Motivic Hall Algebras and Integration Maps

Over the base $X\times Y$, form the vector bundles $V_0={⨁}_{i\in Q_0}\mathrm{Hom}(A_i,B_i),V_1={⨁}_{a\in Q_1}\mathrm{Hom}(A_{s(a)},B_{t(a)})\mathrm{.}$ Their ranks are ${\sum }_i{d}_i{e}_i$ and ${\sum }_{a:i\to j}{d}_i{e}_j$, respectively. For any scheme $T\to X\times Y$, the vertexwise vector-bundle sequences of an extension split Zariski locally on $T$. Their simultaneous splittings form a torsor under $V_0{\mathrm{\mid }}_T$, and the off-diagonal arrow entries give an equivariant map from this torsor to $V_1{\mathrm{\mid }}_T$ for the translation action $\mathrm{\partial }(h{)}_a=B_a{h}_{s(a)}-h_{t(a)}{A}_a$. Conversely, such a torsor with an equivariant map glues the split vertex bundles and block upper-triangular arrow maps into an extension. These constructions are inverse on objects and isomorphisms and commute with base change, giving a global equivalence ${\mathfrak{E}}_{X,Y}\simeq [V_1\mathrm{/}{V}_0]$ over $X\times Y$. No Zariski splitting over the stack base is required.

The rank of $\mathrm{\partial }$ may vary, so stratify $X\times Y$ by its finitely many possible ranks. On each locally closed stratum the kernels and cokernels of $\mathrm{\partial }$ are vector bundles, and the defining vector-stack relation in [F1] gives class ${\mathbb{L}}^{\mathrm{rk}{V}_1-\mathrm{rk}{V}_0}$ times that stratum. This exponent is constant across strata because the two bundle ranks were fixed by $\mathbf{d},\mathbf{e}$. Additivity over the strata yields $[{\mathfrak{E}}_{X,Y}\to X\times Y]={\mathbb{L}}^{{\sum }_{a:i\to j}{d}_i{e}_j-{\sum }_i{d}_i{e}_i}[X\times Y\to X\times Y]\mathrm{.}$

The exponent in step 2.1 is $-{\chi }_Q(\mathbf{d},\mathbf{e})$ by the definition in the Statement. This proves the claimed relative class.

Let $\alpha ,\beta ,\gamma$ be homogeneous classes of degrees $\mathbf{d},\mathbf{e},\mathbf{f}$. The iterated fiber products defining $(\alpha \ast \beta )\ast \gamma$ classify a representation $E$ with a flag $0\subset F_1\subset F_2\subset E$ whose successive factors $F_1$, $F_2\mathrm{/}{F}_1$, and $E\mathrm{/}{F}_2$ have dimensions $\mathbf{f},\mathbf{e},\mathbf{d}$. The maps to the three inputs are the same classes on these three factors.

The iterated fiber products defining $\alpha \ast (\beta \ast \gamma )$ classify the same flags: first $F_2$ is the subrepresentation of $E$ with quotient of dimension $\mathbf{d}$, then $F_1$ is its subrepresentation with quotient of dimension $\mathbf{e}$. For representations of a quiver, kernels, quotients and inverse images are computed vertexwise and inherit the arrow maps, so the two descriptions are inverse equivalences of the flag stacks.

Both parenthesizations therefore pull back the same exterior product of $\alpha ,\beta ,\gamma$ to the same three-step flag stack and push it forward along the same map sending a flag to its middle representation. Functoriality of pullback, pushforward and exterior product in [F2] gives $(\alpha \ast \beta )\ast \gamma =\alpha \ast (\beta \ast \gamma )$. Bilinearity extends this identity from homogeneous classes to all of $H_Q$.

If one factor is the zero representation, the exact sequence with zero subobject or zero quotient is uniquely isomorphic to the identity sequence on the other factor, including its endpoint automorphisms. Thus the degree-zero class acts as a two-sided unit.

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.

By [F1], the finitary carrier is a direct sum of copies of the coefficient ring indexed by object isomorphism classes, and each product coefficient is a natural-number subobject count. By [F2], the motivic carrier consists of relative classes of stacks mapping to representation stacks, and its product pulls those families back to a stack of exact sequences before pushing to the middle-object stack. These are different defining carriers and coefficient operations, even though both use the same quotient-first short-exact-sequence order.

The definitions provide no map from those natural-number counts to the relative stack classes. Any comparison therefore needs a specified realization and normalization; neither associativity proof alone supplies such a comparison.

By [F5], $H_Q$ with its Hall convolution is an associative unital $R$-algebra. Bilinearity of ${\chi }_Q$ gives ${\chi }_Q(\mathbf{d},\mathbf{e})+{\chi }_Q(\mathbf{d}+\mathbf{e},\mathbf{f})={\chi }_Q(\mathbf{e},\mathbf{f})+{\chi }_Q(\mathbf{d},\mathbf{e}+\mathbf{f})$. Hence the two products of $x^{\mathbf{d}},x^{\mathbf{e}},x^{\mathbf{f}}$ have the same coefficient and degree; the zero vector gives the unit. Thus $T_Q$ is also an associative unital $R$-algebra.

Let $\alpha =[X\to {\mathfrak{M}}_{\mathbf{d}}]$ and $\beta =[Y\to {\mathfrak{M}}_{\mathbf{e}}]$ be generators. Their Hall product is the class of the exact-sequence stack over $X\times Y$, pushed to the middle-object stack by [F1]. By [F2], its absolute motive is ${\mathbb{L}}^{-{\chi }_Q(\mathbf{d},\mathbf{e})}[X\times Y]$. Therefore ${\mathcal{I}}_Q(\alpha \ast \beta )={\mathbb{L}}^{-{\chi }_Q(\mathbf{d},\mathbf{e})}[X][Y]x^{\mathbf{d}+\mathbf{e}}={\mathcal{I}}_Q(\alpha ){\mathcal{I}}_Q(\beta )$ by [F3]. The unit from [F5] is the class of ${\mathfrak{M}}_{\mathbf{0}}(Q)=\mathrm{Spec}k$, whose image is $[\mathrm{Spec}k]x^{\mathbf{0}}=x^{\mathbf{0}}$, the target unit.

Isomorphisms, scissor decompositions, and vector-bundle-stack relations in each relative group map to the same relations in $R=\mathcal{K}(\mathrm{Spec}k)$ under $[X\to {\mathfrak{M}}_{\mathbf{d}}]\mapsto [X]x^{\mathbf{d}}$. Thus ${\mathcal{I}}_Q$ is well-defined and additive. For $r\in R$, the coefficient-system product gives ${\mathcal{I}}_Q(r\alpha )=r{\mathcal{I}}_Q(\alpha )$, so it is $R$-linear. Every element of $H_Q$ is a finite sum of homogeneous generators, so bilinearity and step 1.2 extend the multiplication identity to all elements. Together with the unit preservation in step 1.2 and the algebra structures in step 1.1, ${\mathcal{I}}_Q$ is the claimed algebra homomorphism.