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Demazure–Lusztig operators satisfy the affine type-A relations

Statement

Let n≥2, R0=Z[Q,Q−1], A=R0[X1±1,…,Xn±1], and K=Frac⁡(A), with Di and K⋊Sn as in The type-A weight lattice, Laurent ring, and divided differences and The rational skew group algebra for type A. For ri=Xi/Xi+1 define

Ti:=Q[si]+Q−11−ri([1]−[si])∈K⋊Sn.

Then Ti preserves A, satisfies the quadratic and braid relations of The multiplicative Bernstein presentation of the affine Hecke algebra of type A, and obeys

Tif−si(f)Ti=(Q−1)Di(f)(f∈A).

Consequently the assignments Xj↦Xj[1] and Ti↦Ti define an R0-algebra homomorphism from the presented affine Hecke algebra to K⋊Sn. The coefficient of [si] in Ti is nonzero in K.

Facts & Assumptions

Given: The universal coefficient ring R0, Laurent algebra A, fraction field K, type-A action, divided differences Di, and the presented algebra Hnaff.

[F1]

The type-A weight lattice, Laurent ring, and divided differences: A is a domain, Di:A→A, Di(fg)=Di(f)g+si(f)Di(g), and the adjacent transposition interchanges Xi,Xi+1.

[F2]

The rational skew group algebra for type A: K⋊Sn has basis [w] over K and multiplication (a[w])(b[v])=a w(b)[wv].

[F3]

The multiplicative Bernstein presentation of the affine Hecke algebra of type A: the defining relations are commuting invertible Xj, (Ti−Q)(Ti+1)=0, the finite type-A braid relations, and the Bernstein cross-relations on the Xj.

[F4]

Multivariate polynomial and Laurent rings over commutative rings, domains and fraction fields: A has its Laurent monomial basis and embeds in the fraction field K.

Proof

Given: The data of the Statement.

1.1givenF1F2F4

Put c(z):=(Q−1)/(1−z) and d(z):=(1−Qz)/(1−z)=Q−c(z). In the skew group algebra, Ti=c(ri)[1]+d(ri)[si]. Both 1−ri and 1−Qri are nonzero in A, because their two terms are distinct Laurent monomials over R0 by [F4]. Hence c(ri),d(ri) are defined in K and d(ri)≠0.

2.1F1F2step 1.1

For f∈A, multiplication in [F2] gives Tif−si(f)Ti=c(ri)(f−si(f))=(Q−1)Di(f). As an operator on A, Ti(f)=Qsi(f)+(Q−1)Di(f), which lies in A by [F1]. Thus it preserves A and satisfies the full Bernstein relation, not only its values on generators.

2.2F2step 1.1algebra

The rational functions c,d satisfy c(z)+c(z−1)=Q−1, d(z)=Q−c(z), and d(z−1)=1+c(z) by direct simplification. Expanding (c(ri)[1]+d(ri)[si])2 with [F2], its [si] coefficient is d(ri)(c(ri)+c(ri−1))=(Q−1)d(ri), and its [1] coefficient is c(ri)2+d(ri)d(ri−1)=c(ri)2+(Q−c(ri))(1+c(ri))=Q+(Q−1)c(ri). Therefore Ti2=(Q−1)Ti+Q, equivalent to (Ti−Q)(Ti+1)=0.

2.3F1F2step 1.1

If ∣i−j∣>1, the transpositions si,sj commute and each fixes the coefficients c(rj),d(rj) or c(ri),d(ri) of the other operator. Expanding their product by [F2] therefore gives TiTj=TjTi.

3.1F2step 2.2algebra

For adjacent indices put s=si, t=si+1, r=Xi/Xi+1, u=Xi+1/Xi+2, and abbreviate cr=c(r), cu=c(u), cru=c(ru) and similarly for d. The identities in step 2.2 give c(r−1)=Q−1−cr, c(u−1)=Q−1−cu, and cru(Q−1−cr−cu)=−crcu, since the left side is −(Q−1)2/((1−r)(1−u)). The variable permutations give s(u)=t(r)=ru, st(r)=u, and ts(u)=r.

4.1F2step 2.2step 3.1algebra

Expanding TsTtTs and TtTsTt by [F2], their coefficient pairs, in the order [1],[s],[t],[st],[ts],[sts]=[tst], are respectively: [1]: (cr2cu+drcrud(r−1), cu2cr+ducrud(u−1)); [s]: (dr(crcu+cruc(r−1)), cudrcru); [t]: (crducru, du(cruc(u−1)+cucr)); [st]: (drdrucu, cudrdru); [ts]: (crdudru, dudrucr); and [sts]: (drdrudu, dudrudr). The last three pairs agree by commutativity of K. For the [s] pair, substituting c(r−1)=Q−1−cr reduces equality to crcu+cru(Q−1−cr−cu)=0; for [t] the same identity gives equality after substituting c(u−1)=Q−1−cu. For [1], using dr=Q−cr, d(r−1)=1+cr and their u analogues, the difference of the entries is (cr−cu)(crcu+cru(Q−1−cr−cu))=0. Hence the adjacent braid relation holds.

5.1F2F3step 1.1step 2.1step 2.2step 2.3step 4.1∎

The Bernstein, quadratic, distant, and adjacent braid calculations show that the operators satisfy every relation in [F3], including the commuting Laurent variables and their inverses. The universal property of the presented quotient therefore gives the asserted homomorphism. Step 1.1 established that the coefficient d(ri) of [si] is nonzero. No Choice principle or specialization is used.

Depends on

Used by

Cited to discharge well-definedness by The type-A weight lattice, Laurent ring, and divided differences.

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