Spectral sublagebras

Spectral subalgebras

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 18 March 2012

Spectral subalgebras

Then C 0 = μA*| μ xy =μ yx   is a commutative algebra, since, if l 1 , l 2 C 0 and aA then l 2 l 1 a = l 1 l 2 Δ op a = l 1 l 2 Δ a -1 = l 1 l 2 Δ a -1 = l 1 l 2 Δ a= l 1 l 2 a , where the third equality uses the definition of C 0 .

If A is a quasitriangular Hopf algebra the satisfies the quantum Yang-Baxter equation, (QYBE), 12 13 23 = 12 Δid = Δ op id 12 = 23 13 12 .

Since = εidid Δid = εidid 13 23 = εid ., and = ididε id Δ = ididε 13 23 = idε ., and so εid =1and idε =1.

Then, since Sid = mid idSid 13 23 = mid idSid Δid = εid =1, it follows that Sid = -1 . Applying this to the pair A op 21 gives S -1 id 21 = 21 op , and so id S -1 = -1 . Then SS = idS Sid = idS -1 = idS id S -1 =.

The map φ:CZ A in the following proposition is ananalogue of the Harish-Chandra homomorphism.

Let A be a quasitrtiangular Hopf algebra. Then C= λA*| λ xy =λ y S 2 x is a commutative algebra and the map φ: C Z A l idl 21 is a well defined algebra homomorphism.

Proof.
If l 1 , l 2 A* and aA then l 1 l 2 a = l 1 l 2 Δ op a = l 1 l 2 Δ a -1 = l 1 l 2 Δ a -1 S 2 S 2 ,by definiton of C, = l 1 l 2 Δ a -1 = l 1 l 2 Δ a = l 1 l 2 a , and hence C is a commutative algebra.

Let aA. First note that a1 = idε Δ a = idm id S -1 id id Δop Δ a = a a 1 S -1 a 3 a 2 = a 1 S -1 a 2 a 11 a 12 = a 1 S -1 a 2 Δ a , since S -1 is the antipode of A op , and a1 = idε Δ a = idm ididS idΔ Δ a = a a 1 a 2 S a 3 = a a 11 a 12 1S a 2 = a Δ a 1 1S a 2 , .

Then, since 21 Δ a = 21 Δ op a =Δ a 21 , aφ l = a idl 21 = idl a1 21 12 = idl a 1 S -1 a 2 Δ a 1 21 = idl a Δ a 1 21 1S a 2 ,by definition of  C, = idl 21 a Δ a 1 1S a 2 = idl 21 a1 = idl 21 a=φ l a, and so φ l Z A . Since φ l 1 l 2 = id l 1 l 2 21 = id l 1 l 2 idΔ 21 = id l 1 l 2 21 31 13 12 = id l 1 21 φ l 2 1 12 = id l 1 21 φ l 2 1 , since  φ l 2 Z A , = φ l 1 φ l 2 , and so φ is a homomorphism.

Central elements (idqtrV)(a)

The following proposition provides Drinfeld's "second central element construction" [Dr, Prop. 1.2 and Prop 3.3]. The model example is the case that U=Uh𝔤,   x0=ehρ and a=21.

Let (U,) be a quasitriangular Hopf algebra with antipode S. Let x0U and aUU be invertible elements such that x0xx0-1 = S2(x) and aΔ(x) = Δ(x)a,   for   xU, (Ss 2.1) respectively. Let C0 = {lU*  |  l(xy) = l(yx)} and C1 = {U*  |  (xy) = (yS2(x))}. (Ss 2.2) Let Z(U) be the center of U and let Rep be the Grothendieck group of finite dimensional representations of U. Define Rep 0tr0 C0 0x0*0 C1 0fa0 Z(U) l (id)(a) V trV qtrV (idqtrV)(a) where   (u) = l(ux0) (Ss 2.3) for uU. Then

  1. Ck = {U*  |  (xy) = (yS2k(x))} is a commutative subalgebra of U*;
  2. the maps x0* and fa are well defined;
  3. tr:RepC0 is an algebra homomorphism;
  4. if Δ(x0) = x0x0 then x0*: C0C1 is an algebra homomorphism; and
  5. if a=21 then fa: C1Z(U) is an algebra homomorphism.

Proof.
  1. Let 1,2Ck. Then (12)(xy) = (12) (Δ(xy)) = (12) (Δ(x)Δ(y)) = (12) (Δ(y)(S2kS2k)Δ(x)) = (12) (Δ(y)Δ(S2k(x))) = (12) (Δ(yS2k(x))) = (12) (yS2k(x)), where the fourth equality follows from the fact that S2 is a coalgebra automorphism. Thus Ck is a subalgebra of U*.

    Since (SS)() = , (12)(x) = (12) Δ(x) = (21) Δop(x) = (21) (Δ(x)-1) = (21) (Δ(x)-1(S2kS2k)()) = (21) (Δ(x)-1 = (21) Δ(x) = 21(x). So Ck is commutative.
  2. Let = x0*(l). Then (xy) = l(xyx0) = l(yx0x) = l(yx0xx0-1x0) = l(yS2(x)x0) = (yS2(x)), and thus x0*(l)C1 and x0* is well defined.

    The identities ε(x) = xS-1(x(2))x(1) and xx(1)S(x(2)) = ε(x), (Ss 2.4) from the definition of a Hopf algebra are the relations which provide the isomorphisms in [DRV] (A.8). For xU, x(id(a)) = (id) ((x1)a) = (id) x (x(1)ε(x(2)))a = (id) x ( x(1)S-1(x(3))x(2) )a = (id) x (x(1)x(2))a(1S(x(3))) , since   (bc) = (cS2(b)), = (id) xa (x(1)x(2)) (1S(x(3))) , since   Δ(x)a = aΔ(x), = (id) xa (x(1)ε(x(2)))) = (id) (a(x1)) = ((id)(a))x. So (id)(a)Z(U). Hence fa() = (id)(a) is an element of the center of U, and fa is well defined.
  3. Let M,NRep. Let {mi} be a basis of M, {mi} a dual basis in M*, {nj} a basis of N and {nj} a dual basis in N*. Then (trMtrN)(x) = x trM(x(1)) trN(x(2)) = x,i,j x(1)mi x(2)nj, minj = i,j x(minj), minj = trMN(x). So tr:RepC0 is an algebra homomorphism.
  4. Assume Δ(x0) = x0x0. Let l1,l2C0. Then x0* (l1l2)(x) = (l1l2) (xx0) = (l1l2) (Δ(xx0)) = (l1l2) (Δ(x)(x0x0)) = (x0*(l1) x0*(l2)) (Δ(x)) = (x0*(l1) x0*(l2)) (x). So x0*: C0C1 is an algebra homomorphism.
  5. Assume a=21. Let 1,2C1. Then (id12) (21) = (id12) (idΔ) (21) = (id12) (21) (2113) = (id1) (21((id2)(21)1)). Since (id2)(21)Z(U), fa (12) = (id12) (21) = (id1) (21((id2)(21)1)) = (id1) (21) (id2) (21) = fa(1) fa(2). So fa: C1Z(U) is an algebra homomorphism.

Let U=U𝔤 and let a=γ = bbb*. Let 1,2C1. Then fa (12) = (id12) (γ) = (id12) (idΔ) (γ) = (id12) bbb*1 + b1b* = fa (1) 2(1) + 1(1) fa (2). Thus, since (12)(1) = (12) (Δ(1)) = (12) (11) = 1(1) 2(1) , fa(12) (12)(1) = f1(1) 1(1) + fa(2) 2(1) . (Ss 2.5)

(idqtrL(ν))(21) when U=Uh𝔤

In the case that U=Uh𝔤 and x0 = v-1u = ehρ, M and N are U-modules, and 1 = qtrM and 2 = qtrN then (d) and (e) above give that, as elements of U*, (qtrMqtrN)(x) = qtrMN(x), and (idqtrMN) (21) = (idqtrM) (21) (idqtrN) (21) (as explained in [Dr, Prop. 3.3] and [Bau, Prop. 2]). In this case, the computation in the proof of (d) and (e), pictorially, is = = = =

Fix a Cartan subalgebra 𝔥 in 𝔤. For γ𝔥* define the Weyl character sγ = aγ+ρ aρ , where aμ = wW0 det(w)Xwμ. The expressions sγ and aμ are elements of the group algebra of 𝔥*, [𝔥*] = span{Xν  |  ν𝔥*}, and, if wW0 then awμ = det(w)aμ, and swμ = det(w)sμ, where the dot action of W0 on 𝔥* is given by wμ = w(λ+ρ)-ρ, for   wW0,  μ𝔥*. (Ss 3.1) The Weyl denominator formula says that aρ = αR+ (Xα/2-X-α/2) (Ss 3.2) and the Weyl character formula says that if γ is a dominant integral weight, L(λ) is the simple 𝔤-module of highest weight γ, and v1,...,vn is a basis of L(λ) consisting of weight vectors then sλ = i=1n Xwt(vi) = νP Kλν Xν, where Kλν = dim(L(λ)ν), (Ss 3.3) the dimension of the ν weight space of L(λ). For μ𝔥* define an algebra homomorphism evμ: [𝔥*] [𝔥*] by evμ(Xν) = qμ,ν. Then dimq(L(ν)) = ev2ρ(sν), (Ss 3.4) since dimq(L(ν)) = trL(ν) (ehρ) and q=eh2.

[TW, Lemma 3.5.1] Let 𝔤 be a finite dimensional complex semisimple Lie algebra and let Uh𝔤 be the corresponding Drinfeld-Jimbo quantum group. etc etc ???? Let ν be a dominant integral weight so that the irreducible module L(ν) of highest weight ν is finite dimensional. Then (idqtrL(ν)) (21) acts on   L(μ)   by   ev2(μ+ρ) (sν) idL(μ).

Proof.
Let h1,...,hr be an orthonormal basis of 𝔥. By [Dr, §4] (see [LR, (2.13)]) there is an expression = e 1 2 hγ0 + bj+bj-, where γ0 = l=1r hlhl, and bj+U+ and bj-U- are homogeneous elements of degree greater than 0. Let v1,...,vn be a basis of weight vectors of L(ν) and let v1,...,vn be the dual basis in L(ν)*. Let vμ+ be a highest weight vector in L(μ). Since bj+ vμ+ =0 and q=e h 2 , 211 acts on vμ+vivi L(μ)L(ν)L(ν)* by (21id) (vμ+vivi) = qγ0 + j (bj-bj+) qγ0 + j (bj+bj-) (vμ+vi) vi = qγ0 qγ0 + qγ0 j (bj-bj+) (vμ+vi) vi = qγ0 qγ0 (vμ+vivi) + qγ0 j (bj-vμ+bj+vivi) (vμ+vi) vi Since (idqtrL(ν)) (21) is central in U, it acts on vμ+ by a scalar. Therefore, since bj- is a lowering operator, (idqtrL(ν)) (21) = (idqtrL(ν)) (q2γ0), as proved in [Dr, Prop. 5.3]. Then (idqtrL(ν)) (21) = (idqtrL(ν)) (q2γ0) = i q2γ0 (1ehρ) (vμ+vi) |vμ+vi = ivi q 2 l=1r μ(hl) wt(vi) (hl) q 2ρ,wt(vi) (vμ+vi) |vμ+vi = i q2 μ+ρ,wt(vi) = ev2(μ+ρ) (sν) = wW0 det(w) q2μ+ρ,w(ν+ρ) wW0 det(w) q2μ+ρ,wρ .

The Turaev-Wenzl identity almost provides an inverse to the Harish-Chandra homomorphism. In the case of U=Uh𝔰𝔩2, (idqtrL(ω1)) (21) = (idqtrL(ω1)) e h( 1 2 (HH)) = e h 2 (H+1) + e - h 2 (H+1) , and this element acts on L(μω1) by the constant ev2(μ+ρ) (Xω1+X-ω1) = qμ+1 + q-(μ+1) = evμ+ρ (eh2H + e-h2H) = evμ+ρ (K+K-1).

(idqtrL(ν)((21)l)) when U=Uh𝔤

Drinfeld [Dr, last par. of §4] explains the connection between the construction of central elements in (Ss 2.3) and the construction of central elements in [RTF, Theorem 14]. This is expanded by Baumann [Bau, 3rd par. of §3] as follows. Let U=Uh𝔤 and let π:UMn() be a representation of Uh𝔤 and let πij: U be the matrix coefficients of π. As in [RTF, Theorem 16(1) and Theorem 18] set L+ = (lij+) = (πid) (21) so that lij+ = (idπij) (). Let L- = (lij-) = (πid) (idS-1)() = (πid) -1) so that S(lij-) = (πijid) (). Then ehρ (L+S(L-))k is a matrix with entries in U and tr( ehρ ( L+S (L-) )k ) = i1,...,ik j1,...,jk q2ρ,λi1 li1j1+ S(lj1i2-) li1j2+ S(lj2i3-) likjk+ S(ljki1-) = i1,...,ik j1,...,jk q2ρ,λi1 (idπi1j1)() (πj1i2id)() (idπikjk)() (πjki1id)() = i1 q2ρ,λi1 (πi1i1id) ((21)k) = ?????   (idqtrπ) ((21)k).

Examples. Since γ0 = (21)0 = 11 the definition of dim(L(ν)) and dimq(L(ν)) give (idtrL(ν)) (γ0) = dim(L(ν)) and (idqtrL(ν)) ((21)0) = dimq(L(ν)). Since π0( (idtrL(ν)) (γ) ) is a degree 1 element of S(𝔥), S1(𝔥)W0 = 0 and π0: Z(U) σρ( S(𝔥)W0 ) is an isomorphism, it follows that (idtrL(ν)) (γ) = 0.

If f= ν𝔥* fνXν is an element of [𝔥*]W0 then fsμ = f aμ+ρ aρ = 1 aρ f wW0 det(w)w Xμ+ρ = 1 aρ wW0 det(w)wf Xμ+ρ = 1 aρ ν𝔥* wW0 det(w)w fνXν Xμ+ρ = 1 aρ ν𝔥* fν aν+μ+ρ = ν𝔥* fν sν+μ. (Ss 4.1) Two special cases of (Ss 4.1) are sγsμ = νP Kγν sν+μ, (Ss 4.2) and wW0 Xwλ = s wW0 Xwλ = wW0 swλ = wW0 det(w-1) sw-1wλ = wW0 det(w-1) sλ+w-1ρ-ρ. (Ss 4.3) Comparing coefficients of Xν in (Ss 4.3) gives δvλ,ν |(W0)λ| = wW0 Kwλ,ν = wW0 det(w) Kλ+wρ-ρ,ν (Ss 4.4) for some vW0.

Baumann's identity

Identify Rep with [X]W0 = span {sλ  |  λ   dominant integral }. For l0 let Ψl: [X]W0 Z(Uh𝔤) sν (idqtrL(ν)) ((21)l)

[Bau, Thm. 1] For l0 define mλ(l) = wW0 q2l wλ,ρ swλ. Then Ψl (mλ(l)) = Ψ1 (mlλ(1/l)).

Proof.
First note that qtrL(μ) (Ψl(sγ)) = (qtrL(μ)qtrL(γ)) ((21)l) = νP Kγν qtrL(μ+ν) ( ql ( μ+ν,μ+ν+2ρ - μ,μ+2ρ - γ,γ+2ρ ) ) = νP Kγν dimq(L(μ+ν)) ql ( μ+ν,μ+ν+2ρ - μ,μ+2ρ - γ,γ+2ρ ) = νP Kγν ev2ρ(aμ+ν+ρ) ev2ρ(aρ) ql ( μ+ν,μ+ν+2ρ - μ,μ+2ρ - γ,γ+2ρ ) where the second equality uses (Ss 4.2). Taking qtrL(μ) of the LHS of the Baumann identity is qtrL(μ) (Ψl (mλl)) = qtrL(μ) wW0 det(w) q2lλ,wρ Ψl (sλ+wρ-ρ) = wW0 det(w) q2lλ,wρ νP Kλ+wρ-ρ,ν ev2ρ(aμ+ν+ρ) ev2ρ(aρ) ql( μ+ν,μ+ν+2ρ - μ,μ+2ρ - λ+wρ-ρ,λ+wρ-ρ+2ρ ) = νP ev2ρ(aμ+ν+ρ) ev2ρ(aρ) wW0 det(w) Kλ+wρ-ρ,ν q2lλ,wρ ql( μ+ν,μ+ν+2ρ - μ,μ+2ρ - λ+wρ-ρ,λ+wρ-ρ+2ρ ) = νP ev2ρ(aμ+ν+ρ) ev2ρ(aρ) ql( ν,ν - λ,λ +2 ν,μ+ρ ) wW0 det(w) Kλ+wρ-ρ,ν = wW0 ev2ρ(aμ+wλ+ρ) ev2ρ(aρ) q2l wλ,μ+ρ , where the last equality uses (Ss 4.4). Taking qtrL(μ) of the RHS of the Baumann identity is qtrL(μ) ( Ψ1(mlλ(1/l)) ) = qtrL(μ) Ψ1 vW0 det(v) q2 λ,vρ slλ+vρ-ρ = vW0 det(v) q2 λ,vρ dimq (L(μ)) ev2(μ+ρ) (slλ+vρ-ρ), by Turaev-Wenzl. Thus qtrL(μ) (Ψl (mλl)) = vW0 det(v) q2 λ,vρ ev2ρ(aμ+ρ) ev2ρ(aρ) ev2(μ+ρ)(alλ+vρ) ev2(μ+ρ)(aρ) = 1 ev2ρ(aρ) vW0 det(v) q2 λ,vρ ev2(μ+ρ)(alλ+vρ) = 1 ev2ρ(aρ) vW0 q2 v-1λ,ρ ev2(μ+ρ)(alv-1λ+ρ) = wW0 q2 wlλ,μ+ρ ev2ρ (awλ+μ+ρ) ev2ρ (aρ) , since yW0 q2yλ,ρ ev2(μ+ρ) (alyλ+ρ) = ev2(μ+ρ) y,wW0 q2yλ,ρ det(w) Xlwyλ+wρ = y,wW0 q2wyλ,wρ det(w) q2μ+ρ,lwyλ q2μ+ρ,wρ = x,wW0 det(w) q2μ+ρ,lxλ q2xλ+μ+ρ,wρ = xW0 q2μ+ρ,xlλ ev2ρ (axλ+μ+ρ). Thus qtrL(μ) (Ψl(mλl)) = qtrL(μ) (Ψ1(mlλ(1/l))) which completes the proof of the Baumann identity (THOUGH NOT THE STATEMENT THAT THE B L(λ+wρ-ρ) (l) ARE CHARACTERIZED BY THIS IDENTITY.

It follows from (Ss 3.4)??? that the quantum dimension is dimq (L(ν)) = ev2ρ (sν) = ev2ρ ( aν+ρ aρ ) = wW det(w) q w(ν+ρ),2ρ wW det(w) q wρ,2ρ = wW det(w) q 2(ν+ρ),wρ wW det(w) q 2ρ,wρ = ev2(ν+ρ)(aρ) ev2ρ(aρ) = αR+ ev2(ν+ρ) (Xα/2 - X-α/2) ev2ρ (Xα/2 - X-α/2) = αR+ q ν+ρ,α - q -ν+ρ,α q ρ,α - q -ρ,α . Hence (idqtrL(ν)) ((21)0) = αR+ [ν+ρ,α] [ρ,α] , where [k] = qk-q-k q-q-1 , and the Weyl dimension formula is (idqtrL(ν)) (γ0) = dim(L(ν)) = ev0(sν) = limq1 ev2ρ(sν) = αR+ ν+ρ,α ρ,α .

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

[DRV] Z. Daugherty, A. Ram, R. Virk, Appendices to Affine and degenerate affine BMW algebras: Actions on tensor space.

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