Behrend's trace formula

The desire for the formula comes from the fact that it applies to the moduli stack of principal bundles on a curve over a finite field (in some instances indirectly, via the Harder–Narasimhan stratification, as the moduli stack is not of finite type.) See the moduli stack of principal bundles and references therein for the precise formulation in this case.

Pierre Deligne found an example that shows the formula may be interpreted as a sort of the Selberg trace formula.

A proof of the formula in the context of the six operations formalism developed by Yves Laszlo and Martin Olsson is given by Shenghao Sun.

Formulation

By definition, if C is a category in which each object has finitely many automorphisms, the number of points in ${\displaystyle C}$ is denoted by

${\displaystyle \mathrm{\#}C=\sum _p\frac{1}{\mathrm{\#}\mathrm{Aut}(p)},}$

with the sum running over representatives p of all isomorphism classes in C. (The series may diverge in general.) The formula states: for a smooth algebraic stack X of finite type over a finite field ${\displaystyle {\mathbb{F}}_q}$ and the "arithmetic" Frobenius ${\displaystyle {\varphi }^{-1}:X\to X}$, i.e., the inverse of the usual geometric Frobenius ${\displaystyle \varphi }$ in Grothendieck's formula,

${\displaystyle \mathrm{\#}X({\mathbb{F}}_q)=q^{\dim X}\sum _{i=0}^{\mathrm{\infty }}(-1{)}^i\mathrm{tr}\left({\varphi }^{-1};H^i(X,{\mathbb{Q}}_l)\right).}$

Here, it is crucial that the cohomology of a stack is with respect to the smooth topology (not etale).

When X is a variety, the smooth cohomology is the same as etale one and, via the Poincaré duality, this is equivalent to Grothendieck's trace formula. (But the proof of Behrend's trace formula relies on Grothendieck's formula, so this does not subsume Grothendieck's.)