Timeline for Künneth formula for Bredon cohomology theory
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jul 14, 2015 at 0:06 | comment | added | Gustavo Granja | Your question in the comment above is not addressed by my answer as the domain and range of $f$ are not free $G$-spaces. If this is the question you are interested in, I suggest you rephrase your original question. | |
| Jul 13, 2015 at 23:56 | comment | added | Surojit Ghosh | There is a natural map $f : S(\xi)_+ \wedge S^V \rightarrow S(\xi^j)_+ \wedge S^V$. Then for what $n$, $\tilde{H}^n_{Z/n}(f)$ is nonzero?Where $V$ is a representation of $Z/n$ and $f(z,v) = (z^j ,v) , z \in S(\xi) , v \in S^V$. | |
| Jul 13, 2015 at 20:55 | comment | added | Gustavo Granja | Assuming I understand correctly that $G=Z/n$ is acting on the circle $X$ in the standard way, the EMSS seems overkill. In that case $(X\times Y)/G$ is just the mapping torus (see Hatcher Ex. 1.2.11) of the map given by the action of a generator of $G$ on $Y$. The Mayer-Vietoris sequence expresses the cohomology of $(X\times Y)/G$ in terms of the cohomology of $Y$ and the action of a generator of $Z/n$ on $H^*Y$. | |
| Jul 13, 2015 at 2:54 | comment | added | Surojit Ghosh | :What can you say about the convergence when $X$-is $S(\xi)$? Where $\xi$ is the irreducible representation of $Z/n$ given by multiplication by $e^{2i \pi /n}$ | |
| Jul 12, 2015 at 15:32 | history | answered | Gustavo Granja | CC BY-SA 3.0 |