your instinct is true, in the sense that if the graph has no perfect matching, its bridges do not lie on a path. However the example in the beginning of this answer shows that they are not necessarily incident at the same vertex.
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{
"source": [
"https://mathoverflow.net/questions/98385",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3356/"
]
}
|
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