Finite-field groupoid cardinality respects the motivic stack relations

First let $S$ be any of the finite-type Artin stacks with affine stabilizers allowed as a coefficient generator, put $k={\mathbb{F}}_q$, and write $F$ for absolute $q$-power Frobenius. The stack $Z=S{\times }_{S{\times }_{k}S}S$, formed using the diagonal and the graph of $F$, has objects $(s,\alpha :F_T^{\ast }s\sim s)$ over $T$. It is an Artin stack of finite type with finite-type diagonal: its projection to $S$ is the base change of the finite-type representable diagonal of $S$. For a square-zero thickening $i:T_0\hookrightarrow T$, Frobenius factors as $F_T=i\circ f$ with $f:T\to T_0$, since $a\mapsto a^q$ kills the square-zero ideal. A pair $(s_0,{\alpha }_0)$ lifts to $(f^{\ast }{s}_0,f^{\ast }{\alpha }_0)$, whose restriction is identified with $(s_0,{\alpha }_0)$ by ${\alpha }_0$. For pairs $(s,\alpha )$ and $(t,\beta )$ over $T$, a compatible morphism $u_0$ over $T_0$ lifts uniquely to $u=\beta \circ f^{\ast }{u}_0\circ {\alpha }^{-1}$; the compatibility equation $u\alpha =\beta F_T^{\ast }u$ proves uniqueness and, after restriction, proves that this formula is a lift. Thus $Z(T)\to Z(T_0)$ is an equivalence.

If a finite group $G$ acts on a finite groupoid $\mathcal{H}$, then $\mathrm{\mid }\mathcal{H}\mathrm{/}\mathrm{/}G\mathrm{\mid }=\mathrm{\mid }\mathcal{H}\mathrm{\mid }\mathrm{/}\mathrm{\mid }G\mathrm{\mid }$. Indeed, an orbit represented by $h$ has $\mathrm{\mid }G\mathrm{\mid }\mathrm{/}\mathrm{\mid }{\mathrm{Stab}}_G([h])\mathrm{\mid }$ isomorphism classes with the same automorphism order. In the quotient the automorphism order is $\mathrm{\mid }{\mathrm{Aut}}_{\mathcal{H}}(h)\mathrm{\mid }\mathrm{\mid }{\mathrm{Stab}}_G([h])\mathrm{\mid }$, because each stabilizing group element admits exactly $\mathrm{\mid }{\mathrm{Aut}}_{\mathcal{H}}(h)\mathrm{\mid }$ isomorphisms to $h$. Comparing the two contributions proves the formula. In particular, translation of $k^{r_0}$ on $k^{r_1}$ through a linear map gives cardinality $q^{r_1-r_0}$, retaining all kernel automorphisms.

Put $K=\bar{k}$. Every $K$-point $z:\mathrm{Spec}K\to Z_K$ is representable and locally finitely presented: after base change to a finite-type smooth scheme atlas, its source is an algebraic space of finite type over $K$, by the finite-type diagonal. The lifting equivalence of step 1.1 makes this point morphism formally étale, hence étale by [F3] on scheme charts, and open by [F4]. Its image in $\mathrm{\mid }{Z}_K\mathrm{\mid }$ is a single point. Distinct $K$-isomorphism classes have distinct images, since an isomorphism existing over a field extension already exists over the algebraically closed field $K$ (use a finite-type scheme cover of the nonempty isomorphism space). These images are disjoint nonempty opens. A noetherian space has only finitely many such opens, since each contains the generic point of an irreducible component and there are finitely many components. Consequently $Z(K)$ has finitely many isomorphism classes. The natural functor $S(k)\to Z(K)$ is fully faithful: compatibility with the canonical Frobenius identifications says exactly that an isomorphism is Frobenius invariant, so it descends to $k$ by the sheaf condition for isomorphisms and finite-field Galois descent. Finally ${\mathrm{Aut}}_S(s)$ is affine of finite type, so it has finitely many $k$-points. Hence $S(k)$ is a finite groupoid for every allowed coefficient generator. We may therefore use [F1]'s finite-sum formula on all of them, extending the prescribed quiver-stack counts.

Every torsor under an additive vector group $k^r$ is trivial over $k$. To see this, choose a point $t$ over $K$, and write its arithmetic Frobenius image as $F(t)=t+a$. Choose $b\in K^r$ solving $b^q-b=-a$, coordinate by coordinate, which is possible because $K$ is algebraically closed. Then $F(t+b)=t+b$, so $t+b$ descends to a $k$-point and trivializes the torsor. Accordingly, for every $s\in S(k)$, the rational-point groupoid of the vector-stack fiber is equivalent to $[V_1(s)(k)\mathrm{/}{V}_0(s)(k)]$; there are no additional torsor classes. Its cardinality is $q^{r_1-r_0}$ by step 1.2.

Now let $S$ be an arbitrary base allowed in [F2], not necessarily a quiver quotient stack. By step 2.1 its rational-point groupoid is finite. Over a representative $s$, the vector bundles and their map carry the natural linear action of ${\mathrm{Aut}}_S(s)(k)$. Trivial torsors and the origin give an equivariant model for the fiber in step 2.2. Thus the total rational-point groupoid over the isomorphism class $[s]$ is its action groupoid under ${\mathrm{Aut}}_S(s)(k)$, and has cardinality $q^{r_1-r_0}\mathrm{/}\mathrm{\mid }{\mathrm{Aut}}_S(s)(k)\mathrm{\mid }$ by step 1.2. Summing over the finitely many base classes proves $\mathrm{\mid }[V_1\mathrm{/}{V}_0](k)\mathrm{\mid }=q^{r_1-r_0}\mathrm{\mid }S(k)\mathrm{\mid }$ for every defining vector-bundle-stack relation. Finite rank stratification, when needed, gives the same identity stratum by stratum.

Define the count on the free coefficient generators by $[S]\mapsto \mathrm{\mid }S(k)\mathrm{\mid }$ for all $S$ allowed in [F2]. Rational points of a closed/open decomposition form a disjoint union of full groupoids, and rational points of a product form the product groupoid, so the finite-sum formula is additive and multiplicative. Step 3.1 checks every vector-bundle-stack relation in this whole presentation. Since $\mathrm{\mid }{\mathbb{A}}^1(k)\mathrm{\mid }=q\mathrm{\ne }0$, the resulting homomorphism extends across localization at $\mathbb{L}$, with ${\mathbb{L}}^{-1}\mapsto q^{-1}$. Its restriction to the subring generated by the quiver quotient stacks is the claimed realization; verification on the whole coefficient presentation also ensures it respects all ambient equalities between those classes.