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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