Greg Graviton
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 Apr 3 accepted Direct proof that the wedge product preserves integral cohomology classes? Apr 3 comment Direct proof that the wedge product preserves integral cohomology classes? Oh, so there exist orientable manifolds where some submanifolds are not orientable, I hadn't thought of that. Apr 2 comment Direct proof that the wedge product preserves integral cohomology classes? On the issue of non-canonicity, it might be possible to avoid it by arguing that the mapping $\bigoplus_{i+j=k} H_i(M;\mathbb{Z})\otimes H_j(M;\mathbb{Z}) \to H_k(M\times M;\mathbb{Z})$ is "surjective up to torsion" more directly, i.e. to throw away the torsion contributions before the need to exhibit a splitting. Geometrically, I would also be happy to restrict $M$ to be an orientable manifold. Apr 2 comment Direct proof that the wedge product preserves integral cohomology classes? Thanks David! I think this is actually as best an answer as we can hope for. The problem is that the condition for being an integral form uses homology with coefficients $\mathbb{Z}$ in an essential way -- it's not enough to know homology, say, over $\mathbb{C}$ to know what an integral form is. Aug 23 awarded Great Answer Jul 29 awarded Yearling Jan 14 awarded Good Answer Dec 9 awarded Caucus Nov 14 comment Striking applications of integration by parts @JakobH Note that the integration variable $t$ has a minus sign, $f(x-t)$. Oct 1 revised Action of $G/H$ on $H_n(H;M)$ Add concrete calculation with diagrams Sep 30 awarded Explainer Aug 17 awarded Good Answer Aug 6 accepted Action of $G/H$ on $H_n(H;M)$ Aug 6 comment Action of $G/H$ on $H_n(H;M)$ Many thanks for your answer! (I'll make a small edit and add the concrete calculation later.) Brown's book was not available in my university's library. A book by Lang mentioned that the action on the zeroth homology uniquely determines the action on higher homology groups via the long exact sequence / uniqueness of group homology, but a concrete calculation using this method seemed very daunting to me. Aug 6 awarded Custodian Aug 6 reviewed Approve Evaluate the limit as $x$ approaches $3$ : $\displaystyle\frac{x^3 -6x+2}{x^2+2x-3}$ Aug 3 comment Action of $G/H$ on $H_n(H;M)$ @tj_ Apparently, I'm missing the definition of the action. Could you elaborate it into a short answer? That would be much appreciated. Aug 3 comment Action of $G/H$ on $H_n(H;M)$ @tj_ Both are possible, the definition of homology is symmetric with respect to which module you resolve. Aug 2 comment Action of $G/H$ on $H_n(H;M)$ Oh! But wouldn't this mean that the Hochschild-Serre spectral sequence has only limited utility? Right now, it seems to me that in order to calculate the G/H action, I have to look at the bar resolution of $M$ as a $\mathbb{Z}[G]$-module, which I wanted to avoid doing in the first place by decomposing $G$ into $H$ and $G/H$. (Of course, I need some information about how $H$ sits inside $G$.) I was hoping for some "functorial" way of defining the action on $H_n(H;M)$. Aug 1 revised Action of $G/H$ on $H_n(H;M)$ Fixed resolution