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).

 

One thought on “Euler characteristics and perverse sheaves”

  1. The last step using nearby cycle is very interesting. Maybe one can also prove an equivariant version along the line, in particular get a stacky version.

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