One vertex with no arrows gives the quantum dilogarithm

A representation is determined up to isomorphism by its dimension $n$; its automorphism group is ${\mathrm{GL}}_n({\mathbb{F}}_q)$. For $n>0$, $Z(n)=n{z}_1$ and every nonzero subspace has the same phase, so every object is semistable. The zero object supplies the degree-zero term.

An ordered basis of ${\mathbb{F}}_q^n$ can be chosen in $\mathrm{\mid }{\mathrm{GL}}_n({\mathbb{F}}_q)\mathrm{\mid }={\prod }_{j=0}^{n-1}(q^n-q^j)$ ways: after $j$ independent vectors have been chosen, their span contains $q^j$ vectors to exclude. Factoring $q^j$ from the $j$th term and reversing the remaining index gives $\mathrm{\mid }{\mathrm{GL}}_n({\mathbb{F}}_q)\mathrm{\mid }=q^{n(n-1)\mathrm{/}2}{\prod }_{r=1}^n(q^r-1)\mathrm{.}$ For $n=0$ both sides equal $1$.

The Euler form is ${\chi }_Q(n,n)=n^2$. By [F2], the degree-$n$ term of $J(A_{\mathrm{a}\mathrm{l}\mathrm{l}})$ is $\frac{t^{n^2}}{\mathrm{\mid }{\mathrm{GL}}_n({\mathbb{F}}_q)\mathrm{\mid }}{x}^n=\frac{t^n}{{\prod }_{r=1}^n(t^{2r}-1)}{x}^n,$ using [F1], step 1.2 and $q=t^2$. The formula also holds at $n=0$ by the empty-product convention.

Summing the degreewise coefficients in step 2.1 gives the stated formal power series. By [F3] its unweighted coefficient is the finite groupoid cardinality of the semistable stack, and the factor $t^{n^2}$ is precisely the Euler normalization in [F2]. No physical state-counting identification is asserted.