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?


1 Answer 1


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.

  • $\begingroup$ By $R^if_!Y$ you mean $R^if_!\mathbb{Z}/n$ right? $\endgroup$
    – user148212
    Mar 11, 2016 at 12:00
  • $\begingroup$ @user148212 Yes, sorry. $\endgroup$
    – Will Sawin
    Mar 11, 2016 at 14:11

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