Quantum Cohomology of G/P

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

Last update: 18 December 2013

Lecture 11: March 26, 1997

Today we study curves 1G/P.

Fact: Since G/P is projective and thus proper, we have Mor(1,G/P) Mor ( 1\ {a finite set}, G/P ) . In particular Mor(1,G/P) Mor(×,G/P), (×=1\{0,}).

Lemma: Let G be a linear algebraic group. Then every principal G-bundle over 𝔸1 is trivial, so it admits a section.


W.L.O.G., assume that G is connected. Let G E 𝔸1 be a principal G-bundle. Let BG be a Borel subgroup of G. Then have bundle E/B 𝔸1 with fibre G/B which always admits a rational section. Since G/B is proper, we actually have a morphism s:𝔸1E/B. Now form the principal B-bundle Enew = 𝔸1×E/BE 𝔸1 A1×E/BE/B It remains to show that Enew has a section. Consider the normal series of B: BBkBk-1 B0=0 dimBr=r. Br/Br-1 is abelian so is either Ga=(,additive) or Gm=(×,multiplicative).

Case 1 - Gm: Since the only line bundle over 𝔸1 is the trivial one, the associated line bundle over 𝔸1 is trivial.

Case 2 - Ga: Since 𝔸1 is affine, H1(𝔸1,𝒪𝔸1)=0 which is the obstruction for a Ga-bundle to be trivial. (H1(𝔸1,Ga)=H1(𝔸1,𝒪𝔸1)=0).

Recall notation: for a variety X over , X=Mor(×,X).

Theorem 1: The map πP: G G/P = Mor(1,G/P) g(t) g(t)P is surjective.


Given ϕMor(𝔸1,G/P)Mor(1,G/P), form the principal P-bundle over 𝔸1: E= { (t,g)𝔸1×G: ϕ(t)gP } with P acting on the copy of G from the right by right multiplications. By Lemma, E admits a section, ie. s:𝔸1E: s(t)=(t,g(t))E. Thus g(t)Mor(𝔸1,G)G is a lift of ϕ. Similarly, can show that can also lift ϕ to some g(t)Mor(1\{0},G).

Next, we study the degree of the curve πP(g)Mor(1,G/P) for gG=Mor(×,G).

Recall notation

(1) For a variety X over , have X = Mor(×,X) and (X)0 = { ϕX: ϕ|S is trivial inπ1(X) } . For example, for sl(2,), B = { (a(t)b(t)0d(t)) :a,b,d[t,t-1] ,ad=1 } . Now a,d[t,t-1] (Laurent polynomials) and ad=1 a=λtk, d=1λt-k. But must have k=0 in order for g(t)=(a(t)b(t)0d(t))(B)0. Thus (B)0 = { (λb(t)01λ) :λ×,b[t,t-1] } . This is true in general: (B)0 = HU+˜.

Remark: Compare with Baf = { (a(t)b(t)c(t)d(t)) :a,b,c,d[t], ad-bc=1,c(0)=0 } . Very different from (B)0.

(2) πP: Q H2(G/P) πP(αi) = { σriG/P ifriWP, 0 ifriWP.

Theorem 2:

(A) Let w1,w2W and h1,h2Q. If gBaf-w1th1(B)0Bafw2th2(B)0, then ϕπB(g) Mor(1,G/B) satisfies ϕ*[1] = πB(h2-h1), ϕ() B-w1·B, ϕ(0) Bw2·B.
(B) We have two disjoint unions: G=xWaf Baf-x(B)0 =yWafBaf y(B)0. Here, recall Baf = { gMor(1\{},G) :g(0)B } , Baf- = { gMor(1\{0},G) :g(0)B- } .


Proof of (A): Since gBaf-w1tj1(B)0, we can write g(t) = b-(t)n1 t-h1a1 u1(t), t×, where b-(t)Baf-, u1(t)U+, a1H and n1 is a representative of w1 in G. Then by definition ϕ(t) = g(t)·B=b- (t)w1·B, t×. Since b-Mor(1\{0},G) and b-()B-, we have ϕ()B-w1 ·B. Similarly, ϕ(0)Bw2·B. It remains to calculate ϕ*[1]H2(G/B). We do this by calculating ϕ*[1],λ for every dominant integral λh_* considered as an element in H2(G/B). So let λ be a such and let V(λ) be the irreducible highest weight module of G with highest weight λ and highest weight vector vλ+V(λ). Then we have the morphism J: G/B (V(λ)): g·B g·vλ+ and λH2(G/B) is the pullback by J of the standard generator of 𝔸2((V(λ))). Thus ϕ*[1],λ = the degree ofJϕ:1 (V(λ)). Using g(t) = b-(t)n1 t-h1a1 u1(t), t×, we have g(t)·vλ+ = t-λ,h1 a1λb-(t) n1·vλ+,t ×, so in any chosen homogeneous coordinates, we can write (Jϕ)(t) = [ V0(t), V1(t), , V(t) ] where each Vj(t)[t,t-1] and has degree at most -λ,h1 and the degree -λ,h1 occurs. Similarly, using the fact that gBaf+w2 th2(B)0 we see that the minimal degree of the Vj(t)'s is -λ,h2. Thus degree ofJϕ = max. deg-min. deg = λ,h2-h1. Hence ϕ*[1],λ = λ,h2-h1 ϕ*[1] = h2-h1. This finishes the proof of (A).

Proof of (B): First assume we have the unions, ie. G=xWaf Baf-x(B)0 =yWafBaf y(B)0. (*) We prove the disjointness. So assume g(Baf-x1(B)0) (Baf-x1(B)0). Then also gBafy(B)0 for some y. Write x1=w1th1, x1=w1th1, y=w2th2. Then by (A), the curve πB(g)=φ satisfies ϕX[1]= h2-h1= h2-h1 h1=h1. Also ϕ()w1·BB-w1·B. w1=w1. Hence x1=x1. This shows the first union in (*) is disjoint. Similarly is the 2nd. Now we need to show GyWaf Bafy(B)0. Since [U±αi,iIaf] generate G, it suffices to show that yWafBafy(B)0 is stable under the left multiplication by U±αi iIaf. Clearly OK for UαiBaf. Only need to show (U-αi\{id}) yWafBafy (B)0yWaf Bafy(B)0. Now we know: U-αi\{id} BafriUαi.

Case 1: y-1·αi >0 Uy-1·αi (B)0 Uαiy(B)0 y(B)0 (U-αi\{id}) Bafy(B)0 Bafy(B)0OK.

Case 2: y-1·αi<0 Uαi\{id}U-αiriHU-αi Bafri Uαi\{id} y(B)0 BafriU-αi (riH)U-αi y(B)0 Bafri(riH) y(B)0 = Bafy(B)0.

Definition: For w1,w2WP, τH2(G/P), set MG/P,τw1,w2 = the variety of allϕMor (1,G/P)s.t. ϕ*[1]=τ, ϕ()B- w1·P, ϕ(0)Bw2·P. It is a smooth irreducible variety of dimension dimMG/P,τw1,w2 = (w2)-(w1) +τ,c1(TG/P). Thus we have defined a map, for any x1=w1th1, x2=w2th2U πB: Baf-x1(B)0Bafx2(B)0 MG/B,πB(h2-h1)w1,w2. Since πB(h(B)0)=πB(g), we get a well-defined map, still denoted by πB: πB: Baf-x1·(B)0Bafx2·(B)0 MG/B,πB(h2-h1)w1,w2.

Proposition: The map πB: Baf-x1·(B)0Bafx2·(B)0 MG/B,πB(h2-h1)w1,w2 is bijective.


We can in fact prove that πB|Baf-x1·(B)0: Baf-x1·(B)0 MG/B/πB(h2-h1)w1,w2 is injective. Indeed, if g=b-x1 and g=b-x1 where b-,b-Baf- are such that πB(g·(B)0)=πB(g·?????) ie. πB(g)=π(g), then b-(t)x1(t)·B=b-(t)x1(t)·B. Here x1(t) is a representative of x1. Hence b(t)B s.t. b-(t)x1(t)=b-(t)x1(t)b(t). But B=HU+˜ =Γ×HU+˜ =Γ×(B)0. So hΓ, b0(B)0 b-(t)x1(t) = b-(t)x1(t) thb0(t) or b-x1b-x1 th(B)0 or b-x1Baf- x1(B)0 Baf-x1th (B)0. By the disjointness of the union G = xWaf Baf-x(B)0 must have th=id or b(t)(B)0. Hence g·(B)0=g·(B)0. This shows that πB is injective. (Is this argument rigorous enough?) Now suppose ϕMG/B,πB(h2-h1)w1,w2. Let gG be any element such that πB(g)=ϕ. The by Theorem (B), there must exist x1=w1th1 and x2=w2th2Waf s.t. gBaf-x1 (B)0Baf x2(B)0. Let g=gth-h1. Then πB(g)=πB(g)=ϕ but now gBaf-x1(B)0Bafx2th-h1(B)0. But since ϕ*[1] = πB(h2-h1) we must have x2th-h1=x2. Hence gBaf-x1 (B)0Baf x2(B)0 or g·(B)0 (Baf-x1·(B)0) Bafx2·(B)0. This shows that πB is onto. Hence πB is bijective.

Remark: Note that in the definition of MG/P,τw1,w2, we consider a reparametrization of a curve ϕ or a shift of ϕ by an element in H as a new curve.

Connection of MG/P,τw1,w2 to Schubert cells in Gaf/Baf:

Introduce Waf± = { xWaf:βΔ+re ,x·β<0±β>0 } so Waf- is as before the minimal coset representatives of Waf-/W. It is easy to see that Waf-w0Waf+ where w0W is the longest element of W.

Fact: For x=wthWaf±, have (x) = ±s(x) where s(x), the stable length of x, is defined to be s(wth) = (w)+ 2ρ,h.

Theorem 3: Let x1=w1th1, x2=w2th2 be in Waf+. Then we have a natural inverse isomorphism between smooth varieties: Baf-x1·Baf Bafx2·Baf π+π- MG/B,πB(h2-h1)w1,w2 given by π-(g·Baf) = πB(g) ifgBaf-x1, π+(πB(g)) = g·Bafifg Bafx2.

Remark: Note that the intersection Baf-x1·BafBafx2·Baf is smooth and has dimension = (x2)-(x1) = (w2)+ 2ρ,h2- (w1)- 2ρ,h1 = (w2)- (w1)+ 2ρ,h2-h1 = dimMG/B,πB(h2-h1)w1,w2.

Lecture 12: April 8, 1997

Recall last lecture...

The fact G=xWafdisjoint Baf-x(B)0= yWafdisjoint Bafy(B)0 is a special case of the following general fact:

Fact: If V is a subgroup of G such that for each αΔ+re, either UαV or U-αV, then we have two disjoint unions: Gaf=G= xWaf Baf-xV= yWaf BafyV.

Two decompositions for any Kac-Moody group:

xW (of the Kac-Moody group in question) y

0) U-=(U-(x-1B+x))(U-(x-1B-x))
1) (U+xB-x-1)× (B-x·B) xB-·B
2) (U+xB-x-1)× ((B-x·B)By·B) xB-·BBy·B
B-x·B By·B xy, and in this case, B-x·BBy·B is a non-singular irreducible affine variety of dimension =(y)-(x).

In order to prove Theorem 3 stated at the end of last Lecture, we need the following facts. Recall that Waf+ = { xWaf:β Δ+re, x·β<0 β>0 } .

Proposition 1: The following are equivalent:

(0) xWaf+
(1) Bafx-1Baf-x(B)0
(2) xBafx-1Baf-x(B)0x-1
(3) BafxBafBafx(B)0
(1') (B)0x-1BafxBaf
(2') x(B)0x-1BafxBafx-1
(3') Baf-x(B)0Baf-xBaf


The equivalence between (0) and (1) is clear because (2) says that if βΔ+re and xβ<0 then β>0. It is also clear that (1) is equivalent to (2) because x1)x-1=2). Now assume (1). We want to prove (3): It is enough to show that xBafBafx(B)0. Let xbxBaf. Write b=b1b2 where b1Bafx-1Bafx, b2Bafx-1Baf-x. Then xb=xb1b2= (xb1x-1)xb2. Now xb1x-1Baf and b2(B)0 by 1). Hence xbBafx(B)0. This shows that ((0))(1)((2))(3). Now assume 3). We want to prove (1).

Proposition 2: For x1,x2Waf+, the two maps ϕ1: Baf-x1·Baf Baf-x1·(B)0: b-x1·Baf b-x1·(B)0 ϕ2: Bafx2·(B)0 Bafx2·Baf: b+x2·(B)0 b+x2·Baf are both well-defined. Moreover, their restrictions to the following intersections give isomorphisms that are mutually inverses of each other Baf-x1·Baf Bafx2·Baf ϕ2ϕ1 Baf-x1·(B)0 Bafx2·(B)0.


ϕ1 is well-defined because Baf-x1Bafx1-1 x1(B)0x1-1 ((2) in Prop. 1). ϕ2 is well-defined because Bafx2(B)0x2-1 x2Bafx2-1 ((2') in Prop. 1). Since Bafx2BafBafx2(B)0x2-1 ((3) in Prop. 1), we have ϕ1(Baf-x1·BafBafx2·Baf) Baf-x1·(B)0 Bafx2·(B)0. In more details, suppose that m1=b-x1·Baf= b+x2·Baf Baf-x1·Baf Bafx2·Baf where b-Baf, b+Baf. Then bBaf s.t. b-x1=b+x2b. Write b=b1b2 where b1Bafx2-1Bafx2, b2Bafx2-1Baf-x2. Then b-x1=b+(x2b1x2-1)x2b2. We know that x2b1x2-1Baf by definition of b1 and that b2(B)0 by (1) of Prop. 1. Then ϕ(m1)=b-x1 ·(B)0= (b+x2b1x2-1) x2·(B)0 Bafx2·(B)0. Moreover, by the definition of ϕ2, we have ϕ2(ϕ1(m1)) = (b+x2b1x2-1) x2·Baf = b+x2b1·Baf = b+x2·Baf (sinceb1Baf) = m1. This shows that ϕ1 is injective and ϕ2 is onto (when restricted to the intersections). Similarly we can show that ϕ2(Baf-x1·(B)0Bafx2·(B)0) Baf-x1·BafBafx2·Baf and ϕ1(ϕ2(m2))=m2 for m2Baf-x1·(B)0Bafx·(B)0. Let's write out the details again: suppose that m2=b-x1·(B)0 b+x2·(B)0 Baf-x1·(B)0 Bafx2·(B)0 where b-Baf- and b+Baf. Then b0(B)0 s.t. b+x2=b-x1 b0. Write b0=b1b2 where b1(B)0x1-1Baf-x1 and b2(B)0x1-1Bafx1. Then b+x1=b-(x1b1x1-1)x1b2. Now x1b1x1-1Baf- by definition and b2Baf by (1') of Prop. 1. Hence b+x2Baf-x1Baf, or ϕ2(m2)=b+x2·BafBaf-x1·Baf. In other words, ϕ2(Baf-x1·BafBafx2·Baf) Baf-x1·(B)0 Bafx1·(B)0. Moreover, by the definition of ϕ1, we have ϕ1(ϕ2(m2)) = b-(x1b1x1-1) x1·(B)0 = b-x1b1· (B)0 = b-x1·(B)0 (b1(B)0) = m2. This shows that when restricted to the intersections, both ϕ1 and ϕ2 are isomorphisms and that they are the inverses of each other.

We can prove Theorem 3 stated in Lecture 11. We restate

Theorem 3: Let x1=w1th1 and x2w2th2 be in Waf+. Then we have mutually inverse isomorphisms Baf-x1·Baf Bafx2·Baf π+π- MG/B,πB(h2-h1)w1,w2 defined by π-(g·Baf) = πB(g) ifgBaf-x1 s.t.g·Baf Baf-x1·Baf Bafx2·Baf, π+(πB(g)) = g·BafifgBafx2 s.t.πB(g) MG/B,πB(h2-h1)w1,w2.


This is just Proposition 2 and the Proposition at the end of Lecture 11 (page 11-10.5(1)) combined, ie. Baf-x1·Baf Bafx2·Baf Baf-x1·(B)0 Bafx2·(B)0 MG/B,πB(h2-h1)w1,w2.

The Stable Bruhat order s and the stable length s

Say hQ is sufficiently dominant if ρi,h0 for each iI.


1) For x,yWaf, write "xsy" and say "x is y under the stable Bruhat order" if xthyth for sufficiently dominant h.
2) For x=wthWaf, define the stable length of x to be s(x)=(w) +2ρ,h.


1) For any wW and h dominant, have x=wthWaf+.
2) For any given xWaf, have xthWaf+ for sufficiently dominant h.


Clearly 2) follows from 1). We only prove 1). If α<0 is a root for the finite g, then for any n>0, x·(α+nδ)= wα+(n-h,α)δ. Since h,α0, have n-h,αn>0. This always has x·(α+nδ)>0. This shows that if β=α+nδ>0 is such that x·β<0 must have α>0. Thus xWaf+.

Proposition: w1th1||x1 s w2th2||x2 MG/B,πB(h2-h1)w1,w2 Baf-x1·(B)0 Bafx2·(B)0.


We have proved in Lecture 11 that MG/B,πB(h2-h1)w1,w2 Baf-x1·(B)0 Bafx2·(B)0. Now suppose x1sx2. Then sufficiently dominant h st. x1th,x2thWaf+ and x1thx2th. This implies Baf-x1th·Baf Bafx2th·Baf. But by Theorem 3, since x1th,x2thWaf+, we have MG/B,πB(h2-h1)w1,w2 Baf-x1th·Baf Bafx2th·Baf. Conversely, if MG/B,πB(h2-h1)w1,w2, then for h sufficiently dominant so that x1th,x2thWaf+, we have Baf-x1th·Baf Bafx2th·Baf MG/B,πB(h2-h1)w1,w2 . Thus x1thx2th. Hence w1th1sw2th2.

Proposition: For x1,x2Waf+, have x1sx2x1x2.


Suppose x1,x2Waf+. Then x1sx2 MG/B,πB(h2-h1)w1,w2 =Baf-x1·Baf Bafx2·Baf x1x2.

Proposition: For w1,w2W, have w1sw2w1w2.

Proof 1.

Since MG/B,πB(0)w1,w2 =B-w1·BBw2·B we have w1sw2 B-w1·BB w2·B w1w2.

Proof 2.

We first prove that WWaf+.

Suppose β=α+nδ>0 is s.t. w·β<0. Then we must have α>0: for if α<0, then n>0, and thus w·β=wα+nδ>0. Contradiction. Hence α>0. Hence WWaf+.

Question: Given wW, for which xWaf do we have w<sx and s(x)=s(w)+1=(w)+1?

Answer: Iff x is one of the following two forms: either x=wrα where αΔ+ and (x)=(w)+1 (positive roots for the finite 𝔤) or x=wrαtα=wrα+δ where αΔ+ and (wrα)=(w)-2ρ,α+1.



Fact: If wthWaf+ then hQ is dominant.


Proof by contradiction: Suppose h is not dominant. Then i s.t. αi,h<0. Must have αi,h-2. Let β=-αi+δΔ+re. Then x·β=-wαi+(1+αi,h)δ. But β=-αi<0. Contradictory to xWaf+ h dominant.

Lecture 13: April 9, 1997

We first collect some facts about s and s. Then talk about (WP)af and (WP)af.

Proposition s: The following are true about s:

(1) s(w)=w wW.
(2) s(xth)=s(x)+2ρ,h xWaf, hQ.
(3) s(xw0)=(w0)-s(x) xWaf, w0=longest in W.
(4) -(x)s(x)(x) xWaf, s(x)=(x) xWaf+, s(x)=-(x) xWaf-.
(5) For any x,yWaf s(xy) = s(y)+ βΔ+rex·β<0 sign(y-1·β) where signα = { 1 ifαΔ+, -1 ifα-Δ+.

Recall Δ+ = the set of roots of the finite𝔤.


(1) and (2) are clear from the definition.

(3): Write x=wth. Then s(xw0) = (wthw0)= (ww0tw0h) = (ww0)+ 2ρ,w0h = (w0)-(w) -2ρ,h = (w0)-s(x).

(4) We break the proof of (4) into a few parts: We first prove that s(x)=(x) for xWaf+: Assume x=wthWaf+. Then by the definition of Waf+, if α+nδ>0 is s.t. x·(α+nδ)=wα+ (n-α,h) δ<0 we must have α>0. Thus if wα<0, then n can only take values 0,1,,α,h and if wα>0, then n can only take values 0,1,,α,h-1. Thus the set A = { α+nδ>0:x· (α+nδ)<0 } is contained in the set B = { α+nδ:α>0, wα<0,n=0,1, ,α,h } { α+nδ:α>0, wα>0,n=0,1, α,h-1 } . Clearly BA. Thus A=B. Hence (x)=#B = α>0wα<0 (α,h+1)+ α>0wα>0 α,h = α>0α,h+ α>0wα<0 ·1 = 2ρ,h+ (w) = s(x). This shows s(x) = (x)forx Waf+. We have proved (Lecture 8 that) -s(x) = (x)for xWaf-. To prove that s(x) (x)for all xWaf we need the following Lemma:

Lemma: Suppose that h1Q is dominant and regular. Then for all xWaf, we have (xth1) (x)+2ρ,h1. We will prove the Lemma later. Let's assume the Lemma for now. Let xWaf be arbitrary. Let h1 be sufficiently dominant so that xth1Waf+. Then we have s(x) = s(xth1)- 2ρ,h1 = (xth1)- 2ρ,h1 (x)+ 2ρ,h1- 2ρ,h1 (Lemma) = (x). This shows that s(x)(x) for all xWaf. Now if (x)=s(x)=(w)+2ρ,h for x=wthWaf, then since the set B = { α+nδ:α>0, wα<0,n=0,1, α,h } { α+nδ:α>0, wα<0,n=0,1, α,h-1 } (if α,h<0, then the first set in the union is taken to be . Similarly for the 2nd set) is obviously contained in the set A = { α+nδ>0:x· (α+nδ)<0 } . But #B=(w)+2ρ,h B=A. So for every βΔ+reA have β>0. This shows that xWaf+. Similarly we can show s(x)-(x) xWaf and s(x)=-(x) xWaf-. This finishes the proof of (4) (except for the lemma). (Something is not right here).

We now prove (6): x,yWaf (Do not trust this proof!) s(xy) = s(y)+ βΔ+rex·β<0 sign(y-1·β). Write x=w1th1, y=w2th2. Then s(xy) = s(w1w2tw2-1h1+h2) = (w1w2)+ 2ρ,w2-1h1+h2 so s(xy)-s(y) = (w1w2)-(w2) +2w1ρ,h1 so need to show (w1w2)- (w2)+ 2w1ρ,h1 = βΔ+rex·β<0 sign(y-1·β). Notice the special case: x=w1, y=w2, we are saying (w1w2)- (w2) = Δ+β>0w1·β<0 sign(w2-1·β). This is a statement about the finite Weyl group and can be proved by induction on (w2), for example. We assume this. Thus need to show 2w2ρ,h1 = βΔ+rex·β<0 sign(y-1·β)- Δ+α>0w2·α<0 sign(w2-1·α). Let A = { β=α+nδ>0: x·β<0 } = { β=α+nδ>0: w1α+ (n-α,h1) δ<0 } . For β=α+nδA, have y-1·β = w2-1α+ (n+w2-1α,h2) δ so y-1·β = w2-1α. Break A as a disjoint union A = A1A2A3A4 where A1 = { β=α+nδ>0: α>0, w1α>0, w1α+(n-α,h1)δ<0 } , A2 = { β=α+nδ>0: α>0, w1α<0, w1α+(n-α,h1)δ<0 } , A3 = { β=α+nδ>0: α<0, w1α>0, w1α+(n-α,h1)δ<0 } , A4 = { β=α+nδ>0: α<0, w1α<0, w1α+(n-α,h1)δ<0 } , so A1 = { β=α+nδ>0: α>0, w1α>0, n=0,1,,α,h1-1 } , A2 = { β=α+nδ>0: α>0, w1α<0, n=0,1,,α,h1 } , A3 = { β=α+nδ>0: α<0, w1α>0, n=0,1,,α,h1-1 } , A4 = { β=α+nδ>0: α<0, w1α<0, n=0,1,,α,h1 } . Note that αΔ+ sign(w2-1α)= α>0w2-1α>0·1+ α>0w2-1α<0(-1) =2ρ-2(ρ-w2ρ)= 2w2ρ. Similarly, βΔ+rex·β<0 sign(y-1·β) = βA1A2A3A4 sign(y-1·β) so βΔ+rex·β<0 sign(y-1·β)- αΔ+w1·α<0 sign(w2-1·α) = βA1 sign(y-1·β) = 2w2ρ,h1. This shows (5). (This is not a good proof. May not even be correct. Need to come back). This proves the Proposition except for the Lemma.

Lemma: Suppose that h1Q is dominant and regular. Then for all xWaf, we have (xth1) (x)+2ρ,h1.


Set A1 = { α+nδ>0: xth1· (α+nδ)<0 } = { α+nδ>0: x·(α+(n-h1,α)δ) <0 } . Write A1 as A1 = B1B2 where: B1 = A1 { A1 { α+nδ:α+ (n-α,h1) δ>0 } } , B2 = A1 { A1 { α+nδ:α+ (n-α,h1) δ<0 } } . The map B1 A: α+nδ α+(n-α,h1)δ is injective: indeed, if α+(n-α,h1) δ=α+ (n-α,h1) δ α=αand n-α,h1 =n-α,h1 α=α,n= n.Hence#B1 #A=(x). Define the inclusion map B2C = { α+nδ>0:α+ nδ-α,h1 δ<0 } = { α+nδ>0:α>0, n=0,1,, α,h1-1 } . It is clear that #c=αα,h1=2ρ,h1 #B2#C= 2ρ,h1. Hence (xth1)=#A= #B1+#B2 #A+#C=(x)+ 2ρ,h1.

In the next proposition, we collect some facts about s:

Proposition s:

(1) For x,yWaf, we have xsy xthyth for sufficiently dominanth yt-hx t-hfor sufficiently dominanth xthsyth for allh yw0sxw0 wherew0=the longest inW.
(2) For zWaf+, we have
(2a) xz xsz.
(2b) zsy zy.
(3) For zWaf-, we have
(3a) xz zsx.
(3b) ysz zy.
(4) For x,yWaf+, xsy xy.
For x,yWaf-, xsy yx.


(1) Only need to prove that xsy yt-hxt-h for sufficiently dominanth y0sxw0

Lemma 1: If x,yWaf-, then xy xw0yw0.

Lemma 2: xsy w0yw0x.

Proof of Lemma 2.

If ϕMG/B,πB(h2-h1)w1,w2, then ϕ1 defined by ϕ1(t) ϕ(1t)w0·B is in MG/B,πB(h2-h1)w0w2,w0w1. This shows xsy w0yw0x.

I can not prove (1).

Proposition 1: Suppose that hQ. Then hQ+ i +αi idswth wW wsw0th wW xsxth wWaf.

Proposition 2: For βΔ+re and xWaf rβx<sx s(rβx) <s(x) x-1·β<0, x<srβx s(x)s (rβx) x-1·β>0.

Proposition 3: For βΔ+re and β>0, and xWaf xrβ<sx s(xrβ)< s(x) x·β<0, x<sxrβ s(x)< s(xrβ) x·β>0.

Proposition 4: x<sy s(x)<s(y).

Proposition 5: If xsy, then there exists a sequence of the form x=x0<sx1<sx2 <s<sxn=y with n0 and (xk)=s(x)+k for 0kn.

Proposition 6: The following are equivalent: For wW and xWaf,

(a) w<sx and s(x)=(w)+1.
(b) x is one of the following 2 cases:
(1) x=wrα, αΔ+ and (x)=(w)+1.
(2) x=wrαtα=wrα+δ, where αΔ+ and (x)=(w)-2ρ,α+1.
This is related to multiplication by H2 in the quantum cohomology.

We now turn to (WP)af and (WP)af:

Fix a standard parabolic subgroup P of G. Let Δ+(P) = { αΔ+ :𝔤-αP } , QP = αΔ+(P) α. Set (WP)af = { wth:wWP, hQP } . this is the Weylf group of LP, where LP is the Levi-factor of P.


1) P=B, WP=id, (WP)af=id.
2) P=G, WP=W, (WP)af=Waf.
3) P=Pi, WP=1,ri, (WP)af=rαi,rδ-αi.
In general, (WP)af is a Coxeter group; It is a subgroup of Waf, but not a Coxeter subgroup, as seen in the example of P=Pi.

The Length function P(y):

As a Coxeter group, (WP)af has a Length function P(y) = # { β>0:β QP, y·β<0 } . Define (WP)af: (WP)af = { xWaf:β>0, β QPx ·β>0 } .

Proposition: Waf = (WP)af (WP)af ie. each zWaf can be uniquely written as a product z = xy where x(WP)af, y(WP)af.

Define πˆP: Waf (WP)af: z x.

The next proposition gives various properties of πˆP:

(Note: ????? πˆP is what Peterson calls πP in class).

Proposition πˆP:

1) πˆP(W)=WP(WP)af(Waf)P where (Waf)P is the set of minimal representatives for Waf/WP.
2) πˆP(Waf±)Waf±.
3) πP(z)z for all zWaf.
4) For any z,zWaf, hQ, have
  • πˆP(zth)=πˆP(z)πˆP(th).
  • s(πˆP(zth))= s(πˆP(z))+ CP,h where CP=ρ+wPρ= αΔ+w0wP·α<0α (wP=longest inW).
  • zsz πˆP(z)sπˆP(z).
  • πˆP(rβz)<πˆP(z) z-1·βΔ(g.p) (Δ+).
  • πˆP(rβz)=πˆP(z) z-1·βQP+.
  • πˆP(rβz)>πˆP(z) z-1·β-Δ(g/p) (Δ+).

Proposition: For y(WP)af, s,P(y) = s(y) where s,P is the stable length function for (WP)af.

Proposition: For x(WP)af, y(WP)af, s(xy)= s(x)+ s(y), (x)+ s(y) (xy).

Proposition: Any given x(WP)af, can put xthWaf+, xt-hWaf- for sufficiently dominant h(Q)WP, ie. ρi,h0 for all iI such that riWP.

Notation: (P)0 = the identity component ofP, 𝔐P = G/(P)0, *P = (P)0𝔐P, πP: 𝔐B 𝔐P: g*B g·*P. Have action of Γ on 𝔐P: 𝔐P×Γ 𝔐P: (g·*P)·t = gt·*P. This action is trivial if t{th:hQP}. Set, for zWaf, 𝔐P,z± = Baf±z·*P.

Proposition: For zWaf and tΓ 𝔐P,z± = 𝔐P,πˆP(z)±, (𝔐P,z±)·t = 𝔐P,zt±, and for x1,x2(WP)af, 𝔐P,x1- 𝔐P,x2+ x1sx2.

The moduli space 𝔐τ=𝔐τ,P:

Definition: Given a scheme V/ and a morphism f: V×1 G/P, we say that f is of type τ, for τH2(G/P), if for any -valued point v of V, the map fV:1G/P defined by 1 ×1 v×id V×1 f G/P satisfies (fV)* [1] = τ.

The universal property of (𝔐τ,ev):

Proposition: Fix τH2(G/P). There exists a pair (𝔐τ,ev) where 𝔐τ is a reduced scheme of finite type over and ev:𝔐τ×1G/P is a morphism over s.t.

1) ev is of type τ;
2) if V is any reduced scheme of finite type over and f:V×1G/P is a morphism over , then ! morphism fˆ:V𝔐τ over s.t. f = ev (fˆ×id).
Thus (𝔐τ,ev) is unique up to a unique isomorphism. Moreover, 𝔐τ is quasi-projective, and it is either empty or else smooth and of dim dim𝔐τ = dimG/P+ C1TG/P,τ.

Here we outline a proof of the fact that the Zariski tangent space to 𝔐τ at ϕ𝔐τ always has the above dimension: Suppose ϕ: 1 G/P is s.t. ϕ*[1]=τ. Then Tϕ𝔐τ = Γ(1,ϕ*TG/P). Now as sheaves over G/B, we have 0 𝔞 𝔟 TG/P 0 where 𝔟 can be taken as the sheaf of sections of the trivial vector bundle defined by 𝔤, and 𝔞 is the kernel sheaf. Pulling back to 1 by ϕ, we have 0 ϕ*𝔞 ϕ*𝔟 ϕ*TG/P 0. Thus we have the long exact sequence 0 H0(1,ϕ*𝔞) H0(1,ϕ*𝔟) H0(1,ϕ*TG/P) H1(1,ϕ*𝔞) H1(1,ϕ*𝔟) H1(1,ϕ*TG/P) 0. Since 𝔟 is trivial as a vector bundle, have H1(1,ϕ*𝔟) = 0 H1(1,ϕ*TG/P) = 0 dimΓ (1,ϕ*TG/P) = dimH0 (1,ϕ*TG/P) = χ(ϕ*TG/P). Using the general fact that for any vector bundle E over 1, χ(E) = dimE+ C1(E),[1]. We get dimΓ(1,ϕ*TG/P) = dimG/P+ C1(ϕ*TG/P),[1] = dimG/P+ ϕ*C1(TG/P),[1] = dimG/P+ C1(TG/P),ϕ*[1] = dimG/P+ C1(TG/P),τ.

Now for τH2(G/P), v,wWP, set 𝔐τv,w = B-v·P ×G/P 𝔐τ×G/P Bw·P. By a theorem of Kleiman, we have


(1) 𝔐τid,w0wP is open and dense in 𝔐τ,
(2) 𝔐τv,w is quasi-projective, and dim𝔐τv,w = C1(TG/P),τ -(v)+(w).

Kleiman's Theorem: Suppose X is a homogeneous G-space and σY: Y X σZ: Z X are smooth maps. Then for generic g1,g2G, the set g1·Y×X g2·Z = { (y,z):g1 ·σY(y)= g2σZ(z) } is a regular reduced variety of dim=dimY+dimZ-dimX.

Lecture 14: April 15, 1997

Today we introduce two rings for each parabolic P:

1. RP=qHT(G/P)(q): T-equivariant quantum cohomology of G/P with the quantum parameter q inverted,
2. RP=qGT(G/P): T-equivariant quantum cohomology of G/P.

Definition: RP is a free S-module on symbols σP(x), x(WP)af with z-grading deg(sσP(x)) = degs+2s(x).

The A_af module structure on RP

The S-module structure on RP extends to an A_af-module structure on RP by ν(Ai)· σP(x) = { -σP(rix) ifx-1·αi Δ(𝔤_/𝔭_), 0 otherwise where ν is the automorphism of A_af defined at the end of Lecture 10 (page 10-13).

The map ψP:HT(ΩK)RP:

It is the S-module map defined by ψP(σ[x]Ω) = { σP(x) ifx (WP)af, 0 otherwise for all xWaf-.

It should be easy to check that

1) ψP(σ)=j(σ)·σP(id)RP σHT(ΩK),
2) ψP is an A_af-map.

Theorem: There exists a unique commutative S-algebra structure on RP such that

1) σP(id)=1,
2) RP is an Ω-integrable A_af-module with the structure homomorphism SRP: ssσP(id) and the A_af-module structure defined above.

The definition of an Ω-integrable A_af-module is given in Lecture 10. Recall that a proposition in Lecture 10 says that an Ω-integrable A_af-module is equivalent to an affine scheme X over h_=SpecS with a structure morphism πX:S𝒪(X) and

1) an A_-module structure on 𝒪(X),
2) an S-map f:HT(ΩK)𝒪(X)
such that
1) s·p=πX(s)p sS, p𝒪(X),
2) πX is an A_-module map,
4) m:𝒪(X)S𝒪(X)𝒪(X) is an A_-module map,
5) f:HT(ΩK)𝒪(X) is an A_-module map.
Recall that we have used the notation 𝒰 = SpecHT(K/T), 𝒜 = SpecHT(ΩK). Conditions 1)-4) say that X=Spec𝒪(X) is a 𝒰-space, where 𝒰 is a groupoid, and condition 5) says that Specf:X𝒜 is a 𝒰-space morphism.

The geometrical models

The following is from Dale's Lecture at Kac's seminar on April 18, 1997.

The subring ΛPRP:

For hQ, so πP(h)H2(G/P), set qπP(h) = σP(πˆP(th)) RP, and ΛP= {qπP(h):hQ} [H2(G/P)].


Example (πˆP(th) is not necessarily translational): sl3 with extended Dynkin diagram 0 1 2 . Let WP=????? and t=tθ=r0rθ= r0r2r1(WP)af r2(WP)af πˆP(t)=r0r2r1.

Fact: {σP(w):wWP} is a basis of RP over S×ΛP.

Proposition: Formulas for multiplications * in RP and A_af on RP: Ai·(sσ) = (Ai·s)σ+ (ri·s) (Ai·σ) Ai·σP(w) = { -σP(riw) ifw-1·αi Δ(𝔤_/𝔭_), 0 otherwise, (Ai·σ)*σ = Ai· [σ*(ri·σ)] +σ*(Ai·σ).

The operator A0:

Assume that G is simple. α0=δ-θ, Πaf=Π{0}, A0=ν(A0) =-w0A0w0 where w0W is the longest element.

Proposition: A0·(sσ) = -(Aθ·s) σ+(rθ·s) (A0·σ), A0·σP(w) = { -qπP(w-1·θ)-1 σP(πˆP(rθw)) ifw-1·θ Δ(𝔤_/𝔭_), 0 otherwise, (A0·σ)*σ = A0· (σ*(rθ·σ)) -σ*(Aθ·σ).

Theorem: ψP:HT(ΩK)RP is a homomorphism of S-algebras, and ψP(σ)*σ = j(σ)·σ for σHT(ΩK) and σRP.


This is a direct consequence of RP being Ω-integrable.

The structure constants JP,zx,y, x,y,z(WP)af:

For x,y,z(WP)af, define structure constants JP,zx,yS by σP(x)× σP(y) = z(WP)af JP,zx,y σP(z).


(1) degJP,zx,y= 2(s(x)+s(y)-s(z)),
(2) JP,zx,y=JB,zx,y,
(3) JP,zπP(tt)xπˆP(t),yπˆP(t) =JP,zx,y,
(4) For x,y,z(WP)af with xWaf-, JP,zx,y = { ϵ(xyz) jxzy-1 if(yz-1) +s(z)= s(y), 0 otherwise.

Multiplication by H2 in RB:

Theorem: For iI and wW σB(ri)* σB(w) = αΔ+ (wrα)=(w)+1 ρi,α σB(wrα) +αΔ+(wrα)=(w)+1-2ρ,α ρi,α qπB(α) σB(wrα) -(ρi-w·ρi) σB(w).

Remark: One way of looking at the above formula is σB(ri)+ρi = (ρi)R+ αΔ+(rα)=2ρ,α-1 ρi,α qπB(α) Arα, where the left hand side is a multiplication operator on RB and the right hand side is an element in A_af considered as an operator on RB. The right hand side is a commuting family of elements in A_af.

A fact with no classical analogy:

A_af ΛB- ΛB End[RB]W RBwhere ΛB-= hQhdominant q????? [RB]W EndA_afΛB-ΛB RB EndA_ΛB RB A_R ΛB

The ring RP

Define RP = x(WP)afxsid sσP(x). (Recall that x=wthsid hQ+). It is clear from the way A_ acts that RP is an A_-stable submodule of RP.

Fact: For zWaf, x(WP)afxsz sσP(x) is an RP-submodule of RP.

Let ΛP=ΛP RP=τπP(Q+) zqτ. Then RPΛP ΛP RP and {σP(w):wWP} is an SΛP-basis of RP. The augumentation homomorphism is defined to be ε: ΛP : ε(qτ) = δτ,0.

Fact: The map RPΛP HT(G/P) σP(w)1 σP(w) is an isomorphism as A_-modules and S-algebras.

Thus it is reasonable to call RP the T-equivariant quantum cohomology of G/P. It specializes to the T-equivariant cohomology of G/P when the quantum parameters ΛP go to 0.

Poincare Duality

(Compare with the non-quantum case treated in Lecture 7).

Define the SΛP-linear map : RP SΛP by σP(w) = δw,w0wP forwWP.

Theorem: σP(v)*(w0·σP(w0wwP))=δv,w.

Corollary: Have an isomorphism PD: RP HomSΛP(RP,SΛP) defined by PD(φ)(φ) = φ×φ, or concretely PD(σP(w)) = w0· σP(w0wwP).

The Euler Class χG/P:

χG/P =def PD-1(trRP/SΛP) where trRP/SΛP HomSΛP (RP,SΛP) is defined by trRP/SΛP (ϕ) = trace overSΛP of (φ:ϕφ*φ). In other words, trRP/SΛP (ϕ) = ϕ*χG/P. Write σP(v)* σP(w) = uWP buv,w σP(u). Then trRP(SΛP) (σP(v)) = wWP bwv,w = wWP σP(v)* σP(w)* (w0·σP(w0wwP)) χG/P = wWP σP(w)* (w0·σP(w0wwP)).


1) ϕ*χG/P=0 ϕ is nilpotent,
2) χG/P annihilates ΩRP/SΛP.

Example: For sl(3) and P=B, χG/B is invertible q1q2(q1+q2) is invertible.

Lecture 15: April 16, 1997

More facts on RB:

Fact 1: For wW, u,vWuv=w[red] ϵ(u)σB(u-1) *σB(v) = δw,1, u,vWuv=w[red] σB(u)* ϵ(v)σB(v-1) = δw,1.

Remark: Recall from Lecture 7 that similar identities hold for HT(K/T). They can now be considered as a corollary of this fact here about qHT(K/T). Does this follow from any Hopf algebroid structure on qHT(K/T)?

Fact 2: For σRB, σ = wW [ Aw0· (σ·(w0·σB(w0w))) ] *ϵ(w) σB(w). What does this mean? This is not expressing σ in the basis {ϵ(w)σB(w):wW} of RB as an SΛB-module.

Fact 3: RB is a free (RB)A_-module with basis {σB(w):wW}.

Fact 4: (RB)A_ is a polynomial ring on the qπB(αi)'s and the σB(ri)+ρi for iI.

Fact 5: (RB)A_SRB is onto over . (?)

The S-subalgebra RP- of RP:

Define RP-=ImψP= x(WP)afWaf- SσP(x). Then RB- HT(ΩK) but in general HT(ΩK) RP-. We have:

Remark: Working with the case when G is simple, connected but not necessarily simply connected so ΩK is no longer connected, we get the following fact: Assume that ai=1 for all iI in θ=iIaiαi. Let P=Pρi so WP=rjjI,ji. Let w=w0wP. Let Q be a standard parabolic Then σQπˆQ(w) *(wσQ(v)) = qπQ(ρi-v-1ρi) σQ(πˆQ(wv)) for all vWQ. Consequently σ(πˆQ(w)) is invertible in RQ (no clue! How is Q related to P=Pρi?)

Example: G=SL3, (?) WP=r2, G/P=2. σ21×σ21×σ21=q2 (?).

A Filtration:

For hQ, define an A_-submodule FP,h- of RP- (depends only on h mod QP) by FP,h-=RP- q-πP(h) RP= x(WP)afWaf-xsπˆP(t-h) SσP(x) (a finite sum). Then FP,h-* FP,h- FP,h+h-.

Remark: In the geometric models to be given later, elements of FP,h- correspond to trivializing certain line bundles on the (Peterson) variety Y. (a 𝒴).

Fact: When hQ is dominant, FP,h- = ψP(Tt-ω(h)) where Ft-ω(h) is the Bruhat-Filtration in HT(ΩK) in Lecture 9 and ω(h)=-w0·h is the diagram automorphism. Have

More on G/B and G/P:

Fix parabolic P and Q s.t. GPQ. Recall a (classical) fact on H(G/Q): the fibration P/Q G/Q G/P gives rise to a filtration on H(G/Q) such that GrH*(G/Q) H*(P/Q) H*(G/P).

An analogous statement is true for quantum cohomology:

Consider the S-algebra RP,Q = x(WP)afy(WQ)afxsid sσQ(xy). Let RnP,Q = x(WP)af y(WQ)af xsid s(y)n sσQ(xy).

Fact: RmP,Q RnP,Q Rm+nP,Q.

Define RP,Q= grRP,Q= n RnP,Q where RnP,Q = RnP,Q/ R(n-1)P,Q.

Fact: RG/P=RP (SRPQ) RP,Q.

Define R-P,Q = x(WP)af xsid y(WQ)af(WP)af- sσQ(xy), R-nP,Q = x(WP)af xsid y(WQ)af(WP)af- s(y)n sσQ(xy).

Fact: grR-P,Q RG/P=RP [ Im ( H*Ω0 (KP) SRQ ) ] .

Corollary: If SRP/Q and SR?????/Q are reduced, then RG/Q is reduced.


Remark (from informal lecture in the common room after the lecture): Look at the case GPB. The fact grR-P,B RP (Im(H*Ω0(KP)SRB)) (*) has the following meaning in terms of the geometric models: Recall the (Peterson) variety YG/B. It contains 2-T-fixed points {wP:Pparabolic}. Label them by yP. Set YP+ = YB-wP· B, YP- = YB+wP·B (B+=B). Then RP 𝒪(YP+), H*(Ω0(KP)) 𝒪(YP-). Can think of grR-P,B as the subring of 𝒪(YG-YB+) that are regular at yP (not quite sure this is true) so (*) says that near yP, the variety P looks like YP+×YP-.

The quantum cohomology qH(G/P):

What we present here is adequate for G/P but is not the most general case.

For n3, consider the open subscheme Vn() of (1)n: Vn() = { (z1,,zn) (1)n: zizj,ij ,z1=,z2 =0,z3=1 } . For τH2(G/P), let 𝔐τ = {ϕ:1G/P:ϕ*[1]=τ} , 𝔐n,τ = 𝔐τ×Vn() so dim𝔐n,τ = C1(TG/P),τ +dimG/P+n-3. Set ev: 𝔐n,τ (G/P)n: ev(ϕ,z1,,zn) = (ϕ(z1),ϕ(z2),,ϕ(zn)). Roughly speaking, 𝔐n,τ admits a compactification 𝔐n,τ which admits a fundamental class [Mn,τ]. (Manin-Konstevich).

Now for ϕ1ϕnH((G/P)n)H(G/P)n, have 𝔐n,z ev*(ϕ1ϕn) (or?) Using Poincare duality, can regard above as giving a -linear map Jn,τ: n-1H(G/P) H(G/P) of degree =-2[C1(TG/P),τ+(n-3)]. In other words, for any n subvariety X1,,Xn of G/P with i=1n codimXi = 2dim𝔐n,τ we have Jn,τ ( PD-1[X1] PD-1[Xn-1] ) ,[Xn] = # ( 𝔐n,τ ×(G/P)n ( g1X1×× gnXn ) ) () for all (g1,,gn) in a dense open subset of (G())n. These numbers are the Gromov-Witten invariants.

Fact: For ϕH2(G/P) and n4 Jn,τ ( ϕ1 ϕn-2ϕ ) = ϕ,τ Jn,τ ( ϕ1 ϕn-2 ) .

Now let D=[[ε]] with indeterminant ε. Given νε(H*(G/P)D), can make H*(G/P)D into a commutative associative D-algebra with unit σPid with quantum product *ν by σ*νσ = n,τ Jn,τ ( σσ νn-3(n-3)! ) where νn-3=νν ((n-3)-times). In particular, for ϕH2(G/P), define σ*εϕσ = τJ3,τ (σσ) expεϕ,τ. The "potential" function for J3,τ" satisfy WDVV-equation.

The small quantum cohomology:

Make H*(G/P)ΛP into a ΛP-algebra qH*(G/P) by σ*σ = τπP(Q+) qτJ3,τ (σσ).


(1) * is associative.
(2) qH*(G/P) is -graded.
(3) For iI, wW σBri* σBw = αΔ+(wrα)=(w)+1 ρi,α σBwrα +αΔ+(wrα)=(w)+1-2ρ,α ρi,α qπB(α) σBwrα.

The proof of (1) is due to various people. The proof of (3) is a not too hard geometric argument like the one given by Dale in Vogan's seminar.

Relation between qH(G/P) and qH(G/B):

Let τH2(G/P). Then there exists a unique hQ s.t. πP(h) = τ and -1α,h0 for all α-Δ(𝔭_/𝔟_).

Define a standard parabolic P1P by Δ(p1/b) = { αΔ(𝔭_/𝔟_) :α,h=0 } . There have birational morphisms 𝔐πB(h),G/B 𝔐πP(h),G/P1×G/P1G/B 𝔐πP1(h),G/P1 𝔐τ,G/P. This gives a commutative diagram: n-1 H*(G/P) can n-1 H*(G/P1) can n-1 H*(G/B) Jn,τ Jn,πP1(h) Jn,πB(h) H*(G/P) over fibreP/P1integration H*(G/P1) can H*(G/B) This will be used in Lecture 16 to prove SRPqH*(G/P).

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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