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The Rouquier complex of a positive three-strand braid

Example

For the positive word σ1σ2∈B3 the Rouquier complex is F(σ1σ2)=[  B1⊗RB2→d0B1(1)⊕B2(1)→d1R(2)  ] with cohomological degrees 0,1,2 and differentials d0(x⊗y)=(x⋅ε2(y), ε1(x)⋅y),d1(a⊗r, s⊗b)=ε1(a)r−s ε2(b), under the evident identifications B1⊗RR(1)=B1(1), R(1)⊗RB2=B2(1) and R(1)⊗RR(1)=R(2); the example checks d1d0=0 on the four left-basis tensors u1⊗u2, u1⊗(1⊗α2), (1⊗α1)⊗u2, and (1⊗α1)⊗(1⊗α2) of B1⊗RB2, where ui=1⊗1 and records the cohomological and internal degree of every generator.

Facts & Assumptions

Given: The adjacent simple reflections s1,s2 of S3, the bimodules B1,B2 with the generators ui=1⊗1 of degree −1 and 1⊗αi of degree 1, and the complexes F1=[B1→ε1R(1)], F2=[B2→ε2R(1)] of The positive and negative Rouquier generator complexes.

[F1]

Generators and products. Bi has the left R-basis (1⊗1,1⊗αi) of degrees −1 and 1; the multiplication εi sends 1⊗1↦1 and 1⊗αi↦αi, and satisfies εi(r⊗r′)=rr′ (The positive and negative Rouquier generator complexes).

[F2]

Totalization. The signed tensor totalization of F1 and F2 has cohomological degree terms F0⊗G0=B1⊗RB2, F0⊗G1⊕F1⊗G0=B1(1)⊕B2(1) and F1⊗G1=R(2), with Koszul differential d(x⊗y)=dF(x)⊗y+(−1)px⊗dG(y) for x∈Fp (Bounded graded bimodule complexes and signed tensor totalization).

Verification

technique · direct
1.1F1F2

The degree-0 term is B1⊗RB2, of cohomological degree 0; the degree-1 term is F0⊗G1⊕F1⊗G0=B1⊗RR(1)⊕R(1)⊗RB2≅B1(1)⊕B2(1); the degree-2 term is R(1)⊗RR(1)≅R(2). The four basis monomials u1⊗u2, u1⊗(1⊗α2), (1⊗α1)⊗u2, (1⊗α1)⊗(1⊗α2) of B1⊗RB2 have internal degrees −2,0,0,2; the basis ui,1⊗αi of Bi(1) has degrees −2,0; and the generator of R(2) has degree −2.

2.1F2step 1.1

Under the identifications of step 1.1 the Koszul differentials are d0(x⊗y)=(x⋅ε2(y), ε1(x)⋅y)∈B1(1)⊕B2(1) and d1(a,s)=ε1(a)−ε2(s) computed in R(2), i.e. d1(a⊗r,s⊗b)=ε1(a)r−sε2(b) in the notation of the display; the minus sign is the Koszul sign on the differential from bidegree (1,0) to (1,1), where the first factor R(1) sits in cochain degree 1.

3.1F1step 2.1

For every simple tensor x⊗y∈B1⊗RB2 one computes d1d0(x⊗y)=ε1(x)ε2(y)−ε1(x)ε2(y)=0: the first component of d0(x⊗y) contributes ε1(x)ε2(y) and the second contributes the same product with the Koszul sign −1, so the two cancel. Since the differentials are balanced and R-bilinear, this extends to all elements, so d1d0=0.

4.1step 1.1step 3.1∎

The degree bookkeeping of step 1.1 shows that d0 and d1 are homogeneous of internal degree zero on the displayed generators, hence on all elements; this records the cohomological and internal degree of every generator of the three terms and completes the verification of the displayed formula.

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