The generic Jordan-quiver Hall algebra is the symmetric-function ring

If $\lambda =\mathrm{\varnothing }$, then $I_{\lambda }=0$: the coefficient of $[I_{\mathrm{\varnothing }}]$ in $[I_{\mu }]\ast [S^{\oplus r}]$ is $1$ when $r=0$ and $\mu =\mathrm{\varnothing }$, and is $0$ otherwise, so it is a constant polynomial. Now suppose $\lambda =(1^{l_1}{2}^{l_2}\cdots n^{l_n})$ is nonempty, with $n$ its largest part, and let $x$ act on $I_{\lambda }$, with $S=I_1$. By [F1], the coefficient of $[I_{\lambda }]$ in $[I_{\mu }]\ast [S^{\oplus r}]$ counts subobjects isomorphic to $S^{\oplus r}$ with quotient type $\mu$. By [F3], these are exactly the $r$-dimensional subspaces $R\subseteq \ker x$, since $x$ acts as zero on $S^{\oplus r}$.

The Gaussian counts ${\left[\genfrac{}{}{0px}{}{a}{b}\right]}_q$ are polynomial in $q$: for $0<b<a$, fixing a hyperplane gives ${\left[\genfrac{}{}{0px}{}{a}{b}\right]}_q={\left[\genfrac{}{}{0px}{}{a-1}{b}\right]}_q+q^{a-b}{\left[\genfrac{}{}{0px}{}{a-1}{b-1}\right]}_q$, with boundary values ${\left[\genfrac{}{}{0px}{}{a}{0}\right]}_q={\left[\genfrac{}{}{0px}{}{a}{a}\right]}_q=1$. In the second term, after choosing the $(b-1)$-dimensional intersection with the hyperplane, the remaining one-dimensional subspace in the quotient has $q^{a-b}$ choices outside the quotient hyperplane. Induction on $a$ gives polynomials in $\mathbb{Z}[q]$.

Finite-dimensional duality sends $(V,x)$ to $(V^{\ast },x^{\ast })$. If $U\subseteq V$ is $x$-invariant, its annihilator $U^{\perp }\subseteq V^{\ast }$ is $x^{\ast }$-invariant, and dualizing $0\to U\to V\to V\mathrm{/}U\to 0$ exchanges the subobject and quotient. A Jordan block is isomorphic to its transpose by reversing its Jordan-string basis, so [F4] gives $I_{\lambda }^{\ast }\cong I_{\lambda }$. This bijection proves $g_{\mu ,\nu }^{\lambda }(q)=g_{\nu ,\mu }^{\lambda }(q)$; each finite-field Hall algebra is commutative.

For the nonempty partition in step 1.1, put $K_i=\ker x\cap \mathrm{im}{x}^i$ for $i\ge 0$, with $K_0=\ker x$ and $K_n=0$. Each Jordan block of size $j$ contributes one dimension to $K_i$ exactly when $j>i$, so $\dim (K_{i-1}\mathrm{/}{K}_i)=l_i$. For a counted subspace set $r_0=r=\dim R$, $r_i=\dim (R\cap K_i)$, and $r_n=0$. Let $y$ be the operator induced by $x$ on $I_{\lambda }\mathrm{/}R$. Write $x^{-i}(R):=\{v:x^i(v)\in R\}$. The map $x^i:x^{-i}(R)\to R\cap \mathrm{im}{x}^i$ is onto with kernel $\ker x^i$; also $R\subseteq x^{-i}(R)$. Since $R\cap \mathrm{im}{x}^i=R\cap K_i$, quotienting by $R$ gives $\dim \ker y^i=\dim \ker x^i-r+r_i\mathrm{.}$ For a nilpotent Jordan type with block multiplicities $(l_i)$, the number of blocks of size $i$ is the second difference $2\dim \ker x^i-\dim \ker x^{i-1}-\dim \ker x^{i+1}$. Applying this formula to $y$ shows that its multiplicity of blocks of size $i$ is $m_i=l_i+2r_i-r_{i-1}-r_{i+1},$ with all $r_i=0$ beyond $n$.

Let $a_i=r_{i-1}-r_i$. For a valid profile, $0\le a_i\le l_i$. Choosing the graded subspaces $(R\cap K_{i-1})\mathrm{/}(R\cap K_i)\subseteq K_{i-1}\mathrm{/}{K}_i$ contributes ${\prod }_{i=1}^n{\left[\genfrac{}{}{0px}{}{l_i}{a_i}\right]}_q$ possibilities. For fixed graded subspaces, build their lifts from the top of the flag down: at level $i<n$, each lift is the graph of a linear map from an $a_i$-dimensional space to $K_i\mathrm{/}(R\cap K_i)$, giving $q^{a_i(\dim K_i-r_i)}$ choices. Multiplying these independent choices gives $q^{{\sum }_{i=1}^{n-1}{a}_i(\dim K_i-r_i)}=q^{{\sum }_{i=1}^{n-1}(r_{i-1}-r_i)({\sum }_{j>i}{l}_j-r_i)}\mathrm{.}$ Thus this power of $q$ times the displayed product counts all subspaces with the chosen intersection dimensions.

For fixed $\lambda ,\mu ,r$, each profile has integers $r=r_0\ge r_1\ge \cdots \ge r_n=0$ with $0\le r_i\le \dim K_i$, so there are finitely many profiles. Keep exactly those whose block multiplicities $m_i$ from step 2.1 equal $\mathrm{\#}\{j:{\mu }_j=i\}$ for every $i$. These constraints are independent of the field by [F3]–[F4]. Summing the polynomial counts from steps 3.1 and 1.2 over the retained profiles proves that every coefficient in $[I_{\mu }]\ast [S^{\oplus r}]$ is a universal polynomial in $q$.

For the empty partition define $X_{\mathrm{\varnothing }}=[0]$, the empty product; its only nonzero coefficient is the unit coefficient at $[I_{\mathrm{\varnothing }}]$. For a nonempty partition $\lambda =(1^{l_1}{2}^{l_2}\cdots n^{l_n})$, put $a_i=l_i+l_{i+1}+\cdots +l_n$ and form $X_{\lambda }(q)=[S^{\oplus a_n}]\ast [S^{\oplus a_{n-1}}]\ast \cdots \ast [S^{\oplus a_1}]\mathrm{.}$ By step 4.1 and associativity [F2], every coefficient is a polynomial in $q$. Parenthesize the product as $([S^{\oplus a_n}]\ast \cdots \ast [S^{\oplus a_2}])\ast [S^{\oplus a_1}]$. For $n=1$ the coefficient is the single one-step flag counted by the basis coefficient. For $n>1$, the induction hypothesis says the inner coefficient counts flags of length $n-1$ in an intermediate object $Y$. The outer product [F1] chooses $V_1\cong S^{\oplus a_1}$ with $I_{\eta }\mathrm{/}{V}_1\cong Y$; an isomorphism identifies the inner flags with flags in this quotient, and any such isomorphism induces a bijection, so the count is independent of the isomorphism chosen. By [F5] each quotient flag lifts uniquely to a flag containing $V_1$, and its quotient clause identifies each lifted successive quotient with the corresponding quotient in the quotient flag. Thus the coefficient of $[I_{\eta }]$ counts exactly flags $0=V_0\subset V_1\subset \cdots \subset V_n=I_{\eta }$ with $V_i\mathrm{/}{V}_{i-1}\cong S^{\oplus a_i}$.

For every such flag, $x(V_i)\subseteq V_{i-1}$, hence $V_i\subseteq \ker x^i$. Each Jordan block of length $j$ contributes $\min (i,j)$ to $\dim \ker x^i$, so $\dim V_i={\sum }_{h=1}^i{a}_h={\sum }_{j=1}^n\min (i,j)l_j=\dim \ker x_{I_{\lambda }}^i$. For the type $\lambda$ itself, the kernels $V_i=\ker x^i$ form a valid flag: each induced endomorphism on $V_i\mathrm{/}{V}_{i-1}$ is zero and the quotient has dimension $a_i$, so it is isomorphic to $S^{\oplus a_i}$. If all these kernel dimensions agree for another type $\eta$, their block multiplicities agree by second differences, so $\eta =\lambda$; the flag is then forced to be the kernel flag just constructed and is unique. On partitions of a fixed total size, order types by componentwise comparison of these kernel dimensions; antisymmetry follows from the second-difference formula. Therefore $X_{\lambda }(q)=[I_{\lambda }]+{\sum }_{\eta \prec \lambda }{c}_{\lambda ,\eta }(q)[I_{\eta }],$ where $\eta \prec \lambda$ means all kernel dimensions are at least those of $\lambda$, with at least one strict inequality.

In each fixed total dimension there are finitely many partitions. The transition matrix in step 6.1 is triangular with diagonal entries one and polynomial entries from steps 4.1 and 5.1, so it is invertible over $\mathbb{Z}[t]$ by recursive substitution. Thus every $[I_{\lambda }]$ is a polynomial combination of monomials in the semisimple classes. For each finite field, use associativity [F2] and commutativity from step 1.3 to parenthesize and reorder a product of two such monomials so that one basis term is successively multiplied by simple-sum factors on the right. Induction on the number of factors and step 4.1 then show that every structure constant $g_{\mu ,\nu }^{\lambda }(q)$ is the value at $q$ of a polynomial $G_{\mu ,\nu }^{\lambda }(t)\in \mathbb{Z}[t]$.

Each coefficient polynomial is unique: the difference of two candidates vanishes at every field cardinality, and by [F9]–[F10] this includes infinitely many distinct prime cardinalities. A nonzero polynomial in $\mathbb{Z}[t]$ remains nonzero in $\mathbb{R}[t]$ because $\mathbb{Z}\to \mathbb{R}$ is injective; by [F6], it cannot have infinitely many distinct real roots. Thus $H_{\mathrm{g}\mathrm{e}\mathrm{n}}$ is well-defined. Each product is finite because a short exact sequence adds vector-space dimensions, and only finitely many partitions have a fixed size. For a fixed triple of basis classes, the associativity defect is therefore a polynomial in $t$; it vanishes at every field-cardinality specialization by [F2], hence at infinitely many distinct prime cardinalities by [F9]–[F10]. Its coefficients embed injectively from $\mathbb{Z}$ into $\mathbb{R}$, so [F6] implies that the defect is the zero polynomial. The same argument using step 1.3 proves commutativity. The unit equations likewise hold as polynomial identities because they hold at every field-cardinality specialization by [F2], so $u_{\mathrm{\varnothing }}$ is the generic unit.

The monomial $X_{\lambda }$ in step 5.1 uses the generators $u_{(1^{a_i})}$ with exponents $a_i$, which are exactly the parts of the conjugate partition ${\lambda }^{\mathrm{\prime }}$. Since partition conjugation is a bijection, these $X_{\lambda }$ run through every polynomial monomial in the generators $u_{(1^r)}$. Their unitriangular expansions in step 6.1 show that they are also a $\mathbb{Z}[t]$-basis. Together with the coefficient polynomials from step 7.1, this proves the natural map from $\mathbb{Z}[t][u_{(1)},u_{(1^2)},\dots ]$ to $H_{\mathrm{g}\mathrm{e}\mathrm{n}}$ is an isomorphism.

By [F8], the target symmetric-function ring is the polynomial ring $\mathbb{Z}[e_1,e_2,\dots ]$. After adjoining $t^{-1}$, each scalar $t^{-r(r-1)\mathrm{/}2}$ is a unit. Define ${\mathrm{\Phi }}_t(u_{\mathrm{\varnothing }})=1$ and send each free generator $u_{(1^r)}$ to $t^{-r(r-1)\mathrm{/}2}{e}_r$; this defines an isomorphism of polynomial algebras, with inverse sending $e_r$ to $t^{r(r-1)\mathrm{/}2}{u}_{(1^r)}$. Each generator and its image have degree $r$, and $t$ has degree zero, so the isomorphism preserves the grading.