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The rank-one Soergel bimodule square splits

Statement

Let s=si be a simple reflection and Bs=R⊗RsR(1), so that Bs=R⊗RsR{−1} in the internal shift. Then there is an isomorphism of graded (R,R)-bimodules Bs⊗RBs  ≅  Bs(1)⊕Bs(−1), that is Bs⊗RBs≅Bs{−1}⊕Bs{1}; the two summands are the idempotent images of the summands R⊗RsRs⊗RsR and R⊗RsαRs⊗RsR of the middle decomposition, and the two summands are not isomorphic as graded bimodules: they differ by the internal shift {2}, so each is the shift of the other, but their graded left ranks v−1+k+v1+k for k=−1 and k=1, namely v−2+v0 and v0+v2, are different, and an isomorphism of graded bimodules preserves graded ranks.

Facts & Assumptions

Given: A simple reflection s=si, the invariant ring Rs, the element α=αi with s(α)=−α, and the bimodule Bs=R⊗RsR(1)=R⊗RsR{−1}.

[F1]

R=Rs⊕αRs as graded (Rs,Rs)-bimodules, and the inclusion Rs→R is a graded ring map with 2 invertible in k; αRs≅Rs{2} as (Rs,Rs)-bimodules via multiplication by α (The standard type-A reflection realization and its polynomial ring, The Soergel bimodule Bi of a simple reflection).

[F2]

R⊗RsRs⊗RsR≅R⊗RsR and shifts move across a balanced tensor product, so that (R⊗RsR){k}=R⊗Rs(R{k}) (The Bott–Samelson bimodule of a word).

[F3]

Bs is free of rank two on both sides and Bs⊗RBs is free of rank four on both sides, with Δ-flag quotients Rs{2}, R{0} over the piece Rs{1} and Rs{0}, R{−2} over the piece R{−1}; the ∇-flag quotients are R{2}, Rs{0}, R{0}, Rs{−2} (Bott–Samelson bimodules carry delta and nabla support filtrations, Standard graph bimodules, support filtrations and characters).

Proof

1.1

Tensoring the middle factor: by [F1] the (Rs,Rs)-bimodule R decomposes as Rs⊕αRs, so applying R⊗Rs−⊗RsR to it gives R⊗RsR⊗RsR≅(R⊗RsR)⊕(R⊗RsαRs⊗RsR).

F1
2.1

The second summand: multiplication by α is an (Rs,Rs)-bimodule isomorphism Rs{2}→αRs, so R⊗RsαRs⊗RsR≅(R⊗RsR){2} by [F2]; the first summand is R⊗RsR≅(R⊗RsR){0}.

F1F2step 1.1
3.1

Shifts: Bs=R⊗RsR{−1}, so Bs⊗RBs=(R⊗RsR⊗RsR){−2}; applying the decomposition of step 2.1 and distributing the shift gives Bs⊗RBs≅(R⊗RsR){−2}⊕(R⊗RsR){0}=Bs{−1}⊕Bs{1}.

F2step 2.1
4.1

Consistency with the support flags: the two summands Bs{−1} and Bs{1} have Δ-flag quotients Rs{0},R{−2} and Rs{2},R{0} respectively, whose union Rs{0},Rs{2},R{−2},R{0} is the multiset in [F3]; hence the abstract decomposition of step 3.1 realizes the flag computation, and the two summands are the α-divisible and the α-free part of the middle factor.

F3step 3.1
5.1

Non-isomorphism and freeness: Bs{−1} and Bs{1} differ by the shift {2}, and the graded rank of Bs{k} as a left R-module is v−1+k+v1+k, so the two summands have different graded ranks and are not isomorphic; both are free of rank two on each side while Bs⊗RBs is free of rank four, matching step 3.1. ∎

F3step 3.1

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