Euler-Twisted Hall Products, Green Coproducts, and Quantum Groups: Examples and Counterexamples

The acyclic Euler form is $\chi (\alpha ,\beta )={\alpha }_1{\beta }_1+{\alpha }_2{\beta }_2-2{\alpha }_1{\beta }_2$, which is bilinear by [F1]. Define this example’s product by $[M]\star [N]={\nu }^{\chi (\mathrm{deg}M,\mathrm{deg}N)}[M]\ast [N]$, where $\ast$ is the untwisted Hall product of [F3]. This twist preserves associativity, since $\chi (\alpha ,\beta )+\chi (\alpha +\beta ,\gamma )=\chi (\beta ,\gamma )+\chi (\alpha ,\beta +\gamma )$. Thus the composition subalgebra is defined for this quiver.

A representation of dimension $(1,1)$ is a pair $(x,y)\in {\mathbb{F}}_q^2$ of arrow scalars. An isomorphism is a pair of nonzero endpoint scalars $(g_1,g_2)$, acting by common multiplication by $g_2\mathrm{/}{g}_1$. The zero pair is one orbit and gives $S_1\oplus S_2$. Each nonzero orbit is a line in ${\mathbb{F}}_q^2$, giving $q+1$ further orbits. Thus the full Hall component has $q+2$ independent isomorphism-class basis vectors.

A product of simples of total degree $(1,1)$ must use $S_1$ and $S_2$ once each. There are only two possible words, $[S_1]\star [S_2]$ and $[S_2]\star [S_1]$, so the degree component generated by simples has dimension at most two.

Since $q\ge 2$ by [F4], $q+2>2$. The two subspaces have different dimensions, proving strict containment.

Here $\mathrm{Hom}(S,S)\cong k$ and ${\mathrm{YExt}}^1(S,S)=0$, so $⟨S,S⟩=1$. The subobjects of $R=k^2$ isomorphic to $S$ are its $q+1$ lines. Thus $g_{S,S}^R=q+1$, and the twisted product is $[S]\star [S]=\nu (q+1)[R]$. The same count with $S$ as middle object gives ${\mathrm{\Delta }}^{\mathrm{\prime }}([S])=[S]\otimes [0]+[0]\otimes [S]$.

For the mixed coefficient of $[S]\otimes [S]$ in ${\mathrm{\Delta }}^{\mathrm{\prime }}([R])$, an endpoint-labelled exact sequence is an injection $i:k\to k^2$ and a surjection $p:k^2\to k$ with $pi=0$. There are $q^2-1$ choices for $i$; for each, the quotient $k^2\mathrm{/}\mathrm{im}(i)$ is one-dimensional and has $q-1$ isomorphisms to the fixed endpoint $k$. Hence $P_{S,S}^R=(q^2-1)(q-1)$. Also $a_R=\mathrm{\mid }{\mathrm{GL}}_2(k)\mathrm{\mid }=(q^2-1)(q^2-q)$, so $P_{S,S}^R\mathrm{/}{a}_R=1\mathrm{/}q$.

The mixed coproduct coefficient is ${\nu }^{⟨S,S⟩}\mathrm{/}q=\nu \mathrm{/}q={\nu }^{-1}$ because $q={\nu }^2$. The only other possible endpoint pairs are $(R,0)$ and $(0,R)$; each contributes coefficient one. This proves the displayed coproduct.

The acyclic Euler form in [F1] is $\chi (\alpha ,\beta )={\alpha }_1{\beta }_1+{\alpha }_2{\beta }_2-{\alpha }_1{\beta }_2$; it is bilinear. Twisting the associative Hall product in [F3] by ${\nu }^{\chi (\alpha ,\beta )}$ remains associative because $\chi (\alpha ,\beta )+\chi (\alpha +\beta ,\gamma )=\chi (\beta ,\gamma )+\chi (\alpha ,\beta +\gamma )$. In particular, $\chi (e_1,e_2)=-1$, $\chi (e_2,e_1)=0$, and $\chi (e_1,e_1)=1$.

Let $R_0=S_1^{\oplus 2}\oplus S_2$ be the zero-arrow representation of dimension $(2,1)$, and let $R_1=S_1\oplus I$ be the rank-one representation. These are all isoclasses because a map $k^2\to k$ has rank zero or one. In the semisimple one-vertex subcategory, $[S_1{]}^{(2)}={\nu }^2[S_1^{\oplus 2}]$: the ordinary square counts the $q+1$ lines, and dividing $\nu (q+1)$ by $[2{]}_{\nu }=\nu +{\nu }^{-1}$ gives ${\nu }^2$ using [F4].

A representation of dimension $(1,1)$ is one arrow scalar. Its rank is zero, giving $S_1\oplus S_2$, or nonzero, giving the unique isoclass $I$. For each middle object, the unique subobject isomorphic to $S_2$ has quotient $S_1$. Thus both Hall numbers in $[S_1]\star [S_2]$ are one, and [F2] with the Euler value in step 1.1 gives its displayed formula.

For $[S_2]\star [S_1]$, the required subobject is $S_1$ at vertex $1$ with quotient $S_2$. Stability forces the arrow to vanish, so the only middle object is $S_1\oplus S_2$, with coefficient one by [F2] and the Euler value in step 1.1.

The product $[S_1{]}^{(2)}\star [S_2]$ has twist ${\nu }^{⟨2S_1,S_2⟩}={\nu }^{-2}$, cancelled by the coefficient ${\nu }^2$ from step 1.2. In each of $R_0,R_1$, the full vertex-$2$ copy of $S_2$ is the unique subobject with quotient $S_1^{\oplus 2}$. Hence this product is $[R_0]+[R_1]$.

The last product $[S_2]\star [S_1{]}^{(2)}$ has twist ${\nu }^{⟨S_2,2S_1⟩}=1$. The only subobject isomorphic to $S_1^{\oplus 2}$ occurs in $R_0$, where it is the full vertex-$1$ space; no such subobject is stable in $R_1$ because its arrow is nonzero. The coefficient is ${\nu }^2$ from step 1.2, so this product is ${\nu }^2[R_0]$.

By associativity from step 1.1, $[S_1]\star [S_2]\star [S_1]=[S_1]\star [S_1\oplus S_2]$. Its twist is ${\nu }^{⟨S_1,S_1+S_2⟩}=1$. In $R_0$, a subobject $S_1\oplus S_2$ is obtained by choosing any line at vertex $1$, giving $q+1$ choices; in $R_1$, stability forces that line to be the one-dimensional kernel of the arrow, giving one choice. Thus the middle product is $(q+1)[R_0]+[R_1]$.

Subtracting the middle product in step 3.1 from the two outer products in steps 2.3 and 2.4 gives $(1-(q+1)+{\nu }^2)[R_0]+(1-1)[R_1]=0,$ since $q={\nu }^2$. Multiplying by $[2{]}_{\nu }!$ and reindexing the two divided-power terms yields the ordinary symmetric quantum Serre relation from The Hall-normalized extended positive Borel of a loopless quiver.