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A one-vertex Green coproduct and twisted Hall square

Statement

For the one-vertex zero-arrow quiver, with simple $S$ and $R = S \oplus S$ , compute $[S] ⋆ [S] = ν (q + 1) [R], Δ^{'} ([S]) = [S] \otimes [0] + [0] \otimes [S],$ and $Δ^{'} ([R]) = [R] \otimes [0] + [0] \otimes [R] + ν^{- 1} [S] \otimes [S] .$

For this example, $Δ^{'}$ means the coefficient formula $Δ^{'} ([T]) = \sum_{A, B} ν^{\dim A \dim B} P_{A, B}^{T} a_{T}^{- 1} [A] \otimes [B]$ ; it is the one-vertex specialization of Green’s definition. The product is $[A] ⋆ [B] = ν^{\dim A \dim B} [A] * [B]$ .

\node[draw,circle] (v) {$1$};

This is the quiver with one vertex and no arrows.

Facts & Assumptions

Given: $k = F_{q}$ , $ν = \sqrt{q} > 1$ , and $S = k$ as a vector-space representation.

[F1]

The category of finite-dimensional vector spaces is a finite-field hereditary Hall category (Finite-dimensional representations of an acyclic quiver form a hereditary finitary Hall category).

[F2]

The untwisted Hall coefficient counts subspaces and the twisted scalar in this semisimple category is $ν^{\dim A \dim B}$ (The untwisted Hall algebra of a finitary abelian category).

[F3]

$P_{M, N}^{R} = a_{M} a_{N} g_{M, N}^{R}$ (Exact-sequence counts equal endpoint automorphism counts times subobject counts).

Proof

technique · count subspaces and endpoint-labelled sequences

1.1F1F2algebra

Here $Hom (S, S) ≅ k$ and ${YExt}^{1} (S, S) = 0$ , 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 $[S] ⋆ [S] = ν (q + 1) [R]$ . The same count with $S$ as middle object gives $Δ^{'} ([S]) = [S] \otimes [0] + [0] \otimes [S]$ .

1.2F1F3construct

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

2.1F2step 1.2algebra∎

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

Depends on

Used by

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Dependency tree · two levels

17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Olivier Schiffmann, Lectures on Hall Algebras, arXiv:math/0611617v2 (standard reference, not scraped)
Claus Michael Ringel, The Hall Algebra Approach to Quantum Groups, E.L.A.M. Lectures (1993) (standard reference, not scraped)