## Euler characteristics and perverse sheaves

Let $X$ be a variety over a separably closed field $k$, and let $A$ be some object in $D^b_c(X,\mathbf{Q}_{\ell})$. Laumon proved the beautiful result that the usual and compactly supported Euler characteristics $\chi(X,A)$ and $\chi_c(X,A)$ are always equal. Recently while trying to do something else, I found a quick proof of Laumon’s result, as well as a relative version, and I want to sketch this here.

Pick an open immersion into a compactification $j:X \to X'$; after a blowup, we can assume that $Z=X' - X$ is an effective Cartier divisor. Write $i:Z \to X'$ for the inclusion of the boundary. By the usual triangle $j_!A \to Rj_*A \to i_*i^* Rj_*A \to$, we reduce to showing that $\chi(X',i_*i^* Rj_*A)=0$. Filtering $A$ by its perverse cohomology sheaves, we reduce further to the case where also $A$ is perverse. Cover $X'$ by open affines $X_n'$ such that $Z_n= Z \cap X_n'$ is the divisor of a function $f_n$. By an easy Mayer-Vietoras argument, it’s now enough to show that for every open $U$ contained in some $X_n'$, $\chi(U,(i_* i^{\ast}Rj_{\ast}A)|U) = 0$.

But now we win: for any choice of such $U \subset X_n'$, there is an exact triangle $R\psi_{f_n}(A|U \cap X) \to R\psi_{f_n}(A|U \cap X) \to (i_* i^{\ast}Rj_{\ast}A)|U \to$ in $D^b_c(U,\mathbf{Q}_{\ell})$ where $R\psi_{f_n}:\mathrm{Perv}(U \cap X) \to \mathrm{Perv}( U \cap Z)$ is the unipotent nearby cycles functor associated with $f_n$, and the first arrow is the logarithm of the unipotent part of the monodromy. Since $\chi(U, -)$ is additive in exact triangles and the first two terms agree, we’re done.

A closer reading of this argument shows that you actually get the following stronger statement: for any $A$, the class $[i_*i^* Rj_*A] \in K_0\mathrm{Perv}(X')$ is identically zero. From here it’s easy to get a relative version of Laumon’s result.

Theorem. Let $f:X \to Y$ be any map of $k$-varieties. Then for any $A\in D^b_c(X,\mathbf{Q}_\ell)$, there is an equality $[Rf_! A]=[Rf_\ast A]$ in $K_0\mathrm{Perv}(Y)$.