Is the corestriction map from a subgroup H to a group G, on the first Tate cohomology group H^1, dependent on the chosen transversal of H is G?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
No, because the map on cochains is $f\mapsto \sum_{g\in G/H}gfg^{-1}$ where $f$ is $H$-linear, so that any other coset representative $gh$ would not have any effect.
This shows that $cor^G_H$ is independent of the choice of transversal on the level of cohomology. But in that paper of Eunmi Choi, a $\textit{different}$ map on the level of cochains is given (using the bar resolution), which also induces $cor^G_H$ on cohomology, and this $\textit{is}$ dependent on the choice of transversal, as is shown in that paper. But all such images lie in the same cohomology class.
answered Mar 12, 2011 at 19:08
-
$\begingroup$ That is what I thought, but I have a situation where I can't see how it could be independent. I also just found a paper by E. Choi in which he says that it is dependent. I have not studied the paper yet. Thanks $\endgroup$ Commented Mar 12, 2011 at 22:09
-
$\begingroup$ What is the title of the paper (and/or link to it)? $\endgroup$ Commented Mar 13, 2011 at 6:40
-
1$\begingroup$ Nevermind, found it... In that paper he even quotes Eckmann's paper to say that corestriction $\textit{on cohomology groups}$ is independent of transversals. But from what I get from glancing at the paper, Choi looks at a certain map $p$ on the cochains which induces the corestriction map on cohomology groups, and this map $p$ has different values at the cochain-level dependent on transversals. $\endgroup$ Commented Mar 13, 2011 at 9:00