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The rank-two Bott-Samelson bimodule B_sB_t: character, decomposition and the braid relation

Example

Take n=3 and the two simple reflections s=s1, t=s2. The rank-two Bott-Samelson bimodule BsBt has character ch⁡(BsBt)=ch⁡(Bs)ch⁡(Bt)=(Hs+v)(Ht+v)=Hst+vHs+vHt+v2 in the standard basis, so its Δ-multiplicities are 1 at Hst, 1 at Hs and at Ht with shift v, and 1 at the identity with shift v2; in particular all multiplicities are non-negative and dim⁡(BsBt‾)=v2+2+v−2, whose value at v=1 is the ordinary real dimension 4. In S3 the product st has length 2 and Bst has the same character H‾st=Hst+vHs+vHt+v2, so BsBt=Bst is already indecomposable; equivalently the only summand is the top one Bst and all multiplicities my for y<st vanish. Similarly BsBtBs=Bsts⊕Bs with characters H‾sts+H‾s, whereas BtBsBt=Bsts⊕Bt; the two Bott-Samelson products are not isomorphic; the example also verifies the rank-two values of the Kazhdan-Lusztig polynomials Pe,st=Ps,st=Pt,st=1 used on the companion page.

Verification

technique · direct; multiplicativity of the character, the explicit two-letter Hecke computation, and comparison with the rank-two decomposition

Given: The simple reflections s=s1, t=s2 of S3, the one-letter bimodules Bs,Bt and the Bott-Samelson product BsBt (The Bott–Samelson bimodule of a word, The real type-A Soergel category by scalar extension), the character with its Δ-multiplicities (Standard graph bimodules, support filtrations and characters, The type-A standard character is multiplicative), the normalized Hecke algebra with its two bases and H‾s=Hs+v (Existence and uniqueness of the Kazhdan–Lusztig basis), the rank-two decomposition theorem (Rank-two type-A Soergel bimodule decompositions).

[F1] The character is multiplicative and satisfies ch⁡(Bsi)=Hsi+v; the coefficient of Hy in ch⁡(M) is ∑d(M:Δy(d))vd, and the reduced dimension of a layer Δy(d) is one in degree ℓ(y)−d (The type-A standard character is multiplicative, Standard graph bimodules, support filtrations and characters).

[F2] The standard basis of the Hecke algebra multiplies by HuHs=Hus when ℓ(us)=ℓ(u)+1; the Kazhdan-Lusztig basis is the unique basis H‾w∈Hw+∑y<wvZ[v]Hy of bar-invariant elements (Existence and uniqueness of the Kazhdan–Lusztig basis).

[F3] The rank-two decomposition theorem: BsBtBs≅Bsts⊕Bs with no shifts; exchanging s and t gives BtBsBt≅Bsts⊕Bt (Rank-two type-A Soergel bimodule decompositions).

[F4] The polynomial ring has only nonnegative degrees and degree-zero part R. The generator has left basis 1⊗1,1⊗xi for si, since S=Ssi⊕xiSsi; tensor products have the corresponding finite left bases (The real type-A Soergel category by scalar extension).

1.1givenF1F2

The character of the two-letter product. By [F1] and multiplicativity, ch⁡(BsBt)=(Hs+v)(Ht+v)=HsHt+vHs+vHt+v2; since st is reduced, [F2] gives HsHt=Hst, so the standard-basis expansion is Hst+vHs+vHt+v2. Its coefficients give the multiplicities 1,1,1,1 displayed, and reading the layers through [F1] the reduced degrees are 2,0,0,−2, so the graded dimension is v2+2+v−2 and the ordinary real dimension is 4.

2.1step 1.1F2

The Kazhdan-Lusztig element of st. Put A:=v−v−1 and J:=Hst+v(Hs+Ht)+v2; then J∈Hst+vZ[v]Hs+vZ[v]Ht+v2Z[v]. Since Hs−1=Hs+A, Ht−1=Ht+A and Hst‾=Hts−1=Hs−1Ht−1=(Hs+A)(Ht+A)=Hst+A(Hs+Ht)+A2 (the last identity uses HsHt=Hst), applying the semilinear involution gives J‾=Hst‾+v−1(Hs‾+Ht‾)+v−2=Hst+A(Hs+Ht)+A2+v−1(Hs+Ht)+2v−1A+v−2, whose coefficients are A+v−1=v at Hs+Ht and A2+2v−1A+v−2=(v2−2+v−2)+(2−2v−2)+v−2=v2 at He; hence J‾=J, and by uniqueness [F2], J=H‾st.

3.1step 1.1step 2.1F4algebra

Indecomposability of BsBt. The unshifted tensor M=S⊗SsS⊗StS is generated as a bimodule by 1⊗1⊗1. To see this, write a middle polynomial as h0+x1h1 with h0,h1∈Ss, which slide to the first slot. The remaining x1 slides to the last slot because t=s2 fixes x1. Therefore every tensor is a sum of p(1⊗1⊗1)q. After the shift (2) its lowest component is one-dimensional in degree −2, by its finite left tensor basis and S0=R. A degree-zero bimodule endomorphism must send the generator to a scalar multiple of itself, and generation forces that same scalar on all of M. Hence its degree-zero endomorphism algebra is R, with no nontrivial idempotent, so BsBt is indecomposable. The balanced map p⊗q⊗r↦p s(q) st(r) is surjective onto the st graph bimodule, showing that this indecomposable has the top graph of its reduced word; it is the normalized summand Bst. Thus BsBt≅Bst with no lower summands, independently of the Hom formula.

4.1step 2.1step 3.1F3∎

The three-letter braid relation. By multiplicativity and the products computed in steps 1.1 and 2.1, ch⁡(BsBtBs)=H‾sH‾tH‾s=H‾stH‾s; expanding with the Hecke relations gives H‾stH‾s=H‾sts+H‾s Indeed the expansion is J0+Hs+v where J0=Hsts+v(Hst+Hts)+v2(Hs+Ht)+v3. The element J0=H‾stH‾s−H‾s is bar fixed, with triangular lower coefficients in vZ[v], so uniqueness in [F2] identifies it with H‾sts. The ordinary real dimensions of the reductions add to 6+2=8. Exchanging s and t in [F3] gives BtBsBt≅Bsts⊕Bt, whose character differs by H‾t−H‾s=Ht−Hs≠0, so these Bott-Samelson products are not isomorphic; the rank-two Kazhdan-Lusztig data Pe,st=Ps,st=Pt,st=1 are the coefficients displayed in step 2.1. This verifies the example.

Remarks

  • The reduction to rank two is essential here: the computation uses the explicit two- and three-letter decompositions and is an illustration only; the design of the page forbids its use as a supplier inside the general induction, whose base is computed on the axiom page.
  • The two-letter indecomposability is established by its scalar degree-zero endomorphisms; the triple decomposition is the explicit rank-two theorem. Neither computation uses the general character theorem.

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