Finitary Ringel–Hall Algebras

Given a subobject $B\le L\mathrm{/}W$, let $U=q^{\ast }B$ be its inverse image. Since $q_{\ast }W=0\le B$, [F3] gives $W\le U$. The pullback square defining $U$ has a projection $p:U\to B$ that is epic by [F4]; its kernel is the kernel $W$ of $q$. By [F5], its image is canonically $U\mathrm{/}W$. Since $p$ is epic, any cokernel $c$ satisfies $cp=0=0p$, so $c=0$ by cancellation. Its universal property makes its codomain initial, because the only morphism out of $B$ annihilating $p$ is zero; hence the cokernel object is zero. By the image definition, $\mathrm{im}(p)=\ker (\mathrm{coker}p)=B$. Thus $U\mathrm{/}W\cong B$ and $q_{\ast }U=B$.

Conversely let $W\le U\le L$. The restriction $q{\mathrm{\mid }}_U:U\to L\mathrm{/}W$ has kernel $W$, so [F5] identifies its image $q_{\ast }U$ with $U\mathrm{/}W$. By [F6], the quotient map $L\mathrm{/}W\to (L\mathrm{/}W)\mathrm{/}(U\mathrm{/}W)$, composed with $q$, is the quotient map $L\to L\mathrm{/}U$. Its kernel is $U$ by [F1] and [F7], hence the inverse image of $U\mathrm{/}W$ along $q$ is exactly $U$.

Steps 1.1 and 1.2 show that the two assignments are inverse on subobject classes. Both assignments preserve inclusion by the Galois connection [F3], and [F6] gives the stated quotient identification. Thus inverse image along $q$ is the claimed order isomorphism.

Fix a morphism $f$ intertwining two nilpotent endomorphisms. Its kernel and image are invariant, and the induced endomorphism on its cokernel is nilpotent. Intertwining maps are closed under addition, composition is bilinear, and the ordinary finite direct sums with diagonal endomorphisms and the zero object have their vector-space biproduct universal properties. Choose bases for the kernel, image, and cokernel of this one map and transport them to standard $k^d$ objects; their universal properties are preserved. The coimage-to-image isomorphism is an intertwiner because $f$ commutes with the endomorphisms. This verifies the abelian-category clauses in [F3].

For any field $k$, the standard-coordinate objects and matrices between them form sets, since they are a union indexed by pairs of natural numbers of sets of matrices over $k$; hence ${\mathcal{J}}_k$ is small. If $k$ is finite, there are finitely many matrices of each size, and each Hom set is a subset of a finite set of matrices, so Hom sets are finite.

Now suppose $k$ is finite and fix objects $M=(k^m,x_M)$ and $N=(k^n,x_N)$. In a short exact sequence $0\to N\to E\to M\to 0$, the middle space has dimension $m+n$. After identifying it with $k^{m+n}$, only finitely many nilpotent matrices and endpoint maps are possible. Thus the set of such sequences, and its quotient by Yoneda equivalence, are finite.

A monic intertwiner is injective: if $f$ has a nonzero kernel vector $w$, the source nilpotent operator $x$ preserves $\ker f$. Choose the least $j\ge 1$ with $x^{j}w=0$; then $v=x^{j-1}w$ is nonzero, lies in the kernel, and satisfies $xv=0$. The map $S\to \mathrm{dom}(f)$ sending $1$ to $v$ is nonzero and intertwines, but its composite with $f$ is zero, contradicting left cancellation [F4]. Conversely, every injective intertwiner is monic by linear injectivity.

By step 1.4, every monomorphism identifies its domain with an invariant subspace of its target. Conversely, every invariant subspace carries the restricted nilpotent endomorphism; choosing a basis for this single finite-dimensional subspace and transporting the endomorphism to $k^d$ represents its inclusion by a monic map in the standard-coordinate model. Two such monic maps have the same image exactly when they mutually factor: maps with the same image are related by the induced intertwining isomorphism of their domains, and mutual factorisation forces both image inclusions. By [F5], this is precisely equality of subobject classes.

The category is abelian by step 1.1 and small by step 1.2. For finite $k$, its Hom and Yoneda extension sets are finite by steps 1.2 and 1.3, exactly the finitary conditions in [F2]; step 2.1 proves the stated subobject description for every field.

Apply [F1] to $x$. The spans of its Jordan strings are invariant, their concatenation is a basis of $V$, and by [F3] each span is isomorphic as a representation to the block $I_m$ whose length is that string's length. Sorting these finitely many positive lengths weakly decreasing gives a partition $\lambda$ by [F4] and an isomorphism $(V,x)\cong I_{\lambda }$. The zero space gives the empty partition.

If $I_{\lambda }\cong I_{\mu }$, the intertwining isomorphism conjugates their nilpotent matrices, so the ranks of every power agree. By [F2], the multiplicities of blocks of each length agree; hence the multisets of parts and the partitions are equal.

If a string has length $m>1$, the span of its initial vector is a nonzero proper invariant subspace, hence a proper nonzero subobject by [F7]; therefore $I_m$ is not simple by [F6]. The object $I_1=S$ is nonzero and has only the zero and whole subspaces, so it is simple by [F6] and [F7]. Any decomposition in step 1.1 with at least two summands has a proper nonzero invariant summand and hence is not simple. A single summand is simple only when it has length one. Thus $S$ is the unique simple object.

An isomorphism of middle objects transports subobjects and induces an isomorphism of their cokernel quotients; replacing either endpoint by an isomorphic object preserves the two type conditions. Thus the count in [F2] depends only on $[M],[N],[L]$.

For fixed $L$, the equivalence classes counted by [F2] are a quotient of the finite set $\mathrm{Hom}(N,L)$ from [F1]. The subset whose cokernel is isomorphic to $M$ is therefore finite.

Every elementary map in the generating relation for [F1] preserves the middle-object isomorphism class by [F3], and so does its reflexive, symmetric and transitive closure. Thus taking the middle-object class defines a map from the finite set ${\mathrm{YExt}}^1(M,N)$. If $g_{M,N}^L\mathrm{\ne }0$, a counted subobject and quotient give a short exact sequence $0\to N\to L\to M\to 0$ after transporting its endpoints by isomorphisms, so $[L]$ lies in the image. That image is finite, proving finite support without simultaneous choices of extension representatives.

Each coefficient is a natural number and the support is finite by steps 1.2–1.3. Mapping an integer $n$ to $n{1}_R$ therefore gives a well-defined element of the direct sum in [F2], with no automorphism denominator or additional finiteness condition.

Fix basis objects $M,N,P,L$. The coefficient of $[L]$ in $([M]\ast [N])\ast [P]$ is the sum over isomorphism classes $[X]$ of $g_{M,N}^X{g}_{X,P}^L$. By [F1], this counts a subobject $W\le L$ with $W\cong P$ and $L\mathrm{/}W\cong X$, together with a subobject $\bar{U}\le L\mathrm{/}W$ with $\bar{U}\cong N$ and $(L\mathrm{/}W)\mathrm{/}\bar{U}\cong M$. The quotient class $[X]=[L\mathrm{/}W]$ is uniquely determined by $W$, so this sum counts exactly these pairs.

By [F3], each such pair $(W,\bar{U})$ corresponds to exactly one flag $W\le U\le L$ with $W\cong P$, $U\mathrm{/}W\cong N$, and $L\mathrm{/}U\cong M$. The coefficient of $[L]$ in $[M]\ast ([N]\ast [P])$ counts the same flags: first choose $U\le L$ with $U\cong Y$, $L\mathrm{/}U\cong M$, then choose $W\le U$ with $W\cong P$, $U\mathrm{/}W\cong N$, summing over the uniquely determined class $[Y]=[U]$.

The only subobject isomorphic to zero is the zero subobject, and the only subobject with zero quotient is the whole object: in the latter case the inclusion has zero cokernel and hence is an isomorphism by [F4]. Thus $g_{M,0}^L=1$ exactly when $L\cong M$, while $g_{0,M}^L=1$ exactly when $L\cong M$. The zero-object symbol is a two-sided unit.

The two parenthesizations therefore have the same coefficient at every basis vector $[L]$, so $([M]\ast [N])\ast [P]=[M]\ast ([N]\ast [P])$ for basis elements.

Basis products have finite support by [F2]. Every element of $H_{\mathcal{A},R}$ is a finite linear combination of basis vectors, so bilinearity extends the basis calculation in step 2.1 to all triples and preserves finite support. Step 1.3 supplies the unit, proving the statement.

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.

For the Jordan-string basis $v_1,\dots ,v_r$ of $I_r$, we have $T{v}_1=0$ and $T{v}_j=v_{j-1}$ for $j>1$. Give $I_r$ the $k[T]$-module structure in which $T$ acts by its nilpotent operator. The linear map $k[T]\mathrm{/}(T^r)\to I_r$ sending $T^i$ to $v_{r-i}$ for $0\le i<r$ is a bijection commuting with $T$, hence a module isomorphism. Intertwining linear maps are exactly module homomorphisms. A module homomorphism $k[T]\mathrm{/}{T}^r\to k[T]\mathrm{/}{T}^h$ is determined by the image of $1$, which may be any vector killed by $T^r$; that space has dimension $\min (r,h)$. Thus the endomorphism algebra of $I_{\lambda }$ has dimension $D={\sum }_{r,h}{m}_r{m}_h\min (r,h)={\sum }_j({\lambda }_j^{\mathrm{\prime }}{)}^2$. Counting each column's ordered pairs of boxes gives ${\sum }_j({\lambda }_j^{\mathrm{\prime }}{)}^2=\mathrm{\mid }\lambda \mathrm{\mid }+2{\sum }_i(i-1){\lambda }_i$: in a column of height $c$, its square is $c+2{\sum }_{i=1}^c(i-1)$.

Consider the induced map of an endomorphism $f$ on $I_{\lambda }\mathrm{/}T{I}_{\lambda }$. Ordered by block length, its matrix is block triangular: a homomorphism from length $r$ to a larger length $h$ has image of $1$ divisible by $T^{h-r}$ and hence has zero reduction in the target head. Its diagonal blocks are arbitrary matrices $B_r\in {\mathrm{Mat}}_{m_r}(k)$; all choices occur independently by constant matrices between blocks of the same length. The linear map from the $D$-dimensional endomorphism space to these diagonal blocks is onto, with every fiber of size $q^{D-{\sum }_r{m}_r^2}$.

The map $f$ is invertible exactly when all $B_r$ are invertible. Necessity follows from invertibility on the quotient. Conversely those blocks make the triangular head map surjective, so $I_{\lambda }=f(I_{\lambda })+T{I}_{\lambda }$. Since $f$ commutes with $T$, iteration gives $I_{\lambda }=f(I_{\lambda })+T^N{I}_{\lambda }$ for every $N$, and nilpotence makes the last summand zero for large $N$. Thus $f$ is onto and, on a finite-dimensional vector space, bijective; its inverse commutes with $T$.

Successively choosing linearly independent columns gives $\mathrm{\mid }{\mathrm{GL}}_m(k)\mathrm{\mid }={\prod }_{i=0}^{m-1}(q^m-q^i)=q^{m^2}{\prod }_{j=1}^m(1-q^{-j})$, including $m=0$. Hence the two preceding steps count the automorphisms as $q^{D-{\sum }_r{m}_r^2}{\prod }_r\mathrm{\mid }{\mathrm{GL}}_{m_r}(k)\mathrm{\mid }=q^D{\prod }_r{\prod }_{j=1}^{m_r}(1-q^{-j})$. Substitute $D$ from step 1.1. For $\lambda =\mathrm{\varnothing }$ the space is zero and its sole automorphism is the identity.