Finitary Ringel–Hall Algebras: Examples and Counterexamples

Use the small model with objects $k^n$ and all matrices between them; every finite-dimensional vector space is isomorphic to one of these. The zero object and finite direct sums exist, and each Hom set is an abelian group under pointwise addition with bilinear composition. For a linear map $f:V\to W$, its kernel and cokernel $W\mathrm{/}\mathrm{im}f$ are finite-dimensional. Choose bases of these two spaces for this single map and transport them to standard $k^d$ objects; the transported inclusion and projection retain the universal properties, so the model has kernels and cokernels. The canonical map $V\mathrm{/}\ker f\to \mathrm{im}f$ is injective because $f(v)=0$ implies $v\in \ker f$, and surjective by the definition of image, hence is an isomorphism. Thus [F1] makes the model abelian. Hom sets are finite sets of matrices over $k$. Any short exact sequence $0\to U\stackrel{i}{\to }E\stackrel{p}{\to }Q\to 0$ splits: lift a finite basis of $Q$ to $E$ and extend linearly to a section $s:Q\to E$. The map $U\oplus Q\to E$, $(u,q)\mapsto i(u)+s(q)$, is injective by applying $p$ and then using monicity of $i$; it is surjective because $e-s(p(e))$ lies in $\ker p=\mathrm{im}i$. It is an isomorphism fixing the endpoints, so every Yoneda extension is equivalent to the split one and its Yoneda extension set has one class. Therefore the model satisfies [F2].

If $g_{S,S}^L\mathrm{\ne }0$, an exact sequence $0\to S\to L\to S\to 0$ gives ${\dim }_{k}L=2$, so the only possible isomorphism class is $k^2$. Every one-dimensional subspace of $k^2$ is counted by [F3], and its quotient is one-dimensional. The nonzero vectors of $k^2$ are partitioned by their spans, each span contains $q-1$ nonzero vectors, and $k^2$ has $q^2-1$ nonzero vectors. Hence the number of lines is $(q^2-1)\mathrm{/}(q-1)=q+1$.

There are no other middle-object classes by step 1.2, and the coefficient of $[k^2]$ is the line count from step 1.2. This proves the displayed untwisted product.

Kernels and cokernels of homomorphisms between finite abelian $p$-groups are finite abelian $p$-groups, and finite direct sums remain finite abelian $p$-groups. Transporting each kernel or cokernel along a bijection of its underlying finite set with a standard set gives the required object in the chosen model, with the same universal property. Thus the model is abelian using [F1]. For fixed finite groups $M,N$, Hom is a finite set. Any short exact sequence with endpoints $M,N$ has a middle group of cardinality $\mathrm{\mid }M\mathrm{\mid }\mathrm{\mid }N\mathrm{\mid }$; there are finitely many group laws on a finite set of that size and finitely many endpoint maps. Therefore its Yoneda extension set is finite. The stated group-table model is a set of objects, so [F3] applies.

If $0\to C_p\to L\to C_p\to 0$ is exact, then $\mathrm{\mid }L\mathrm{\mid }=p^2$. If $L$ contains an element of order $p^2$, then $L\cong C_{p^2}$. Otherwise every nonzero element has order $p$, so scalar multiplication by ${\mathbb{F}}_p$ makes $L$ a two-dimensional vector space and $L\cong C_p\oplus C_p$. Thus these are the only possible middle groups.

In $C_{p^2}$, the unique subgroup of order $p$ is $p\mathbb{Z}\mathrm{/}{p}^2\mathbb{Z}$, and its quotient is $C_p$, giving coefficient one by [F2]. In $C_p\oplus C_p$, every one-dimensional subspace is a subgroup isomorphic to $C_p$, and its quotient is $C_p$. The number of such subspaces is $(p^2-1)\mathrm{/}(p-1)=p+1$, by partitioning nonzero vectors into lines.

The extension classification and counts of step 2.1 give the formula. Moreover $\mathrm{End}(C_{p^2})\cong \mathbb{Z}\mathrm{/}{p}^2\mathbb{Z}$ additively, which has an element of order $p^2$; it is not an ${\mathbb{F}}_p$-vector space. Thus the category is not ${\mathbb{F}}_p$-linear.

The category is finitary by [F3], so its Hall product is defined. Any middle object contributing to $[S]\ast [S]$ has vector-space dimension two. By [F1], its type is either $(1,1)$ or $(2)$.

In $I_{(1,1)}=S\oplus S$, the nilpotent operator is zero, so every one-dimensional subspace is an invariant subobject by [F3], and its quotient is $S$. There are $(q^2-1)\mathrm{/}(q-1)=q+1$ such lines.

In $I_2$, an invariant one-dimensional subspace must be annihilated by the nilpotent operator, so it is contained in $\ker x$. This kernel is one-dimensional, hence it is the unique such subobject by [F3], and the quotient is $S$.

The two possible middle types and their counts from steps 1.1–1.3 yield the displayed Hall product.

The normalized generator map gives ${\mathrm{\Phi }}_q(u_{(1)})=e_1=m_{(1)}$ and ${\mathrm{\Phi }}_q(u_{(1,1)})=q^{-1}{e}_2=q^{-1}{m}_{(1,1)}$. Since $e_1^2=m_{(2)}+2m_{(1,1)}$, applying it to [F1] yields ${\mathrm{\Phi }}_q(u_{(2)})=e_1^2-(1+q^{-1})e_2=m_{(2)}+(1-q^{-1})m_{(1,1)}$. The exponents are $n((1))=n((2))=0$, $n((1,1))=1$, so [F2] identifies exactly the four asserted $P$ functions, including the unit.

In the power-sum basis, $m_{(2)}=p_2$ and $m_{(1,1)}=(p_1^2-p_2)\mathrm{/}2$, hence $P_{(2)}=((1-s)p_1^2+(1+s)p_2)\mathrm{/}2$. The diagonal Hall–Littlewood norms are $⟨p_1^2,p_1^2{⟩}_s=2\mathrm{/}(1-s{)}^2$ and $⟨p_2,p_2{⟩}_s=2\mathrm{/}(1-s^2)$. Substitution gives $⟨P_{(2)},P_{(1,1)}{⟩}_s=\frac{1}{4}((1-s)2\mathrm{/}(1-s{)}^2-(1+s)2\mathrm{/}(1-s^2))=0$, $⟨P_{(2)},P_{(2)}{⟩}_s=1\mathrm{/}(1-s)$ and $⟨P_{(1,1)},P_{(1,1)}{⟩}_s=1\mathrm{/}((1-s)(1-s^2))$. These independently check the theorem's degree-two scalar convention.

The automorphism factors from [F2] are $a_{(2)}=q^2(1-q^{-1})$ and $a_{(1,1)}=q^4(1-q^{-1})(1-q^{-2})$. Since ${\mathrm{\Phi }}_q(u_{(1,1)})=q^{-1}{P}_{(1,1)}$, the rule $B_q=q^{-2}⟨,{⟩}_{q^{-1}}$ gives exactly $a_{(1,1)}^{-1}$ on that image, and gives $a_{(2)}^{-1}$ on the other image. In particular $B_q(p_2,p_2)=q^{-2}2\mathrm{/}(1-q^{-2})=2\mathrm{/}(q^2-1)$, whereas [F3] gives the ordinary Hall value $2$. The factors $q^{-1}$ on the column class and $q^{-2}$ on the degree-two form are both necessary.