User craig westerland - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T08:40:21Z http://mathoverflow.net/feeds/user/4649 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118865/spaces-parametrizing-ramified-covers-of-surfaces/118912#118912 Answer by Craig Westerland for Spaces parametrizing ramified covers of surfaces Craig Westerland 2013-01-14T19:13:44Z 2013-01-14T19:13:44Z <p>As Jason Starr mentions, the set of components of this space is somewhat difficult to pin down. I will work over the Riemann sphere $\Sigma =S^2$ for simplicity (it's already complicated enough). If you restrict your focus to the subspace $CF(\Sigma, G,n)$ consisting of branched covers $G$-covers which are connected, then for large $n$, there is a description of the set of components in terms of a certain quotient of the $H_2(G)$ by commutators. When $G = S_n$ (and monodromy is constrained to be given by transpositions), this case was studied by Clebsch and Hurwitz, with the result being that the space is connected. For more general $G$, this is an unpublished result of Conway-Parker, which was expanded upon by Fried-Völklein in the appendix to their paper "The inverse Galois problem and rational points on moduli spaces."</p> <p>Computing the cohomology of this space is quite difficult. It follows from your description above that the cohomology is the same as the cohomology of the mapping class group of the surface being fibred over, with (nontrivial twisted) coefficients in the free abelian group generated by $Hom(\pi_1(\Sigma \setminus { s_1, \dots, s_n}), G)$; i.e.,</p> <p>$$H^\ast(F(\Sigma, G, n)) \cong H^\ast(mcg(\Sigma); \mathbb{Z}Hom(\pi_1(\Sigma \setminus { s_1, \dots, s_n}), G)).$$</p> <p>Note that you can identify the cohomology of individual components as the group cohomology with coefficients in the subrepresentation generated by a given orbit of homomorphisms.</p> <p>Abstractly, this is nice, but it doesn't actually tell you what the groups are. One thing that you could hope for is a form of homological stability (i.e., that the homology of a single component stabilizes as $n \to \infty$), as it does for the configuration space (here I'm using the unordered configuration space, not the ordered one). Then, if you figure out the limiting homology, you can at least compute some of the homology groups of $F(\Sigma, G, n)$ (those to which your stability theorem applies). </p> <p>To advertise some recent work of Ellenberg, Venkatesh, and myself, in "Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields, I and II," we have proven a rational homological stability result for certain choices of $G$ (also, we often restrict the sort of monodromy around branch points that is allowed to lie in a given collection of conjugacy classes), as well as giving a method of determining the limiting homology.</p> <p>It's far from an optimal answer: we can give you partial, rational information only under certain assumptions on $G$. Also, if you're really after the ordered moduli space (we treat the unordered), our results don't quite answer your question (though I think that you can bootstrap them to get some information at least). Any improvements in the computations of the cohomology of these spaces would be very interesting indeed.</p> http://mathoverflow.net/questions/117794/rational-morava-e-theory-of-cyclic-groups Rational Morava E-theory of cyclic groups Craig Westerland 2013-01-01T18:12:05Z 2013-01-01T19:46:43Z <p>I'm confused about a seeming contradiction that is probably just a reflection of ignorance on my part. Let's try to compute the Morava E-theory of $B \mathbb{Z}/p$ in two different ways.</p> <ol> <li><p>First, following <a href="http://www.math.rochester.edu/people/faculty/doug/mypapers/emspace.pdf" rel="nofollow">Ravenel-Wilson</a> (see also <a href="http://www.math.rochester.edu/people/faculty/doug/mypapers/barc.pdf" rel="nofollow">Hopkins-Kuhn-Ravenel</a>, <a href="http://journals.cambridge.org/action/displayAbstract;jsessionid=6121AED0FBA78F8A056023186B2EB949.journals?fromPage=online&amp;aid=2092580" rel="nofollow">Hunton</a>, etc.), one computes (via knowledge of $K(n)^\ast(\mathbb{C} P^\infty)$ and a Gysin sequence) that the Morava K-theory of $B \mathbb{Z}/p$ is $$K(n)^\ast (B \mathbb{Z}/p) = K(n)_* [x] / x^{p^n}$$ where $x$ is of degree $2$. In particular, it is concentrated in even degrees and free of rank $p^n$ over $K(n)_\ast$. By Bockstein arguments, the Morava E-theory is then also free of the same rank.</p></li> <li><p>Second, using the knowledge that $E_n^\ast (B \mathbb{Z}/p)$ is free over $E_n^\ast$, it must embed into its rationalisation, $E_n^\ast (B \mathbb{Z}/p) \otimes \mathbb{Q}$. Let's try to compute the rank of that rationalisation using the group cohomological Atiyah-Hirzebruch spectral sequence: $$H^*(\mathbb{Z} / p, E_n^\ast \otimes \mathbb{Q}) \implies (E_n \otimes \mathbb{Q})^\ast (B \mathbb{Z}/p)$$ However, the $E_2$ term of the spectral sequence is rank 1 over $E_n^\ast \otimes \mathbb{Q}$ since $\mathbb{Z} / p$ is a finite group, and must collapse there. Therefore it certainly doesn't contain a sublattice of rank $p^n$.</p></li> </ol> <p>I've indicated my misgivings with the second argument by notating the target of the spectral sequence as $(E_n \otimes \mathbb{Q})^\ast (B \mathbb{Z}/p)$, that is, the value on $B \mathbb{Z}/p$ of the cohomology theory which is the rationalisation of $E_n$ (whose homotopy groups are $E_n^\ast \otimes \mathbb{Q}$). My guess is that this is NOT the same as the rationalisation of the E-theory of $B \mathbb{Z}/p$.</p> <p>So my real question is: is it easy to see why this is the case? More importantly, can one compute the correct answer (i.e., coming from argument 1) through methods based upon argument 2? I'm envisioning some sort of spectral sequence that knows how rationalisation and cohomology may fail to commute.</p> http://mathoverflow.net/questions/116894/cohomology-ring-of-bg/116903#116903 Answer by Craig Westerland for Cohomology ring of BG Craig Westerland 2012-12-20T20:03:18Z 2012-12-20T20:03:18Z <p>Notice that there is a sequence of homomorphisms $T \to N \to G$, where $N$ is the maximal torus normaliser (so $W = N/T$). $W$ acts on $BT$ (because it acts on $T$ by conjugation through group homomorphisms), and there is an equivalence from the classifying space of $N$ to the Borel construction for this action:</p> <p>$$BN \simeq EW \times_W BT.$$</p> <p>Consequently, we can compute the cohomology of $BN$ from the Leray-Serre spectral sequence</p> <p>$$H^\ast(W; H^\ast(BT)) \implies H^\ast(BN).$$</p> <p>Taking rational cohomology, this spectral sequence is concentrated in group-cohomological degree $0$, since $W$ is a finite group. Therefore the spectral sequence collapses at $E_2$, which is $H^0(W, H^\ast BT) = H^\ast(BT)^W$.</p> <p>It therefore suffices to show that the map $BN \to BG$ is an isomorphism in rational cohomology. If we write $BN$ as $EG / N$, this map is a fibre bundle with fibre $G / N$, so it's enough to show that $G/N$ has the rational homology of a point. </p> <p>For instance, if $G = SU(2)$, $N = \mathbb{Z} / 2 \ltimes T$, and $T = S^1$. Then $G/T = \mathbb{C} P^1$, and the action of $\mathbb{Z} / 2$ is antipodal, giving $G / N = \mathbb{R} P^2$, which is indeed rationally a point. I don't remember the argument in general, but I think this is always true.</p> <p>Hopefully this indicates how the corresponding integral statement can fail - there can be torsion contributions from the higher group cohomology of $W$, which needs to be exactly cancelled (via a differential in the second spectral sequence above) with a torsion cohomology class from $G/N$.</p> http://mathoverflow.net/questions/116507/can-i-compute-k-theory-in-serre-fibrations/116517#116517 Answer by Craig Westerland for Can I compute K theory in Serre fibrations? Craig Westerland 2012-12-16T09:30:49Z 2012-12-16T09:30:49Z <p>Here are a couple possible answers:</p> <p>(1). You could follow Dylan's suggestion about the Atiyah-Hirzebruch spectral sequence for </p> <p>$$H^{\ast}(B, K_{\ast}) \implies K^{\ast}(B)$$ </p> <p>and use the fact that you know the K-theory of $B$ to conclude something about the differentials in this spectral sequence. Then plug this information in to the Atiyah-Hirzebruch-Serre spectral sequence for the fibration.</p> <p>(2). You can't (easily) use the Eilenberg-Moore spectral sequence</p> <p>$$Tor_{K^{\ast} (B)} (K^{\ast} (E), K_{\ast}) \implies K^{\ast}(F),$$</p> <p>since it won't generally converge (because K-theory is not connective). To my understanding, <a href="http://arxiv.org/abs/0803.3798" rel="nofollow">this</a> paper by Tilman Bauer is the state of the art on these sorts of convergence questions.</p> <p>(3). This is pretty distant from your desired answer, but it's too tempting not to share. In the case that $F =G$ is a group and $E \to B$ is a principal $G$-bundle, there is a bar spectral sequence</p> <p>$$Tor_{K_{\ast}(G)} (K_{\ast} (E), K_{\ast}) \implies K_{\ast}(B),$$</p> <p>at least in the case that $K_{\ast}(G)$ is flat over $K_{\ast}$, since $B$ is the Borel construction for the action of $G$ on $E$ (When the flatness assumption doesn't hold, you can either make do with a simplicial construction that's trying to be the bar complex; i.e., $K_{\ast}(G^{\times n} \times E)$, or reduce mod $p$ and do it one prime at a time). This of course requires that you know $K_{\ast}(G)$, and how it acts on $K_{\ast}(E)$, but can be fantastically successful when you do (see, e.g., Ravenel-Wilson's <a href="http://www.jstor.org/discover/10.2307/2374093?uid=3737536&amp;uid=2129&amp;uid=2&amp;uid=70&amp;uid=4&amp;sid=21101451910923" rel="nofollow">paper</a> on the Morava K-theories of Eilenberg-MacLane spaces).</p> http://mathoverflow.net/questions/101497/formal-group-laws-arising-from-localizations-of-mu/101500#101500 Answer by Craig Westerland for Formal group laws arising from localizations of MU Craig Westerland 2012-07-06T14:12:07Z 2012-07-06T14:34:20Z <p>Regarding (1), since the localisation map from $\mathbb{Z}$ to <code>${\mathbb{Z}}_{(p)}$</code> is injective, and $MU_\ast$ is free over $\mathbb{Z}$, localisation $MU \to MU_{(p)}$ will be injective, and so the image of the localisation map will again be $MU$.</p> <p>Regarding $(2)$, localisation does preserve complex orientations -- a class $u \in E^2(\mathbb{C} P^\infty)$ is an orientation if its restriction to $S^2$ is a unit in $E^2(S^2) = E_0$. Since the map from a ring into a localisation of the ring carries units to units, the same will hold for any localisation of $E$.</p> <p>This procedure doesn't in some sense make the cohomology theory more interesting, but rather less interesting, as localisation usually kills some of the difficulties in the algebra of the original cohomology theory. But that in itself is perhaps interesting.</p> http://mathoverflow.net/questions/101011/homology-of-homotopy-fixed-point-spectra/101039#101039 Answer by Craig Westerland for Homology of homotopy fixed point spectra Craig Westerland 2012-07-01T03:25:08Z 2012-07-01T03:25:08Z <p>A good reference for the sort of spectral sequence that you're looking for is an article by Bruner-Rognes, "<a href="http://arxiv.org/abs/math.AT/0406081" rel="nofollow">Differentials in the homological homotopy fixed point spectral sequence</a>." It treats some of the convergence questions that you're worried about.</p> http://mathoverflow.net/questions/99663/homotopy-groups-of-on/99666#99666 Answer by Craig Westerland for homotopy groups of O(n) Craig Westerland 2012-06-15T01:40:54Z 2012-06-15T01:46:33Z <p>In practice, this isn't really possible. There are fibre sequences</p> <p>$$O(n-1) \to O(n) \to S^{n-1}$$</p> <p>which allow you to inductively compute the homotopy groups of $O(n)$ in terms of the homotopy of $S^{k}$, for $k &lt; n$. But the latter is one of the main open questions in homotopy theory.</p> <p>Of course, real Bott periodicity tells you the homotopy groups of $O = \lim_{n\to \infty} O(n)$. By the previous fibre sequence, this is the same as $\pi_k(O(n))$ for $n>k+1$ -- the homotopy groups stabilise at that point -- since $\pi_k(S^{n-1}) = 0$ in that range. But the higher homotopy of $O(n)$ for a fixed $n$ is less tractable.</p> <p>A good reference for what you <em>can</em> do with the fibre sequence above (and others like it) is Mimura-Toda's <em>Topology of Lie Groups, I and II</em>.</p> http://mathoverflow.net/questions/99032/roots-of-unity-in-algebraic-k-theory Roots of unity in algebraic K-theory Craig Westerland 2012-06-07T12:20:22Z 2012-06-07T12:20:22Z <p>For any commutative ring $R$, the tensor product of (finitely generated, projective) $R$-modules equips the algebraic K-theory $K(R) = K_0(R)$ with the structure of a commutative ring with unit.</p> <p>For $n>2$, I'm wondering what assumptions one might need to impose upon $R$ to ensure that $K(R)$ contains primitive $n^{\rm th}$ roots of unity (i.e., a cyclic subgroup of $K(R)^{\times}$ of order $n$). I'd even be happy with a good family of examples.</p> http://mathoverflow.net/questions/94566/equivariant-cohomology-of-the-complement-to-the-arrangment-cup-i-neq-joverrig/94738#94738 Answer by Craig Westerland for equivariant cohomology of the complement to the arrangment $\cup_{i\neq j}overrightarrow{x_i} = overrightarrow{x_j}$? Craig Westerland 2012-04-21T13:58:08Z 2012-04-21T13:58:08Z <p>When $V$ is a complex vector space, I computed the equivariant homology of the configuration space in a paper called <a href="http://front.math.ucdavis.edu/1005.2002" rel="nofollow">Operads of moduli spaces of points in $\mathbb{C}^d$</a>, not with respect to $U(V)$, but its restriction to $U(1)$, along the diagonal embedding. I'm not sure, but I think that extending to $U(V)$ just tensors this result with $H*(BU(\dim(V)-1))$, since the answer that I got for $U(1)$ doesn't seem to allow much room for nonzero operations coming from $H_*(U\dim(V)-1)$. </p> <p>The methods are closely related to those that Mark Grant sketches above, but are nicely packaged using the language of operads. If you're only interested in rational computations, I think a similar answer is obtainable in the real setting.</p> http://mathoverflow.net/questions/92422/to-which-extent-can-one-recover-a-manifold-from-its-group-of-homeomorphisms/92426#92426 Answer by Craig Westerland for To which extent can one recover a manifold from its group of homeomorphisms Craig Westerland 2012-03-28T00:30:17Z 2012-03-28T00:30:17Z <p>I suppose that the answer to this question depends upon how much information you are willing to allow yourself to extract from $G$. Since your manifold is connected, $G$ acts transitively upon it, and so if $x \in M$ is any point, and $G_x$ the stabiliser of $x$ in $G$, then there is a homeomorphism $G/G_x \cong M$. So $M$ can be completely reconstructed from $G$.</p> <p>To be fair, though, this presupposes that you have a very good understanding of $G$ and its subgroups, perhaps more than is reasonable. Furthermore, this is not the sort of information that is preserved by passing to the mapping class group (e.g., a surface is clearly not homeomorphic a quotient of its mapping class group).</p> http://mathoverflow.net/questions/89687/dualizable-classifying-spaces/90352#90352 Answer by Craig Westerland for Dualizable classifying spaces Craig Westerland 2012-03-06T11:12:16Z 2012-03-06T11:12:16Z <p>This example is undoubtedly covered by John Klein's general response, but is somewhat illustrative in its own right.</p> <p>By the Cartan-Hadamard theorem, the universal cover of every connected, non-positively curved manifold $M$ is homeomorphic to $\mathbb{R}^n$. So if $\Gamma = \pi_1(M)$ is the fundamental group of $M$, then $M = \mathbb{R}^n / \Gamma = B \Gamma$ is the classifying space of $\Gamma$. If $M$ is itself compact, it is dualisable.</p> <p>One tends to consider the case when $\Gamma$ is a discrete subgroup of $Isom(\mathbb{H}^n)$, the group of isometries of hyperbolic space. If $\Gamma$ acts freely and cocompactly on $\mathbb{H}^n$, then $B\Gamma$ is stably dualisable.</p> http://mathoverflow.net/questions/88993/alternate-proofs-of-quillens-theorem-on-formal-group-laws-and-mu Alternate proofs of Quillen's theorem on formal group laws and MU Craig Westerland 2012-02-20T04:20:44Z 2012-02-20T14:14:51Z <p>This is related to this posting:</p> <p><a href="http://mathoverflow.net/questions/5166/complex-cobordism-from-formal-group-laws" rel="nofollow">http://mathoverflow.net/questions/5166/complex-cobordism-from-formal-group-laws</a></p> <p>but not entirely the same. I'd like to know if there are any proofs of Quillen's theorem that $\pi_* (MU)$ is the Lazard ring (home of the universal formal group law) other than Quillen's. Specifically, is it possible to prove the result without a deep understanding of the structure of $H^*(MU)$ as a module over the Steenrod algebra?</p> http://mathoverflow.net/questions/88491/stable-triviality-of-fiber-bundles Stable triviality of fiber bundles Craig Westerland 2012-02-15T06:10:32Z 2012-02-20T02:51:00Z <p>This is probably has an obvious proof or a straightforward counterexample, but I'm having trouble finding either.</p> <p>Let $p:E \to B$ be a fibre bundle, with fibre $F$. Assume that there is a spectrum $X$ and a homotopy equivalence of spectra</p> <p>$$f: X \wedge \Sigma^{\infty} B_+ \to \Sigma^{\infty} E_+$$</p> <p>(in particular, there is a stable section to $p$). </p> <p>Can we conclude that $X \simeq \Sigma^{\infty} F_+$? If so, we may conclude that the bundle is stably trivial; that is, </p> <p>$$\Sigma^{\infty} E_+ \simeq \Sigma^{\infty} (B\times F)_+.$$</p> <p>If this is not the case, what sort of conditions do we need to demand of the fibration to make it hold? I'm happy to localize everything in site at your favorite prime, or for that matter, cohomology theory...</p> <p><strong>Edit:</strong> There is an immediate type of counterexample, gotten by taking $E = B \times F$ to be trivial, but where $\Sigma^\infty F_+ \simeq X \vee Y$, where $Y$ is $\Sigma^\infty B_+$-acyclic; i.e., </p> <p>$$\pi_*(Y \wedge \Sigma^\infty B_+) = 0$$</p> <p>One can define this problem away by assuming that $X$ and $\Sigma^\infty F_+$ are $\Sigma^\infty B_+$-local, which I'm happy to do for now.</p> http://mathoverflow.net/questions/88090/the-connected-components-of-the-free-loop-space/88092#88092 Answer by Craig Westerland for The connected components of the free loop space Craig Westerland 2012-02-10T11:50:27Z 2012-02-10T11:50:27Z <p>The different components are, indeed, not all homotopy equivalent, and you are quite right in noting that the argument that works for $\Omega M$ (via concatenation of loops) does not hold here.</p> <p>This is best illustrated when $M = BG = K(G,1)$ is an Eilenberg-MacLane space with precisely one (non-abelian) homotopy group $G$ in dimension 1. Geometrically, we are simply assuming $M$ is aspherical. Admittedly, not all manifolds fit this description (nor are all of these spaces manifolds), but this case captures the important part of the failure of these components to be homotopy equivalent.</p> <p>As you say, the set of components of $\Lambda M$ are indexed by conjugacy classes in $\pi_1(M) = G$. This statement may be promoted to the general claim that $\Lambda M$ is homotopy equivalent to the Borel construction </p> <p>$$\Lambda M \simeq G^{ad} \times_G EG,$$</p> <p>where $G^{ad}$ is $G$, regarded as a $G$-space via the conjugation action, $EG$ is a contractible space with a free $G$ action (e.g., the universal cover of $M$), and the notation indicates the quotient by the diagonal action of $G$ on the cross product. </p> <p>Since $G$ is discrete, one may write it as a disjoint union of orbits. These are, of course, just conjugacy classes of elements in $G$. I'll write $(g)$ for the conjugacy class of $g \in G$, so</p> <p>$$\Lambda M \simeq \coprod_{(g)} (g) \times_G EG.$$</p> <p>Then an individual component of $\Lambda M$ is of the form $(g) \times_G EG$. What is the topology of this space? Well, for one, it has fundamental group given by the centralizer of $g$ in $G$, $C(g)$. This is because it is the quotient of $(g)$ copies of the universal cover of $M$ by an action which permutes the copies via conjugation (transitively, by assumption), and the stabilizer of a given copy (say the one indexed by $g$) is simply the set of elements that commute with $g$. </p> <p>In fact, since $M$ was aspherical (i.e., $EG$ was contractible), this is the <em>only</em> homotopy group of this space. We conclude:</p> <p>$$\Lambda M \simeq \coprod_{(g)} K(C(g), 1) = \coprod_{(g)} BC(g).$$</p> <p>Now, as long as $G$ is not abelian, the centralizers of elements of $G$ will not all be isomorphic. Consequently, these components are not homotopy equivalent, as they have different fundamental groups.</p> http://mathoverflow.net/questions/87207/can-one-do-without-a-classifying-space-when-showing-vanishing-of-cohomology/87420#87420 Answer by Craig Westerland for Can one do without a classifying space when showing vanishing of cohomology Craig Westerland 2012-02-03T11:58:50Z 2012-02-03T11:58:50Z <p>To highlight the assumption of high connectivity in Johannes' answer, I should point out that for any $G$, you can always take $Y = G$ (a simplicial complex with only vertices, seeing as $G$ is discrete). Then you have a simplicial, cocompact, free action, and it is true that $H^n(Y/G, A) = 0$ for $n>0$. But that tells us nothing about $H^n(G, A)$.</p> http://mathoverflow.net/questions/68498/cohomology-of-finite-quotients-of-lie-groups/68688#68688 Answer by Craig Westerland for Cohomology of finite quotients of Lie groups Craig Westerland 2011-06-24T02:17:34Z 2011-06-24T02:17:34Z <p>Here's another approach using the Leray-Serre spectral sequence. Using the fact that $\Gamma$ acts freely on $G$, the map $G \to G/\Gamma$ is a covering space, and so the cohomology of $G/\Gamma$ may be computed by the spectral sequence</p> <p>$$E_2^{s, t} = H^s(\Gamma, H^t(G)) \implies H^{s+t}(G/\Gamma).$$</p> <p>Here $H^s(\Gamma, M)$ is the s'th group cohomology of $\Gamma$ with coefficients in the $\Gamma$-representation $M$. We record three facts to start:</p> <ol> <li>If $\Gamma \cong \mathbb{Z} / p$, then $H^s(\Gamma, \mathbb{Z})$ is 0 for $s$ odd, and $\mathbb{Z} / p$ for $s$ even.</li> <li>For $G=Spin(n)$, $H^t(G) = \mathbb{Z}, 0, 0, \mathbb{Z}$, for $t=0,1,2,3$.</li> <li>For $p$ odd, $\mathbb{Z} / p$ cannot act nontrivially on $\mathbb{Z}$, since $Aut(\mathbb{Z}) = \mathbb{Z} / 2$.</li> </ol> <p>Let's compute part of the $E_2$-term of the spectral sequence. Fact 2 implies that the $t=1$ and $2$ rows vanish entirely. Fact 3 implies that $H^0(G) = \mathbb{Z}$ and $H^3(G) = \mathbb{Z}$ are trivial $\Gamma$-modules, so the $t=0$ and $3$ rows are the group cohomology described in Fact 1.</p> <p>Thus the only possible term that can contribute to $H^3(G/\Gamma)$ is $E_2^{0, 3} = H^3(G) = \mathbb{Z}$. There is, however, a possiblity of a single differential in this region of the spectral sequence, namely</p> <p>$$d_4: E_2^{0, 3} = \mathbb{Z} \longrightarrow E_2^{4, 0} = H^4(\Gamma, \mathbb{Z}) = \mathbb{Z} / p.$$</p> <p>Therefore, $H^3(G/\Gamma)$ surjects onto $H^3(G)$ precisely when this differential $d_4= 0$. We note that if it's not 0, it is surjective; thus at worst $H^3(G/\Gamma)$ may be identified with an index $p$ subgroup of $H^3(G)$.</p> <p>So, how do we compute $d_4$? I claim that it's given as:</p> <p>$$H^3(G) \cong H^4(BG) \to H^4(B\Gamma) = H^4(\Gamma)$$</p> <p>where the map is the restriction in cohomology, induced by the (inclusion) homomorphism $\Gamma \subseteq G$. This can be seen, for instance, by comparing this with the spectral sequence for the (rather dumb) fibration $G \to G/G=pt$.</p> <p>So a long winded answer to your question is: The map is surjective if and only if none of the $H^4(BG) = \mathbb{Z}$ is supported on $H^4(\Gamma) = \mathbb{Z} /p$. I would imagine that determining when that is the case is highly dependent upon the subgroup in question.</p> http://mathoverflow.net/questions/61029/equivariant-cohomology-with-respect-to-a-loop-group/61103#61103 Answer by Craig Westerland for equivariant cohomology with respect to a loop group Craig Westerland 2011-04-08T23:40:19Z 2011-04-08T23:40:19Z <p>You can compute $H^\ast(LBG)$ as the Hochschild cohomology </p> <p>$$HH^\ast(C_\ast(G), C^\ast(G)),$$</p> <p>where $C_\ast(G)$ is the singular chain complex of $G$, equipped with the Pontrjagin product, and $C^\ast(G)$ are the cochains, with the $C_\ast(G)$-module structure dual to the obvious one on $C_\ast(G)$. If you like, you can then replace $C^\ast(G)$ with the Chevalley-Eilenberg complex $K$ for the Lie algebra, and $C_\ast(G)$ with its dual $K^\ast$. At this point, though, it is perhaps not so obvious how to recover the ring structure on $K^*$ which lets you define the Hochschild cohomology.</p> http://mathoverflow.net/questions/46521/formality-of-classifying-spaces/60680#60680 Answer by Craig Westerland for Formality of classifying spaces Craig Westerland 2011-04-05T12:53:06Z 2011-04-05T12:53:06Z <p>I think that you outlined the proof. In more detail, let $W$ be the Weyl group of $G$, and $T$ its maximal torus. Pick $p$ coprime to $|W|$; this allows to ignore higher $W$ group cohomology in the computation </p> <p>$$H^*(BG, \mathbb{F}_p) \cong H^*(BT, \mathbb{F}_p)^W$$</p> <p>Since $W$ is a reflection group, $H^*(BT, \mathbb{F}_p)^W$ is a polynomial algebra, say on $d$ generators. Pick <em>cocycle</em> representatives $x_1, \dots, x_d \in C^*(BG, \mathbb{F}_p)$. Now let $R = \mathbb{F}_p[y_1, \dots, y_d]$ be the free graded commutative algebra on generators $y_i$ in the same degree as $x_i$, and equip $R$ with the $0$ differential. By freeness (and the fact that $d(x_i) = 0$), you get a map $R \to C^*(BG, \mathbb{F}_p)$ of DGA's which sends $y_i$ to $x_i$. You know (because you constructed it that way) that it induces an isomorphism in cohomology, and so $BG$ is formal at the prime $p$.</p> <p>If you have some other mechanism for ensuring that $H^*(BG, \mathbb{F}_p)$ is a polynomial algebra (e.g., the statement is known integrally, as for $G = U(n)$, $Sp(n)$), the same argument works.</p> http://mathoverflow.net/questions/58389/homology-of-homotopy-fixed-point-spaces Homology of homotopy fixed point spaces Craig Westerland 2011-03-14T06:57:36Z 2011-03-16T21:48:24Z <p>This is a general question for the homotopy theory crowd: How does one go about computing the homology and homotopy groups of homotopy fixed point spaces $X^{hG}:= Map^G(EG, X)$ for the action of a group $G$ on a space $X$? There seem to be some tools:</p> <ol> <li>Lannes' theory: which allows you to compute (or at least say <em>something</em> about) $H_*(X^{hG}, \mathbb{F}_p)$ when $G$ is a $p$-group.</li> <li>Homotopy fixed point spectral sequences, which allow you to compute the stable homotopy groups of homotopy fixed point <em>spectra</em>.</li> </ol> <p>Are there other tools out there? I feel like (1.) should be the harder version of a fact that I'm missing about computing $H_*(X^{hG}, \mathbb{F}_p)$ when $|G|$ is coprime to $p$. Regarding (2.), is there any hope of an unstable homotopy fixed point spectral sequence?</p> http://mathoverflow.net/questions/58379/does-there-exist-a-contractible-fiber-bundle-with-fiber-g-infty-and-base-su/58392#58392 Answer by Craig Westerland for Does there exist a contractible fiber bundle with fiber $G(\infty)$ and base $SU(\infty)$? Craig Westerland 2011-03-14T07:24:15Z 2011-03-14T07:24:15Z <p>This is a lousy answer, but it gets you part of the way. There is an $E_\infty$ H-space structure on $G(\infty)$ given by direct sum of subspaces of $\mathbb{C}^\infty$. Consequently one is entitled to form the classifying space $B(G(\infty))$. </p> <p>Now, one can rectify $G(\infty)$ into a group; that is, there is a group $G$ and a homotopy equivalence $G \to G(\infty)$. For instance, one could take $G$ to be the Kan loop group on the singular complex of $B(G(\infty))$, a (very large) group model for $\Omega B(G(\infty)) \simeq G(\infty)$. Then one gets a homotopy equivalence $BG \simeq BG(\infty)$.</p> <p>Bott periodicity implies that there is a homotopy equivalence $SU \to B(G(\infty))$. Replace this with a map $f: SU \to BG$. The associated principal $G$-bundle $E \to SU$ is contractible, since $f$ is a homotopy equivalence. So you get a contractible fiber bundle over $SU$ with fibre homotopy equivalent to $G(\infty)$.</p> <p>That's not quite what you want. If you could show that the homotopy equivalence $G \to G(\infty)$ was via an action that mimicked left translation in $G$, then you can form the simultaneous quotient $E \times_G G(\infty)$; this would be precisely the sort of object that you had in mind.</p> <p>I'm also ignoring the issue about whether the universal bundle $EG \to BG$ is in fact a fibre bundle; $G$ is potentially sufficiently exotic that the standard theorems about Lie groups don't necessarily hold. Again, though, if you could show that $G$ acts on $G(\infty)$ nicely, then you could realize $G$ (as you say above) as a subgroup of $Diff(G(\infty))$, where I believe that the result is known to hold.</p> http://mathoverflow.net/questions/56363/list-of-classifying-spaces-and-covers/56367#56367 Answer by Craig Westerland for List of Classifying Spaces and Covers Craig Westerland 2011-02-23T07:38:54Z 2011-02-23T07:38:54Z <p>If $G$ is linear topological group, then a model for $EG$ may be taken to be an infinite Stiefel manifold. More precisely, if there is an faithful representation of $G$ into $GL_n(\mathbb{C})$, that gives a free action of $G$ on the space $V_n$ of $n$-frames in $\mathbb{C}^\infty$. $V_n$ is contractible, so $BG$ may be taken to be the quotient $BG = V_n / G$. If it's helpful to think of it this way, this is a fibre bundle over $BGL_n(\mathbb{C}) = G_n(\mathbb{C}^\infty)$, with fibre $GL_n(\mathbb{C}) / G$.</p> <p>Another favorite example comes from spaces of embeddings: if $M$ is a compact manifold without boundary, then it is a consequence of Whitney's embedding theorem that $Emb(M, \mathbb{R}^\infty)$ is contractible. The group $G=Diff(M)$ of diffeomorphisms of $M$ acts freely on $Emb(M,\mathbb{R}^\infty)$ by precomposing an embedding with a diffeomorphism. Therefore a model for $BG$ is the quotient $Emb(M, \mathbb{R}^\infty) / Diff(M)$, which is often thought of as the space of subspaces of $\mathbb{R}^\infty$ diffeomorphic to $M$. One can combine this idea with the previous idea for subgroups of diffeomorphism groups.</p> <p>Lastly, if there is a homomorphism $G \to H$ which is a homotopy equivalence, then of course there is a homotopy equivalence $BG \to BH$. So in the previous examples, one can for instance replace $GL_n(\mathbb{C})$ with $U(n)$, and $Diff^{+}(\Sigma)$ with the mapping class group $\Gamma(\Sigma) = \pi_0(Diff^{+}(\Sigma))$ for closed surfaces $\Sigma$.</p> http://mathoverflow.net/questions/22299/what-are-some-examples-of-colorful-language-in-serious-mathematics-papers/22343#22343 Answer by Craig Westerland for What are some examples of colorful language in serious mathematics papers? Craig Westerland 2010-04-23T12:42:44Z 2010-04-23T12:42:44Z <p>From Tilman Bauer's "p-compact groups as framed manifolds:"</p> <p><em>For our purposes, it is enough to work in the category of so-called naive G-spectra. I will drop the word “naive” since it will make this work appear so puny.</em></p> <p>And in Tilman's paper with Natalia Castellana, "Adjoint spaces and flag varieties of p-compact groups:"</p> <p><em>This comment is only meant to intimidate the reader and is insubstantial for what follows.</em></p> http://mathoverflow.net/questions/21742/knot-complement-diffeomorphism-groups-and-embedding-spaces Knot complement diffeomorphism groups and embedding spaces Craig Westerland 2010-04-18T13:32:23Z 2010-04-18T19:53:15Z <p>I'm interested in the following collection of questions: Let $S^n_k = \sqcup_k S^n$ be a disjoint union of $k$ distinct $n$-dimensional spheres. Write $Emb(S_k^n, S^{n+2})$ for the space of embeddings of these spheres into $S^{n+2}$. Pick your favorite embedding $e: S_k^n \to S^{n+2}$, and let $X_e = S^{n+2} \setminus im(e)$ be the complement of the image of the embedding.</p> <ol> <li><p>What is $\pi_1(Emb(S_k^n, S^{n+2}), e)$? Since this is probably unknown, what is known?</p></li> <li><p>How is this related to the mapping class group $\pi_0(Diff(X_e))$ of $X_e$?</p></li> </ol> <p>I ask #2 because in dimension $n=0$, they are the same: the space of embeddings is the configuration space of points in the sphere. Its fundamental group is the (spherical) braid group, which is the same as the mapping class group of the punctured sphere. My guess is that life is not so simple in higher dimensions. Lastly, does any of this simplify out when you get into the range of dimensions where surgery theory starts working well?</p> http://mathoverflow.net/questions/21121/dyer-lashof-based-spectral-sequence-for-homotopy-classes-of-maps-between-infinite/21305#21305 Answer by Craig Westerland for Dyer-Lashof based spectral sequence for homotopy classes of maps between infinite loop spaces (spectra). Craig Westerland 2010-04-14T04:43:03Z 2010-04-14T07:33:54Z <p>This might not quite be what you're looking for, Dev, but you should check out Paul Goerss and Mike Hopkins' "Multiplicative ring spectra project," on Paul's <a href="http://www.math.northwestern.edu/~pgoerss/" rel="nofollow">webpage</a>. They construct such a spectral sequence using Andre-Quillen cohomology in "Moduli spaces of commutative ring spectra," and "Andre-Quillen (co-)homology for simplicial algebras over simplicial operads." A relevant theorem would be 4.3 in the first reference, which gives the spectral sequence.</p> <p>Though this doesn't use Dyer-Lashof operations, they appear in section 6 (especially Prop 6.4) where Goerss and Hopkins give a second spectral sequence which computes the $E_2$ term of the original spectral sequence. The new $E_2$ term is given in terms of an $Ext$ functor in the category of unstable modules over the Dyer-Lashof algebra.</p> <p>They use this machinery to show in section 7 that the space of $E_\infty$ maps between Lubin-Tate spectra is homotopically discrete. If you're looking for computations using these spectral sequences, that's a great place to start.</p> http://mathoverflow.net/questions/21207/applications-of-fabers-conjecture Applications of Faber's conjecture Craig Westerland 2010-04-13T14:09:49Z 2010-04-13T17:20:47Z <p>Faber's perfect pairing conjecture states that the tautological ring $R^*$ of the moduli space $\mathcal{M}_g$ of curves of genus $g$ behaves as if it were the rational cohomology of a closed, oriented manifold of dimension $g-2$. Specifically, $R^{g-2}$ is rank one, and multiplication into this degree gives a perfect pairing between $R^k$ and $R^{g-2-k}$.</p> <p>My understanding is that it is known (through work of Looijenga, Faber, and Pandharipande) that $R^{g-2} = \mathbb{Q}$, but the perfect pairing part hasn't been proven (though it has been verified in low genus cases). I'd like to know:</p> <ol> <li><p>Why might Faber have conjectured this to be the case? What is it about $R^*$ that suggests that it might satisfy Poincare duality?</p></li> <li><p>If true, what sort of applications does this have (to our understanding of $\mathcal{M}_g$, for instance)?</p></li> </ol> http://mathoverflow.net/questions/20671/what-is-the-equivariant-cohomology-of-a-group-acting-on-itself-by-conjugation/21006#21006 Answer by Craig Westerland for What is the equivariant cohomology of a group acting on itself by conjugation? Craig Westerland 2010-04-11T13:32:31Z 2010-04-11T13:32:31Z <p>With regards to some of the comments towards the bottom from David Ben-Zvi and Tim Perutz: you can get around some finite dimensional restrictions. Specifically, there is an interpretation of the Hochschild cohomology of $C^*(BG)$ in terms of string topology. Namely, it's an inverse limit of the homology of a pro-object that approximates the free loop space of $BG$ by finite dimensional manifolds. As hinted at by Dan Ramras' comments, a lot of this comes from Kate Gruher's work. This comment is explained in detail in a paper of hers and mine: arXiv:0710.1445.</p> http://mathoverflow.net/questions/16468/topologists-loops-versus-algebraists-loops/18256#18256 Answer by Craig Westerland for Topologists loops versus algebraists loops Craig Westerland 2010-03-15T08:08:49Z 2010-03-15T08:08:49Z <p>Just a metacomment on Bhargav's answer: it's not always true that $\pi_0(LX) = \pi_1(X)$, namely when $X$ is not simply connected (as is certainly the case in this example). In general $\pi_0(LX)$ is the set of conjugacy classes of elements of $\pi_1(X)$ -- think about the change-of-basepoint isomorphism in $\pi_1$.</p> <p>However, this certainly doesn't break the argument: there are lots of nontrivial conjugacy classes in $\pi_1(X)$.</p> http://mathoverflow.net/questions/122309/bar-construction-for-spectra Comment by Craig Westerland Craig Westerland 2013-02-20T11:29:35Z 2013-02-20T11:29:35Z Why is it uninteresting for $S$ to split off of $R$? Doesn't this happen in the case that $R = \Sigma^\infty G_+$ for a group $G$? The associated Tor spectrum $Tor^R(S, S)$ (or derived smash product of $S$ with itself over $R$) will then compute the suspension spectrum of $BG$, certainly an interesting object. http://mathoverflow.net/questions/121832/analogue-of-cyclic-homology-for-e-n-algebras Comment by Craig Westerland Craig Westerland 2013-02-14T21:40:30Z 2013-02-14T21:40:30Z This is not a great answer, but you can try to take the factorization homology of the $e_n$-algebra over $S^n$ (perhaps you need to assume that the algebra is framed). Then, since $S^n$ has an action of $SO(n+1)$, you could take the homotopy orbits (or fixed points) of the result. When $n=1$, this returns the usual definition of cyclic homology (or negative cyclic homology). But it's not obvious that this can be expressed in the terms that you're asking for above. http://mathoverflow.net/questions/121808/in-the-cohomology-of-thom-spectrum-over-loops2-and-p-adic-characteristic-class Comment by Craig Westerland Craig Westerland 2013-02-14T21:26:17Z 2013-02-14T21:26:17Z What is the group $G_3$? http://mathoverflow.net/questions/118865/spaces-parametrizing-ramified-covers-of-surfaces/118912#118912 Comment by Craig Westerland Craig Westerland 2013-01-15T11:23:14Z 2013-01-15T11:23:14Z Yeah, sorry, that's right. The setup in my response is appropriate for computing the homology of the moduli space of these branched covers; your setup (fibering over the configuration space) will only use the braid group. And indeed, computing the homology of these spaces even when $\Sigma = \mathbb{R}^2$ is really quite difficult for non-abelian $G$. http://mathoverflow.net/questions/118865/spaces-parametrizing-ramified-covers-of-surfaces/118912#118912 Comment by Craig Westerland Craig Westerland 2013-01-14T19:18:12Z 2013-01-14T19:18:12Z Also, compactifications of these spaces do exist; see, for instance, the work of Abramovich-Corti-Vistoli, amongst others. http://mathoverflow.net/questions/117818/computing-homotopy-groups-of-x-such-that-pi-1x-has-solvable-word-problem Comment by Craig Westerland Craig Westerland 2013-01-02T00:36:33Z 2013-01-02T00:36:33Z This is probably missing the point, but of course if $\pi_1(X)$ is finite, the universal cover of $X$ is a finite simply connected simplicial complex with the same higher homotopy groups as $X$; now invoke Brown's result. http://mathoverflow.net/questions/117794/rational-morava-e-theory-of-cyclic-groups/117803#117803 Comment by Craig Westerland Craig Westerland 2013-01-01T21:19:17Z 2013-01-01T21:19:17Z Nice, thank you! http://mathoverflow.net/questions/116894/cohomology-ring-of-bg/116903#116903 Comment by Craig Westerland Craig Westerland 2012-12-20T20:27:51Z 2012-12-20T20:27:51Z You can avoid the first spectral sequence if you have another way of talking yourself into believing that the cohomology of $BN$ is the same as the $W$-invariants of $H^*(BT)$, e.g., using the transfer. For the latter spectral sequence, you can use the fact that the Euler class of $G/T$ is nonzero, as Chris Gerig indicates in his answer. http://mathoverflow.net/questions/116894/cohomology-ring-of-bg/116904#116904 Comment by Craig Westerland Craig Westerland 2012-12-20T20:23:45Z 2012-12-20T20:23:45Z Oh, maybe this is nonsense -- the 0 dimensional class is the trivial factor in the regular representation, and the 2 dimensional factor is the reduced regular representation. What is subtle here is that how the regular representation knows which summands correspond to which cohomological degrees. I suppose that this should be related to the Bruhat order on the Weyl group. http://mathoverflow.net/questions/116894/cohomology-ring-of-bg/116904#116904 Comment by Craig Westerland Craig Westerland 2012-12-20T20:16:42Z 2012-12-20T20:16:42Z Chris, this was my memory, too, but it's hard to imagine that it's exactly true: for the case $G=SU(2)$, $N=\mathbb{Z}/2$ and $G/T=P^1$, whose cohomology is indeed free of rank two. However there's no way that $W$ can act on it by the regular representation, since one generator is in dimension 0, and the other in dimension 2. http://mathoverflow.net/questions/116894/cohomology-ring-of-bg Comment by Craig Westerland Craig Westerland 2012-12-20T20:04:51Z 2012-12-20T20:04:51Z The differences between the power series and polynomial rings in this case depend upon your choice to define $H^\ast(X)$ as either the product or sum over all $n$ of $H^n(X)$. http://mathoverflow.net/questions/116507/can-i-compute-k-theory-in-serre-fibrations/116517#116517 Comment by Craig Westerland Craig Westerland 2012-12-16T11:43:54Z 2012-12-16T11:43:54Z This may be particularly useful in the case that $B = BG$ is the classifying space of a group whose group ring is easily presented. http://mathoverflow.net/questions/116507/can-i-compute-k-theory-in-serre-fibrations/116517#116517 Comment by Craig Westerland Craig Westerland 2012-12-16T09:56:27Z 2012-12-16T09:56:27Z An addendum about the last suggestion: if it's not $K_{\ast}(B)$ that you're after, but rather $K_{\ast}(E)$, you can back up the fibre sequence by one to $\Omega B \to F \to E$, and run the indicated SS on that. http://mathoverflow.net/questions/116490/why-do-we-use-the-diagonal-for-diagonal-approximations/116500#116500 Comment by Craig Westerland Craig Westerland 2012-12-16T01:26:59Z 2012-12-16T01:26:59Z Actually, the map $g \mapsto (g, 1)$ induces a product which carries $\alpha \otimes \beta$ to $\alpha$ only when $\beta \in H^0(G)$; otherwise the &quot;product&quot; is $0$. http://mathoverflow.net/questions/110568/on-variable-and-primitive-cohomology-of-a-hypersurface-in-a-projective-space Comment by Craig Westerland Craig Westerland 2012-10-24T23:04:05Z 2012-10-24T23:04:05Z What is $X$ here?