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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 suggested edit on 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
Aug
1
asked Action of $G/H$ on $H_n(H;M)$
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29
awarded  Yearling
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Dec
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awarded  Nice Answer
Jul
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awarded  Yearling
Jan
6
accepted Compact operators: why is the image of the unit ball only assumed to be relatively compact?
Jan
6
comment Compact operators: why is the image of the unit ball only assumed to be relatively compact?
Thanks a lot! So the qualification "relative" is apparently meant for the case of Banach spaces that are not reflexive. My argument would fail there as the (closed) unit ball would no longer be weakly compact.
Jan
5
accepted How to draw a complex line bundle?