Craig Westerland
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Registered User
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Mar 15 |
awarded | ● Yearling |
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Feb 20 |
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Bar construction for spectra 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. |
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Feb 14 |
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Analogue of cyclic homology for e_n-algebras? 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. |
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Feb 14 |
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In the cohomology of Thom spectrum over LoopS^{2} and p-adic characteristic classes What is the group $G_3$? |
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Jan 15 |
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Spaces parametrizing ramified covers of surfaces 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 $G$. |
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Jan 14 |
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Spaces parametrizing ramified covers of surfaces Also, compactifications of these spaces do exist; see, for instance, the work of Abramovich-Corti-Vistoli, amongst others. |
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Jan 14 |
answered | Spaces parametrizing ramified covers of surfaces |
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Jan 2 |
awarded | ● Nice Question |
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Jan 2 |
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Computing homotopy groups of X such that pi_1(X) has solvable word problem 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. |
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Jan 1 |
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Rational Morava E-theory of cyclic groups Nice, thank you! |
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Jan 1 |
asked | Rational Morava E-theory of cyclic groups |
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Dec 20 |
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Cohomology ring of BG 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. |
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Dec 20 |
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Cohomology ring of BG 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. |
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Dec 20 |
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Cohomology ring of BG 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. |
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Dec 20 |
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Cohomology ring of BG 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 $H^n(X)$. |
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Dec 20 |
answered | Cohomology ring of BG |
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Dec 18 |
accepted | Can I compute K theory in Serre fibrations? |
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Dec 16 |
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Can I compute K theory in Serre fibrations? This may be particularly useful in the case that $B = BG$ is the classifying space of a group whose group ring is easily presented. |
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Dec 16 |
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Can I compute K theory in Serre fibrations? 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. |
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Dec 16 |
answered | Can I compute K theory in Serre fibrations? |
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Dec 16 |
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Why do we use the diagonal for diagonal approximations ? 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 "product" is $0$. |
