Quantum Cohomology of G/P

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 16 December 2013

Lecture 6: February 26, 1997

(The following is the beginning of Lecture 4 given on Feb. 19).

Schubert Cells in G/P

Recall that a closed subgroup P of G is called a standard parabolic subgroup if PB.

Let PG be a standard parabolic subgroup. Then subset JI s.t. P=BWJB where WJ=rjjJ is the subgroup of W by {rj:jJ}. Set WP = WJ WP = { uW:u<uvfor all vWP,vid } . Thus WP is the set of minimum representatives of the coset space W/WP. We have G/P=wWP BwP ????? Bwp(w) ????? is called the Schubert Cell corresponding to w.

Each Ewp is T-stable and G/P=wWP BwP takes G/P into a CW-complex.

????? XwP= closure ofBwPinG/P ????? is a complex projective variety called the Schubert variety ????? have XwP= vWPvw BvP ????? wWP, let iwP: XwP G/P ????? [XwP]H2(w)(XwP,Z). Set σwP= (iwP)* [XwP] H2(w) (G/P).

Schubert Basis for H(G/P,) and H(G/P,)

Fact: {σwP:wW} is a basis for H(G/P,).

Notation: The dual basis of H(G/P,) dual to {σwP:wW} is denoted by {σPw:wW} .

Remark: Hodd(G/P)=0.

(Here starts Lecture 6)

Schubert Basis for HomS(HT(G/P),S) and HT(G/P):

Definition: For wWP, put σ(w)P= (iwP)* [XwP] HomS(HT(G/P),S). Then {σ(w)P:wW} is a basis for HomS(HT(G/P),S). There is then a unique basis { σP(w):wW } of HT(G/P) (over S) s.t. σP(v),σ(w)P =δv,w. Bot {σ(w)P} and {σP(w)} are called Schubert basis.

????? basis {σP(w):wW} of HT(G/P) is characterized by ????? properties:

(1) deg(σP(w))=2(w)
(2) Under evaluation at 0: SHT(G/P) H*(G/P) we have σP(w) σPw.
(3) (iwP:XwPG/P)*(σP(v))=0 if vw.

?????, we look at

Another set of elements {ψwP:wWP} in HomS(HT(G/P),S):

For wW, consider the T-equivariant map jwP: pt G/P: pt wP set ψwP= (jwP)* HomS(HT(G/P),S). Of course ψwP=ψw1P if ww1Wp. We think of ψwP as localizing at the T-fixed pt wP.

Warning: {ψwP:wWP} is NOT an S-basis for HomS(HT(G/P)) because σ(ri)P= 1αi ψidP-1αi ψriP.

Remark: Expressing ψwP as a linear combination over S of the σ(v)P we get the D-matrix in Kostant-Kumar. Will do this later.

Properties: Consider the G-equivariant map πP: G/B G/P: gB gP. Then (πP)* ψwB = ψwP wW (πP)* ψP(w) = ψB(w) wWP.

Action of A_ on HomS1(HT(G/P),S) in the basis {σ(w)P:wWP}

Proposition 1: Ai·σ(w)P= { σ(riw)P ifw<riw andriwWP 0 otherwise


Let iwP: XwP G/P. Recall that σ(w)P= (iwP)* [XwP] From the fact stated at the end of last lecture, Ai·σ(w)P= μ*σi*[XwP] HomS(G/P,S) where μ: Ki×TXwP G/P: (ki,x) kix. It follows that (?) Ai·σ(w)P= { σ(riw)P ifw<riwand riwWP, 0 otherwise.

?????ave: for v,wW, v·ψwP= ψvwP

Action of A_ on HT(G/P) in the basis {σP(w):wWP}

Proposition 2: For vW, wWP, Av·σP(w)= { ϵ(v)σP(vw) if(v-1)+ (vw)=(w) vw?????, 0 otherwise.


Let's first check that Ai·σP(w)= { -σP(riw) if1+(riw) =(w)(ie. riw?????, 0 otherwise. From the previous Proposition 1, if riw<w ( ri????? Ai·σ(riw)P =σ(w)P. But (Aif)(z)= Ai·f(z)-ri ·f(Ai·z) zHT(G/P) by definition, so by letting f=σ(riw)P and zσP(v), we get δw,v=0-ri· σ(riw)P (Ai·σP(v)) or ( Ai·σP(v), σ(riw)P ) =-δw,v Ai· σP(w)=- σP(riw) otherwise follows.

Remark: Recall that ε=ψidB= σ(id)B HomS(HT(G/B),S). We can identify A_HomS (HT(G/B),S) (*) by a fa: fa(z) = ε(a·z). Then this is an identification of S-modules, and from Proposition 2, fAW=ϵ(w) σ(w-1)B ie. Aw ϵ(w)σ(w-1)B Thus by Proposition 1, we see that under the identification (*), the (left) A_-action on HomS(HT(G/B),S) becomes the (left) action A_ on A_ by a·b=b(*a) where, recall from lecture 2, that *S = S, *w = w-1, *Aw = ϵ(w) Aw-1. (The * in Lecture 2 is defined to be *S=S, *w=ϵ(w)w-1 and *Aw=Aw-1).

The ring A_ˆ of characteristic operators again

Set ε = ψidB HomS(HT(G/B),S) = σB(id). So ε(σB(w))= δw,idwW.


(1) Every characteristic operator aAˆ_ can be uniquely written as a=wW swAwsw S. In fact, sw=ε (a·(ϵ(w)σB(w-1))) (Recall ϵ(w)=(-1)(w)).
(2) a is compactly supported iff only finitely many w's occur in the sum. (ie. at f only finitely many sw's are ?????


(1) For any aAˆ, write a=a-wWε (a·(ϵ(w)σB(w-1))) Aw Then aAˆ_. Thus to show a=0 it is enough to show that ε(a·z)=0 for any zHT(G/B). (See Lecture 5). Since both a and ε are S-linear, it is enough to show that ε(a·σB(v))=0 for all vW. Now a-σB(v) = a·σB(v)- wWε (a·ϵ(w)σB(w-1)) Aw·σB(v) = a·σB(v)- wW(w-1)+(wv)=(v)ε (a·ϵ(w)σB(w-1)) ϵ(w)σB(wv) = a·σB(v)- wW(w-1)+(wv)=(v)ε (a·σB(w-1)) σB(v) * ΔσB(v) = u,wW uw=v (v)=(u)+(w) σB(u) σB(w) a = ((εa)id) ΔX ?????e Corollary 3 in Lecture 5). a·σB(v) = 0 a 0.

Uniqueness is clear.

If a has compact support, since any compact subset of K is contained in some Kw where Kw=Ki1Ki2Kir if w=ri1ri2rir [red], we see that there are only finitely many w's involved in the expression a=wWswAw.

Note: The following remark was crossed out in the scanned notes.

Remark: We can think of A_ as HomS(HT(K/T),S), or the ????? of HT(K/T) via the pairing: (a,z) =def ε(a·z). Let's check then that A_ action on HomS(HT(K/T),S) becomes the A_-action on A_ by left multiplications: For aA_, use faHomS(HT(K/T),S) to denote the element given by fa(z)= (a,z)=ε (a·z). For iI. we want to check Ai·fa = fAia.

The Hopf Algebroid Structure on HT(K/T)

Recall from Lecture 5 that HT(K/T) is a Hopf algebroid over S. We now express the structure maps for this Hopf algebroid in the basis {σB(w):wW}.

First, recall that we have ring homomorphisms πL: S HT(K/T) πR: S HT(K/T). This gives two S-module structures on HT(K/T). The map πL is nothing but the characteristic homomorphism ch in Lecture 5. The map πR is a little more mysterious. It gives the 2nd S-module structure on HT(K/T) in Lecture 4.

Proposition: The elements {σB(w):wW} is also a basis for the second S-module on HT(K/T) defined by πR.

Remark: I (Lu) suspect that πR has a lot to do with the Bruhat-Poisson structure on K/T.

The next theorem expresses the structure maps for the Hopf algebroid structure on HT(K/T) in the basis {σB(w):wW}.

Theorem: (Recall notation from Lecture 5):

1) For λh*, πR(λ)= πL(λ)+ iI λ,αi σB(ri)
2) ε(σB(w))=δw,id
3) ΔσB(w)= u,vWw=uv[red] σB(u) σB(v) (w=uv [red] means w=uv ?????
4) c(σB(w))= ϵ(w)σB(w-1)
5) For any K-space X and σHT(X) ΔX(σ)= wWϵ(w) σB(w-1) (Aw·σ) HT(K/T)S HT(X)


We first prove 5). 5) is due to the general fact if an algebra A acts on a space M, then using a basis a1,an, of A and the dual basis ξ1,ξn, of A*, the co-module map is nothing but ΔM: M A*M: ΔM(m) = iξiai·m ????? in our example, we are identifying HT(K/T) with A_* ????? the pairing (a,z)=ε(a·z) aA_,z HT(K/T). ????? this pairing; we have {Aw:wW} as a basis for A_ ????? dual basis in HT(K/T) is {ϵ(w)σB(w-1):wW} (see page 6-8). Thus for any σHT(X) ΔX(σ)= wWϵ(w) σB(w-1) (Aw·σ) Peterson gave the following proof in class:

Since {ϵ(w)σB(w-1):wW} is a basis for HT(K/T), we know ΔX(σ)= wWϵ(w) σB(w-1) ϕw for some ϕwHT(X) for each wW. Need to show ϕw=Aw·σ. To do this, let vW and calculate Av·σ. We have Av·σ = (εid)ΔX (Av·σ) = (εAvid) ΔX(σ) (see Lecture 5, Corollary 1) = wWε (Av·ϵ(w)σB(w-1)) ϕw = ε(Av·ϵ(v)σB(w-1)) ϕv = ϕv. This finishes the proof of 5)

Remark: What is quoted as Corollary 1 in Lecture 5 is the fact that the action of A_ on HT(X) is obtained by the comodule map ΔX: HT(X) HT(K/T)SHT(X) by a·σ=(a,σ(1)) σ(2)ifΔX σ=σ(1)σ(2) and (a,z)=ε(a·z) is the pairing between HT(K/T) and A_. This is just like in the Hopf algebra case.

We now prove 3). This is just a special case of 5) for X=K/T. Indeed, ????? 5), we get ΔσB(w)= u1Wϵ (u1)σB(u1-1) Au1·σB(w). But Au1·σB(2)= { ϵ(u1)σB(u1w) if(u1-1) +(u1w)= (w), 0 otherwise. ????? ΔσB(w) = u1Ww=u1-1·(u1w)[red]ϵ (u1)σB(u1-1) ϵ(u1) σB(u1w) = u=u1-1w v=u1wW w=uv[red] σB(u) σB(v). This finishes the proof of 3).

2) is is clear from definition since ε=σ(id)B.

It remains to prove 1) and 4).

To prove 1), we need the following Lemma:

Lemma: For any σHT(K/T), σ=wWπR (ε(Aw·σ)) ϵ(w)σB(w-1).


Write σ=wWπR (sw)ϵ(w) σB(w-1) for some swS for each wW. Using επR=idS and (Av,ϵ(w)σB(w-1)) (=ε(Av·ε(w)σB(w-1))) =δv,w we get ε(Av·σ) || (Av,σ) = wWε πR(sw) (Av·ϵ(w)σB(w-1)) =επR(sv) = sv σ = wWπR(ε(Av·σ))ϵ(w)σB(w-1).

This proves the Lemma.

Remark: In proving the Lemma, we used the fact that S-valued ????? the pairing () between A_ and HT(K/T) defined by (a,σ) = ε(a·σ) satisfies (πR(s)a,σ)= ε(πR(s)) (a,σ)=s(a,σ) and ε(πR(s))=s sS. It says that ε:HT(K/T)S is not only an S-map for the first S-module structure on HT(K/T), (defined by πL) but also for the 2nd S-module structure HT(K/T) defined by πR.

Is this really true? Recall that πR:SHT(K/T) is the pullback of the map (Eu×K)/(T×T) E/T [e,k] e·k. It is not clear why ε:HT(K/T)S is πR(S)-linear.

Now we prove 1): By Lemma πL(λ)=wW πR(ε(Aw·πL(λ))) ϵ(w)σB(w-1) But Aw·πL(λ)= πL(Aw·λ)= { πL(λ) w=id, λ,αi w=ri, 0 otherwise πL(λ) = πR(ε(πL(λ))) +iIπR (ε(λ,αi)) (-1)σB(ri) = πR(λ)- iI λ,αi σB(ri) πR(λ) = πL(λ)+ iI λ,αi σB(ri).

Remark: This is an interesting formula. Understand what this says for Kostant's Harmonic form sri later.

It remains to prove 4), ie. c(σB(w))= ϵ(w) σB(w-1).

The following is the proof given by Peterson. It is kind of strange.

W>e first prove that c(σB(w))= ±σB(w-1) We'll determine the sign later. For wW, let Ew(2)= {(e,ek):eEu,kKw}. Then H*(Eu(2)/T×T) HT(XwB). (Why? This is saying that we do not distinguish XwB and its Bott-Samelson resolution?)

Recall that t: E(2) E(2): (e1,e2) (e2,e1). So t(E2(2)) = Ew-1(2). But Rw= { σHT(E(2)/T×T) :degσ=2(w)and σ|Ev(2)/T×T =0forvWs.t.vw } . We know Rw = σB(w) c(Rw) = Rw-1 c(σB(w)) = ±σB(w-1). Now show that c(σB(w))=ϵ(w)σB(w-1).

w=id OK.

w=ri OK.

For (w)2, assume sign =·ϵ(v) for (v)<(w). Since (cc)·TΔ= Δ·c where T(σσ)= σσ we get, from Δ(σB(w))= w=uv[red] σB(u) σB(v) that Δ(cσB(w)) = vu?????[red] c(σB(v)) c(σB(u)) = c(σB(w))1+ 1c(σB(w))+ w=uv[red]u1v1 ϵ(u)ϵ(v) σB(v-1) σB(u-1). But Δ(ϵ(w)σB(w))= ϵ(w)σB(w-1)1+ 1ϵ(w)σB(w-1)+ same sum0 must have Δ(σB(w))= ϵ(w) σB(w-1). This proves 4).

This completes the proof of the theorem.

?????rable A_-modules ( actions of 𝒰=SpecHT(K/T))

Definition: Let X be an affine scheme over h_=SpecS with structure homomorphism πX:S𝒪(X). An A_-module structure on 𝒪(X) is said to be integrable if for all sS and p𝒪(X).

1) s·p=πX(s)p
2) πX:s𝒪(X) and m:𝒪(X)S𝒪(X)𝒪(X)
are both A_-module maps
3) For each p𝒪(X), Aw·p=0 for all but finitely many wW.

Example: 𝒰 as a scheme over h_=SpecS with structure homomorphism πL (?) Is this an example? Maybe not, because HT(K/T)SHT(K/T) we use πR to define the S-module structure on the first copy of HT(K/T). (OK. Because in the multiplication of HT(K/T)SHT(K/T), even the S-structure on the first copy is defined by πL). IntegrableA_-module structure on 𝒪(X) actionϕ: 𝒰×h_XX. One way:

If ϕ:𝒰×h_XX is an action, have ϕ*: 𝒪(X) HT(K/T)𝒪(X) Then for aA_, define pP a·p=m· ( πX(ε(a·p(1))) p(2) ) ifϕ*p= p(1)p(2). The other way, given A_-action on 𝒪(X), define ϕ*(p)=wW c(σG/B(w)) (Aw·p) This is the map giving the action ϕ: 𝒰×h_X X.

Next, we look at the 2nd action of A_ on HT(K/T).

Notation: The action of A_ on HT(K/T) that we have been talking about way along will from now on be denoted by aL·. the second action that we will now introduce now will be denoted by aR·.

The second action of A_ on HT(K/T)

Define a second action of A_ on HT(K/T) by aR·= c·(aL·)c


1) aLbR= bRaL a,bA_
2) ΔaL=(aLid)Δ
3) for sS, aA_ and zHT(K/T)
sL·z = πL(s)z sR·z = πR(s)z and aL·πL(s) = πL(a·s) aR·πR(s) = πR(a·s) a·ε(z) = ε(a(1)La(2)R·z) ifΔa= a(1)a(2) w ·ε(z) = ε(wLwR·z)
Thus, in the basis {σB(w):wW}.

Lecture 7: March 4, 1997

Recall formulas from last time: AvR· σB(w)= { σ(wv-1) if(wv-1) +(v)=(w), 0 otherwise. For any aAˆ_ a=wWε (aR·σB(w)) Aw ????? zHT(K/T) z=wWπL (ε(AwR·z)) σB(w) ????? εAwR = σ(w)B εwR = ψwB Given wW, du,wS????? of degree (u) for each uw s.t. w=uwdu,w Au moreover dw,w= αΔ+redw-1α<0 (-α)=ϵ(w) αΔ+redw-1α<0α


Induction on (w): (w)=0 w=id id=id. (w)=1 w=ri ri=1-αiAi OK. Assume w=riw1>w1. Assume w1=uw1 du,w1Au du,w1 S(u) (h*) Then w = riw1= (1-αiAi) uw1 du,w1 Au = uw1 du,w1 Au- uw1 αi(Aidu,w1) Au Since Aidu,w1= (ri·du,w1) Ai+Ai· du,w1 w = uw1 du,w1 Au-uw1 αi(ri·ru,w1) AiAu+αi (Ai·du,w1) = uw1 ( du,w1-αiAi ·du,w1 ) Au-uw1 αi(ri·du,w1) AiAu = uw1 (ri·du,w1) Au- uw1riu>u αi(ri·du,w1) Ariu du,w = ri·du,w1 ifuw1 dri·u,w = -αi(ri·du,w1) ifuw1,riu>u shows that du,wS(u)(h*) for any uw.

Moreover, driw1,w= -αi(ri·dw1,w1). Assume dw1,w1=ϵ (w1) αΔ+redw1-1α<0 α Then dw,w = -α1 (ri·dw1,w1) = ϵ(w) αΔ+redw?????-1α<0α

Remark: Sara Billey's formula gives an express for each du,w. Will come back to this later.


1) ψwB=uwdu,wσ(w)B
2) wWkerψwB=0.
3) HT(K/T) is reduced, ie. the only nilpotent elemtn
4) HT(G/P)(HT(K/T))(WP)R is also reduced.


1) follows from εAwR = σ(w)B εwR = ψ(w)B 2) If zwWkerψwB then ψwB(z)=0 w. Since the matrix D=(du,w) is upper-triangular it is invertible σ(w)B(z)=0. But {σ(w)B} is a basis for HomS(HT(K/T),S) z=0. If zHT(K/T) is s.t. zm=0 for some m1 then for each wW ε(wR·zm)=0 But wR·zm= (wR·z)m ε((wR·z)m) = 0 (ε(wn·z))m = 0 ε(wR·z) = 0 ie. zkerψwB=0 w z=0. Clear.

Proposition: The action aR· of A_ on HT(K/T) descends to an action on H(K/T) via the map SHT(K/T) H(K/T) where the S-module structure on HT(K/T) is defined by πL.


This is because the S action defined by πL commutes with aR· for any aA_.

Remark: The incuded action of AwA_ on H(K/T) is by the BGG-operators.

?????ne Constants" for the multiplication on HT(K/T)

For u,v,wW, define awu,vS by ΔAw= u,vW awu,v AuAv (Δ cocommutative awuv=awvu)

Proposition: σB(u) σB(v)= wW πL (awu,v) σB(w)


We know that σB(u) σB(v)= wWπL (ε(AwR·σB(u)σB(v))) σB(w) Then AwR· (σB(u)σB(v))= u,vW πR (awu,v) (AuR·σB(u)) (AvR·σB(v)) and ε (AwR·(σB(u)σB(v))) = u,vW awu,vε (AuR·σB(u)) ε(AvR·σB(v)) = u,vW awu,vε (σ(u)B,σB(u)) (σ(v)B,σB(v)) = u,vW awu,vε δu,u δv,v = awu,v σB(u) σB(v)= wW πL (awu,v) σB(w).

Special properties of the awu,v's:

(1) awu,v=0 unless uw, vw


This is seen from the definition: ΔAi = Ai1+riAi = Ai1+ (1-αiAi)Ai = 1Ai+Ai1- AiαiAi = 1Ai+Ai1- AiαiAi ΔAiAj = (1Ai+Ai1-AiαiAi) (1Aj+Aj1-AjαjAj) = 1AiAj+ AjAi+ AiAj+ AiAj1 -AiαiAiAj -AiAjαiAi -AiAjαiAiαjAj -AjAiαjAj -AiAjαjAj +AiAjαiAiαjAj so clear from induction on (w).

Proposition: awu,v is a homogeneous polynomial of degree (u)+(v)- (w)inS.


degσB(u) σB(v) = deg(πLawu,vσB(w)) 2(u)+2(v) = 2(degawu,vinS) +2(w). deg(awu,vinS) = (u)+(v)-(w).

Proposition: For w,vW, vw dv,w=awv,w where, recall dv,wS are defined by w=vw dv,wAv ?????e w=vw awv,wAv.


Write ww=u1,u2w Swu1,u2 Au1Au2. wε (wR·σB(w)) = u1,u2w Swu1,u2 au1ε (Au2R·σB(w)) = u1,u2w Swu1,u2 Au1δu2,w = u1w Swu1,w Au1. But ε(wR·σB(w)) =dw,w. dw,ww = u1w Swu1,w Au1 Swu1w = dw,wdu1,w On the other hand, w = du,wAu. ww = u1,u2 (vdv,wavu1,u2) Au1Au2 Swu1u2 = vdv,w avu1,u2 Swu1,w = vdv,w avu1,w= dw,w awu1,w. By Su1w=dw,wdu1,w and dw,w0 get du1,w= awu1,w

(Very strange proof).

Proposition: For wW, wuv[red] ϵ(u)σB(u-1) σB(v)= δw,id (1) wuv[red] σB(u) ϵ(v) σB(v)= δw,id (2)


(1) m (cid)Δ =ε, (2) m (idc)Δ =ε.

Remark: This will also be true for quantum cohomology.

Remark: Fix e0Eu. Define i: K/T Eu/T: kT e0kT. Then i×i: K/T×K/T Eu(2)/T×T. Consequently, (i×i)*: HT(K/T) HT(K/T)HT(K/T). We have (i×i)* σB(w) = wuv[red] ϵ(u) σβu-1 σBv.

The Finite Case

Proposition: In the finite case, we have A_L = EndA_R (HT(K/T)) A_R = EndA_L (HT(K/T)). HT(K/T) is a free A_L (as well as A_R) module with one generator σB(w0), where w0 is the longest element in W. If ϕEndA_L(HT(K/T)) then aA_ s.t. ϕ(σB(w0))= aR·σB(w0).

Claim: zHT(K/T), ϕ(z)=aR·z.


For any zHT(K/T), bA_ s.t. z=bL·σB(w0) ϕ(z) = ϕ(bL·σB(w0)) = bL·ϕ(σB(w0)) (ϕEndA_L) = bL·aR· σB(w0) = aR·bL· σB(w0) = aR·z.

The space HT(K) with K acting on K by conjugations

Consider now K as a K-space by conjugations. The map p: K K/T is T-equivariant (but not K-equivariant). Thus p*: H(K/T) HT(K) is an S-module map: p*(πL(s)z) =π(s)p*(z) where π = [kpt]*: S HT(K). Now A_ acts on both HT(K) and HT(K/T) by characteristic operations. But since p is not a K-map, p* does not intertwine the A_-actions on HT(K) and on HT(K/T). We have, nevertheless, the following:

Proposition: For aA_ with Δa=a(1)a(2), and for all zHT(K/T) a·p*(z)= p*(a(1)La(2)R·z) In particular, for sS and wW π(s)p*(z) = p*(πL(s)z) =p*(πR(s)z) w·p*(z) = p*(wLwR·z) Aw·p*(z) = p* ( uwvw πL(awuv) AuLAvR·z )

Proposition: For any K-space X with action map μX: K×X X the pullback μX*: HT(X) HT(K×X) is the composition HT(X) ΔX HT(K/T)SHT(X) p*id HT(K)SHT(X) HT(K×X).

The Pontrayagin action of the ring H*(K):

μK: K×K K: (k1,k2) k1k2 gives a map μK*: H*(K)H*(K) H*(K). This defines a ring structure on H*(K). Now for any K-space X with μX: K×X X get μX*: H*(K)H*(X) H*(X) which defines an action of H*(K) on H*(X).

????? at the special case X=K/T with μX = μK/T: K×K/T K/T. A_R acts on H*(K/T), and this action commutes with the Pontryagin action of H*(K) on H*(K/T).

Define a ring structure on H*(K/T) by σvσw= { σvw if(v)+ (w)= (vw), 0 otherwise. Then μK/T* ( σ H*(K) × σ H*(K/T) ) = p*(σ)σ. Consequently, p*: H*(K) H*(K/T) is a ring homomorphism.

Theorem (Peterson-Kac): Over any field 𝔽

1) p*(H*(K/T),𝔽) is a Hopf subalgebra of H*(K𝔽).
2) p*(H*(K/T),𝔽)= H*(K/T,𝔽)S= {σ:λσ=0λh*}.
3) If mij= for all ij, then p*(H*(K/T),) the dual of a tensor algebra as a Hopf algebra.

Poincare Duality in the finite case

Define A_-module homomorphism PD: HT(G/P) HomS(HT(G/P),S), PD(z)(y)= [G/P]yzS Consider the case P=B: [G/B]=ε Aw0R. In general, [G/P] σP(w)= δw,w0wP where wP is the longest element in WP, so w0wP is the longest element in WP.

Recall that (from Lecture 2) ΔAw0= wWAw w0Aw0w PD(σP(w))= w0· σ(w0wwP)P Also w0Lw0R· σB(w)=ϵ (w)σB(w0ww0). It follows that PD is an S-module isomorphism.

The Euler Class:

For zHT(G/P), consider the operator Mz on HT(G/P) by yzy. The Euler Class χG/PHT(G/P) is defined by the property: traceMz= [G/P] χG/P·z.

Proposition: χG/P= wWP σP(w) [w0·σP(w0wwP)].


By the definition of trace and using the "dual basis" {σ(w)P} of {σP(w)}, we have Mz = wWP ( σ(w)P,z σP(w) ) σ(w)P = PD(w0·σP(w0wwP)) ????? ?????Mz = wWP ( PD(w0·σP(w0wwP)) ,zσP(w) ) = wWP [G/P]z σP(w) (w0·σP(w0wWP)) ????? χG/P=wWP σP(w) (w0·σP(w0wwP)).

We will use PD to denote its inverse as well.

Lemma: For v,wWP σP(v)PD (σ(w)P)=0 unless vw.

So χG/P is the trace of a rank 1 upper triangular matrix. Also σP(w)PD (σ(w)P)= w·PD(σ(id)P).


1) χG/P has image α>0w0wP>0αR·1 in HT(K/T)
2) χG/P is W-invariant under the left action
3) Image of χG/P in H*(G/P) is |WP|σPw0wP

Some facts on the classifcying spaces

H*(BT) πR HT(G/B) H*(BT)wP HT(G/B)(wP)R HT(G/P) ????? , we have H*(BT)wP H*(BKP) HT(G/P). In fact

1) H*(BKP,) (HT(G/P))W (?)
2) SH*(BK) H*(BKP) HT(G/P) ZSHT(G/P) H*(G/P)

Open Problems

(1) In what sense does the diagonal map K K×K: k (k,k) correspond to the co-product Δ: A_ A_SA_ Given homomorphism K1K2 with T1T2, N1N2 can easily calculate HT2(K2/T2) HT1(K1/T1)
(2) Conjecture: For each u,v,wW, the ϵ(uvw)awu,v is a polynomial in the αj's iI with +-coefficient.

True for:
(1) (u)+(v)=(w) - Kumar
(2) v=w or u=w - Sara Billey.
(3) Similar models for K-theory (done?). Cobordism: HT(G/P) KT(G/P). BGG-operators Demazure operators
(4) Find combinatorial interpretation of the coefficients of ϵ(uvw)awu,v
(5) Find combinatorial interpretation of the structure constants of HS1(Grass(k,n)) with S1 acting by exp(tρ).
(6) Prove Little-Richardson Rule for σ where σ is a diagram automorphism of f dim G and σ is admissible, re. ασk(i),αi0σk(i)=1. (In this case Gσ has the structure of a Kac-Moody group. λh* σ(λ)=λ λ minuscule αΔ+ 0λ,α1 H*(G/Pλ)H*(Gσ/(GσP)) ?)
(7) Study more of the Bruhat Graph (G/B)T W Vertices: W, edges wwrα α>0 T-stable curves (P) in G/P

Full subgraphs correspond to XwP with vertices vw. vvrα iff v,vrαw.
(8) Theorem (Carrell-Peterson): The Kazdan-Lusztig Polynomial Pv,w=1 for this graph, have the same # of edges emanate from each point.
(9) Study directed Bruhat graphs: wαwrα ifw<wrα.

Lecture 8: March 11, 1997

Recall picture for the next two lectures

Let K: compact simple Lie group
ΩK: base preserving algebraic loops in K
Then TK acts on ΩK by conjugation: (t·k)(z)=tk (z)t-1. Roughly, the diagonal embedding ΩK ΩK×ΩK gives a co-product HT(ΩK) HT(ΩK)SHT(ΩK) and the multiplication map for the group structure on ΩK: ΩK×ΩK ΩK gives a product HT(ΩK)SHT(ΩK) HT(ΩK) In fact, HT(ΩK) is a commutative and cocommutative Hopf algebra over S. We will identify this Hopf algebra structure using A_af. In fact, we have a map ΩK Gaf/Baf which gives HT(ΩK) HT(Gaf/Baf) = A_af. Under this, we will identify HT(ΩK) Zaf(S) (centralizer ofSin A_af) and describe Zaf(S) using the affine Weyl group Waf.

Notation: For a variety X over , use X=Mor(×,X). Let G be a finite dimensional connected simple algebraic group over . We then have the finite root datum I, αi,h, αih. Δ+, Π, W, g_, h_, b_, Let θ be the highest root. From these we form the following Kac-Moody root datum:

Corresponding to this root datum, we have the following Kac-Moody Lie algebra 𝔤_af: 𝔤_af = 𝔤_ [t,t-1]= 𝔤_ ei = ei1 fi = fi1 e0 = e-θt f0 = fθt-1 [e0,f0] = [e-θ1,eθ1]=-????? Roots are in Qaf. They are all those in Qaf of the form α+nδn,αΔ, orα=0. The root spaces are (𝔤_af)α+nδ= { 𝔤αtn ifαΔ,n, h_tn α=0,n so Δre= { αnδ:αΔ,n } and all nδ's n, are "imaginary roots". They have multiplicity =dimh_.

The positive roots are (Δaf)+ = { α+nδ:n>0or n0αΔ+ } , (Δaf)+re = { α+nδ:n0αΔ+ } .

The affine Weyl group Waf:

By definition, Waf=WΓ the semi-direct product, where ΓQ with Q Γ: h th. wthw-1 = tw·h thth = th+h The reason why this is the same as the group generated by the reflections r0,ri, iI is because tθ= r0rθ For wW, w·(α+nδ) = w·α+nδ (wδ=δ) th· (α+nδ) = α+nδ- α,hδ (soth·α=α- α,hδ, th·(nδ)= nδ-α,h δ).

The Kac-Moody group: Gaf=G=Mor (×,G) (Laurent series int) set P0 = Mor(,G) (power series int) Baf = {gMor(,G):g(0)B} P0 Uaf+ = {gMor(,G):g(0)U+} Kaf = {gGaf:g(S1)K} ΩK = {kKaf:k(1)=id} Taf = T G const. loopsGaf Kaf acts on ΩK by k·k=kkk (1)-1 Then iΩ: ΩK Gaf/P0: k k·* *=P0 is a Kaf-equivariant map. This map is also a home????? because Gaf = KafBaf KafBaf=T = (ΩK)K Baf=(ΩK) P0

The compact involution on Gaf:

(wKaf)(g) (t)=wK (g(t-1)) gMor(×,G) =Gaf where wk:GG is the compact involution on G corresponding to K.

The normalizer Naf of Haf=H in Gaf is Naf=N=Mor (×,N) where, recall, N is the normalizer of H in G, so also have Naf = semi-direct product ofNandΩT ΩT = {gΩK:g(S1)T} so gΩT must be a homomorphism from S1 to T. Thus ΩTΓQ where Q ΩT: h hˆ: hˆ(z)=zh,z×. This way we also see WΓ Waf: (w,th) (W,hˆ-1H) Naf/H.

The nil-Hecke rings A_ and A_af:

Since (h)af=h, we have Saf=S. Let A_ be the nil-Hecke ring defined by W. Let A_af be the nil-Hecke ring defined by Waf. Then we have the embedding A_ A_af: s s Ai Ai, iI,(i0). Recall that if β=waiϵΔre with iI, then we define Aβ = wAαiw-1= wAiw-1 rβ = 1-βAβ (It is not obvious how to write Aβ in terms of the Ai's).

Define a ring homomorphism ev: A_af A_: ev|S = id ev(Aβ) = Aβ ev(wth) = w where if β=α+n, β=α. This is well-defined.

The embedding A_A_af is a section of ev.

Now identify ΩK iΩ Gaf/P0 we see that ΩK is a Kac-Moody G/P, so we have all we discussed before, namely:

In the Schubert basis Ax·σ(y)Ω= { σ(xy)Ω ifxyWaf-, (xy)=(x)+ (y), 0 otherwise. Define HT(ΩK)=S-span of {σ(x)Ω:x?????} HomS(HT(ΩK),S) In our special case at hand, not only do we have Gaf/BafGaf/????? but also: ΩKGaf/Baf. Thus have HT(ΩK) A_af. Next time, write the images of σ(x)Ω, for xWaf-, in A_af under the above embedding and identify HT(ΩK) as a subalgebra of A_af.

About Waf- and Waf/W

Recall that Waf-=WafP0 is the set of minimal representatives of the coset space Waf/WP0=Waf/W. (WP0=W). xWaf- x<xriiI, (i0), x·αi>0iI. Write x=wt-h. Then x·αi = w·t-h·αi = w· (αi+h,αiδ) = wαi+ h,αiδ xWaf- wαi+h,αi δ>0iI h,αi0 and whenh,αi =0must havewαi>0 his dominant and when h,αi=0 must havew<wri. Now for h dominant, set Wh = the subgroup ofWgenerated by ri:h,αi=0 = {wW:wh=h}. Set Ph=BWhBB parabolic. Then Wh=WPh. Let Wh=WPh be the set of minimal representatives of the coset space W/Wh, ie. wWh w<wriri Wh so wWh For eachiwith h,αi=0 havew<wri. Thus we have proved Waf- = { wt-h:hdominant (ie. h,αi0 iIandwWh } = { wt-h:hdominant and if h,αi=0 foriImust have wαi>0 } . The map Waf- Waf/W wt-h wt-h/W is of course a bijection.

Now another model for Waf/W is ΓQ: Γ Waf/W th th/W. In other words, each coset Waf/W has a unique translation element t-h in it, namely wt-h/W=w t-hw-1/W =t-w·h/W.


(1) each coset in Waf/W has a unique minimal representative.
(2) each coset in Waf/W has a unique translation element as a representative.
(3) Let xWaf-. Then x is the minimal representative for the coset xW. We know that x must be of the form x=wt-h where h is dominant and wWh. The translation element in this coset is t-w-h, so wt-ht-w·h.
(4) When h is dominant and regular, we have wt-hWaf- for all wW. So for different w1,w2W, the two elements w1t-h and w2t-h lie in two different cosets in Waf/W.
(5) A special case is when xWaf-Γ. This is the case iff the minimal representative for xW, namely x itself, coincides with the translational representative of xW. Write x=wt-h where h is dominant and wWh. Then x=t-w·h wt-h=t-w·h w=1 so Waf-P = {t-h:his dominant}.
(6) Let's now calculate the length (t-h) when h is dominant. Recall that α+nδ>0 either n>0 or n=0,α>0. Now we need to see for α+nδ>0, when do we have t-h·(α+nδ)<0. Now t-h·(α+nδ) =α+(n+h,α) δ If n>0,α<0, then t-h·(α+nδ)<0 for n=0,1,,h,α-1.
If n>0,α=0, then t-h·(α+nδ)nδ<0.
If n>0,α>0, then t-h·(α+nδ)<0.
If n=0,α>0, then t-h·(α+nδ)<0.
Then the only case when α+nδ>- and t-h·(α+nδ)<0 is when α=-β<0 (so β>0) n=0,1,,h,β-1 The number of such element is β>0h,β=h,2ρ. Hence (t-h)= h,2ρ= β>0 h,β for h dominant. Let's notice that the sum of all {α+nδ>0;t-h(α+nδ)<0} = β>0 ( -β-β+δ+ (-β+2δ) ++ (-β+(h,β-1)δ) ) = β>0 ( -h,ββ +12h,β (h,β-1) δ ) .
(7) For any x=wt-hWaf-, t=t-h1Γ-=Waf-Γ we have xt=wt-(h+h1)Waf- and (xt)=(x)+(t).
(8) Can prove that for x=wt-hWaf-, α+nδ>0 be st. x·(α+nδ)=wα+(n+h,α)δ<0 either α<0,wα>0 and n=1,2,,-α,h-1 or α<0,wα<0 and n=1,2,,-α,h. In other words { α+nδ>0:wt-h ·(α+nδ)<0 } = { -β+nδ:β>0,wβ<0, n=1,,β,h-1 } { -β+nδ:β>0, wβ>0,n=1,2, β,h } . Consequently, (wt-h)= 2ρ,h- (w).

Lecture 9: March 12, 1997

Recall the A_af-action on HomS(HT(ΩK),S): Ax·σ(y)Ω = { σ(xy)Ω ifxyWaf- (x)+(y)= (xy), 0 otherwise, w·ψt = ψwt t·ψt = ψtt t,tΓ,wW.

Define HT(ΩK)= xWaf-s σ(x)Ω as the A_af-submodule of HomS(HT(ΩK),S) spanned over S by {σ(x)Ω:xWaf-}. For xWaf-, set Fx= yWaf-yx sσ(y)Ω. Then ixΩ: X_xΩ ΩK gives HomS(HT(X_xΩ),S) Fx HomS(Fx,S) HT(X_xΩ).

Structure on Fx

(1) {1ψt:tΓ,txw0} is a free S-basis for Frac(S)SFx where Frac(S) = the fractional field ofS xΓ- the minimal rep. ofxWcoincides with the translational representative ofxW.
(2) Set Γ-=ΓWaf-= {t-h:hh_dominant} see end of Lecture 8 onWaf- Waf/WΓ. Then:
  • X_tΩ is K-stable, so Ft is an A_-submodule of HT(ΩK),
  • σ(t)Ω[HT(ΩK)]A_ ie. σ(t)Ω is A_-invariant.


To show that X_tΩ is K-stable, it is enough to show P0t·P0 X_tΩ t-1B-tP0 But for any αΔ+ t-h·α=α+ h,αδ Δ(P0/baf) (ie. a root forP0) t-1B-tP0 X_tΩ isJ-stable Ft isA_-submod. of HT(ΩK). Next, we need to show that iI, (rit)<(t)+1, Ai·σ(t)Ω=0 ?????. But Ai·σ(t)Ω=0 unless ritWaf-. So just need to show that ritWaf-. for any iI. This is not possible. Suppose ritWaf- for some i. Then ri must satisfy "h,αj=0 for some jI riαj>0". Since riαi<0, must have h,αi>0. If (rit)=(t)+1, then t<rit or t-1<t-1ri t-1αi>0. But t-1·αi=th·αi=αi-h,αiδ, Since h,αi>0 th·αi<0. Contractiction. Hence Ai·σ(t)Ω=0 iI.

Hopf algebra structure on HT(ΩK)

Proposition: HT(ΩK) is a Hopf algebra over S, commutative and cocommutative.

Proof (outline) and structure maps.

  • The T-equivariant multiplication map m: ΩK×ΩK ΩK induces the product map: μ: HT(ΩK)HT(ΩK) HT(ΩK). Since m(X_xΩ×X_tΩ) X_xtΩ we actually have μ: FxFt Fxt.
  • The diagonal imbedding ΩK ΩK×ΩK induces the co-product: Δ: HT(ΩK) HT(ΩK)HT(ΩK). Clearly ΔFxFxFx
  • Co-commutativity is clear. As for commutativity of μ, one can give a couple of reasons. One reason is that over Frac(S), Fx has a basis {1ψt:tΓ,txw0} and ψtψt=ψtψt=ψtt. Another reason is because ΩK is a double loop space so its (at least ordinary) homology is commutative.
  • unit: ψid
  • antipode: c(Ft)=Fω(t) where ω is the diagram automorphism defined by ω·αi=-αω(i), i0,ω(0)=0 ( ω(w)=w0ww0 for wW and ω(th)=t-w0·h). In terms of the ψt's, the Hopf algebra structure is easier to express. ε(ψt) = 1, c(ψt) = ψt-1, Δψt = ψtψt, ψtψt = ψtt, ψid = 1.

In the following, we describe a model for HT(ΩK).

The map j:HT(ΩK)A_af

First, we have the general fact that if X is a T-space and ϕ: ΩK×X X is a T-equivariant map (with T acting on ΩK by conjugations and on ΩK×X by the diagonal action), then each σHT(ΩK)HomS(HT(ΩK),S) defines the following composition map HT(X) ϕ* HT(ΩK)SHT(X) c(σ)id SSHT(X) HT(X). If ϕ defines an action of ΩK on X, then these composition maps define an HT(ΩK)-module structure on HT(X).

Now assume that X is a Kaf-space. By restriction to T and ΩK, it is both a T-space and an ΩK-space and the action map ϕ: ΩK×X X is T-equivariant. Thus each σHT(ΩK) defines an operator on HT(X). This is functorial in X, so we get a characteristic operator. In other words, we have a map j: HT(ΩK) Aˆ_af. A calculation shows that j(ψt)=t. Thus j(σ) is compactly supported, (?) so ie j(σ)A_af. It is obvious that j is a ring homomorphism. Since HT(ΩK) is commutative and since j is an s-map, (?) we have j(HT(ΩK)) ZA_af(s), centralizer ofsin A_af(tWaf A_afcommutes withs). Set A_Ω= ZA_af(s). It is a commutative S-algebra. Thus we have an S-algebra homomorphism j: HT(ΩK) A_Ω = ZA_af(S) Will show that it is in fact an isomorphism.

Connection between j:HT(ΩK)A_af and jΩ:ΩKGaf/Baf:kkBaf.

Have commutative diagram HT(ΩK) j A_af HomS(HT(ΩK),S) (jΩ)X HomS(HT(Gaf/Baf),S) a ε·aR = ε·c(a)L Before we find j(σ(x)Ω), we collect some facts about the action of HT(ΩK) on HT(X) for a Kaf-space X.

Note: The following Lemma 1 had a large cross through it in the scanned copy.

Lemma 1: For any Kaf-space X, the action of A_af on HT(X) factors through A_ via the map (Is this right?) ev: A_af A_ where, recall, ev|S = id, ev|Aβ = Aβ, ev|wth = w.

Lemma 2: For σHT(ΩK), (idev) Δ·j(σ)= j(σ)1. ?


This is roughly due to the fact that ΩK Kaf: k (k,1).

Now for any A_af-module M and A_-module N, set M*SN=MSev*N, an A_af-module. Then by Lemma 2, j(σ)·(mn)= j(σ)·mn. Apply this to the action map F: HT(ΩK)SHT(X) HT(X).

Proposition: The above action map is an A_af-module map.


For σHT(ΩK) and zHT(X), we know F(σz)= j(σ)·z ? so for wW w·F(σz)= w·j(σ)·z. In particular w·F(ψtz)= w·t·z=wtw-1 ·w·z=(w·t)· (w·z). On the other hand F(w·(ψtz))= F(w·ψtw·z)= F(w·(ψtz)). Also t·F(ψtt)= tt·z=F (ψttz)=F (t·(ψtz)).

Proposition: The multiplication map HT(ΩK)SHT(ΩK) HT(ΩK) is an A_af-map.


This is because σσ=j(σ)· σ.

More generally, for any A_af-module M, the map ϕ: HT(ΩK)SM M σm j(σ)·m is always an A_af-module map.

We now look at j(σ(x)Ω),

Introduce the ideal IA_af: (left ideal) I=wWwid A_afAw. This is the ideal of annihilators of 1HT(ΩK) for the action of A_af on HT(ΩK).

Proposition: For xWaf- j(σ(x)Ω) =AxmodI.


j(σ(x)Ω)·1= σ(x)Ω1= σ(x)Ω= Ax·σ(id)Ω= Ax·1. j(σ(x)Ω) -AxI.

Corollary 1: Axw0=j(σ(x)Ω)Aw0 where w0= longest in W.


j(σ(x)Ω) Aw0= (Ax+a)Aw0= AxAw0= Axw0(aI).

Corollary 2: For any xWaf-, tΓ- σ(x)Ω σ(t)Ω = σ(xt)Ω, FxFt = Fxt. ((x)+(w0)=(xw0) holds for all xWaf-). This is due to the following general fact: For any parabolic P, xWP,yWP, (xy)=(x)+(y).


Since σ(t)Ω[HT(ΩK)]A_, have σ(x)Ω σ(t)Ω = j(σ(x)Ω) ·σ(t)Ω = (Ax+a)· σ(t)Ω aI = Ax· σ(t)Ω (a·(σ(t)Ω=0)) = σ(xt)Ω. (We are saying (x)+(t)=(xt)? automatically?)

Proposition: HT(ΩK)SA_ A_af: σa j(σ)a is an A_af-module isomorphism, where A_af acts on A_ via ev:A_afA_.


?????or: j:HT(ΩK)A_Ω is an isomorphism.

Thus we have a direct sum decomposition A_afA_Ω +I as an A_Ω-module.

Structures on A_Ω

Proposition: For sS, aA_Ω, wW, tΓ and βΔre s·a = sa=as wt·a = wtaw-1 Aβ·a = Aβa- rβaAβ (β=α+nδ,β=α) = Aβarβ +aAβ (Aα0· a0 = -θ). The proof of this proposition is not trivial. Need calculat?????

Introduce Hopf algebra (over S) structure on A_Ω: π(s) = s, ε(t) = 1, c(t) = t-1, Δ(t) = tt.

Theorem: The map j: HT(ΩK) A_Ω is an isomorphism of both A_af-modules and Hopf algebra modules.

Lecture 10: March 19, 1997

Ω-integrable A_af-modules

We first recall the definition of the integrable A_-modules where A_ is A_af or A_finite, that was given at the end of Lecture 6:

An integrable A_-module is an A_-module structure on 𝒪(X), where X is an affine scheme over h_=SpecS with structure homomorphism πX:S𝒪(X) such that

(1) s·p=πX(s)p sS,p𝒪(X).
(2) πX:S𝒪(X) is an A_-module map.
(3) m:𝒪(X)S𝒪(X)𝒪(X) is an A_-module map.
(4) For each p𝒪(X), Aw·p=0 for all but finitely many wW.

Now back to our notation where A_ denotes the nil-Hecke ring for the finite Wely group W. Then condition (4) is not needed.

Definition: An Ω-integrable A_af-module is by definition an affine scheme X over h_=SpecS, with structure homomorphism πX:S𝒪(X), and an A_af-module structure on 𝒪(X) such that

(1) X is an integrable A_-module by restricting the action of A_af to A_;
(2) m:𝒪(X)*S𝒪(X)𝒪(X) is an A_af-map.
(Part of the requirement for (1) is in (2) as well).

Question: Is (2) weaker than asking m:𝒪(X)S𝒪(X)𝒪(X) being an A_af-map? This seems to be just a different requirement. So the notion of Ω-integrable A_af-module seems different from that of an integrable A_af-module.

Set 𝒜=SpecHT(ΩK). Then 𝒜 is an integrable A_-module. We know from Lecture 9 (page 9-9) that m:HT(ΩK)*SHT(ΩK)HT(ΩK) is an A_af-module map, so 𝒜 is an Ω-integrable A_af-module.

Proposition: An Ω-integrable A_af-module structure on 𝒪(X) is equivalent to

(1) an integrable A_-module structure 𝒪(X); and
(2) an A_-module map g:HT(ΩK)𝒪(X).
More explicitly, given an A_af-module structure on 𝒪(X), by restriction to A_ we get an integrable A_-module structure on 𝒪(X), and the map g: HT(ΩK) 𝒪(X): g(σ) = j(σ)·1. Conversely, given (1) and (2), the A_af-module structure on 𝒪(X) is defined by (j(σ)a)·p= g(σ)(a·p).


Assume that the A_af-module structure on 𝒪(X) is given. We need to show that the map g is an A_-map, i.e., for aA_ and σHT(ΩK), need to show g(a·σ) = a·g(σ). Now g(a·σ) = j(a·σ)·1, a·g(σ) = a·j(σ)·1= (aj(σ))·1. Thus we need to show ( j(a·σ)- aj(σ) ) ·1=0𝒪(X). But we know that the action of A_ on HT(ΩK) is characterised by the fact that j(a·σ)- aj(σ)I= wWwid A_afAw. Since for any iI, Ai·1=Ai· πX(1)=πX (Ai·1)=0 𝒪(X) we see that b·1=0 for any bI. Thus ( j(a·σ)- aj(σ) ) ·1=0 or g:HT(ΩK)𝒪(X) is an A_-map.

Conversely, assume that we are given an integrable A_-module structure on 𝒪(X) and an A_-map g:HT(ΩK)𝒪(X). Define, for σHT(ΩK) and aA_, p𝒪(X) (j(σ)a)·p= g(σ)(a·p). Need to show that this gives an Ω-integrable A_af-module structure on 𝒪(X). First need to show that this is indeed and action of A_af. This must follow from the fact that HT(ΩK)*SA_ A_af: σa j(σ)a is an A_af-module map. (?) In order to show m: 𝒪(X)S𝒪(X) 𝒪(X) is an A_af-module map, only need to show m(j(σ)·(p1p2)) =j(σ)·(p1p2). But j(σ)·(p1p2)= g(σ)p1p2 and (Remark after Lemma 2 in Lecture 9 on page 9-7) m(j(σ)·(p1p2)) = m(j(σ)·p1p2) =m(g(σ)p1p2) = g(σ)p1p2 so m(j(σ)·(p1p2)) =j(σ)·(p1p2).

Need to fill in the proof of why (j(σ)a)·p=defg(σ)(a·p) defines an A_af-action.

In more geometrical terms, let 𝒰=SpecHT(K/T). We said in Lecture 6 that an integrable A_-module should be thought of as an action ϕ:𝒰×A_XX. In this language, an Ω-integrable A_af-module structure on 𝒪(X) pairs (ϕ,f) where ϕ is an action of 𝒰 on X and f:X𝒜 is a 𝒰-equivariant map.

The polynomials jxy, xWaf-, yWaf

For xWaf-, introduce jxyS, yWaf, by j(σ(x)Ω)= yWafjxy Ay. In terms of the map jΩ: Ωx Gaf/Baf we have jΩ* σGaf/Baf(y) =xWaf- jxyσΩ(x).

Immediate properties of the polynomial jxy's, xWaf-, yWaf:

Property 1: degjxy=2 ((y)-(x)).

This is because deg(σ(x)Ω) = -2(x), degAy = -2(y).

Property 2: jxy=δxy ifyWaf-.

Property 3: jxy=0unless ytxw0for some tΓ. ( degjxy = 2((y)-(x)) 2((xw0)-(x)) = 2((x)+(w0)-(x)) =2(w0) ) .


Since jΩ(X_xΩ) πp-1 (X_xGaf/Baf) =X_xw0Gaf/Baf and since jΩ*(ψt)=ψt by definition, we have jΩ*(z)=0 in HTX_xΩ if ψt(z)=0 for all tΓ with t ?????. Property 3 now follows from this.

Proposition: For x,zWaf- σ(x)Ω σ(z)Ω= yWafyzWaf-(y)+(z)=(yz) jxyσ(xz)Ω.


σ(x)Ω σ(z)Ω = j(σ(x)Ω) ·σ(z)Ω = yWaf jxyAy· σ(z)Ω = yWafyzWaf-(y)+(z)=(yz) jxyσ(xz)Ω.

Conjecture: The jxy's are polynomials in the αi's with coefficients in +={0,1,2,}.

Remark 1: Can show jxy+ when (y)=(x) by making connection with quantum cohomology: these are the Gromov-Witten invariants.

Remark 2: We proved last time that xWaf- and tΓ-=Waf-Γ, σ(x)Ω σ(t)Ω= σ(xt)Ω. On the other hand, since HT(ΩK) is commutative, we have σ(x)Ω σ(t)Ω= σ(t)Ω σ(x)Ω= j(σ(t)Ω) ·σ(x)Ω. It follows that, for h dominant j(σ(t-h)Ω) =wWAt-w·h since σ(t-h)Ω is A_-invariant, we know that j(σ(t-h)Ω) is in the center of A_af.

An integral formula

Define ev1: Kaf/T K/T ev1(kT) = k(1)T.

Proposition: For x,yWaf- and wW, jxyω(w-1) = σGaf/Baf(yw0) ev1* (w0L·σG/B(w)) ,σ(xw0)Gaf/Baf = [X_Ωyw0X_xw0Ω] ev1*(w0L·σG/B(w)) where X_Ωyw0= Baf-yw0·Baf, X_xw0Ω= Bafxw0·Baf and ω(w)=w0w w0-1 is the diagram automorphism.


1. w0L·σG/B(w) restricts to σG/Bw under the restriction map HT(K/T) H(K/T).
2. The formula for (w)=1 will be used later to show that H*(ΩK) qH*(G/B).


The proof uses various formulas we have proved so far. σGaf/Baf(yw0) ev1* (w0L·σG/B(w)), σ(xw0)Gaf/Baf = ε ( (Axw0)R· ( σGaf/Baf(yw0) ev1* (w0L·σG/B(w)) ) ) (definition of) = ε ( j(σ(x)Ω)R ·Aw0R· ( σGaf/Baf(yw0) ev1* (w0L·σG/B(w)) ) ) (Axw0=j (σ(x)Ω) Aw0from Lecture?????) = ε ( j(σ(x)Ω)R· ( vW ( (Aw,v)R· σGaf/Baf(yw0) ) ( (w0Av)R· ev1*????? ) ) ) (Δλw0= vWAw0v w0Avfrom Lecture?????) = ε ( j(σ(x)Ω)R · ( vW σGaf/Baf(yw0v-1w0) ev1* ( w0RAvR w0L· σG/B(w) ) ) ) (yw0v-1w0) +(w0v)=(yw0) (w-v-1w0) +(w0v)= (w0) automatically satisfied. (Formula forAw0vR· from Lecture 6 and beginning of Lecture 7), ev1*comm????? = ε ( j(σ(x)Ω)R ·vW(wv-1)+(v)=(w) σGaf/Baf(yω(v)-1) ev1* ( w0Rw0L· σG/B(wv-1) ) ) = ε ( vW(wv-1)+(v)=(w) ( j(σ(x)Ω)R ·σGaf/Baf(yω(v)-1) ) ev1* ( w0Rw0L· σG/B(wv-1) ) ) ( (idev)Δ· j(σ)=j(σ)1 in Lecture 9) = vW(wv-1)+(v)=(w) ε ( jσ(x)ΩR· σGaf/Baf(yω(v)-1) ) ε ( ev1* ( w0Rw0L· σG/B(wv-1) ) ) (εis a homomorphism). = vW(wv-1)+(v)=(w) ε ( jσ(x)ΩR· σGaf/Baf(yω(v)-1) ) δv,w ( ε ( ev1* ( w0Rw0K· σG/B(wv-1) ) ) =ε(σG/B(wv-1)) =δv,wWhy?) = ε ( j(σ(x)Ω)R· σGaf/Baf(yω(w)-1) ) = j(σ(x)Ω), σGaf/Baf(yω(w)-1) = jxyω(w)-1 . The fact that this is then equal to the integral is almost by definition of the Schubert basis and of the pairing .

Remark: X_xΩ is rational and irreducible (?).

The basis {σ[x]:xWaf-} for HT(ΩK)

For xWaf-, set σ[x]=ϵ(x) c(σ(x)Ω) HT(ΩK). This is an S-basis for HT(ΩK).

The automorphism ν of A_af is used to obtain properties for this basis: ν|Δ=id |Δ, ν|A_Ω =c. Can check that ν(a)=(-1)12dega w0ω(a)w0, aWaf where, recall, ω(w)=w0ww0, ω(th)=?tω(h)=????? Also have ν(a)·c(σ)= c(a·σ).

Fact 1: xWaf- σ[x]=w0· σ(ω(x))Ω.


σ[x] = ϵ(x)c(σ(x)Ω) = ϵ(x)c(Ax·1) = ϵ(x)ν(Ax)·1 = ϵ(x) (-1)(x) w0ω(Ax)w0 ·1 = ϵ(x) (-1)(x) w0Aω(x) ·1 = w0· (σ(ω(x))Ω).

Fact 2: For xWaf, yWaf- ν(Ax)· σ[y]= { ϵ(x) σ[xy] ifxyWaf-, (x)+(y)= (xy), 0 otherwise.


Follows from σ[x]=ϵ(x)ν(Ax)·1.

Fact 3: For tΓ-, x,zWaf- σ[t]= σ(ω(t))Ω.

Fact 4: For x,zWaf- σ[x]σ[z] yWafyzWaf-(y)+(z)=(yz) ϵ(xy)jxy σ[yz].

Ideals in HT(ΩK) and A_af

Proposition: If M is an A_af-submodule of HT(ΩK), then

1) M is an ideal of HT(ΩK) which is stable under A_;
2) j(M)A_=A_j(M) is a 2-sided ideal of A_af.


Assume that M is an A_af-submodule of HT(ΩK). Then it is automatically A_-stable. If σHT(ΩK) and mM, we have σm=j(σ)·m. Since M is A_af-stable, j(σ)·mM σmM. Hence MHT(ΩK) is an ideal. Now for iI and mM, Aij(m) = j(m) Ai+j(Ai·m) ri A_j(m) j(M)A_. Also have j(m)Ai = Aij(m)-ri j(Ai·m) j(M)A_ A_j(M) j(M)A_ = A_j(M). Thus j(M) is stable under both left and right multiplications by elements in both j(HT(ΩK)) and A_. Hence j(M) is a 2-sided ideal of A_af.

Examples of ideals of HT(ΩK):

For βΔ+re, let K(β)= xWaf-x·β<0 sσ[x]. Since (zx)=(z)+(x) and x·β<0 (zx)·β<0, the formula in Fact 2 implies that K(β) is an A_af-stable submodule of HT(ΩK). Hence it is an A_-stable ideal of HT(ΩK). The sum of these things will be the kernel of the map from HT(ΩK) to qH(G/B).

Future Lectures:

Notes and references

This is a typed version of Lecture Notes for the course Quantum Cohomology of G/P by Dale Peterson. The course was taught at MIT in the Spring of 1997.

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