1
$\begingroup$

Let $V_k(\mathbb{R}^n)$ be Stiefel manifolds.

In the paper The cohomology rings of real Stiefel manifolds with integer coefficients, Martin Čadek, Mamoru Mimura, and Jiří Vanžura, J. Math. Kyoto Univ. Volume 43, Number 2 (2003), 411-428., the cohomology rings $$ H^*(V_k(\mathbb{R}^n);\mathbb{Z}_2), $$ $$ H^*(V_k(\mathbb{R}^n);\mathbb{Z}), $$ are obtained.

Let $\Sigma_k$ be permutation grup of order $k$. For any $\sigma\in \Sigma_k$, let $\sigma$ act on $V_k(\mathbb{R}^n)$ by $$\sigma(v_1,\cdots,v_k)=(v_{\sigma(1)},\cdots,v_{\sigma(k)}).$$

I want to know the induced homomorphism on cohomology ring $$ \sigma^*: H^*(V_k(\mathbb{R}^n);\mathbb{Z}_2)\to H^*(V_k(\mathbb{R}^n);\mathbb{Z}_2), $$ $$ \sigma^*: H^*(V_k(\mathbb{R}^n);\mathbb{Z} )\to H^*(V_k(\mathbb{R}^n);\mathbb{Z} ). $$ Is it possible or difficult? How to solve it?

Could you illustrate the example how to obtain the action of $\mathbb{Z}_2$ on $H^*(V_2(\mathbb{R}^n);\mathbb{Z})$?

$\endgroup$
7
$\begingroup$

The space $V_k(\mathbb{R}^n)$ can be identified with the space $L(\mathbb{R}^k,\mathbb{R}^n)$ of linear isometric inclusions $\mathbb{R}^k\to\mathbb{R}^n$. From this point of view, it is clear that the action of $\Sigma_k$ extends to an action of the orthogonal group $O(k)$. Now $SO(k)$ is connected so it acts by maps that are homotopic to the identity, and so the action in cohomology is trivial. Thus, you only need to calculate the effect of a single transposition to get the full action of $\Sigma_k$. I don't remember how that works out, but it can't be hard.

$\endgroup$
  • $\begingroup$ Dear Prof. Neil, could you give the example how to obtain the action of $\mathbb{Z}_2$ on $H^*(V_2(\mathbb{R}^n);\mathbb{Z})$? $\endgroup$ – QSH Mar 7 '15 at 5:06
  • $\begingroup$ @RenShiquan: what have you tried? $\endgroup$ – Neil Strickland Mar 7 '15 at 12:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.