User aaron mazel-gee - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T08:17:39Z http://mathoverflow.net/feeds/user/303 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114647/what-is-the-universal-deformation-of-the-formal-additive-group-widehat-mathbb What is the universal deformation of the formal additive group $\widehat{\mathbb{G}}_a$ over $\mathbb{F}_p$? Aaron Mazel-Gee 2012-11-27T10:53:50Z 2012-11-27T19:54:10Z <p>Lubin and Tate show in their paper <em>Formal moduli for one-parameter formal Lie groups</em> that for any formal group over a field $k$ of characteristic $p>0$ with height $h&lt;\infty$, the functor of deformations is represented by a formal scheme isomorphic to $\mbox{Spf } \mathbb{W}(k)[[u_1,\ldots,u_{h-1}]]$. Modulo lower terms, $u_i$ is the coefficient of $x^{p^i}$ in the $p$-series of the universal deformation. (We take $p=u_0$.)</p> <p>Does this carry over for the additive group? Certainly there is an evident deformation to $\mbox{Spf } \mathbb{W}(k)[[u_1,u_2,\ldots]]$. In the paper, the finite height assumption is present in various results that they cite from elsewhere, so without being intimately familiar with the whole theory it's kind of hard to tell if this assumption is essential.</p> http://mathoverflow.net/questions/110285/on-the-natural-bigraded-homotopy-groups-of-a-simplicial-object-in-a-model-categ On the natural (bigraded) homotopy groups of a simplicial object in a model category Aaron Mazel-Gee 2012-10-22T00:44:02Z 2012-10-22T18:21:11Z <p>$\def\mc{\mathcal} \def\sm{\wedge}$ This question stems from the Goerss-Hopkins paper <a href="http://www.math.northwestern.edu/~pgoerss/spectra/obstruct.pdf" rel="nofollow">Moduli Problems for Structured Ring Spectra</a>. Let me begin by attempting to summarize the relevant framework -- this comes from the beginning of Chapter 3, on page 92.</p> <blockquote> <p>Let $\mc{C}$ be a model category, and let $\mc{P}$ be a set of projectives defining the $\mc{P}$-resolution model stucture on $s\mc{C}$. (See Question 1 below.) Given $X \in s\mc{C}$ and for any $P\in \mc{P}$, define the $(n,P)$th <em>natural homotopy group</em> of $X$ by $\pi_{n,P}(X)=\pi_n(\mbox{map}(P,X))$ (using the derived mapping space). This is corepresentable in $\mbox{Ho}(s\mc{C})$ by $P \sm \Delta^n/\partial\Delta^n$, which is defined as the pushout of the corner $\emptyset \otimes \ast \leftarrow P \otimes \ast \rightarrow P \otimes \Delta^n / \partial \Delta^n$. This smash product construction actually fits into an adjunction $$(-) \sm K : \mc{C} / \emptyset \leftrightarrows s\mc{C} : C_K,$$ where we define the right adjoint as follows. First, there is an adjunction $$(-)\otimes K:\mc{C} \leftrightarrows s\mc{C}:\mbox{hom}(K,-)_0 \stackrel{def}{=} M_K(-)$$ (where the right adjoint is the $0$th object in the "hom object" of $s\mc{C}$ coming its simplicially contensored structure), and then we define $C_K$ via the pullback diagram <code>$$\begin{array}{ccc} C_KX &amp; \rightarrow &amp; M_KX \\ \downarrow &amp; &amp; \downarrow \\ \emptyset &amp; \rightarrow &amp; M_*X = X_0. \end{array}$$</code> In particular, write $C_nX=C_{\Delta^n / \Lambda^n_0}X$ and $Z_nX=C_{\Delta^n / \partial \Delta^n}X$. The "inclusion of the boundary" <code>$\Delta^{n} /\partial \Delta^{n} \rightarrow \Delta^{n+1}/\Lambda^{n+1}_0$</code> induces the second map in a fibration sequence $$Z_{n+1} X \rightarrow C_{n+1}X \rightarrow Z_{n}X$$ in the category $\mc{C}/\emptyset$ whenever $X \in s\mc{C}$ is Reedy fibrant. The neat result is that in this case, this in fact gives a "presentation" of the natural homotopy group $\pi_{n,P}(X)$ via the exact sequence $$[P,C_{n+1}X] \rightarrow [P,Z_nX] \rightarrow \pi_{n,P}(X) \rightarrow 0.$$</p> </blockquote> <p>I have a few questions about the details and heuristics here.</p> <p><strong>Question 1</strong>: I'm confused about the extra conditions on $\mc{C}$ that might be necessary. They claim that the objects $P\in \mc{P}$ will be h-cogroup objects; this is what in particular gives a canonical map $P \rightarrow \emptyset$, which endows $\mbox{map}(P,X)$ with a basepoint. On the one hand, the definition of the $\mc{P}$-resolution model structure (see pages 24-26) starts out with the axiom that $\mc{P}$ be closed under suspension and desuspension, which suggests strongly that we're in a stable model category. But then it's silly to point out that the objects of $\mc{P}$ corepresent homotopy functors valued in abelian groups, since this is true of all objects. Even more, a stable model category is automatically pointed, which would mean that $\mc{C}/\emptyset$ is a totally vacuous thing to discuss; I'd consider this to be strong evidence that somehow $\mc{C}$ <em>isn't</em> meant to necessarily be stable. Of course the final application will be for $\mc{C}$ a category of spectra so this doesn't matter, but I've been pondering this long enough that the question has become of independent interest.</p> <p><strong>Question 2</strong>: Slightly further down (3.1.3 on page 95), they claim that the case $n=0$ of the result I cited yields an isomorphism $\pi_0([P,X]) \cong \pi_{0,P}(X)$ (where $[P,X]$ is a simplicial abelian group -- "simplicial" via <code>$[P,X]_\bullet = [P,X_\bullet]$</code>, and "abelian group" since supposedly $P$ is an (abelian, I guess) h-cogroup). However, unless I'm making a totally stupid mistake, I'm pretty sure that $Z_0X=C_{\Delta^0/\partial \Delta^0}X = C_*X = \emptyset$. I don't see how to make sense of this.</p> <p><strong>Question 3</strong> (assuming $\mc{C}$ doesn't need to be stable): How should I understand taking a pullback over the initial object? If $\emptyset=*$, then I could consider $M_KX$ as "maps from $K$ to $X$" and $C_KX$ as "based maps from $K$ to $X$". This is really appealing; then I could view the "presentation" as saying that $\pi_{n,P}(X)$ is just $[P,-]$ applied to "based maps $S^n \rightarrow X$ mod based maps $D^{n+1} \rightarrow X$"! (It also makes sense that "based maps" should be adjoint to "smash product".) But as things stand, this isn't really honest. In fact, I don't think I've <em>ever</em> seen anyone take a pullback over an initial object (besides in the pointed case, of course). I can't seem to wrap my head around what $\mbox{lim}(\emptyset \rightarrow A \leftarrow B)$ should mean, besides "the last object $C$ over $B$ such that $C \rightarrow B \rightarrow A$ is trivial" (i.e. factors through $\emptyset$, obviously). This sounds an awful lot like the fiber of a map, but I'm not convinced that this analogy makes any sense. I'd be grateful to hear any geometric intuition that anyone can give here.</p> http://mathoverflow.net/questions/107030/relationships-between-properties-of-model-categories relationships between properties of model categories Aaron Mazel-Gee 2012-09-12T16:49:45Z 2012-09-13T11:54:12Z <p>I've recently found myself running up against all sorts of adjectives that can describe a model category: cofibrantly generated, combinatorial, tractable, stable, locally (finitely) presentable, (left and/or right) proper, simplicial, admits Bousfield localizations, et al. I'm hoping for a reference whose explicit goal is to give an intuitive explanation of these concepts; presumably, such a reference would also tell me what each of these adjectives can buy me.</p> <p>(As a bonus, this reference might even have a diagram analogous to the one on pp. 576-7 of Gortz &amp; Wedhorn's book <a href="http://www.math.ucdavis.edu/~blnli/buildings/bag.pdf" rel="nofollow">Algebraic Geometry</a>, which summarizes the relationships between different properties that a morphism of schemes may satisfy. But that'd just be for fun.)</p> http://mathoverflow.net/questions/86985/if-a-in-mboxrings-subset-e-infty-mbox-rings-what-is-the-equivalence-betw If $A\in \mbox{Rings}\subset E_\infty\mbox{-rings}$, what is the equivalence between objects of $\mathcal{D}(\mbox{Mod}_A)$ and $A$-module spectra? Aaron Mazel-Gee 2012-01-29T20:34:07Z 2012-01-30T20:33:36Z <p>In Lurie's "A Survey of Elliptic Cohomology", he writes on page 14 that if $A$ is an ordinary commutative ring considered as an $E_\infty$-ring, then $A$-module spectra are the same thing as objects of the derived category of $A$-modules. This is mysterious to me. On the one hand, to an $A$-module spectrum $M$ we might associate the $A$-modules $\pi_n(M)$, but I don't know of any interesting maps between these; perhaps this will just end up being the homology of any representative chain complex of $A$-modules. But then, I certainly don't see a natural way of getting from an object of $\mathcal{D}(\mbox{Mod}_A)$ to an $A$-module spectrum.</p> <p>Incidentally, what does this induce on the level of categories? The obvious first guess is that $A$-module spectra actually form a topological category and that passing to $\mathcal{D}(\mbox{Mod}_A)$ applies $\pi_0$.</p> http://mathoverflow.net/questions/57082/is-there-a-unified-reason-that-there-are-an-infinite-number-of-geodesics-between Is there a unified reason that there are an infinite number of geodesics between nonconjugate points on a compact manifold? Aaron Mazel-Gee 2011-03-02T07:20:08Z 2012-01-22T12:18:29Z <p>The proof of this statement seems to break into two really different arguments. So, I'm wondering if there is a better argument that can explain them both, or whether it's really just two theorems that happen to be easy to say at the same time. Both rely on a bit of Morse theory, namely that (assuming $p$ and $q$ are nonconjugate) we get a CW-complex for the space of paths $\Omega (M;p,q)$ from $p$ to $q$ which has one cell for each geodesic from $p$ to $q$, whose dimension is the index of the geodesic (i.e. the number of points along it that are conjugate to the starting point, with multiplicity).</p> <p><strong>Case 1 ($|\pi_1(M)|&lt;\infty$):</strong> When $\pi_1(M)=0$, applying the Serre spectral sequence to the path fibration $\Omega M \rightarrow \mathcal{P}M \rightarrow M$ shows that it would be a contradiction if we ever had $H_m(\Omega M)=0$ for all $m\geq N$. So the statement follows from cellular homology. If $0&lt;|\pi_1(M)|&lt;\infty$, pull back the metric on $M$ to its universal cover $\tilde{M}$, which is also compact. Choose $\tilde{p}\in \pi^{-1}(p)$ and $\tilde{q}\in \pi^{-1}(q)$. Then from what we have just said, there are an infinite number of geodesics on $\tilde{M}$ from $\tilde{p}$ to $\tilde{q}$, and these project to geodesics from $p$ to $q$. (I don't want to use the Serre spectral sequence when the base isn't simply connected, if I can help it!)</p> <p><strong>Case 2 ($|\pi_1(M)|=\infty$):</strong> Note that $\pi_1(M)=\pi_0(\Omega(M))$, so any CW-decomposition of $\Omega (M)$ must have an infinite number of cells.</p> <p>I'm pretty sure that since my manifold is complete, in Case 2 I could also have just said "lift a representative of each element of $\pi_1$ (concatenated with some fixed path from $p$ to $q$), homotope it to a geodesic, and project back down" but I'm not positive I can do that. In any case, that still feels kind of different from the argument in Case 1, but maybe there's something here I'm just not seeing.</p> http://mathoverflow.net/questions/69967/titles-composed-entirely-of-math-symbols/85896#85896 Answer by Aaron Mazel-Gee for Titles composed entirely of math symbols Aaron Mazel-Gee 2012-01-17T09:36:24Z 2012-01-17T09:36:24Z <p><a href="http://www.nullplug.org/publications/e-vs-h-infinity.pdf" rel="nofollow">Here</a> is $H_\infty\not= E_\infty$, wherein Justin Noel gives an example of an $H_\infty$-structure on a ring spectrum which does not descend from an $E_\infty$-structure.</p> http://mathoverflow.net/questions/80372/reference-request-john-baez-on-1-and-2-categories-and-propertiesstructur reference request: John Baez on (-1)- and (-2)-categories and properties+structure+stuff Aaron Mazel-Gee 2011-11-08T10:37:21Z 2011-11-08T10:49:30Z <p>I vaguely recall reading a long time ago a 50-or-so page paper, either by John Baez or linked from his page (I think the former), which among other things gave a justification for his table of n-categories and a very cute explanation of the realization that really it should start at n=-2. I believe this paper also introduced (to me) the ideas of properties, structure, and stuff. I was just thinking about how I'd love to look back at this, but among the wealth of resources available on Baez's webpage I can't seem to find what I'm looking for. Googling hasn't turned it up either. Does anyone know what paper I'm talking about? And if it doesn't talk about properties, structure, and stuff, can anyone recommend a friendly and polished exposition of these ideas?</p> http://mathoverflow.net/questions/71523/how-can-i-see-that-h-mathbbz-doesnt-admit-a-bousfield-complement How can I see that $H\mathbb{Z}$ doesn't admit a Bousfield complement? Aaron Mazel-Gee 2011-07-28T22:33:17Z 2011-07-29T10:53:26Z <p>From Ravenel's article "Localization and Periodicity in Homotopy Theory":</p> <blockquote> <p>Two spectra $E$ and $F$ are said to be <em>Bousfield equivalent</em> when they give the same localization functor, or equivalently when $E_\ast (X)=0$ iff $F_\ast (X)=0$. The equivalence class of $E$ is denoted by $\langle E \rangle$. There is a partial ordering on the set of Bousfield classes. We say that $\langle E \rangle \geq \langle F \rangle$ if $E_\ast (X)=0$ implies that $F_\ast (X)=0$. Thus $\langle S^0 \rangle$ is the biggest class and $\langle pt \rangle$ is the smallest. Smash products and wedges are well defined on Bousfield classes. A class $\langle F \rangle$ is the <em>complement</em> of $\langle E \rangle$ if $\langle E \rangle \vee \langle F \rangle = \langle S^0 \rangle$ and $\langle E \rangle \wedge \langle F \rangle = \langle pt \rangle$. A class may or may not have a complement. It is easy to find examples of classes (e.g., that of an integer Eilenberg-Mac Lane spectrum) that do not.</p> </blockquote> <p>I was trying to figure out why this last statement is true, and at first I wanted to apply cohomotopy to a hypothetical equivalence $H\mathbb{Z} \vee F \simeq S^0$, but then I realized that of course there's no reason that we should have such an equivalence. Is there some other easy approach?</p> http://mathoverflow.net/questions/70022/sh-sh-map-represents-the-category-of-sheaves-on-a-stack (Sh,Sh-map) represents the category of sheaves on a stack. Aaron Mazel-Gee 2011-07-11T15:35:20Z 2011-07-29T10:37:33Z <p>I'm trying to understand the following theorem, but I don't think I'm reading it correctly.</p> <p>Let $(\mathcal{C},J)$ be a site (with a subcanonical topology). Write $\mathcal{C}/X$ for the groupoid of objects over $X\in \mathcal{C}$. Let $\mbox{Sh}:\mathcal{C}^{op} \rightarrow \mbox{Gpds}$ be the functor taking $X$ to the category of sheaves on $\mathcal{C}/X$ and isomorphisms of sheaves, and let $\mbox{Sh-map}:\mathcal{C}^{op} \rightarrow \mbox{Gpds}$ be the functor taking $X$ to the category whose objects are sheaf morphisms $\mathscr{F} \rightarrow \mathscr{G}$ and whose morphisms are commuting squares of sheaves determined by isomorphisms $\mathscr{F}_1 \stackrel{\sim}{\rightarrow} \mathscr{F}_2$ and $\mathscr{G}_1 \stackrel{\sim}{\rightarrow} \mathscr{G}_2$. These are in fact both stacks on $\mathcal{C}$, and moreover they determine a category-object $(\mbox{Sh},\mbox{Sh-map})$ in the category of stacks.</p> <blockquote> <p><strong>Theorem</strong>: The category of sheaves on a stack $\mathscr{M}$ is equivalent to the category of morphisms of stacks $\mathscr{M} \rightarrow (\mbox{Sh,Sh-map})$. That is, the objects are the 1-morphisms and the morphisms are the 2-morphisms.</p> </blockquote> <p>I'd like to interpret this to mean that the objects of $Shv(\mathscr{M})$ are associated to 1-morphisms $\mathscr{M} \rightarrow \mbox{Sh}$, and that the morphisms of $Shv(\mathscr{M})$ are associated to 2-morphisms in $Hom_{Stacks}(\mathscr{M},\mbox{Sh})$, which in turn should be the same as 1-morphisms $\mathscr{M} \rightarrow \mbox{Sh-map}$. But there a number of problems with this.</p> <p>First, given a sheaf $\mathcal{F} \in Shv(\mathscr{M})$ I'm having trouble constructing a natural transformation $\mathscr{M} \rightarrow \mbox{Sh}$. Perhaps I shouldn't, but to check this I'm using a test object $X\in \mathcal{C}$. By Yoneda, an object of $\mathscr{M}(X)$ is the same as a 1-morphism of stacks $f:X\rightarrow \mathscr{M}$, and so I obtain an object of $Sh(X)$ (i.e. a sheaf on $\mathcal{C}/X$) via $(\alpha:Y\rightarrow X) \mapsto \mathcal{F}(f\alpha:Y \rightarrow X \rightarrow \mathscr{M})$. That's natural enough. Again by Yoneda, a morphism in $\mathscr{M}(X)$ is a 2-morphism between maps $f,g:X\rightarrow \mathscr{M}$ of stacks, i.e. a section $s:X\rightarrow X\times_\mathscr{M} X$ of the projection from the 2-category fiber product. Out of this, I'm supposed to construct a natural transformation from the sheaf $(\alpha:Y\rightarrow X) \mapsto \mathcal{F}(f\alpha:Y \rightarrow X \rightarrow \mathscr{M})$ to the sheaf $(\alpha:Y\rightarrow X) \mapsto \mathcal{F}(g\alpha:Y \rightarrow X \rightarrow \mathscr{M})$. But the only structure in place to give me such a thing is a morphism in $Stacks/\mathscr{M}$ between $f\alpha$ and $g\alpha$, and I don't see how to construct this.</p> <p>Second, a 2-morphism between 1-morphisms $f,g\in Hom_{Stacks}(\mathscr{M},\mbox{Sh})$ is a section $s:\mathscr{M} \rightarrow \mathscr{M} \times_{\mbox{Sh}} \mathscr{M}$. Thus for any $(\alpha:X\rightarrow \mathscr{M})\in \mathscr{M}(X)$, we get an object $(\alpha,\beta:X \rightarrow \mathscr{M},\varphi:f\alpha \stackrel{\sim}{\rightarrow} g\alpha)\in (\mathscr{M}\times_{\mbox{Sh}}\mathscr{M})(X)$. On the other hand, a 1-morphism $\mathscr{M} \rightarrow \mbox{Sh-map}$ is for each $\alpha:X \rightarrow \mathscr{M}$ an arbitrary morphism on sheaves on $\mathcal{C}/X$. These can't be the same.</p> <p>By the way, I've tried to do (what I think is) the right thing and work out the sheaf in $Shv(\mbox{Sh})$ associated to the 1-morphism $\mbox{Id}:\mbox{Sh} \rightarrow \mbox{Sh}$, following Yoneda and all. From the above, it's easy to see what this sheaf should do to morphisms $X\rightarrow \mbox{Sh}$ from a representable stack. But it appears that I need to make choices if I want to say what it does to arbitrary morphisms of stacks $\mathscr{N} \rightarrow \mbox{Sh}$. Perhaps instead I should take a limit or colimit over its application to the full subcategory of representable stacks over $\mathscr{N}$?</p> http://mathoverflow.net/questions/69085/can-one-use-atiyah-singer-to-prove-that-the-chern-weil-definition-of-chern-classe Can one use Atiyah-Singer to prove that the Chern-Weil definition of Chern classes are $\mathbb{Z}$-cohomology classes? Aaron Mazel-Gee 2011-06-29T02:46:17Z 2011-06-30T13:28:21Z <p>In Chern-Weil theory, we choose an arbitrary connection $\nabla$ on a complex vector bundle $E\rightarrow X$, obtain its curvature $F_\nabla$, and then we get Chern classes of $E$ from the curvature form. A priori it looks like these live in <code>$H^*(X;\mathbb{C})$</code>, but by an argument that I don't really understand they're actually in the image of <code>$H^*(X;\mathbb{Z})$</code> (which is where they're usually considered to live). Meanwhile, I've heard people say that whenever I see a arbitrary real constants that end up having to be integers I should wonder whether the Atiyah-Singer index theorem is lurking in there somewhere. Is there anything to this wild guess?</p> http://mathoverflow.net/questions/67324/homological-algebra-with-spectra homological algebra with spectra Aaron Mazel-Gee 2011-06-08T23:49:25Z 2011-06-09T18:52:47Z <p>I'm reading Mike Hopkins' <a href="http://www.math.rochester.edu/u/faculty/doug/otherpapers/coctalos.pdf" rel="nofollow">COCTALOS</a> notes and having trouble with some pretty basic statements about (naive) spectra. Basically I'm nervous doing homological algebra with them, although in these problems I'm having I couldn't imagine the proofs being anything other than simple diagram chases. The relevant definitions are in section 4.</p> <p>First, I'm having trouble with the proof of Prop. 4.12, which claims that every E-Adams resolution arises from an Adams tower. The proof is based on the diagram at the bottom of page 13 (sorry, I'm not sure how to put diagrams in an MO problem) and shows how to build the first level of the tower. To continue the tower, we need that two composite maps (at the top of page 14) are null, and this is what's confusing me. For (i), I have no idea, although $h$ must be E-mono since it's the first map in a factorization of an E-mono. For (ii), the problem is that we know both $\Sigma^{-1} I_0\rightarrow \Sigma^{-1}I_3$ and $\Sigma^{-1}I_1\rightarrow \Sigma^{-1}I_3$ are null, but the cofiber $X_1 = \Sigma^{-1}I_1/\Sigma^{-1}I_0$ may not map E-nully to $\Sigma^{-1}I_3$. In fact, there isn't necessarily a single choice of map anyways; it depends on the choice of nullhomotopy of $\Sigma^{-1}I_0\rightarrow \Sigma^{-1}I_3$ (factoring through $\Sigma^{-1}I_2$). All I know is that I need to find a map $C (E\wedge X_1) \simeq I\wedge E\wedge I_0 \rightarrow E\wedge \Sigma^{-1}I_3$ extending the map from $E\wedge X_1$. (Here $I$ is the pointed interval.)</p> <p>Second, I don't understand the statement that "these maps are null after smashing with E and hence must actually be null as [their images] are E-injective". Suppose $A\rightarrow J$ is E-null and $J$ is E-injective. The only thing to do here is start with an E-mono $A\rightarrow B$ and obtain a lift $B\rightarrow J$ in the diagram $B\leftarrow A \rightarrow J$. Then the lift will have to be E-null too. But I don't see how this would imply that $A\rightarrow J$ is actually null. And there's no obvious E-mono I've got to work with anyways, besides maybe ${Id}_A$ (which doesn't help at all).</p> <p>And lastly, I'm having trouble proving the $\Leftarrow$ direction of Lemma 5.2, that if for $E$ a ring spectrum the map $J=S\wedge J\rightarrow E\wedge J$ is the inclusion of a retract then $J$ is E-injective. I'd prove this by taking an E-mono $X\rightarrow Y$ and trying to find a lift in $Y\leftarrow X \rightarrow J$ for any $X\rightarrow J$. The only thing to do here is smash with $E$ and obtain $E\wedge B\leftarrow E\wedge A \rightarrow E\wedge J \rightarrow J$ (where the first map is mono), but I can't see what this buys me. There's one fact we've got about monos (Lemma 4.3), but not sure whether it helps here: $A \rightarrow B$ is mono iff there is some $C\rightarrow B$ such that $A\vee C\rightarrow B$ is a weak equivalence.</p> <p>Since these seem like they should be such routine verifications, I'm feeling that there must be some basic technique I'm missing. Does anyone have any suggestions?</p> http://mathoverflow.net/questions/63439/how-can-we-detect-the-existence-of-almost-complex-structures How can we detect the existence of almost-complex structures? Aaron Mazel-Gee 2011-04-29T16:46:39Z 2011-05-03T20:45:39Z <p>Any smooth $k$-manifold $M$ comes with a well-defined map $f:M\rightarrow BGL_{k}(\mathbb{R})$ (up to homotopy) classifying its tangent bundle. Since $GL_{k}(\mathbb{R})$ deformation-retracts onto $O_k$, then $BGL_{k}(\mathbb{R})\simeq BO_k$, which is a cute way (though it's certainly overkill) of proving that every smooth manifold admits a Riemannian metric. An almost-complex structure, on the other hand, is equivalent to a reduction of the structure group from $GL_{2n}(\mathbb{R})$ to $GL_n(\mathbb{C})$, which is the same as asking for a lift of the classifying map through $BU_n\simeq BGL_n(\mathbb{C})\rightarrow BGL_{2n}(\mathbb{R})$.</p> <blockquote> <p>Can we detect the nonexistence of a lift entirely using characteristic classes? If not, what else goes into the classification?</p> </blockquote> <p>I'd imagine these don't suffice themselves. I know that $w_{2n}(TM) \equiv_2 c_n(TM)$, so this holds in the universal case $H^\ast(BO_{2n};\mathbb{Z}/2) \rightarrow H^\ast(BU_n;\mathbb{Z}/2)$. And certainly there are necessary conditions like $w_1(TM)=0$ (which of course just means that $TM$ is an orientable bundle, which is the same as asking that $M$ be an orientable manifold). But I have no idea of what sufficient conditions would look like. I've heard that this problem is indeed solved. Maybe it takes some characteristic class &amp; cohomology operation gymnastics, or maybe it even needs extraordinary characteristic classes. Or maybe there's yet another ingredient in the classification?</p> <p><strong>Edit</strong>: Apparently I misquoted my source, and this is only known stably (which makes sense, in light of Joel's answer and Tom's comments on it).</p> http://mathoverflow.net/questions/60035/holomorphic-k-theory holomorphic K-theory Aaron Mazel-Gee 2011-03-30T05:07:53Z 2011-03-30T10:30:49Z <p>Topological K-theory is usually defined by setting $K(X)$ to be the groupification of the monoid $Vect_\mathbb{C}(X)$ of complex vector bundles over $X$ (with addition given by Whitney sum). However, we can alternatively declare that $[B]\sim [A]+[C]$ whenever $0\rightarrow A \rightarrow B\rightarrow C\rightarrow 0$ is a short exact sequence of vector bundles over $X$ (morphisms are required to have locally constant rank): certainly if $B\cong A\oplus C$ then we have such a sequence, and in the other direction we can take a metric on $B$ and identify $C$ with $A^\perp \subseteq B$.</p> <p>You can take the $K$-theory of any abelian category using this second definition. So, I'm curious to know if people do this for the category of holomorphic vector bundles over a complex manifold. The above splitting construction no longer works since it uses partitions of unity, so assuming we use this more general definition we'd get more equivalence relations. On the other hand, there's all this funny business going on with vector bundles topologically but not holomorphically isomorphic, which means that $K_{hol}(X)$ wouldn't just be a subquotient of $K(X)$. So in the end, I'm not sure whether I should expect this to be a more or less tractable sort of object.</p> <p>I'm told that the Chow ring might have something to do with this, but the wikipedia page seems to indicate that it's more analogous to singular cohomology than anything else.</p> http://mathoverflow.net/questions/52492/localization-at-a-homology-theory-and-the-adams-spectral-sequence localization at a homology theory and the Adams spectral sequence Aaron Mazel-Gee 2011-01-19T09:23:10Z 2011-01-19T15:04:42Z <p>I'm reading Switzer's "Algebraic Topology", which talks about the (homology) ASS in chapter 19. His ASS (where he puts $S^0$ in the first slot) either converges to $\pi_n(Y)$, or to $\pi_n(Y)/\cap_{s\geq 0}F^{s,n+s}$ (which happens e.g. when our theory $E$ is the spectrum $H\mathbb{F}_p$ representing $H^*(-;\mathbb{F}_p)$).</p> <p>Switzer doesn't talk about this at all, but I've seen/heard before that one can also present the ASS as converging to something denoted $[L_EX, L_EY]$. I don't know much about the functor $L_E:Top\rightarrow Top$, but I believe it is characterized as follows: If $E_*X=0$ then we say that $X$ is <em>$E$-acyclic</em>. Now, a space $Y$ is called <em>$E$-local</em> if whenever $X$ is $E$-acyclic, $[X,Y]=0$. Then, the <em>$E$-localization</em> $L_EY$ of $Y$ is an initial object in the category of $E$-local spaces under $Y$.</p> <p>So here is what I'm wondering. The next result after the ASS is the following Proposition 19.11: If $\iota_*:\pi_q(S^0)\rightarrow \pi_q(E)$ is an isomorphism for $q\leq 0$ and an epimorphism for $q=1$ and if $\pi_q(Y)=0$ for $q\lt N$ for some $N\in \mathbb{Z}$, then $A^{s,t}=D^s=0$ for all $s,t$, and hence the Adams spectral sequence converges to $\pi_*(Y)$. (Here $\iota:S^0\rightarrow E$ is the unit map.) This seems very much like a fact about localization in disguise -- presumably it'd be saying something about $L_EY$, perhaps that $Y=L_EY$. Except that I don't know what $L_ES^0$ is, which I'd imagine should be the first thing anyone would try to figure out.</p> <p>So I guess my questions are:</p> <ol> <li>Do I have the right characterization of localization?</li> <li>Assuming the answer to (a) is yes (or I guess even if it isn't), what is this proposition saying about localization?</li> <li>Replacing $S^0$ with an arbitrary $X$, in Switzer's setup we can hope for the ASS to give us $[X,Y]/\cap_{s\geq 0}F^{s,*+s}$ -- how can we pass between this quotient and $[L_EX,L_EY]$? Is there anything more geometric to say than that the denominator is the subgroup of maps inducing 0 in homology?</li> </ol> http://mathoverflow.net/questions/49784/the-z-2-cohomology-functor-from-top-to-grvecspaces-factors-through-unstable-a-mo "The Z/2-cohomology functor from Top to GrVecSpaces factors through Unstable-A-Mod, and this is the largest algebraic category through which it factors." Aaron Mazel-Gee 2010-12-18T08:07:19Z 2010-12-18T20:45:12Z <p>(Here A is the Steenrod algebra.)</p> <p>I have it on <a href="http://math.berkeley.edu/~ericp/latex/haynes-notes/haynes-notes.pdf" rel="nofollow">good authority</a> (p. 23) that this is true, but I can't quite make sense of it. The crux of the matter is something I've been wondering for a while:</p> <blockquote> <p>Why exactly is it that cohomology operations are "the best we can do" (i.e./more generally, why is it that the finest structure we can impose on $E$-cohomology is that it be a module over $E^*E$)?</p> </blockquote> <p>There might be something tied up in the qualification that we're talking about <em>algebraic</em> categories, which are defined <a href="http://ncatlab.org/nlab/show/algebraic+category" rel="nofollow">here</a>. I've never heard of these before. Presumably their properties show up in GrVecSpaces as manifestations of actual things on the topological side, which makes it seem like we're only restricting ourselves to categories that are going to preserve some information that we obviously (?) want to preserve, but probably there's more to it than that. And in any case I don't understand why those should be exactly the properties of GrVecSpaces that we care about.</p> <p>(I'd love to hear about category theory stuff of course, but my main goal is to understand the boxed question.)</p> http://mathoverflow.net/questions/43762/toy-examples-of-equivariant-homotopy-theory toy examples of equivariant homotopy theory Aaron Mazel-Gee 2010-10-27T05:25:06Z 2010-10-27T14:57:34Z <p>I've heard a little recently about equivariant homotopy theory, and so I decided to try out some baby examples just to get a feel for it. I'm not even sure if these are the right thing to look at, and I'm sure I'm butchering the notation, but I've attempted to compute $\pi_2^{C_n}(S^2)$ and $\pi_3^{C_3}(S^2)$; here, $C_n$ acts by rotations on the plane and $C_3$ acts by the standard representation on 3-space.</p> <p>For the first one, there's the obvious equivariant cell structure on $S^2$, which has fixed vertices at 0 and at the basepoint $\infty$, one orbit of $n$ edges, and one orbit of $n$ faces. My maps are based and I need to map fixed points to fixed points, so either both vertices go to $\infty$ or else my map fixes them.</p> <ul> <li>Suppose both go to $\infty$. The image of any one edge is an arbitrary based loop, and of course the other edges' images are determined by this. Assuming my faces are going to able to be mapped in at all, the inclusion of an edge into a face is a cofibration, so I may as well just homotope the edges to $\infty$ now. From here, the map is determined by the image of one face, i.e. an element of (nonequivariant) $\pi_2(S^2)$.</li> <li>In the case where $0$ and $\infty$ are both fixed, the image of an edge is an arbitrary path between $0$ and $\infty$, and then we just get to decide where a face goes, which I think is just an element of the preimage of 1 in the connecting map $\pi_2(S^2,S^1)\rightarrow \pi_1(S^1)$. There's no extension problem in the lexseq, so this is just a projection onto one factor $\mathbb{Z}\times \mathbb{Z}\rightarrow \mathbb{Z}$, so this looks like $\mathbb{Z}\times {1}$.</li> </ul> <p>So as a set, $\pi_2^{C_n}(S^2)=\mathbb{Z} \sqcup \mathbb{Z}$.</p> <p>For the second one, I have a 1-cell of fixed points at $(x,y,z)=(t,t,t)$, my three 2-cells are the half-planes through that 1-cell and one of the three axes, and my three 3-cells fill in the rest. The 1-cell needs to be sent to fixed points, and these are still just $0$ and $\infty$, so by continuity it's all sent to $\infty$. So the image of a 2-cell is just an element of $\pi_2(S^2)$. Once we choose that (which determines the images of the other 2-cells, of course) we can always extend to the 3-cells, since these are just homotopies from a map to itself. Once we've chosen one, any other gives an element of $\pi_3(S^2)=\mathbb{Z}$. So as a set, $\pi_3^{C_3}(S^2)=\mathbb{Z} \times \mathbb{Z}$.</p> <p>So, my first question is: Are these right? Also, I've learned to compute (usual) homotopy groups of spheres by making a Postnikov tower, and I'm wondering if there's a sufficiently easy example for the equivariant case where I can do the analogous calculations by hand without the full generality of slice cells or whatever's going on (those could be the wrong words -- I don't think I understand what these are well enough to know whether this is a decent request, either). In any case, I'd love suggestions of better/more instructive examples. Lastly, I'm wondering if there are actually group structures here. In the first example, it looks like I can reasonably hope to add guys that are both in the same copy of $\mathbb{Z}$, but not otherwise. I think it's easy to show that there's no $C_n$-equivariant coproduct on $S^2$. On the other hand, quotienting by the plane $x+y+z=0$ in $\mathbb{R}^3$ looks like it gives a $C_3$-equivariant coproduct on $S^3$. If two elements came from the same choice of $\pi_2(S^2)$, then there's an obvious origin for the $\pi_3(S^2)$-torsor, namely using the trivial homotopy to extend the map over the 3-cells. I think this agrees with my equivariant coproduct, just because it matches up with the usual picture you draw of how to add elements of $\pi_2$ (two squares sitting on top of each other). However, I can't tell whether this makes any sense when the elements came from different choices of $\pi_2(S^2)$. It seems like if it does work at all, there might be something funnier than the obvious group structure...</p> http://mathoverflow.net/questions/38931/how-can-i-show-that-the-map-l-k-pi-nl-n-representing-the-fundamental-class How can I show that the map L-->K(\pi_n(L),n) representing the fundamental class of an (n-1)-connected space is an isomorphism on \pi_n? Aaron Mazel-Gee 2010-09-16T05:46:47Z 2010-10-15T21:15:13Z <p>As an exercise, I'm trying to show that for an $(n-1)$-connected space $L$ with $\pi=\pi_n(L)$, the map $\iota_L:L\rightarrow K(\pi,n)$ associated to the fundamental class $\iota_L\in H^n(L;\pi)$ induces an isomorphism $\pi_n(L)\rightarrow \pi_n(K(\pi,n))$. After playing with this for a while, I've found that this feels so tautological that it <i>has</i> to be true, although that doesn't quite constitute a proof. I'm supposed to use only basic definitions / first principles. Presumably, these are:</p> <ol> <li><p>The universal coefficient theorem yields $H^n(L;\pi)\cong Hom(H_n(L),\pi)$. The Hurewicz homomorphism $h:\pi \rightarrow H_n(L)$ is an isomorphism, and its inverse $h^{-1}\in Hom(H_n(L),\pi)$ corresponds to $\iota_L\in H^n(L;\pi)$.</p></li> <li><p>We have a canonical bijection $H^n(L;\pi) \cong [L,K(\pi,n)]$. Any $[f]\in [L,K(\pi,n)]$ corresponds to $f^*\iota\in H^n(L;\pi)$, where $\iota\in H^n(K(\pi,n);\pi)$ is the fundamental class of $K(\pi,n)$ (which is associated to its identity map). This is actually a group isomorphism if we add maps on the right side by using the fact that $K(\pi,n)=\Omega K(\pi,n+1)$ (or at least $\simeq$, although what does $K(\pi,n)$ even mean really). I'd imagine that this is canonical too, but I don't know for sure.</p></li> <li><p>We have a map $[L,K(\pi,n)]\rightarrow Hom(\pi,\pi)$ given by <code>$[f]\mapsto f_\#$</code>. Presumably the idea is to show that the image of $\iota_L$ is an isomorphism, but I can't tell if I'm just complicating the question by phrasing it in these terms.</p></li> </ol> <p>I think my problem is that I don't really understand the defining way Eilenberg-MacLane spaces work. I have an intuitive picture of the Hurewicz homomorphism, and so I guess I have an intuitive picture of its inverse: it takes a homology class in degree $n$ and realizes it as the image of a bunch of based $n$-spheres (which is made possible by the Hurewicz theorem). We can look at this more or less as a cochain in $C^n_{cell}(L;\pi)$, and this is the element $\iota_L\in H^n(L;\pi)$. But then I have no idea how to actually turn this into a map $\iota_L : L\rightarrow K(\pi,n)$, or whether I'm even supposed to.</p> <p>Because I'd still like to work this out myself to whatever extent I can, a good hint (if one exists) is worth more to me than a straight-up answer. Of course, I'm happy with either. Thanks!</p> http://mathoverflow.net/questions/38931/how-can-i-show-that-the-map-l-k-pi-nl-n-representing-the-fundamental-class/42328#42328 Answer by Aaron Mazel-Gee for How can I show that the map L-->K(\pi_n(L),n) representing the fundamental class of an (n-1)-connected space is an isomorphism on \pi_n? Aaron Mazel-Gee 2010-10-15T21:15:13Z 2010-10-15T21:15:13Z <p>Here is a solution, based on Jeffrey's comment:</p> <p>Let $f:L\rightarrow K(\pi,n)$ be the map classifying $\iota$. We will show that the induced map $f_*:H_i(L)\rightarrow H_i(K(\pi,n))$ is isomorphic for $i \lt n+1$ and epimorphic for $i=n+1$. This will imply by Whitehead's theorem that the same is true for <code>$f_\# :\pi_i(L)\rightarrow \pi_i(K(\pi,n))$</code> (assuming $n \gt 1$).</p> <p>First, $f_*$ is automatically an isomorphism for $i \lt n$, since in that case by the Hurewicz theorem $H_i(L)$ and $H_i(K(\pi,n))$ are both trivial.</p> <p>Next, in dimension $n$, we use the fact that <code>$f^*:H^n(K(\pi,n);\pi)\rightarrow H^n(L;\pi)$</code> satisfies $f^*\iota_n = \iota$, where $\iota_n$ is the fundamental class of $K(\pi,n)$. By the universal coefficient and Hurewicz theorems, we may regard $H^n(K(\pi,n))=Hom(H_n(K(\pi,n)),\pi)$ and $H^n(L)=Hom(H_n(L),\pi)$, and the fundamental classes both correspond to isomorphisms (namely, the inverses of the respective Hurewicz homomorphisms). In general, if a group homomorphism $\varphi:B\rightarrow A$ induces <code>$\varphi^\#:Hom(A,C)\rightarrow Hom(B,C)$</code> and <code>$\varphi^\#$</code> takes an isomorphism $\psi_1:A\rightarrow C$ to an isomorphism $\psi_2:B \rightarrow C$, then $\varphi$ itself must be an isomorphism, because <code>$\psi_2 = \varphi^\# (\psi_1)=\psi_1\varphi$</code> and so $\varphi = \psi_1^{-1}\psi_2$. Here, since $f^*:Hom(H_n(K(\pi,n)),\pi)\rightarrow Hom(H_n(L),\pi)$ is induced from $f_*:H_n(L)\rightarrow H_n(K(\pi,n))$ and takes an isomorphism to an isomorphism, then $f_*:H_n(L)\rightarrow H_n(K(\pi,n))$ itself is an isomorphism.</p> <p>Lastly, by the Hurewicz theorem, since $K(\pi,n)$ is $n$-connected, the Hurewicz homomorphism $h:\pi_{n+1}(K(\pi,n))\rightarrow H_{n+1}(K(\pi,n))$ is an epimorphism, but $\pi_{n+1}(K(\pi,n))=0$ by definition so $H_{n+1}(K(\pi,n))=0$ as well. Hence $f_*: H_{n+1}(L)\rightarrow H_{n+1}(K(\pi,n))$ is automatically epimorphic.</p> http://mathoverflow.net/questions/41701/how-does-the-cokernel-of-the-j-homomorphism-count-exotic-spheres How does the cokernel of the J-homomorphism count exotic spheres? Aaron Mazel-Gee 2010-10-10T18:05:06Z 2010-10-10T18:05:06Z <p>The <a href="http://en.wikipedia.org/wiki/J-homomorphism" rel="nofollow">wikipedia article on the J-homomorphism</a> says that "the cokernel of the J-homomorphism is of interest for counting exotic spheres". I'd like to think this makes some sort of philosophical sense; as I understand it, the homomorphism comes from a Hopf construction, which isn't really a smooth sort of thing, but on the other hand it still feels like constructions using the (stable) (special) orthogonal group should somehow keep us in the non-exotic world. Is this at all close to the right intuition?</p> http://mathoverflow.net/questions/41525/how-should-i-think-about-delooping How should I think about delooping? Aaron Mazel-Gee 2010-10-08T17:12:01Z 2010-10-09T21:13:27Z <p>When talking about the Eilenberg-Maclane space $K(G,n)$, we usually restrict our attention to the situation where $G$ is abelian. In that case, we get $\Omega K(G,n)=K(G,n-1)$, so we can call $K(G,n)$ a <em>delooping</em> of $K(G,n-1)$.</p> <p>Since $\pi_n$ is always abelian for $n>1$, it only makes sense to talk about $K(G,1)=BG$ for $G$ nonabelian anyways. So there definitely shouldn't be delooping of this space, because then it would have $\pi_2=G$, which is impossible. From the previous paragraph, it seems like we should therefore be able to say that the nonabelianity of $G$ (i.e., the nontriviality of the commutator $[G,G]$) is the obstruction to delooping $BG$. But this isn't very satisfying, because I can't quite see what's going on with the actual space.</p> <p>All of which motivates my (slightly open-ended/up-to-interpretation) question:</p> <p><strong>How should I think about delooping? Is it nothing more than thing like "for the space $X$ that we care about, it just so happens that we've got $Y$ with $\Omega Y\simeq X$", or is there a definite way to measure obstructions? In the cases where a delooping exists, is there an explicit method for its construction?</strong></p> http://mathoverflow.net/questions/39447/is-there-an-easy-way-to-find-the-minimum-dimensions-of-representations-for-these Is there an easy way to find the minimum dimensions of representations for these R-algebras? Aaron Mazel-Gee 2010-09-20T23:06:58Z 2010-09-21T01:55:47Z <p>I'm working with Clifford algebras, of which the first few are $C_0 = \mathbb{R}$, $C_1 = \mathbb{C}$, $C_2 = \mathbb{H}$, $C_3= \mathbb{H}^2$, $C_4 = M_{2,2}(\mathbb{H})$, $C_5= M_{4,4}(\mathbb{C})$, $C_6=M_{8,8}(\mathbb{R})$, $C_7 = M_{8,8}(\mathbb{R})^2$, $C_8 = M_{16,16}(\mathbb{R})$, etc. (The operations on the product of two algebras are componentwise.)</p> <p>The notes I'm working with claim without proof that the minimum dimensions of representations are 1, 2, 4, 4, 8, 8, 8, 8, 16. (Sorry I probably could've formatted this better.) I assume that they mean faithful representations, and that they're representations as $\mathbb{R}$-algebras (and not just as rings), i.e. they're $\mathbb{R}$-linear maps $C_k \rightarrow End(V)$ (or equivalently maps $C_k \otimes_\mathbb{R} V \rightarrow V$, I think). I have no idea why these numbers are right, though. The first few I can see by ad hoc arguments, but that's not really that satisfying. I know no representation theory beyond a few definitions, so this might be really obvious and I just don't know the right theorems/arguments?</p> http://mathoverflow.net/questions/37423/is-there-an-easy-way-to-describe-the-sheaf-of-smooth-functions-on-a-product-manif Is there an easy way to describe the sheaf of smooth functions on a product manifold? Aaron Mazel-Gee 2010-09-01T20:53:36Z 2010-09-03T11:24:00Z <p>A smooth structure on a manifold $M$ can be given in the form of a sheaf of functions $\mathcal{F}$ such that there is an open cover $\mathcal{U}$ of $M$ with every $U\in \mathcal{U}$ isomorphic (along with $\mathcal{F}|_U$) to an open subset $V$ of $\mathbb{R}^n$ (along with $\mathcal{O}|_V$, where $\mathcal{O}$ is the sheaf of smooth functions on $\mathbb{R}^n$). I think we might also need to say that this satisfies a smooth-coordinate-change axiom, although maybe that's already tied up in the definition of a sheaf. In any case, here is my question:</p> <blockquote> <p>Given two smooth manifolds $(M,\mathcal{F})$ and $(N,\mathcal{G})$, is there an easy way to write the sheaf of functions on $M\times N$ without reference to coordinate neighborhoods?</p> </blockquote> <p>I'm wondering this because in one of my classes we defined smooth manifolds in this way (and we defined analytic and holomorphic manifolds similarly). It seems like some people are very fond of this alternative definition because it doesn't refer to an atlas, which at first seems like it's an inherent part of the structure of the manifold. So okay fine, everyone loves a canonical definition. However, this is only going to be useful as long as we can tell our whole story in this canonical language. In class, the only way the professor was able to give the sheaf on the product was by breaking down and using coordinates. (Admittedly, he was on the spot and presumably unprepared for the question.)</p> <p>This also suggests the broader, more open-ended question:</p> <blockquote> <p>Are there longer-run advantages to the above definition (compared to the usual definition involving an atlas and perhaps a maximal atlas)?</p> </blockquote> http://mathoverflow.net/questions/25792/the-hopf-invariant-of-the-hopf-construction the hopf invariant of the hopf construction Aaron Mazel-Gee 2010-05-24T19:07:45Z 2010-08-14T13:21:19Z <p>I'm having some trouble with a problem about the Hopf construction, in the exercises for Ch. 4 of <a href="http://www.amazon.com/Cohomology-Operations-Applications-Homotopy-Theory/dp/0486466647/ref=sr_1_1?ie=UTF8&amp;s=books&amp;qid=1274726521&amp;sr=8-1-spell" rel="nofollow">Mosher &amp; Tangora</a>. Given a map $g : S^{n-1} \times S^{n-1} \rightarrow S^{n-1}$, we get a map $h(g) : S^{2n-1} \rightarrow S^n$ by considering</p> <p>$S^{2n-1} = S^{n-1} * S^{n-1} = S^{n-1} \times I \times S^{n-1} / \sim$</p> <p>$S^n = S(S^{n-1}) = S^{n-1} \times I / \sim$</p> <p>and by putting $h(g)(a,t,b) = (g(a,b),t)$. There's an easy homotopy invariant $(a,b)$ of the map $g$, the degree of the map when restricted to either factor (times any point of the other $S^{n-1}$). The problem asks me to show that the Hopf invariant of the map $h(g)$ is $H(h(g))=ab$. This is defined by $s^2=H(h(g))t$ for generators $s\in H^n(X)$ and $t\in H^{2n}(X)$, where $X=S^n \cup_{h(g)} S^{2n}$ (here $h(g)$ is the attaching map).</p> <p>I'm trying to mimic a proof in the chapter, which constructs a map with Hopf invariant 2 for even $n$ which is really just the Whitehead square of the identity map. They use the following diagram ($f$ is the folding map, $F$ is induced from $f$, $g$ is the attaching map to get from $S^n \vee S^n$ to $S^n \times S^n$, the vertical maps are inclusions, and $K=S^n \cup_{fg} e^{2n}$ is the complex in which we need to calculate cup products):</p> <pre><code> g f S^{2n-1} ----&gt; S^n V S^n ------------&gt; S^n | | | | | | i | | V F V S^n x S^n = ( S^n V S^n ) U_g e^{2n} -----&gt; K </code></pre> <p>[<strong>Edit</strong>: Uh oh, how do I get it to put stuff in a uniformly spaced font? I'm too illiterate to understand the "formatting reference" -- any help??]</p> <p>Note that $F$ exists because the composition $ifg:S^{2n-1} \rightarrow K$ is nullhomotopic. We know that $f^*s=s_1 + s_2$, where $s$ and $s_i$ represent the obvious cohomology generators. Denoting 2n-dim. cohomology generators of $K$ and $S^n \times S^n$ by $t$ and $r$ respectively, since $F$ induces an isomorphism in $H^{2n}$, (we can choose orientations so that) $F^* t=r$. And by the Kunneth formula, in $H^*(S^n \times S^n)$ we have $s_1 s_2=r$ and $s_1^2=s_2^2=0$. So now we calculate: $F^*(s^2) = (F^*s)^2 = (f^*s)^2 = (s_1 + s_2)^2 = s_1^2 + s_1 s_2 + s_2 s_1 + s_2^2 = 2r$ (assuming $n$ is even; otherwise $F^*(s^2)=0$).</p> <p>So somehow they're using things they know about the cohomology of $S^n \times S^n$ to compute cup products in $H^*(K)$. This seems like it's possible because the attaching map $fg$ factors through $S^n \vee S^n$. In the problem I'm stuck on, I can't figure out what should be the analogous factorization. Alternatively, I was thinking that I could possibly use Poincare duality and just try to find the self-intersection of the copy of $S^n$ inside of $K$, but that seem too silly/gimmicky. Another fact I know is that $A * B \simeq \Sigma(A \wedge B)$. But suspension doesn't preserve cup products, so I don't think this could help either...</p> http://mathoverflow.net/questions/2890/cohomology-and-eilenberg-maclane-spaces cohomology and Eilenberg-MacLane spaces Aaron Mazel-Gee 2009-10-27T20:41:56Z 2010-08-01T17:13:26Z <p>This question is related to <a href="http://mathoverflow.net/questions/1346/representablity-of-cohomology-ring" rel="nofollow">this question</a> from Dinakar, which I found interesting but don't yet have the background to understand at that level.</p> <p>Unless I'm mistaken, the rough statement is that H<sup>n</sup>(X;G) (the n-dimensional cohomology of X with coefficients in G) should somehow correspond to (free?) homotopy classes of maps X --> K(G,n). I want to understand this better, in relatively elementary terms. Here are some questions which (I hope) will point me in the right direction.</p> <ol> <li>What category are we working in? My guess is that X should just be a topological space, the cohomology is singular cohomology, and our maps X --> K(G,n) just need to be continuous.</li> <li>Does this carry over if we give X a smooth structure, take de Rham cohomology, and require our maps X --> K(G,n) to be smooth?</li> <li>How does addition in H<sup>n</sup>(X;G) carry over?</li> <li>How does the ring structure on H*(X;G) carry over? (This has probably been adequately answered to Dinakar already.)</li> </ol> http://mathoverflow.net/questions/33589/stably-isomorphic-groups Stably isomorphic groups Aaron Mazel-Gee 2010-07-28T00:07:59Z 2010-07-28T09:15:12Z <p>The original question is: For groups, $A$, $B$, and $C$, can we have $A$ and $B$ be nonisomorphic but still have $A\oplus C = B\oplus C$?</p> <p>Clearly the answer is yes; for example, we can take $A=D$, $B=D\oplus D$, and $C=\oplus_\mathbb{N} D$. A spicier example is when $A$ and $B$ are the groups of sections of vector bundles over the same base space that are stably isomorphic but not isomorphic. But in general, both of these are going to involve non-finitely generated groups. So perhaps we should restrict the question to finite groups. Or we could even ask: Exactly what conditions on the groups give positive and negative answers?</p> <p><strong>Edit:</strong> To clarify what sorts of answers I'm hoping exist, recall that the definition of stable isomorphism of vector bundles requires that you add <em>trivial</em> bundles. So, what should be the algebraic analogue?</p> http://mathoverflow.net/questions/33250/references-general-idea-of-kervaire-invariant-problem references / general idea of kervaire invariant problem Aaron Mazel-Gee 2010-07-25T05:40:56Z 2010-07-26T23:38:19Z <p>There's a workshop at MSRI in a couple months on the Kervaire invariant problem that I'd really like to attend. I saw Hopkins speak about it a while back without understanding much of the talk, but I'm thinking that maybe by now I'm more prepared to see what it's all about. At least, I'd like to be able to get something out of the workshop, even if I don't understand everything that's going on. I don't know much more about the problem than the <a href="http://en.wikipedia.org/wiki/Kervaire_invariant" rel="nofollow">wikipedia article</a> can tell me. So, I'm looking for:</p> <p>(a) recommendations for reading to acquaint myself with the necessary background information, and</p> <p>(b) a sketch of the story -- I can somewhat follow the intro to the <a href="http://arxiv.org/PS_cache/arxiv/pdf/0908/0908.3724v1.pdf" rel="nofollow">paper</a> itself, but seeing some of the details spelled out would help me start to actually understand what's going on.</p> <p>(Or, unfortunately possibly: (c) a warning that it's probably too difficult to pick up all the material in just a few months and that it's not worth my time to try.)</p> <p>By the way, wikipedia says that Kervaire used the invariant to create a 10-dimensional PL manifold with no differentiable structure, but I thought that the invariant is zero in dimension 10. What's the deal there?</p> http://mathoverflow.net/questions/25792/the-hopf-invariant-of-the-hopf-construction/31933#31933 Answer by Aaron Mazel-Gee for the hopf invariant of the hopf construction Aaron Mazel-Gee 2010-07-15T00:52:11Z 2010-07-15T16:47:45Z <p><em>This is my attempt at an answer. I think I might be off in justifying the second diagram. It's based on the proof in Bredon, but I think it might be a little simpler. Corrections are welcome, of course!</em></p> <hr> <p>For convenience, we introduce some notation and conventions. We define subsets of <code>$$S^n= \{ (c,t):c\in S^{n-1},t\in I \} /\sim$$</code> by <code>$X=\{(c,t):t\leq 1/2\}$</code>, <code>$Y=\{(c,t):t\geq 1/2\}$</code>, and <code>$Z=\{(c,1/2)\}$</code>. We define subsets of <code>$$S^{2n-1}=\{ (a,t,b):a,b\in S^{n-1},t\in I\}$$</code> by <code>$A=\{(a,t,b)\in S^{2n-1}:t\leq 1/2\}$</code> and <code>$B=\{(a,t,b)\in S^{2n-1}:t\geq 1/2\}$</code>. We define subsets of <code>$A\cap B=\{(a,1/2,b)\}$</code> by <code>$S^{n-1}_A = \{(a,1/2,b_0)\}$</code> and <code>$S^{n-1}_B = \{(a_0,1/2,b)\}$</code> for fixed $a_0,b_0\in S^{n-1}$. We consider $S^{2n-1}$ as the boundary of $e^{2n}$; in this cell, $S^{n-1}_A$ is the boundary of a cell $e^n_A$ and $S^{n-1}_B$ is the boundary of a cell $e^n_B$. We write $K=S^n\cup_{h(g)}e^{2n}$ for the complex in which we must compute $H(h(g))$. Throughout, all cohomology is integral.</p> <p>To compute cup products in $K$, we use the commutative diagram</p> <pre><code>H^n(K) x H^n(K) ---------&gt; H^{2n}(K) ^ cup ^ | | | cup | H^n(K,X) x H^n(K,Y) -----&gt; H^{2n}(K,S^n) </code></pre> <p>which follows from the map $(K,\emptyset,\emptyset)\rightarrow (K,X,Y)$ and naturality; the vertical maps are isomorphisms by the long exact sequence in cohomology for pairs. Let the generator $x$ of $H^n(K)$ correspond to the generator $x_Y$ of $H^n(K,X)$ and the generator $x_X$ of $H^n(K,Y)$, so that $x_Y\cup x_X$ in $H^{2n}(K,S^n)$ corresponds to $x^2$ in $H^{2n}(K)$.</p> <p>To understand the image of the cup product, we use the evident map $j:(e^{2n},A,B)\rightarrow (K,X,Y)$. We have the diagram</p> <pre><code>H^n(K,X) = H^n(S^n,X) = H^n(Y,Z) = H^{n-1}(Z) | | | j^* | j^* | | V V H^n(e^{2n},A) = H^n(e^n_A,S^{n-1}_A) = H^{n-1}(S^{n-1}_A) </code></pre> <p>which commutes because of the naturality of the isomorphisms involved (namely, along the top: cellular cohomology, excision and homotopy invariance, lexseq for pairs; along the bottom: homotopy invariance, lexseq for pairs). By definition of degree, the map $j^*:H^{n-1}(Z)\rightarrow H^{n-1}(S^{n-1}_A)$ is multiplication by $\alpha$. Hence, so is the map $j^* : H^n(K,X)\rightarrow H^n(e^{2n},A)$. Similarly, the map $j^*:H^n(K,Y)\rightarrow H^n(e^{2n},B)$ is multiplication by $\beta$. Now, $j^*:H^{2n}(K,S^n)\rightarrow H^{2n}(e^{2n},S^{2n-1})$ is an isomorphism, and $j^*(x_Y\cup x_X)$ is $\alpha\beta$ times a generator of $H^{2n}(e^{2n},S^{2n-1})$. Thus $x_Y\cup x_X$ is $\alpha\beta$ times a generator of $H^{2n}(K,S^n)$. So by the first diagram, $H(h(g))=\alpha\beta$.</p> http://mathoverflow.net/questions/15322/visualizing-whats-going-on-in-based-homotopy-theory-et-al visualizing what's going on in based homotopy theory, et al. Aaron Mazel-Gee 2010-02-15T09:26:30Z 2010-07-07T00:49:54Z <p>I'm reading J.P. May's <a href="http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf" rel="nofollow" title="Concise Course in Algebraic Topology">Concise Course in Algebraic Topology</a>, and I'm having a lot of trouble visualizing how things work in Chapter 8, "Based cofiber and fiber sequences". Of course this is pretty basic stuff, but it's really cool to me that there are such clear topological analogues to the usual exact sequences in homological algebra. Still, I can't even get a clear picture of what a smash product looks like for any but the most basic of spaces, and based cones/suspensions/loopspaces make my head hurt.</p> <p>a) Will I be alright if in my head I just sort think of a smash product as a usual product (for example), with the understanding that I need to tack on an extra condition that I really shouldn't think too hard about?</p> <p>b) Why all the fuss about based homotopy theory, anyways?</p> <p>c) While I'm at it, can anyone suggest a book that is less terse? I feel like this one rarely gives the motivation and visual intuition that I'd like...</p> http://mathoverflow.net/questions/30597/how-can-i-prove-that-the-derived-couple-of-the-homotopy-exact-couple-is-an-invari How can I prove that the derived couple of the homotopy exact couple is an invariant? Aaron Mazel-Gee 2010-07-05T08:56:09Z 2010-07-05T19:18:35Z <p>I'm working on (yet) an(other) exercise from Mosher &amp; Tangora's "Cohomology Operations and Applications to Homotopy Theory". This one is about the homotopy exact couple, which is defined for a complex $K$ by $D_{p,q}=\pi_{p+q}(K^p)$ and $E_{p,q}=\pi_{p+q}(K^p,K^{p-1})$. So that we have relative Hurewicz, we assume K to be simply connected. As stated in the title, the object of the exercise is to show that this is not a homotopy invariant but that its derived couple is.</p> <p>The motivating example I've got in my head (let me know if you've got a better one) is $S^2$ realized either with 1 vertex, 1 edge, and 2 faces, or with 1 edge and 1 face. This already easily proves that the homotopy exact couple itself is not an invariant. For the harder part, I've drawn the (presumably standard) grid with rows like $\cdots \rightarrow D_{p,q} \rightarrow E_{p,q} \rightarrow D_{p-1,q} \rightarrow \cdots$ connected by vertical inclusion maps $D_{p,q} \rightarrow D_{p+1,q-1}$, and I can see how these both give the same derived couple, but I'm having trouble figuring out exactly how to make this into a general argument. I begin with a homotopy equivalence $f:K \rightarrow L$, $g:L \rightarrow K$, and I can assume these maps are cellular so I get induced maps between all corresponding groups of the homotopy exact couples associated to $K$ and $L$. But what can I say about these maps? Clearly from my motivating example the restrictions to skeleta need not be homotopy equivalences, or even anything close. I'm pretty sure they commute with the intra-couple maps, but I haven't had any success pushing through the commutative algebra with that fact alone. It smells like obstruction theory should be involved here since in general you'll need to move $K^p$ through $K^{p+1}$ to realize the homotopy $gf\simeq 1_K$ (consider it as a map $K\times I \rightarrow K$, which can be assumed to be cellular), but I don't think I understand it well enough to see how (or if that's even true, I guess). Am I headed in the right direction?</p> <hr> <p>P.S. I'm camping right now so I typed all of this on my phone. Might this be a first for MO? Or have people been asking math questions from their phones since before I was born...</p> http://mathoverflow.net/questions/23654/pontrjagin-square-and-possible-typo-in-mosher-tangora Pontrjagin square (and possible typo in Mosher & Tangora?) Aaron Mazel-Gee 2010-05-06T01:13:20Z 2010-05-06T12:20:45Z <p>There's an exercise at the end of Ch. 2 of Mosher &amp; Tangora's "Cohomology Operations and Applications in Homotopy Theory", which says:</p> <p>Suppose the cocycle $u\in C^{2p}(X;Z)$ satisfies $\delta u=2a$ for some $a$.</p> <p>i. Show that $u \cup_0 u + u\cup_1 u$ is a cocycle mod 4.</p> <p>ii. Define a natural operation, the Pontrjagin square, $P_2:H^{2p}(-;Z_2)\rightarrow H^{4p}(-;Z_4)$.</p> <p>iii. Show that $\rho P_2(u)=u\cup u$, where $\rho:H^*(-;Z_4)\rightarrow H^*(-;Z_2)$ denotes reduction mod 2.</p> <p>iv. Show that $P_2(u+v)=P_2(u)+P_2(v)+u\cup v$, where $u\cup v$ is computed with the non-trivial pairing $Z_2 \otimes Z_2\rightarrow Z_4$.</p> <hr> <p>First of all, I'm confused by the first sentence. Shouldn't it just say $u$ is a cochain, not a cocycle? Probably more importantly, I don't really see what's supposed to be going on in part (i), since those two addends aren't in the same cohomological dimensions. I'm getting that</p> <p>$\delta (u\cup_0 u+u\cup_1 u) = 4(a\cup u)+2(a\cup_1 u+u\cup_1 a)$,</p> <p>and I'm not sure how to cancel the last terms. The addends in the RHS came straight from the addends on the LHS, i.e. there's no interaction between the two terms as far as I can tell, which makes me doubt myself. This could be just a lot of silliness on my part, but I'd appreciate it if someone could clear this up for me. And of course I'd love to hear about fun or unexpected applications of this particular operation...</p> <p><strong>EDIT</strong></p> <p>Following Tyler's suggestion, I've found that changing the formula to $u\cup_0 u+u\cup_1 \delta u$ does wonders.</p> <p>Now that I've reached it, I'm having difficulty with part (iv). I have that</p> <p>$P_2(u+v)=P_2(u)+P_2(v)+u\cup_0 v+v\cup_0 u+u\cup_1 \delta v+v\cup_1 \delta u$.</p> <p>So somehow those last four terms are supposed to collapse into $u\cup v$ as computed using the non-trivial pairing. How exactly should this work? I have a vague sense that a representing cochain should only be spitting out the values 0(mod 4) and 2(mod 4), the latter only when the cup guys both spat out 1(mod 2)'s. So I'm slightly inclined to believe that the first two loose terms, which are equal since $\deg(u)=\deg(v)=2p$ and $\cup_0=\cup$ (the usual cup product), might sum to the desired "$u\cup v$" thing. Assuming we represent $u$ and $v$ by the same cochains throughout the equation, which I'm pretty sure is a valid thing to do(???), then if they evaluate to an even number then that sum will be 0(mod 4) while if they evaluate to an odd number then that sum will be 2(mod 4). So I'm thinking that somehow those last two terms should vanish. But why? And is there a concise way of writing everything I've just said, e.g. using notation $\rho:Z_2\otimes Z_2\rightarrow Z_4$ for the pairing, etc.? <strong><em>And probably more importantly, is there any significance to the fact that the cup product measures the failure of the Pontrjagin square to be a group homomorphism?</em></strong></p> http://mathoverflow.net/questions/124459/is-there-an-analog-of-directed-graph-for-topological-spaces Comment by Aaron Mazel-Gee Aaron Mazel-Gee 2013-03-13T21:27:30Z 2013-03-13T21:27:30Z <a href="http://ncatlab.org/nlab/show/directed+topological+space" rel="nofollow">ncatlab.org/nlab/show/directed+topological+space</a> http://mathoverflow.net/questions/124166/are-loop-spaces-of-homotopically-equivalent-spaces-homotopically-equivalent/124167#124167 Comment by Aaron Mazel-Gee Aaron Mazel-Gee 2013-03-10T19:12:44Z 2013-03-10T19:12:44Z CW complexes are the opposite of &quot;pathological&quot;. http://mathoverflow.net/questions/13638/which-popular-games-are-the-most-mathematical/13641#13641 Comment by Aaron Mazel-Gee Aaron Mazel-Gee 2013-01-13T09:42:45Z 2013-01-13T09:42:45Z One way to measure the complexity of a game is by declaring that Person 1 is someone who knows only the rules of the game, and that Person n+1 is someone who will beat Person n 90% of the time; then, the complexity is the maximal such chain. I've heard (though I can't remember where) that chess is 5 and go is 9. I'd imagine that this measure of complexity is correlated with how well we can program computers to play the game, although (presumably interestingly, from a psych point of view) there are probably cases where this doesn't match up perfectly. http://mathoverflow.net/questions/114647/what-is-the-universal-deformation-of-the-formal-additive-group-widehat-mathbb/114695#114695 Comment by Aaron Mazel-Gee Aaron Mazel-Gee 2012-12-02T16:48:06Z 2012-12-02T16:48:06Z Thanks, Eric! I'd say this certainly means there isn't a deformation *space*; if anything, there might be an Artin stack. If we could present it as $X/\!/G$ carrying a universal deformation $\widetilde{\mathbb{G}}$, then perhaps &quot;Morava $E_n$ at $n=\infty$&quot; could be reasonably taken to mean the Lubin-Tate spectrum associated to $\widetilde{\mathbb{G}}(X)$ along with its group action... or maybe the fixed points thereof. http://mathoverflow.net/questions/69086/lawvere-theories-versus-classical-universal-algebra/95773#95773 Comment by Aaron Mazel-Gee Aaron Mazel-Gee 2012-11-30T15:54:43Z 2012-11-30T15:54:43Z Ah, so I think the point is that there's no a priori reason that up-to-homotopy models (in $\mbox{Top}$) for the abelian group Lawvere theory should give rise to $E_\infty$-spaces. And of course, this result immediately implies that not all $E_\infty$-spaces can be written as homotopy models. http://mathoverflow.net/questions/114647/what-is-the-universal-deformation-of-the-formal-additive-group-widehat-mathbb Comment by Aaron Mazel-Gee Aaron Mazel-Gee 2012-11-27T10:54:32Z 2012-11-27T10:54:32Z Cross-posted to here from math.se due to lack of activity: <a href="http://math.stackexchange.com/questions/242061/what-is-the-universal-deformation-of-widehat-mathbbg-a-over-mathbbf-p" rel="nofollow" title="what is the universal deformation of widehat mathbbg a over mathbbf p">math.stackexchange.com/questions/242061/&hellip;</a> http://mathoverflow.net/questions/69086/lawvere-theories-versus-classical-universal-algebra/95773#95773 Comment by Aaron Mazel-Gee Aaron Mazel-Gee 2012-11-27T00:47:30Z 2012-11-27T00:47:30Z I just looked briefly at Badzioch's paper and wondered about the exact same thing that you bring up in your last paragraph. Did you ever sort this out? http://mathoverflow.net/questions/111025/monoidal-model-categories-with-suspension-functor Comment by Aaron Mazel-Gee Aaron Mazel-Gee 2012-10-29T19:57:51Z 2012-10-29T19:57:51Z In any model category $M$ you have $\Sigma X = \mbox{hocolim}(\ast \leftarrow X \rightarrow \ast)$. Then you can define a category $S^M$ of internal spectra in the same way as you do for spaces, and this comes with a stabilization functor $\Sigma^\infty:M \rightarrow S^M$. http://mathoverflow.net/questions/110285/on-the-natural-bigraded-homotopy-groups-of-a-simplicial-object-in-a-model-categ/110354#110354 Comment by Aaron Mazel-Gee Aaron Mazel-Gee 2012-10-24T00:40:32Z 2012-10-24T00:40:32Z I see; thanks for the clarification. And yes, I suppose that convention $X/\emptyset = X_+$ makes sense: the rule is to take the subspace and identify it all with a (new) point. I'll leave this open for a few days in case anyone has anything to say about my third question. http://mathoverflow.net/questions/110285/on-the-natural-bigraded-homotopy-groups-of-a-simplicial-object-in-a-model-categ Comment by Aaron Mazel-Gee Aaron Mazel-Gee 2012-10-22T01:28:51Z 2012-10-22T01:28:51Z Thanks for the suggestion. I've already checked out Dwyer-Kan-Stover briefly, but it didn't seem to resolve these questions. And no, I don't care strongly about the unstable version -- my hands are already pretty full with trying to understand this paper! -- I've just become curious about it in the course of things. http://mathoverflow.net/questions/107030/relationships-between-properties-of-model-categories Comment by Aaron Mazel-Gee Aaron Mazel-Gee 2012-09-17T10:57:52Z 2012-09-17T10:57:52Z David -- if you go ahead and copy your comment as an answer, I'll accept it. http://mathoverflow.net/questions/107030/relationships-between-properties-of-model-categories Comment by Aaron Mazel-Gee Aaron Mazel-Gee 2012-09-14T22:08:50Z 2012-09-14T22:08:50Z Ah, so there's no magic pill that I can take and suddenly understand all this stuff? ;o) I'll give this another day or so, and if nobody else responds then I'll accept your comment as an answer. Already Hirschhorn's introduction (as well as his introductions to the various Parts and Chapters) has been quite helpful. http://mathoverflow.net/questions/107030/relationships-between-properties-of-model-categories Comment by Aaron Mazel-Gee Aaron Mazel-Gee 2012-09-13T11:33:08Z 2012-09-13T11:33:08Z Thanks for the feedback. The reason I asked is that I've seen these sorts of adjectives running around before, but only recently have I begun to actually wrestle with them. I wasn't sure a diagram would be helpful, but the one in Gortz-Wedhorn popped into my head when I was thinking about it. I suppose the more important part is the last sentence. Certainly nLab has been a good resource, but I think I'd benefit from something a little more narrative. I'll edit. http://mathoverflow.net/questions/106963/a-question-on-an-intuitive-way-to-look-at-stacks Comment by Aaron Mazel-Gee Aaron Mazel-Gee 2012-09-12T00:56:44Z 2012-09-12T00:56:44Z No, this should still be true in the world of stacks; saying the triangle commutes really means there's a 2-morphism there, which is an isomorphism of elliptic curves over $S$. You should write to one of the authors, I guess. http://mathoverflow.net/questions/57082/is-there-a-unified-reason-that-there-are-an-infinite-number-of-geodesics-between/86345#86345 Comment by Aaron Mazel-Gee Aaron Mazel-Gee 2012-05-16T23:27:33Z 2012-05-16T23:27:33Z I like this answer; the Morse relations certainly seem to play a fundamental role here. As for your remark at the end though, if $\Omega_{p,p}(M) \simeq *$ then shouldn't its $0^{th}$ Betti number be 1 and all others be 0? Although this seems to contradict your observation for $\mathbb{R}^n$, since it says that $\sum k_m = 2Q(1)$, so I must be missing something...