2
$\begingroup$

Suppose everything below is defined over $k=\overline{\mathbb{F}}_q$.

Let $H$ be a connected algebraic group acting on a separated variety $Y$. Denote the morphism $H\times Y\rightarrow H\times Y; (h,y)\mapsto (h,hy)$ by $f$, and denote by $\pi$ the left projection of $H\times Y$ to $H$. By applying proper base change to the Cartesian diagram \begin{equation*} \require{AMScd} \begin{CD} H\times Y @>>> Y\\ @V \pi V V @VV V\\ H @>>> \mathrm{Spec} (k) \end{CD} \end{equation*} we see $R^i\pi_!\mathbb{Z}/n$ is the constant sheaf $H^i_c(Y,\mathbb{Z}/n)$ on $H$.

In the argument of Proposition 6.4 of Deligne--Lusztig's seminal paper, they assert that $f$ gives an endomorphism on this constant sheaf, in the way that at each stalk at $h$ the endomorphism at $H^i_c(Y,\mathbb{Z}/n)_h=H^i_c(Y,\mathbb{Z}/n)$ is given by the induced map of $h$.

I understand that at each stalk at $h$, there is an induced action of $h$ on the cohomology group, but why they constitute a sheaf endomorphism?

$\endgroup$
2
$\begingroup$

It follows from functoriality. For three varieties $X, Y, Z$ with maps $f:X \to Z$ and $g: Y\to Z$, for every isomorphism $h: X \to Y$ forming a commutative triangle with $f$ and $g$, we get an induced isomorphism between $R^i f_! Y$ and $R^i g_! Y$.

Apply this to your morphism $f$ and you get the desired action.

You can see it's the desired action because this isomorphism agrees with base change - because everything in base change is defined canonically so has to commute with isomorphisms.

$\endgroup$
2
  • $\begingroup$ By $R^if_!Y$ you mean $R^if_!\mathbb{Z}/n$ right? $\endgroup$
    – user148212
    Mar 11 '16 at 12:00
  • $\begingroup$ @user148212 Yes, sorry. $\endgroup$
    – Will Sawin
    Mar 11 '16 at 14:11

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.