The A₂ quiver gives the quantum-dilogarithm pentagon

A representation is one linear map $f:V_1\to V_2$. Choose a complement to $\ker f$ and a complement to $\mathrm{im}f$. The restriction of $f$ from the first complement to $\mathrm{im}f$ is an isomorphism. In bases adapted to these spaces, the representation is $S_1^{\oplus a}\oplus E^{\oplus r}\oplus S_2^{\oplus b},r=\mathrm{rank}f,a=\dim \ker f,b=\dim (V_2\mathrm{/}\mathrm{im}f)\mathrm{.}$ Thus these three isomorphism types account for every representation.

The Euler form gives ${\chi }_Q({\mathbf{e}}_1,{\mathbf{e}}_2)=-1$ and ${\chi }_Q({\mathbf{e}}_2,{\mathbf{e}}_1)=0$, so $⟨{\mathbf{e}}_1,{\mathbf{e}}_2⟩=-1$ and $⟨{\mathbf{e}}_2,{\mathbf{e}}_1⟩=1$. By [F3], $x_2\ast x_1=t{x}^{(1,1)},x_1\ast x_2=t^{-1}{x}^{(1,1)},x^{(1,1)}=t{x}_1\ast x_2\mathrm{.}$ In particular $x_2\ast x_1=t^2{x}_1\ast x_2$.

A subrepresentation of $E$ is a pair of subspaces $U_1,U_2\subseteq {\mathbb{F}}_q$ with $U_1\subseteq U_2$. Its only proper nonzero possibility is $S_2$. The phase of $z_1+z_2$ lies between ${\varphi }_1$ and ${\varphi }_2$ by [F4]. The sequence $0\to S_2\to E\to S_1\to 0$ is nonsplit because $E$ has no subrepresentation isomorphic to $S_1$; every nonzero arrow scalar is changed to the identity by vertex base changes, so this is the unique nontrivial extension of these simples. Hence $E$ is stable when ${\varphi }_1>{\varphi }_2$, unstable when ${\varphi }_1<{\varphi }_2$, and strictly semistable when ${\varphi }_1={\varphi }_2$.

Suppose ${\varphi }_1>{\varphi }_2$. In the decomposition of step 1.1, if two different summand types occur, the direct sum of all summands of greatest phase is a proper subobject. Repeated application of [F4] shows the total phase is strictly below that greatest phase, because at least one summand has smaller phase; hence the representation is unstable. Pure sums $S_1^{\oplus n}$ and $S_2^{\oplus n}$ are semistable because all their nonzero subobjects have the same phase. A subrepresentation of $E^{\oplus n}$ is a pair $U_1\subseteq U_2$; writing $r=\dim U_1$ and $s=\dim U_2$ gives $r\le s$. The determinant calculation $\mathrm{Im}(\overline{(z_1+z_2)}(r{z}_1+s{z}_2))=(s-r)\mathrm{Im}({\bar{z}}_1{z}_2)\le 0$ shows its phase is at most that of $E^{\oplus n}$. Thus the semistable objects are exactly pure sums of $S_1$, $E$, or $S_2$, in descending phase order.

Suppose ${\varphi }_1<{\varphi }_2$. Every summand in the decomposition of step 1.1 has phase at most ${\varphi }_2$, and an $E$ summand or an $S_1$ summand has smaller phase. Repeated application of [F4] therefore makes the phase of any object containing a lower-phase summand strictly less than ${\varphi }_2$. An object with an $E$ summand has the nonzero subobject $S_2$, and an object with both $S_1$ and $S_2$ summands has an $S_2$ summand; each is a destabilizing subobject. Hence only pure sums of $S_2$ and pure sums of $S_1$ are semistable, in that descending order. If ${\varphi }_1={\varphi }_2$, all nonzero dimension vectors have the same phase, so every representation is semistable and the HN product has one phase factor containing both $S_1\oplus S_2$ and $E$.

Each $A\in \{S_1,S_2,E\}$ has $\mathrm{End}(A)={\mathbb{F}}_q$. For $E$, an endomorphism is a pair of scalars at the vertices that must agree because its arrow is the identity. Also ${\chi }_Q(\mathbf{d}(A),\mathbf{d}(A))=1$ for each of these three objects. [F5] gives ${\mathrm{YExt}}^1(A^{\oplus m},A^{\oplus n})=0$: the Hom space has dimension $mn$ and ${\chi }_Q(m\mathbf{d}(A),n\mathbf{d}(A))=mn$. The semistable subcategory at each stable phase in step 3.1 is consequently equivalent to ${\mathrm{V}\mathrm{e}\mathrm{c}\mathrm{t}}_{{\mathbb{F}}_q}$, with the displayed object as its one-dimensional simple.

For $A$ one of these three stable objects, put $\gamma =\mathbf{d}(A)$. Then ${\chi }_Q(n\gamma ,n\gamma )=n^2$ and $\mathrm{\mid }\mathrm{Aut}(A^{\oplus n})\mathrm{\mid }=\mathrm{\mid }{\mathrm{GL}}_n({\mathbb{F}}_q)\mathrm{\mid }$. Counting ordered bases directly gives $\mathrm{\mid }{\mathrm{GL}}_n({\mathbb{F}}_q)\mathrm{\mid }={\prod }_{j=0}^{n-1}(q^n-q^j)=q^{n(n-1)\mathrm{/}2}{\prod }_{r=1}^n(q^r-1)\mathrm{.}$ Therefore $\frac{t^{n^2}}{\mathrm{\mid }{\mathrm{GL}}_n({\mathbb{F}}_q)\mathrm{\mid }}=\frac{t^n}{{\prod }_{r=1}^n(t^{2r}-1)}$ using $q=t^2$. Thus the $J$-image of the semistable series at that phase is $E_{t^2}(x^{\gamma })$. The empty product gives the same result for $n=0$.

If ${\varphi }_1>{\varphi }_2$, the descending semistable phases are $S_1,E,S_2$; if ${\varphi }_1<{\varphi }_2$, they are $S_2,S_1$. Apply [F2] to these two chamber assignments and substitute the phase images from steps 5.1 and 1.2. This gives the stated pentagon over $\mathbb{Q}(\sqrt{q})$.

Each fixed-bidegree coefficient of the pentagon is a rational function of $t$. Specializing to $q=2^m$ for every $m\ge 1$ gives infinitely many distinct $t=2^{m\mathrm{/}2}>1$, where every denominator $t^{2r}-1$ is nonzero and step 6.1 proves equality. The numerator of the difference therefore has infinitely many roots and is the zero polynomial. This proves the identity coefficientwise in $\mathbb{Q}(t)$.