## Quantum Cohomology of $G/P$

Last update: 18 December 2013

## Lecture 11: March 26, 1997

Today we study curves ${ℙ}^{1}\to G/P\text{.}$

Fact: Since $G/P$ is projective and thus proper, we have $Mor(ℙ1,G/P) ≃ Mor ( ℙ1\ {a finite set}, G/P ) .$ In particular $Mor(ℙ1,G/P) ≃ Mor(ℂ×,G/P), (ℂ×=ℙ1\{0,∞}).$

Lemma: Let $G\prime$ be a linear algebraic group. Then every principal $G\prime \text{-bundle}$ over ${𝔸}^{1}\supset ℂ$ is trivial, so it admits a section. Proof. W.L.O.G., assume that $G\prime$ is connected. Let $G′ ⟶ E ↓ 𝔸1$ be a principal $G\prime \text{-bundle.}$ Let $B\prime \subset G\prime$ be a Borel subgroup of $G\text{.}$ Then have bundle $E/B′ ↓ 𝔸1$ with fibre $G/B\prime$ which always admits a rational section. Since $G/B\prime$ is proper, we actually have a morphism $s:{𝔸}^{1}\to E/B\prime \text{.}$ Now form the principal $B\prime \text{-bundle}$ $Enew = 𝔸1×E/B′E ↓ 𝔸1 ≃ A1×E/B′E/B′$ It remains to show that ${E}_{\text{new}}$ has a section. Consider the normal series of $B\prime \text{:}$ $B′⊃Bk⊃Bk-1 ⊃⋯⊃B0=0 dim Br=r.$ ${B}_{r}/{B}_{r-1}$ is abelian so is either ${G}_{a}=\left(ℂ,\text{additive}\right)$ or ${G}_{m}=\left({ℂ}^{×},\text{multiplicative}\right)\text{.}$ Case 1 $-$ ${G}_{m}\text{:}$ Since the only line bundle over ${𝔸}^{1}$ is the trivial one, the associated line bundle over ${𝔸}^{1}$ is trivial. Case 2 $-$ ${G}_{a}\text{:}$ Since ${𝔸}^{1}$ is affine, ${H}^{1}\left({𝔸}^{1},{𝒪}_{{𝔸}^{1}}\right)=0$ which is the obstruction for a ${G}_{a}\text{-bundle}$ to be trivial. $\text{(}{H}^{1}\left({𝔸}^{1},{G}_{a}\right)={H}^{1}\left({𝔸}^{1},{𝒪}_{{𝔸}^{1}}\right)=0\text{).}$ $\square$

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

Theorem 1: The map $π∼P: G∼ ⟶ G/P∼ = Mor(ℙ1,G/P) g(t) ⟼ g(t)P$ is surjective. Proof. Given $\varphi \in \text{Mor}\left({𝔸}^{1},G/P\right)\simeq \text{Mor}\left({ℙ}^{1},G/P\right),$ form the principal $P\text{-bundle}$ over ${𝔸}^{1}\text{:}$ $E= { (t,g)∈𝔸1×G: ϕ(t)≃gP }$ with $P$ acting on the copy of $G$ from the right by right multiplications. By Lemma, $E$ admits a section, ie. $\exists$ $s:{𝔸}^{1}\to E:$ $s\left(t\right)=\left(t,g\left(t\right)\right)\in E\text{.}$ Thus $g\left(t\right)\in \text{Mor}\left({𝔸}^{1},G\right)\in \stackrel{\sim }{G}$ is a lift of $\varphi \text{.}$ Similarly, can show that can also lift $\varphi$ to some $g\prime \left(t\right)\in \text{Mor}\left({ℙ}^{1}\\left\{0\right\},G\right)\text{.}$ $\square$

Next, we study the degree of the curve ${\stackrel{\sim }{\pi }}_{P}\left(g\right)\in \text{Mor}\left({ℙ}^{1},G/P\right)$ for $g\in \stackrel{\sim }{G}=\text{Mor}\left({ℂ}^{×},G\right)\text{.}$

### Recall notation

(1) For a variety $X$ over $ℂ,$ have $X∼ = Mor(ℂ×,X)$ and $(X∼)0 = { ϕ∈X∼: ϕ|S′ is trivial in π1(X) } .$ For example, for $sl\left(2,ℂ\right),$ $B∼ = { (a(t)b(t)0d(t)) :a,b,d∈ℂ[t,t-1] ,ad=1 } .$ Now $a,d\in ℂ\left[t,{t}^{-1}\right]$ (Laurent polynomials) and $ad=1⇒ a=λtk, d=1λt-k.$ But must have $k=0$ in order for $g\left(t\right)=\left(\begin{array}{cc}a\left(t\right)& b\left(t\right)\\ 0& d\left(t\right)\end{array}\right)\in {\left(\stackrel{\sim }{B}\right)}_{0}\text{.}$ Thus $(B∼)0 = { (λb(t)01λ) :λ∈ℂ×,b∈ℂ[t,t-1] } .$ This is true in general: $(B∼)0 = H⋉U+˜.$

Remark: Compare with $Baf = { (a(t)b(t)c(t)d(t)) :a,b,c,d∈ℂ[t], ad-bc=1, c(0)=0 } .$ Very different from ${\left(\stackrel{\sim }{B}\right)}_{0}\text{.}$

(2) $πP: Q∨ ⟶ H2(G/P) πP(αi∨) = { σriG/P if ri∉WP, 0 if ri∈WP.$

Theorem 2:

 (A) Let ${w}_{1},{w}_{2}\in W$ and ${h}_{1},{h}_{2}\in {Q}^{\vee }\text{.}$ If $g\in {B}_{\text{af}}^{-}{w}_{1}{t}_{{h}_{1}}{\left(\stackrel{\sim }{B}\right)}_{0}\cap {B}_{\text{af}}{w}_{2}{t}_{{h}_{2}}{\left(\stackrel{\sim }{B}\right)}_{0},$ then $ϕ≔π∼B(g)∈ Mor(ℙ1,G/B)$ satisfies $ϕ*[ℙ1] = πB(h2-h1), ϕ(∞) ∈ B-w1·B, ϕ(0) ∈ Bw2·B.$ (B) We have two disjoint unions: $G∼=⨆x∈Waf {B}_{\text{af}}^{-}x(B∼)0 =⨆y∈WafBaf y(B∼)0.$ Here, recall $Baf = { g∈Mor(ℙ1\{∞},G) :g(0)∈B } , Baf- = { g∈Mor(ℙ1\{0},G) :g(0)∈B- } .$ Proof. Proof of (A): Since $g\in {B}_{\text{af}}^{-}{w}_{1}{t}_{{j}_{1}}{\left(\stackrel{\sim }{B}\right)}_{0},$ we can write $g(t) = b-(t)n1 t-h1a1 u1(t), t∈ℂ×,$ where ${b}_{-}\left(t\right)\in {B}_{\text{af}}^{-},$ ${u}_{1}\left(t\right)\in {\stackrel{\sim }{U}}_{+},$ ${a}_{1}\in H$ and ${n}_{1}$ is a representative of ${w}_{1}$ in $G\text{.}$ Then by definition $ϕ(t) = g(t)·B=b- (t)w1·B, t∈ℂ×.$ Since ${b}_{-}\in \text{Mor}\left({ℙ}^{1}\\left\{0\right\},G\right)$ and ${b}_{-}\left(\infty \right)\in {B}_{-},$ we have $ϕ(∞)∈B-w1 ·B.$ Similarly, $ϕ(0)∈Bw2·B.$ It remains to calculate ${\varphi }_{*}\left[{ℙ}^{1}\right]\in {H}_{2}\left(G/B\right)\text{.}$ We do this by calculating $⟨ ϕ*[ℙ1],λ ⟩$ for every dominant integral $\lambda \in {\underset{_}{h}}^{*}$ considered as an element in ${H}^{2}\left(G/B\right)\text{.}$ So let $\lambda$ be a such and let $V\left(\lambda \right)$ be the irreducible highest weight module of $G$ with highest weight $\lambda$ and highest weight vector ${v}_{\lambda }^{+}\in V\left(\lambda \right)\text{.}$ Then we have the morphism $J: G/B ⟶ ℙ(V(λ)): g·B ⟼ ℂg·vλ+$ and $\lambda \in {H}^{2}\left(G/B\right)$ is the pullback by $J$ of the standard generator of ${𝔸}^{2}\left(ℙ\left(V\left(\lambda \right)\right)\right)\text{.}$ Thus $⟨ϕ*[ℙ1],λ⟩ = the degree of J∘ϕ:ℙ1 ⟶ℙ(V(λ)).$ Using $g(t) = b-(t)n1 t-h1a1 u1(t), t∈ℂ×,$ we have $g(t)·vλ+ = t-⟨λ,h1⟩ a1λb-(t) n1·vλ+,t ∈ℂ×,$ so in any chosen homogeneous coordinates, we can write $(J∘ϕ)(t) = [ V0(t), V1(t), ⋯, Vℓ(t) ]$ where each ${V}_{j}\left(t\right)\in ℂ\left[t,{t}^{-1}\right]$ and has degree at most $-⟨\lambda ,{h}_{1}⟩$ and the degree $-⟨\lambda ,{h}_{1}⟩$ occurs. Similarly, using the fact that $g∈Baf+w2 th2(B∼)0$ we see that the minimal degree of the ${V}_{j}\left(t\right)\text{'s}$ is $-⟨\lambda ,{h}_{2}⟩\text{.}$ Thus $degree of J∘ϕ = max. deg-min. deg = ⟨λ,h2-h1⟩.$ Hence $⟨ϕ*[ℙ1],λ⟩ = ⟨λ,h2-h1⟩ ⇒ϕ*[ℙ1] = h2-h1.$ This finishes the proof of (A). Proof of (B): First assume we have the unions, ie. $G∼=⋃x∈Waf Baf-x(B∼)0 =⋃y∈WafBaf y(B∼)0. (*)$ We prove the disjointness. So assume $g∈(Baf-x1(B∼)0) ∩(Baf-x1′(B∼)0).$ Then also $g∈Bafy(B∼)0$ for some $y\text{.}$ Write $x1=w1th1, x1′=w1′th1′, y=w2th2.$ Then by (A), the curve ${\stackrel{\sim }{\pi }}_{B}\left(g\right)=\phi$ satisfies $ϕX[ℙ1]= h2-h1= h2′-h1′$ $⇒$ ${h}_{1}={h}_{1}^{\prime }\text{.}$ Also $\varphi \left(\infty \right)\in {w}_{1}·B\cap {B}_{-}{w}_{1}^{\prime }·B\text{.}$ $⇒$ ${w}_{1}={w}_{1}^{\prime }\text{.}$ Hence ${x}_{1}={x}_{1}^{\prime }\text{.}$ This shows the first union in $\left(*\right)$ is disjoint. Similarly is the 2nd. Now we need to show $G∼⊂⨆y∈Waf Bafy(B∼)0.$ Since $\left[{U}_{±{\alpha }_{i}},i\in {I}_{\text{af}}\right]$ generate $\stackrel{\sim }{G},$ it suffices to show that $\underset{y\in {W}_{\text{af}}}{⨆}{B}_{\text{af}}y{\left(\stackrel{\sim }{B}\right)}_{0}$ is stable under the left multiplication by ${U}_{±{\alpha }_{i}}$ $\forall$ $i\in {I}_{\text{af}}\text{.}$ Clearly OK for ${U}_{{\alpha }_{i}}\subset {B}_{\text{af}}\text{.}$ Only need to show $(U-αi\{id}) ⨆y∈WafBafy (B∼)0⊂⨆y∈Waf Bafy(B∼)0.$ 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: $\stackrel{‾}{{y}^{-1}·{\alpha }_{i}}<0$ $⇒$ ${U}_{{\alpha }_{i}}\\left\{\text{id}\right\}\subset {U}_{-{\alpha }_{i}}{r}_{i}H{U}_{-{\alpha }_{i}}$ $⇒Bafri Uαi\{id} y(B∼)0 ⊂ BafriU-αi (riH)U-αi y(B∼)0 ⊂ Bafri(riH) y(B∼)0 = Bafy(B∼)0.$ $\square$

Definition: For ${w}_{1},{w}_{2}\in {W}^{P},$ $\tau \in {H}_{2}\left(G/P\right),$ set $MG/P,τw1,w2 = the variety of all ϕ∈Mor (ℙ1,G/P) s.t. ϕ*[ℙ1]=τ, ϕ(∞)∈B- w1·P, ϕ(0)∈Bw2·P.$ It is a smooth irreducible variety of dimension $dim MG/P,τw1,w2 = ℓ(w2)-ℓ(w1) +⟨τ,c1(TG/P)⟩.$ Thus we have defined a map, for any ${x}_{1}={w}_{1}{t}_{{h}_{1}},$ ${x}_{2}={w}_{2}{t}_{{h}_{2}}\in U$ $π∼B: Baf-x1(B∼)0∩Bafx2(B∼)0 ⟶ MG/B,πB(h2-h1)w1,w2.$ Since ${\stackrel{\sim }{\pi }}_{B}\left(h{\left(\stackrel{\sim }{B}\right)}_{0}\right)={\stackrel{\sim }{\pi }}_{B}\left(g\right),$ we get a well-defined map, still denoted by ${\stackrel{\sim }{\pi }}_{B}\text{:}$ $π∼B: Baf-x1·(B∼)0∩Bafx2·(B∼)0 ⟶ MG/B,πB(h2-h1)w1,w2.$

Proposition: The map $π∼B: Baf-x1·(B∼)0∩Bafx2·(B∼)0 ⟶ MG/B,πB(h2-h1)w1,w2$ is bijective. Proof. We can in fact prove that $π∼B|Baf-x1·(B∼)0: Baf-x1·(B∼)0 ⟶ MG/B/πB(h2-h1)w1,w2$ is injective. Indeed, if $g={b}^{-}{x}_{1}$ and $g\prime ={b}^{-}\prime {x}_{1}$ where ${b}^{-},{b}^{-}\prime \in {B}_{\text{af}}^{-}$ are such that ${\stackrel{\sim }{\pi }}_{B}\left(g·{\left(\stackrel{\sim }{B}\right)}_{0}\right)={\stackrel{\sim }{\pi }}_{B}\left(g\prime ·\text{?????}\right)$ ie. ${\stackrel{\sim }{\pi }}_{B}\left(g\right)=\stackrel{\sim }{\pi }\left(g\prime \right),$ then ${b}^{-}\left(t\right){x}_{1}\left(t\right)·B={b}^{-}\prime \left(t\right){x}_{1}\left(t\right)·B\text{.}$ Here ${x}_{1}\left(t\right)$ is a representative of ${x}_{1}\text{.}$ Hence $\exists$ $b\left(t\right)\in \stackrel{\sim }{B}$ s.t. ${b}^{-}\left(t\right){x}_{1}\left(t\right)={b}^{-}\prime \left(t\right){x}_{1}\left(t\right)b\left(t\right)\text{.}$ But $B∼=H∼⋉U+˜ =Γ×H⋉U+˜ =Γ×(B∼)0.$ So $\exists$ $h\in \Gamma ,$ ${b}_{0}\in {\left(\stackrel{\sim }{B}\right)}_{0}$ $b-(t)x1(t) = b-′(t)x1(t) thb0(t)$ or $b-x1∈b-′x1 th(B∼)0$ or $b-x1∈Baf- x1(B∼)0∩ Baf-x1th (B∼)0.$ By the disjointness of the union $G∼ = ⨆x∈Waf Baf-x(B∼)0$ must have ${t}_{h}=\text{id}$ or $b\left(t\right)\in {\left(\stackrel{\sim }{B}\right)}_{0}\text{.}$ Hence $g·{\left(\stackrel{\sim }{B}\right)}_{0}=g\prime ·{\left(\stackrel{\sim }{B}\right)}_{0}\text{.}$ This shows that ${\stackrel{\sim }{\pi }}_{B}$ is injective. (Is this argument rigorous enough?) Now suppose $\varphi \in {M}_{G/B,{\pi }_{B}\left({h}_{2}-{h}_{1}\right)}^{{w}_{1},{w}_{2}}\text{.}$ Let $g\prime \in \stackrel{\sim }{G}$ be any element such that ${\stackrel{\sim }{\pi }}_{B}\left(g\prime \right)=\varphi \text{.}$ The by Theorem (B), there must exist ${x}_{1}^{\prime }={w}_{1}{t}_{{h}_{1}^{\prime }}$ and ${x}_{2}^{\prime }={w}_{2}{t}_{{h}_{2}^{\prime }}\in {W}_{\text{af}}$ s.t. $g′∈Baf-x1′ (B∼)0∩Baf x2′(B∼)0.$ Let $g=g′th-h1′.$ Then ${\pi }_{B}\left(g\right)={\pi }_{B}\left(g\prime \right)=\varphi$ but now $g\prime \in {B}_{\text{af}}^{-}{x}_{1}{\left(\stackrel{\sim }{B}\right)}_{0}\cap {B}_{\text{af}}{x}_{2}^{\prime }{t}_{h-{h}_{1}^{\prime }}{\left(\stackrel{\sim }{B}\right)}_{0}\text{.}$ But since $ϕ*[ℙ1] = πB(h2-h1)$ we must have ${x}_{2}^{\prime }{t}_{h-{h}_{1}^{\prime }}={x}_{2}\text{.}$ Hence $g′∈Baf-x1 (B∼)0∩Baf x2(B∼)0$ or $g′·(B∼)0∈ (Baf-x1·(B∼)0) ∩Bafx2·(B∼)0.$ This shows that ${\pi }_{B}$ is onto. Hence ${\pi }_{B}$ is bijective. $\square$

Remark: Note that in the definition of ${M}_{G/P,\tau }^{{w}_{1},{w}_{2}},$ we consider a reparametrization of a curve $\varphi$ or a shift of $\varphi$ by an element in $\stackrel{\sim }{H}$ as a new curve.

### Connection of ${M}_{G/P,\tau }^{{w}_{1},{w}_{2}}$ to Schubert cells in ${G}_{\text{af}}/{B}_{\text{af}}\text{:}$

Introduce $Waf± = { x∈Waf:β∈Δ+re ,x·β<0 ⇒ ±β‾>0 }$ so ${W}_{\text{af}}^{-}$ is as before the minimal coset representatives of ${W}_{\text{af}}^{-}/W\text{.}$ It is easy to see that $Waf-w0⊂Waf+$ where ${w}_{0}\in W$ is the longest element of $W\text{.}$

Fact: For $x=w{t}_{h}\in {W}_{\text{af}}^{±},$ have $ℓ(x) = ±ℓs(x)$ where ${\ell }_{s}\left(x\right),$ the stable length of $x,$ is defined to be $ℓs(wth) = ℓ(w)+ ⟨2ρ,h⟩.$

Theorem 3: Let ${x}_{1}={w}_{1}{t}_{{h}_{1}},$ ${x}_{2}={w}_{2}{t}_{{h}_{2}}$ be in ${W}_{\text{af}}^{+}\text{.}$ 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 $π-(g·Baf) = π∼B(g) if g∈Baf-x1, π+(π∼B(g)) = g·Bafif g∈ Bafx2.$

Remark: Note that the intersection ${B}_{\text{af}}^{-}{x}_{1}·{B}_{\text{af}}\cap {B}_{\text{af}}{x}_{2}·{B}_{\text{af}}$ is smooth and has dimension $=$ $ℓ(x2)-ℓ(x1) = ℓ(w2)+ ⟨2ρ,h2⟩- ℓ(w1)- ⟨2ρ,h1⟩ = ℓ(w2)- ℓ(w1)+ ⟨2ρ,h2-h1⟩ = dim MG/B,πB(h2-h1)w1,w2.$

## Lecture 12: April 8, 1997

Recall last lecture...

The fact $G∼=⨆x∈Wafdisjoint Baf-x(B∼)0= ⨆y∈Wafdisjoint Bafy(B∼)0$ is a special case of the following general fact:

Fact: If $V$ is a subgroup of $\stackrel{\sim }{G}$ such that for each $\alpha \in {\Delta }_{+}^{\text{re}},$ either ${U}_{\alpha }\subset V$ or ${U}_{-\alpha }\subset V,$ then we have two disjoint unions: $Gaf=G∼= ⨆x∈Waf Baf-xV= ⨆y∈Waf BafyV.$

### Two decompositions for any Kac-Moody group:

$\forall$ $x\in W$ (of the Kac-Moody group in question) $\ni y$

 0) ${U}_{-}=\left({U}_{-}\cap \left({x}^{-1}{B}_{+}x\right)\right)\left({U}_{-}\cap \left({x}^{-1}{B}_{-}x\right)\right)$ 1) $\left({U}_{+}\cap x{B}_{-}{x}^{-1}\right)×\left({B}_{-}x·B\right)\stackrel{\sim }{⟶}x{B}_{-}·B$ 2) $\left({U}_{+}\cap x{B}_{-}{x}^{-1}\right)×\left(\left({B}_{-}x·B\right)\cap By·B\right)\stackrel{\sim }{⟶}x{B}_{-}·B\cap By·B$
$⇒B-x·B∩ By·B≠∅⟺ x≤y,$ and in this case, ${B}_{-}x·B\cap By·B$ is a non-singular irreducible affine variety of dimension $=\ell \left(y\right)-\ell \left(x\right)\text{.}$

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

Proposition 1: The following are equivalent:

 (0) $x\in {W}_{\text{af}}^{+}$ (1) ${B}_{\text{af}}\cap {x}^{-1}{B}_{\text{af}}^{-}x\subset {\left(\stackrel{\sim }{B}\right)}_{0}$ (2) $x{B}_{\text{af}}{x}^{-1}\cap {B}_{\text{af}}^{-}\subset x{\left(\stackrel{\sim }{B}\right)}_{0}{x}^{-1}$ (3) ${B}_{\text{af}}x{B}_{\text{af}}\subset {B}_{\text{af}}x{\left(\stackrel{\sim }{B}\right)}_{0}$ (1') ${\left(\stackrel{\sim }{B}\right)}_{0}\cap {x}^{-1}{B}_{\text{af}}x\subset {B}_{\text{af}}$ (2') $x{\left(\stackrel{\sim }{B}\right)}_{0}{x}^{-1}\cap {B}_{\text{af}}\subset x{B}_{\text{af}}{x}^{-1}$ (3') ${B}_{\text{af}}^{-}x{\left(\stackrel{\sim }{B}\right)}_{0}\subset {B}_{\text{af}}^{-}x{B}_{\text{af}}$ Proof. The equivalence between (0) and (1) is clear because (2) says that if $\beta \in {\Delta }_{+}^{\text{re}}$ and $x\beta <0$ then $\stackrel{‾}{\beta }>0\text{.}$ It is also clear that (1) is equivalent to (2) because $x \text{1)} {x}^{-1}=\text{2).}$ Now assume (1). We want to prove (3): It is enough to show that $xBaf⊂Bafx(B∼)0.$ Let $xb\in x{B}_{\text{af}}\text{.}$ Write $b={b}_{1}{b}_{2}$ where $b1∈Baf∩x-1Bafx, b2∈Baf∩x-1Baf-x.$ Then $xb=xb1b2= (xb1x-1)xb2.$ Now $x{b}_{1}{x}^{-1}\in {B}_{\text{af}}$ and ${b}_{2}\in {\left(\stackrel{\sim }{B}\right)}_{0}$ by 1). Hence $xb\in {B}_{\text{af}}x{\left(\stackrel{\sim }{B}\right)}_{0}\text{.}$ This shows that $\left(\text{(0)}⇔\right)\text{(1)}\left(⇔\text{(2)}\right)⇒\text{(3).}$ Now assume 3). We want to prove (1). $\square$

Proposition 2: For ${x}_{1},{x}_{2}\in {W}_{\text{af}}^{+},$ 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 $Baf-x1·Baf∩ Bafx2·Baf ⇄ϕ2ϕ1 Baf-x1·(B∼)0∩ Bafx2·(B∼)0.$ Proof. ${\varphi }_{1}$ is well-defined because ${B}_{\text{af}}^{-}\cap {x}_{1}{B}_{\text{af}}{x}_{1}^{-1}\subset {x}_{1}{\left(\stackrel{\sim }{B}\right)}_{0}{x}_{1}^{-1}$ ((2) in Prop. 1). ${\varphi }_{2}$ is well-defined because ${B}_{\text{af}}\cap {x}_{2}{\left(\stackrel{\sim }{B}\right)}_{0}{x}_{2}^{-1}\subset {x}_{2}{B}_{\text{af}}{x}_{2}^{-1}$ ((2') in Prop. 1). Since ${B}_{\text{af}}{x}_{2}{B}_{\text{af}}\subset {B}_{\text{af}}{x}_{2}{\left(\stackrel{\sim }{B}\right)}_{0}{x}_{2}^{-1}$ ((3) in Prop. 1), we have $ϕ1(Baf-x1·Baf∩Bafx2·Baf) ⊂Baf-x1·(B∼)0∩ Bafx2·(B∼)0.$ In more details, suppose that $m1=b-x1·Baf= b+x2·Baf∈ Baf-x1·Baf∩ Bafx2·Baf$ where ${b}^{-}\in {B}_{\text{af}},$ ${b}^{+}\in {B}_{\text{af}}\text{.}$ Then $\exists$ $b\in {B}_{\text{af}}$ s.t. $b-x1=b+x2b.$ Write $b={b}_{1}{b}_{2}$ where ${b}_{1}\in {B}_{\text{af}}\cap {x}_{2}^{-1}{B}_{\text{af}}{x}_{2},$ ${b}_{2}\in {B}_{\text{af}}\cap {x}_{2}^{-1}{B}_{\text{af}}^{-}{x}_{2}\text{.}$ Then ${b}^{-}{x}_{1}={b}^{+}\left({x}_{2}{b}_{1}{x}_{2}^{-1}\right){x}_{2}{b}_{2}\text{.}$ We know that ${x}_{2}{b}_{1}{x}_{2}^{-1}\in {B}_{\text{af}}$ by definition of ${b}_{1}$ and that ${b}_{2}\in {\left(\stackrel{\sim }{B}\right)}_{0}$ 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 ${\varphi }_{2},$ we have $ϕ2(ϕ1(m1)) = (b+x2b1x2-1) x2·Baf = b+x2b1·Baf = b+x2·Baf (since b1∈Baf) = m1.$ This shows that ${\varphi }_{1}$ is injective and ${\varphi }_{2}$ is onto (when restricted to the intersections). Similarly we can show that $ϕ2(Baf-x1·(B∼)0∩Bafx2·(B∼)0) ⊂Baf-x1·Baf∩Bafx2·Baf$ and ${\varphi }_{1}\left({\varphi }_{2}\left({m}_{2}\right)\right)={m}_{2}$ for ${m}_{2}\in {B}_{\text{af}}^{-}{x}_{1}·{\left(\stackrel{\sim }{B}\right)}_{0}\cap {B}_{\text{af}}x\prime ·{\left(\stackrel{\sim }{B}\right)}_{0}\text{.}$ 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 ${b}^{-}\in {B}_{\text{af}}^{-}$ and ${b}^{+}\in {B}_{\text{af}}\text{.}$ Then $\exists$ ${b}_{0}\in {\left(\stackrel{\sim }{B}\right)}_{0}$ s.t. $b+x2=b-x1 b0.$ Write ${b}_{0}={b}_{1}{b}_{2}$ where ${b}_{1}\in {\left(\stackrel{\sim }{B}\right)}_{0}\cap {x}_{1}^{-1}{B}_{\text{af}}^{-}{x}_{1}$ and ${b}_{2}\in {\left(\stackrel{\sim }{B}\right)}_{0}\cap {x}_{1}^{-1}{B}_{\text{af}}{x}_{1}\text{.}$ Then ${b}^{+}{x}_{1}={b}^{-}\left({x}_{1}{b}_{1}{x}_{1}^{-1}\right){x}_{1}{b}_{2}\text{.}$ Now ${x}_{1}{b}_{1}{x}_{1}^{-1}\in {B}_{\text{af}}^{-}$ by definition and ${b}_{2}\in {B}_{\text{af}}$ by (1') of Prop. 1. Hence ${b}^{+}{x}_{2}\in {B}_{\text{af}}^{-}{x}_{1}{B}_{\text{af}},$ or ${\varphi }_{2}\left({m}_{2}\right)={b}^{+}{x}_{2}·{B}_{\text{af}}\in {B}_{\text{af}}^{-}{x}_{1}·{B}_{\text{af}}\text{.}$ In other words, $ϕ2(Baf-x1·Baf∩Bafx2·Baf) ⊂Baf-x1·(B∼)0∩ Bafx1·(B∼)0.$ Moreover, by the definition of ${\varphi }_{1},$ 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 ${\varphi }_{1}$ and ${\varphi }_{2}$ are isomorphisms and that they are the inverses of each other. $\square$

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

Theorem 3: Let ${x}_{1}={w}_{1}{t}_{{h}_{1}}$ and ${x}_{2}\in {w}_{2}{t}_{{h}_{2}}$ be in ${W}_{\text{af}}^{+}\text{.}$ Then we have mutually inverse isomorphisms $Baf-x1·Baf∩ Bafx2·Baf ⇄π+π- MG/B,πB(h2-h1)w1,w2$ defined by $π-(g·Baf) = π∼B(g) if g∈Baf-x1 s.t. g·Baf∈ Baf-x1·Baf∩ Bafx2·Baf, π+(π∼B(g)) = g·Bafif g∈Bafx2 s.t. π∼B(g) ∈MG/B,πB(h2-h1)w1,w2.$ Proof. This is just Proposition 2 and the Proposition at the end of Lecture 11 (page 11-10.5(1)) combined, ie. $Baf-x1·Baf∩ Bafx2·Baf≃ Baf-x1·(B∼)0 ∩Bafx2·(B∼)0 ≃MG/B,πB(h2-h1)w1,w2.$ $\square$

### The Stable Bruhat order $\stackrel{s}{\le }$ and the stable length ${\ell }_{s}$

Say $h\in {Q}^{\vee }$ is sufficiently dominant if $⟨{\rho }_{i},h⟩\gg 0$ for each $i\in I\text{.}$

Definition:

 1) For $x,y\in {W}_{\text{af}},$ write $\text{"}x\stackrel{s}{\le }y\text{"}$ and say $\text{"}x$ is $\le y$ under the stable Bruhat order" if $x{t}_{h}\le y{t}_{h}$ for sufficiently dominant $h\text{.}$ 2) For $x=w{t}_{h}\in {W}_{\text{af}},$ define the stable length of $x$ to be $ℓs(x)=ℓ(w) +⟨2ρ,h⟩.$

Facts:

 1) For any $w\in W$ and $h$ dominant, have $x=w{t}_{h}\in {W}_{\text{af}}^{+}\text{.}$ 2) For any given $x\in {W}_{\text{af}},$ have $x{t}_{h}\in {W}_{\text{af}}^{+}$ for sufficiently dominant $h\text{.}$ Proof. Clearly 2) follows from 1). We only prove 1). If $\alpha <0$ is a root for the finite $g,$ then for any $n>0,$ $x·(α+nδ)= wα+(n-⟨h,α⟩)δ.$ Since $⟨h,\alpha ⟩\le 0,$ have $n-⟨h,\alpha ⟩\ge n>0\text{.}$ This always has $x·\left(\alpha +n\delta \right)>0\text{.}$ This shows that if $\beta =\alpha +n\delta >0$ is such that $x·\beta <0$ must have $\alpha >0\text{.}$ Thus $x\in {W}_{\text{af}}^{+}\text{.}$ $\square$

Proposition: $w1th1| |x1 ≤s w2th2| |x2 ⟺ MG/B,πB(h2-h1)w1,w2 ≠∅ ⟺ Baf-x1·(B∼)0 ∩Bafx2·(B∼)0≠∅.$ Proof. We have proved in Lecture 11 that $MG/B,πB(h2-h1)w1,w2 ≃ Baf-x1·(B∼)0 ∩Bafx2·(B∼)0.$ Now suppose ${x}_{1}\stackrel{s}{\le }{x}_{2}\text{.}$ Then $\exists$ sufficiently dominant $h$ st. ${x}_{1}{t}_{h},{x}_{2}{t}_{h}\in {W}_{\text{af}}^{+}$ and ${x}_{1}{t}_{h}\le {x}_{2}{t}_{h}\text{.}$ This implies $Baf-x1th·Baf ∩Bafx2th·Baf≠∅.$ But by Theorem 3, since ${x}_{1}{t}_{h},{x}_{2}{t}_{h}\in {W}_{\text{af}}^{+},$ we have $MG/B,πB(h2-h1)w1,w2 ≃ Baf-x1th·Baf∩ Bafx2th·Baf≠∅.$ Conversely, if ${M}_{G/B,{\pi }_{B}\left({h}_{2}-{h}_{1}\right)}^{{w}_{1},{w}_{2}}\ne \varnothing ,$ then for $h$ sufficiently dominant so that ${x}_{1}{t}_{h},{x}_{2}{t}_{h}\in {W}_{\text{af}}^{+},$ we have $Baf-x1th·Baf∩ Bafx2th·Baf ≃ MG/B,πB(h2-h1)w1,w2 ≠∅.$ Thus ${x}_{1}{t}_{h}\le {x}_{2}{t}_{h}\text{.}$ Hence ${w}_{1}{t}_{{h}_{1}}\stackrel{s}{\le }{w}_{2}{t}_{{h}_{2}}\text{.}$ $\square$

Proposition: For ${x}_{1},{x}_{2}\in {W}_{\text{af}}^{+},$ have ${x}_{1}\stackrel{s}{\le }{x}_{2}⇔{x}_{1}\le {x}_{2}\text{.}$ Proof. Suppose ${x}_{1},{x}_{2}\in {W}_{\text{af}}^{+}\text{.}$ Then $x1≤sx2 ⟺ MG/B,πB(h2-h1)w1,w2 =Baf-x1·Baf∩ Bafx2·Baf≠∅ ⟺ x1≤x2.$ $\square$

Proposition: For ${w}_{1},{w}_{2}\in W,$ have ${w}_{1}\stackrel{s}{\le }{w}_{2}⇔{w}_{1}\le {w}_{2}\text{.}$ Proof 1. Since ${M}_{G/B,{\pi }_{B}\left(0\right)}^{{w}_{1},{w}_{2}}={B}_{-}{w}_{1}·B\cap B{w}_{2}·B$ we have $w1≤sw2 ⟺ B-w1·B∩B w2·B≠∅ ⟺ w1≤w2.$ $\square$ Proof 2. We first prove that $W\subseteq {W}_{\text{af}}^{+}\text{.}$ Suppose $\beta =\alpha +n\delta >0$ is s.t. $w·\beta <0\text{.}$ Then we must have $\alpha >0\text{:}$ for if $\alpha <0,$ then $n>0,$ and thus $w·β=wα+nδ>0.$ Contradiction. Hence $\alpha >0\text{.}$ Hence $W\subset {W}_{\text{af}}^{+}\text{.}$ $\square$

Question: Given $w\in W,$ for which $x\in {W}_{\text{af}}$ do we have $w\stackrel{s}{<}x$ and ${\ell }_{s}\left(x\right)={\ell }_{s}\left(w\right)+1=\ell \left(w\right)+1\text{?}$

Answer: Iff $x$ is one of the following two forms: either $x=w{r}_{\alpha }$ where $\alpha \in {\stackrel{‾}{\Delta }}_{+}$ and $\ell \left(x\right)=\ell \left(w\right)+1$ (positive roots for the finite $𝔤\text{)}$ or $x=w{r}_{\alpha }{t}_{{\alpha }^{\vee }}=w{r}_{\alpha +\delta }$ where $\alpha \in {\stackrel{‾}{\Delta }}_{+}$ and $\ell \left(w{r}_{\alpha }\right)=\ell \left(w\right)-⟨2\rho ,{\alpha }^{\vee }⟩+1\text{.}$ Proof. Later. $\square$

Fact: If $w{t}_{h}\in {W}_{\text{af}}^{+}$ then $h\in {Q}^{\vee }$ is dominant. Proof. Proof by contradiction: Suppose $h$ is not dominant. Then $\exists$ $i$ s.t. $⟨{\alpha }_{i},h⟩<0\text{.}$ Must have $⟨{\alpha }_{i},h⟩\le -2\text{.}$ Let $\beta =-{\alpha }_{i}+\delta \in {\Delta }_{+}^{\text{re}}\text{.}$ Then $x·\beta =-w{\alpha }_{i}+\left(1+⟨{\alpha }_{i},h⟩\right)\delta \text{.}$ But $\stackrel{‾}{\beta }=-{\alpha }_{i}<0\text{.}$ Contradictory to $x\in {W}_{\text{af}}^{+}$ $⇒$ $h$ dominant. $\square$

## Lecture 13: April 9, 1997

We first collect some facts about ${\ell }_{s}$ and $\stackrel{s}{\le }\text{.}$ Then talk about ${\left({W}_{P}\right)}_{\text{af}}$ and ${\left({W}^{P}\right)}_{\text{af}}\text{.}$

Proposition ${\ell }_{s}\text{:}$ The following are true about ${\ell }_{s}\text{:}$

 (1) ${\ell }_{s}\left(w\right)=w$ $\forall$ $w\in W\text{.}$ (2) ${\ell }_{s}\left(x{t}_{h}\right)={\ell }_{s}\left(x\right)+⟨2\rho ,h⟩$ $\forall$ $x\in {W}_{\text{af}},$ $h\in {Q}^{\vee }\text{.}$ (3) ${\ell }_{s}\left(x{w}_{0}\right)=\ell \left({w}_{0}\right)-{\ell }_{s}\left(x\right)$ $\forall$ $x\in {W}_{\text{af}},$ ${w}_{0}=\text{longest}$ in $W\text{.}$ (4) $-\ell \left(x\right)\le {\ell }_{s}\left(x\right)\le \ell \left(x\right)$ $\forall$ $x\in {W}_{\text{af}},$ $ℓs(x)=ℓ(x) ⟺ x∈Waf+, ℓs(x)=-ℓ(x) ⟺ x∈Waf-.$ (5) For any $x,y\in {W}_{\text{af}}$ $ℓs(xy) = ℓs(y)+ ∑β∈Δ+rex·β<0 sign(y-1·β‾)$ where $sign α = { 1 if α∈Δ‾+, -1 if α∈-Δ‾+.$

Recall $Δ‾+ = the set of roots of the finite 𝔤.$ Proof.

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

(3): Write $x=w{t}_{h}\text{.}$ Then $ℓs(xw0) = ℓ(wthw0)= ℓ(ww0tw0h) = ℓ(ww0)+ ⟨2ρ,w0h⟩ = ℓ(w0)-ℓ(w) -⟨2ρ,h⟩ = ℓ(w0)-ℓs(x).$

(4) We break the proof of (4) into a few parts: We first prove that ${\ell }_{s}\left(x\right)=\ell \left(x\right)$ for $x\in {W}_{\text{af}}^{+}\text{:}$ Assume $x=w{t}_{h}\in {W}_{\text{af}}^{+}\text{.}$ Then by the definition of ${W}_{\text{af}}^{+},$ if $\alpha +n\delta >0$ is s.t. $x·(α+nδ)=wα+ (n-⟨α,h⟩) δ<0$ we must have $\alpha >0\text{.}$ Thus if $w\alpha <0,$ then $n$ can only take values $0,1,\dots ,⟨\alpha ,h⟩$ and if $w\alpha >0,$ then $n$ can only take values $0,1,\dots ,⟨\alpha ,h⟩-1\text{.}$ 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 $B\subset A\text{.}$ Thus $A=B\text{.}$ Hence $ℓ(x)=#B = ∑α>0wα<0 (⟨α,h⟩+1)+ ∑α>0wα>0 ⟨α,h⟩ = ∑α>0⟨α,h⟩+ ∑α>0wα<0 ·1 = ⟨2ρ,h⟩+ ℓ(w) = ℓs(x).$ This shows $ℓs(x) = ℓ(x)for x∈ Waf+.$ We have proved (Lecture 8 that) $-ℓs(x) = ℓ(x)for x∈Waf-.$ To prove that $ℓs(x) ≤ ℓ(x)for all x∈Waf$ we need the following Lemma:

Lemma: Suppose that ${h}_{1}\in {Q}^{\vee }$ is dominant and regular. Then for all $x\in {W}_{\text{af}},$ we have $ℓ(xth1)≤ ℓ(x)+⟨2ρ,h1⟩.$ We will prove the Lemma later. Let's assume the Lemma for now. Let $x\in {W}_{\text{af}}$ be arbitrary. Let ${h}_{1}$ be sufficiently dominant so that $x{t}_{{h}_{1}}\in {W}_{\text{af}}^{+}\text{.}$ Then we have $ℓs(x) = ℓs(xth1)- ⟨2ρ,h1⟩ = ℓ(xth1)- ⟨2ρ,h1⟩ ≤ ℓ(x)+ ⟨2ρ,h1⟩- ⟨2ρ,h1⟩ (Lemma) = ℓ(x).$ This shows that ${\ell }_{s}\left(x\right)\le \ell \left(x\right)$ for all $x\in {W}_{\text{af}}\text{.}$ Now if $\ell \left(x\right)={\ell }_{s}\left(x\right)=\ell \left(w\right)+⟨2\rho ,h⟩$ for $x=w{t}_{h}\in {W}_{\text{af}},$ 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 $⟨\alpha ,h⟩<0,$ then the first set in the union is taken to be $\varnothing \text{.}$ Similarly for the 2nd set) is obviously contained in the set $A = { α+nδ>0:x· (α+nδ)<0 } .$ But $\text{#}B=\ell \left(w\right)+⟨2\rho ,h⟩$ $⇒$ $B=A\text{.}$ So for every $\beta \in {\Delta }_{+}^{\text{re}}\in A$ have $\stackrel{‾}{\beta }>0\text{.}$ This shows that $x\in {W}_{\text{af}}^{+}\text{.}$ Similarly we can show ${\ell }_{s}\left(x\right)\ge -\ell \left(x\right)$ $\forall$ $x\in {W}_{\text{af}}$ and ${\ell }_{s}\left(x\right)=-\ell \left(x\right)$ $⇔$ $x\in {W}_{\text{af}}^{-}\text{.}$ This finishes the proof of (4) (except for the lemma). (Something is not right here).

We now prove (6): $\forall$ $x,y\in {W}_{\text{af}}$ (Do not trust this proof!) $ℓs(xy) = ℓs(y)+ ∑β∈Δ+rex·β<0 sign(y-1·β‾).$ Write $x={w}_{1}{t}_{{h}_{1}},$ $y={w}_{2}{t}_{{h}_{2}}\text{.}$ Then $ℓs(xy) = ℓs(w1w2tw2-1h1+h2) = ℓ(w1w2)+ ⟨2ρ,w2-1h1+h2⟩$ so $ℓs(xy)-ℓs(y) = ℓ(w1w2)-ℓ(w2) +⟨2w1ρ,h1⟩$ so need to show $ℓ(w1w2)-ℓ (w2)+ ⟨2w1ρ,h1⟩ = ∑β∈Δ+rex·β<0 sign(y-1·β‾).$ Notice the special case: $x={w}_{1},$ $y={w}_{2},$ we are saying $ℓ(w1w2)- ℓ(w2) = ∑Δ‾+∋β>0w1·β<0 sign(w2-1·β).$ This is a statement about the finite Weyl group and can be proved by induction on $\ell \left({w}_{2}\right),$ for example. We assume this. Thus need to show $⟨2w2ρ,h1⟩ = ∑β∈Δ+rex·β<0 sign(y-1·β‾)- ∑Δ‾+∋α>0w2·α<0 sign(w2-1·α).$ Let $A = { β=α+nδ>0: x·β<0 } = { β=α+nδ>0: w1α+ (n-⟨α,h1⟩) δ<0 } .$ For $\beta =\alpha +n\delta \in A,$ have $y-1·β = w2-1α+ (n+⟨w2-1α,h2⟩) δ$ so $y-1·β‾ = w2-1α.$ Break $A$ as a disjoint union $A = A1∪A2∪A3∪A4$ 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 $∑α∈Δ‾+ sign(w2-1α)= ∑α>0w2-1α>0·1+ ∑α>0w2-1α<0(-1) =2ρ-2(ρ-w2ρ)= 2w2ρ.$ Similarly, $∑β∈Δ+rex·β<0 sign(y-1·β‾) = ∑β∈A1∪A2∩A3∩A4 sign(y-1·β‾)$ so $∑β∈Δ+rex·β<0 sign(y-1·β‾)- ∑α∈Δ‾+w1·α<0 sign(w2-1·α) = ∑β∈A1 sign(y-1·β‾) = ⟨2w2ρ,h1⟩.$ 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 ${h}_{1}\in {Q}^{\vee }$ is dominant and regular. Then for all $x\in {W}_{\text{af}},$ we have $ℓ(xth1)≤ ℓ(x)+⟨2ρ,h1⟩.$ Proof. Set $A1 = { α+nδ>0: xth1· (α+nδ)<0 } = { α+nδ>0: x·(α+(n-⟨h1,α⟩)δ) <0 } .$ Write ${A}_{1}$ as $A1 = B1∪B2$ 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 $α+(n-⟨α,h1⟩) δ=α′+ (n′-⟨α′,h1⟩) δ ⇒α=α′and n-⟨α,h1⟩ =n′-⟨α′,h1⟩ ⇒α=α′,n= n′.Hence#B1 ≤#A=ℓ(x).$ Define the inclusion map $B2⟶C = { α+nδ>0:α+ nδ-⟨α,h1⟩ δ<0 } = { α+nδ>0:α>0, n=0,1,…, ⟨α,h1⟩-1 } .$ It is clear that $\text{#}c=\sum _{\alpha }⟨\alpha ,{h}_{1}⟩=⟨2\rho ,{h}_{1}⟩$ $⇒#B2≤#C= ⟨2ρ,h1⟩.$ Hence $ℓ(xth1)=#A= #B1+#B2≤ #A+#C=ℓ(x)+ ⟨2ρ,h1⟩.$ $\square$

$\square$

In the next proposition, we collect some facts about $\stackrel{s}{\le }\text{:}$

Proposition $\stackrel{s}{\le }\text{:}$

(1) For $x,y\in {W}_{\text{af}},$ we have $x≤sy ⟺ xth≤yth for sufficiently dominant h ⟺ yt-h≤x t-hfor sufficiently dominant h ⟺ xth≤syth for all h ⟺ yw0≤sxw0 where w0=the longest in W.$
(2) For $z\in {W}_{\text{af}}^{+},$ we have
 (2a) $x\le z$ $⇒$ $x\stackrel{s}{\le }z\text{.}$ (2b) $z\stackrel{s}{\le }y$ $⇒$ $z\le y\text{.}$
(3) For $z\in {W}_{\text{af}}^{-},$ we have
 (3a) $x\le z$ $⇒$ $z\stackrel{s}{\le }x\text{.}$ (3b) $y\stackrel{s}{\le }z$ $⇒$ $z\le y\text{.}$
(4) For $x,y\in {W}_{\text{af}}^{+},$ $x\stackrel{s}{\le }y$ $⇔$ $x\le y\text{.}$
For $x,y\in {W}_{\text{af}}^{-},$ $x\stackrel{s}{\le }y$ $⇔$ $y\le x\text{.}$ Proof.

(1) Only need to prove that $x≤sy ⟺ yt-h≤xt-h for sufficiently dominant h ⟺ y0≤sxw0$

Lemma 1: If $x,y\in {W}_{\text{af}}^{-},$ then $x\le y$ $⇔$ $x{w}_{0}\le y{w}_{0}\text{.}$

Lemma 2: $x\stackrel{s}{\le }y$ $⇔$ ${w}_{0}y\le {w}_{0}x\text{.}$ Proof of Lemma 2. If $\varphi \in {M}_{G/B,{\pi }_{B}\left({h}_{2}-{h}_{1}\right)}^{{w}_{1},{w}_{2}},$ then ${\varphi }_{1}$ defined by $ϕ1(t)≔ ϕ(1t)w0·B$ is in ${M}_{G/B,{\pi }_{B}\left({h}_{2}-{h}_{1}\right)}^{{w}_{0}{w}_{2},{w}_{0}{w}_{1}}\text{.}$ This shows $x\stackrel{s}{\le }y$ $⇔$ ${w}_{0}y\le {w}_{0}x\text{.}$ $\square$

I can not prove (1).

$\square$

Proposition 1: Suppose that $h\in {Q}^{\vee }\text{.}$ Then $h∈Q+∨≔ ∑i∈ℤ ℤ+αi∨ ⟺ id≤swth∀ w∈W ⟺ w≤sw0th∀ w∈W ⟺ x≤sxth∀ w∈Waf.$

Proposition 2: For $\beta \in {\Delta }_{+}^{\text{re}}$ and $x\in {W}_{\text{af}}$ $rβx0.$

Proposition 3: For $\beta \in {\Delta }_{+}^{\text{re}}$ and $\stackrel{‾}{\beta }>0,$ and $x\in {W}_{\text{af}}$ $xrβ0.$

Proposition 4: $x\stackrel{s}{<}y$ $⇒$ ${\ell }_{s}\left(x\right)<{\ell }_{s}\left(y\right)\text{.}$

Proposition 5: If $x\stackrel{s}{\le }y,$ then there exists a sequence of the form $x=x0 with $n\ge 0$ and $\ell \left({x}_{k}\right)={\ell }_{s}\left(x\right)+k$ for $0\le k\le n\text{.}$

Proposition 6: The following are equivalent: For $w\in W$ and $x\in {W}_{\text{af}},$

(a) $w\stackrel{s}{<}x$ and ${\ell }_{s}\left(x\right)=\ell \left(w\right)+1\text{.}$
(b) $x$ is one of the following 2 cases:
 (1) $x=w{r}_{\alpha },$ $\alpha \in {\stackrel{‾}{\Delta }}_{+}$ and $\ell \left(x\right)=\ell \left(w\right)+1\text{.}$ (2) $x=w{r}_{\alpha }{t}_{{\alpha }^{\vee }}=w{r}_{\alpha +\delta },$ where $\alpha \in {\stackrel{‾}{\Delta }}_{+}$ and $\ell \left(x\right)=\ell \left(w\right)-⟨2\rho ,{\alpha }^{\vee }⟩+1\text{.}$
This is related to multiplication by ${H}^{2}$ in the quantum cohomology.

### We now turn to ${\left({W}_{P}\right)}_{\text{af}}$ and ${\left({W}^{P}\right)}_{\text{af}}\text{:}$

Fix a standard parabolic subgroup $P$ of $G\text{.}$ Let $Δ+(P) = { α∈Δ‾+ :𝔤-α∈P } , QP∨ = ∑α∈Δ+(P) ℤα∨.$ Set $(WP)af = { wth:w∈WP, h∈QP∨ } .$ this is the Weylf group of $\stackrel{\sim }{{L}_{P}},$ where ${L}_{P}$ is the Levi-factor of $P\text{.}$

Examples:

 1) $P=B,$ ${W}_{P}=\text{id},$ ${\left({W}_{P}\right)}_{\text{af}}=\text{id.}$ 2) $P=G,$ ${W}_{P}=W,$ ${\left({W}_{P}\right)}_{\text{af}}={W}_{\text{af}}\text{.}$ 3) $P={P}_{i},$ ${W}_{P}=⟨1,{r}_{i}⟩,{\left({W}_{P}\right)}_{\text{af}}=⟨{r}_{{\alpha }_{i}},{r}_{\delta -{\alpha }_{i}}⟩\text{.}$
In general, ${\left({W}_{P}\right)}_{\text{af}}$ is a Coxeter group; It is a subgroup of ${W}_{\text{af}},$ but not a Coxeter subgroup, as seen in the example of $P={P}_{i}\text{.}$

### The Length function ${\ell }_{P}\left(y\right)\text{:}$

As a Coxeter group, ${\left({W}_{P}\right)}_{\text{af}}$ has a Length function $ℓP(y) = # { β>0:β‾∨ ∈QP∨, y·β<0 } .$ Define ${\left({W}^{P}\right)}_{\text{af}}\text{:}$ $(WP)af = { x∈Waf:β>0, β‾∨∈ QP∨ ⇒ x ·β>0 } .$

Proposition: $Waf = (WP)af (WP)af$ ie. each $z\in {W}_{\text{af}}$ 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 ${\stackrel{ˆ}{\pi }}_{P}\text{:}$

(Note: ????? ${\stackrel{ˆ}{\pi }}_{P}$ is what Peterson calls ${\pi }_{P}$ in class).

Proposition ${\stackrel{ˆ}{\pi }}_{P}\text{:}$

 1) ${\stackrel{ˆ}{\pi }}_{P}\left(W\right)={W}^{P}\subset {\left({W}^{P}\right)}_{\text{af}}\subset {\left({W}_{\text{af}}\right)}^{P}$ where ${\left({W}_{\text{af}}\right)}^{P}$ is the set of minimal representatives for ${W}_{\text{af}}/{W}_{P}\text{.}$ 2) ${\stackrel{ˆ}{\pi }}_{P}\left({W}_{\text{af}}^{±}\right)\subset {W}_{\text{af}}^{±}\text{.}$ 3) ${\stackrel{\vee }{\pi }}_{P}\left(z\right)\le z$ for all $z\in {W}_{\text{af}}\text{.}$ 4) For any $z,z\prime \in {W}_{\text{af}},$ $h\in {Q}^{\vee },$ have ${\stackrel{ˆ}{\pi }}_{P}\left(z{t}_{h}\right)={\stackrel{ˆ}{\pi }}_{P}\left(z\right){\stackrel{ˆ}{\pi }}_{P}\left({t}_{h}\right)\text{.}$ ${\ell }_{s}\left({\stackrel{ˆ}{\pi }}_{P}\left(z{t}_{h}\right)\right)={\ell }_{s}\left({\stackrel{ˆ}{\pi }}_{P}\left(z\right)\right)+⟨{C}_{P},h⟩$ where $CP=ρ+wPρ= ∑α∈Δ‾+w0wP·α<0α (wP=longest in W).$ $z\stackrel{s}{\le }z\prime$ $⇒$ ${\stackrel{ˆ}{\pi }}_{P}\left(z\right)\stackrel{s}{\le }{\stackrel{ˆ}{\pi }}_{P}\left(z\prime \right)\text{.}$ ${\stackrel{ˆ}{\pi }}_{P}\left({r}_{\beta }z\right)<{\stackrel{ˆ}{\pi }}_{P}\left(z\right)$ $⟺$ $\stackrel{‾}{{z}^{-1}·\beta }\in \Delta \left(g\text{.}p\right)$ $\left(\subset {\stackrel{‾}{\Delta }}_{+}\right)\text{.}$ ${\stackrel{ˆ}{\pi }}_{P}\left({r}_{\beta }z\right)={\stackrel{ˆ}{\pi }}_{P}\left(z\right)$ $⟺$ $\stackrel{‾}{{z}^{-1}·\beta }\in {Q}_{P}^{+}\text{.}$ ${\stackrel{ˆ}{\pi }}_{P}\left({r}_{\beta }z\right)>{\stackrel{ˆ}{\pi }}_{P}\left(z\right)$ $⟺$ $\stackrel{‾}{{z}^{-1}·\beta }\in -\Delta \left(g/p\right)$ $\left(\subset {\stackrel{‾}{\Delta }}_{+}\right)\text{.}$

Proposition: For $y\in {\left({W}_{P}\right)}_{\text{af}},$ $ℓs,P(y) = ℓs(y)$ where ${\ell }_{s,P}$ is the stable length function for ${\left({W}_{P}\right)}_{\text{af}}\text{.}$

Proposition: For $x\in {\left({W}^{P}\right)}_{\text{af}},$ $y\in {\left({W}_{P}\right)}_{\text{af}},$ $ℓs(xy)= ℓs(x)+ ℓs(y), ℓ(x)+ ℓs(y) ≤ℓ(xy).$

Proposition: Any given $x\in {\left({W}^{P}\right)}_{\text{af}},$ can put $xth∈Waf+, xt-h∈Waf-$ for sufficiently dominant $h\in {\left({Q}^{\vee }\right)}^{{W}_{P}},$ ie. $⟨{\rho }_{i},h⟩\gg 0$ for all $i\in I$ such that ${r}_{i}\notin {W}_{P}\text{.}$

Notation: $(P∼)0 = the identity component of P∼, 𝔐P = G∼/(P∼)0, *P = (P∼)0∈𝔐P, π∘P: 𝔐B ⟶ 𝔐P: g*B ⟼ g·*P.$ Have action of $\Gamma$ on ${𝔐}_{P}\text{:}$ $𝔐P×Γ ⟶ 𝔐P: (g·*P)·t = gt·*P.$ This action is trivial if $t\in \left\{{t}_{h}:h\in {Q}_{P}^{\vee }\right\}\text{.}$ Set, for $z\in {W}_{\text{af}},$ $𝔐P,z± = Baf±z·*P.$

Proposition: For $z\in {W}_{\text{af}}$ and $t\in \Gamma$ $𝔐P,z± = 𝔐P,πˆP(z)±, (𝔐P,z±)·t = 𝔐P,zt±,$ and for ${x}_{1},{x}_{2}\in {\left({W}^{P}\right)}_{\text{af}},$ $𝔐P,x1-∩ 𝔐P,x2+≠∅ ⟺x1≤sx2.$

### The moduli space ${𝔐}_{\tau }={𝔐}_{\tau ,P}\text{:}$

Definition: Given a scheme $V/ℂ$ and a morphism $f: V×ℂℙ1 ⟶ G/P,$ we say that $f$ is of type $\tau ,$ for $\tau \in {H}_{2}\left(G/P\right),$ if for any $ℂ\text{-valued}$ point $v$ of $V,$ the map ${f}_{V}:{ℙ}^{1}\to G/P$ defined by $ℙ1 ≃ ℂ×ℂℙ1 ⟶v×id V×ℂℙ1 ⟶f G/P$ satisfies $(fV)* [ℙ1] = τ.$

The universal property of $\left({𝔐}_{\tau },\text{ev}\right)\text{:}$

Proposition: Fix $\tau \in {H}_{2}\left(G/P\right)\text{.}$ There exists a pair $\left({𝔐}_{\tau },\text{ev}\right)$ where ${𝔐}_{\tau }$ is a reduced scheme of finite type over $ℂ$ and $\text{ev}:{𝔐}_{\tau }{×}_{ℂ}{ℙ}^{1}\to G/P$ is a morphism over $ℂ$ s.t.

 1) ev is of type $\tau \text{;}$ 2) if $V$ is any reduced scheme of finite type over $ℂ$ and $f:V{×}_{ℂ}{ℙ}^{1}\to G/P$ is a morphism over $ℂ,$ then $\exists !$ morphism $\stackrel{ˆ}{f}:V\to {𝔐}_{\tau }$ over $ℂ$ s.t. $f = ev∘ (fˆ×id).$
Thus $\left({𝔐}_{\tau },\text{ev}\right)$ is unique up to a unique isomorphism. Moreover, ${𝔐}_{\tau }$ is quasi-projective, and it is either empty or else smooth and of dim $dim 𝔐τ = dim G/P+ ⟨C1TG/P,τ⟩.$

Here we outline a proof of the fact that the Zariski tangent space to ${𝔐}_{\tau }$ at $\varphi \in {𝔐}_{\tau }$ always has the above dimension: Suppose $ϕ: ℙ1 ⟶ G/P$ is s.t. ${\varphi }_{*}\left[{ℙ}^{1}\right]=\tau \text{.}$ Then $Tϕ𝔐τ = Γ(ℙ1,ϕ*TG/P).$ 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 ${ℙ}^{1}$ by $\varphi ,$ we have $0 ⟶ ϕ*𝔞 ⟶ ϕ*𝔟 ⟶ ϕ*TG/P ⟶ 0.$ Thus we have the long exact sequence $0 ⟶ H0(ℙ1,ϕ*𝔞) ⟶ H0(ℙ1,ϕ*𝔟) ⟶ H0(ℙ1,ϕ*TG/P) ⟶ ⟶ H1(ℙ1,ϕ*𝔞) ⟶ H1(ℙ1,ϕ*𝔟) ⟶ H1(ℙ1,ϕ*TG/P) ⟶ 0.$ Since $𝔟$ is trivial as a vector bundle, have $H1(ℙ1,ϕ*𝔟) = 0 ⇒H1(ℙ1,ϕ*TG/P) = 0 ⇒dim Γ (ℙ1,ϕ*TG/P) = dim H0 (ℙ1,ϕ*TG/P) = χ(ϕ*TG/P).$ Using the general fact that for any vector bundle $E$ over ${ℙ}^{1},$ $χ(E) = dim E+ ⟨C1(E),[ℙ1]⟩.$ We get $dim Γ(ℙ1,ϕ*TG/P) = dim G/P+ ⟨C1(ϕ*TG/P),[ℙ1]⟩ = dim G/P+ ⟨ϕ*C1(TG/P),[ℙ1]⟩ = dim G/P+ ⟨C1(TG/P),ϕ*[ℙ1]⟩ = dim G/P+ ⟨C1(TG/P),τ⟩.$

Now for $\tau \in {H}_{2}\left(G/P\right),$ $v,w\in {W}^{P},$ set $𝔐τv,w = B-v·P ×G/P 𝔐τ×G/P Bw·P.$ By a theorem of Kleiman, we have

Proposition:

 (1) ${𝔐}_{\tau }^{\text{id},{w}_{0}{w}_{P}}$ is open and dense in ${𝔐}_{\tau },$ (2) ${𝔐}_{\tau }^{v,w}$ is quasi-projective, and $dim 𝔐τv,w = ⟨C1(TG/P),τ⟩ -ℓ(v)+ℓ(w).$

Kleiman's Theorem: Suppose $X$ is a homogeneous $G\text{-space}$ and $σY: Y ⟶ X σZ: Z ⟶ X$ are smooth maps. Then for generic ${g}_{1},{g}_{2}\in G,$ the set $g1·Y×X g2·Z = { (y,z):g1 ·σY(y)= g2σZ(z) }$ is a regular reduced variety of $\text{dim}=\text{dim} Y+\text{dim} Z-\text{dim} X\text{.}$

## Lecture 14: April 15, 1997

Today we introduce two rings for each parabolic $P\text{:}$

 1 ${R}_{P}^{\prime }=q{H}^{T}{\left(G/P\right)}_{\left(q\right)}\text{:}$ $T-equivariant$ quantum cohomology of $G/P$ with the quantum parameter $q$ inverted, 2 ${R}_{P}=q{G}_{T}\left(G/P\right)\text{:}$ $T\text{-equivariant}$ quantum cohomology of $G/P\text{.}$

Definition: ${R}_{P}^{\prime }$ is a free $S\text{-module}$ on symbols ${\sigma }_{P}^{\left(x\right)},$ $x\in {\left({W}^{P}\right)}_{\text{af}}$ with $z\text{-grading}$ $deg(sσP(x)) = deg s+2ℓs(x).$

### The ${\underset{_}{A}}_{\text{af}}$ module structure on ${R}_{P}^{\prime }$

The $S\text{-module}$ structure on ${R}_{P}^{\prime }$ extends to an ${\underset{_}{A}}_{\text{af}}\text{-module}$ structure on ${R}_{P}^{\prime }$ by $ν(Ai)· σP(x) = { -σP(rix) if x-1·αi‾ ∈Δ(𝔤_/𝔭_), 0 otherwise$ where $\nu$ is the automorphism of ${\underset{_}{A}}_{\text{af}}$ defined at the end of Lecture 10 (page 10-13).

### The map ${\psi }_{P}:{H}_{T}\left(\Omega K\right)\to {R}_{P}^{\prime }\text{:}$

It is the $S\text{-module}$ map defined by $ψP(σ[x]Ω) = { σP(x) if x∈ (WP)af, 0 otherwise$ for all $x\in {W}_{\text{af}}^{-}\text{.}$

It should be easy to check that

 1) ${\psi }_{P}\left(\sigma \right)=j\left(\sigma \right)·{\sigma }_{P}^{\left(\text{id}\right)}\in {R}_{P}^{\prime }$ $\forall$ $\sigma \in {H}_{T}\left(\Omega K\right),$ 2) ${\psi }_{P}$ is an ${\underset{_}{A}}_{\text{af}}\text{-map.}$

Theorem: There exists a unique commutative $S\text{-algebra}$ structure on ${R}_{P}^{\prime }$ such that

 1) ${\sigma }_{P}^{\left(\text{id}\right)}=1,$ 2) ${R}_{P}^{\prime }$ is an $\Omega \text{-integrable}$ ${\underset{_}{A}}_{\text{af}}\text{-module}$ with the structure homomorphism $S\to {R}_{P}^{\prime }:$ $s↦s{\sigma }_{P}^{\left(\text{id}\right)}$ and the ${\underset{_}{A}}_{\text{af}}\text{-module}$ structure defined above.

The definition of an $\Omega \text{-integrable}$ ${\underset{_}{A}}_{\text{af}}\text{-module}$ is given in Lecture 10. Recall that a proposition in Lecture 10 says that an $\Omega \text{-integrable}$ ${\underset{_}{A}}_{\text{af}}\text{-module}$ is equivalent to an affine scheme $X$ over $\underset{_}{h}=\text{Spec} S$ with a structure morphism ${\pi }_{X}:S\to 𝒪\left(X\right)$ and

 1) an $\underset{_}{A}\text{-module}$ structure on $𝒪\left(X\right),$ 2) an $S\text{-map}$ $f:{H}_{T}\left(\Omega K\right)\to 𝒪\left(X\right)$
such that
 1) $s·p={\pi }_{X}\left(s\right)p$ $\forall$ $s\in S,$ $p\in 𝒪\left(X\right),$ 2) ${\pi }_{X}$ is an $\underset{_}{A}\text{-module}$ map, 4) $m:{𝒪}_{\left(X\right)}{\otimes }_{S}{𝒪}_{\left(X\right)}\to 𝒪\left(X\right)$ is an $\underset{_}{A}\text{-module}$ map, 5) $f:{H}_{T}\left(\Omega K\right)\to 𝒪\left(X\right)$ is an $\underset{_}{A}\text{-module}$ map.
Recall that we have used the notation $𝒰 = Spec HT(K/T), 𝒜 = Spec HT(ΩK).$ Conditions 1)-4) say that $X=\text{Spec} 𝒪\left(X\right)$ is a $𝒰\text{-space,}$ where $𝒰$ is a groupoid, and condition 5) says that $\text{Spec} f:X\to 𝒜$ is a $𝒰\text{-space}$ morphism.

### The geometrical models

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

• $𝒰=\text{Spec}{H}^{T}\left(K/T\right)\text{:}$ $G∨, 𝔤∨, h_=(h_∨)*⊂(𝔤_∨)* etc.$ For each $i\in I,$ let ${{f}_{i}^{\vee }}^{*}\subset {\left({\underset{_}{𝔤}}^{\vee }\right)}^{*}$ be such that $⟨{{f}_{i}^{\vee }}^{*},{f}_{j}^{\vee }⟩={\delta }_{\text{?????}}$ and with weight ${\alpha }_{i}^{\vee }\text{.}$ Let $E=∑i∈I fi∨*∈ (𝔤∨)*.$ Set $𝒰 = { (E+h,u)∈ (E+h_) ×U-∨: u-1· (E+h)∈ (n_+∨)⊥ } .$ Notice that $𝒰$ can be identified with the following subset of $\left(E+\underset{_}{h}\right)×{U}_{-}×\left(E+\underset{_}{h}\right)\text{:}$ $𝒰 = { (E+h1,u,E+h2) :u-1·(E+h1) =E+h2 } .$ It thus has a groupoid structure as a subgroupoid of the direct product groupoid $\left(E+\underset{_}{h}\right)×{U}_{-}×\left(E+\underset{_}{h}\right)\text{:}$
• $𝒜=\text{Spec} {H}_{T}\left(\Omega K\right)={{B}^{\vee }}^{E+\underset{_}{h}}=\left\{\left(E+h,b\right)\in \left(E+\underset{_}{h}\right)×{B}^{\vee }:b·\left(E+h\right)=E+h\right\}\text{.}$
• Action of $𝒰$ on ${{B}^{\vee }}^{E+\underset{_}{h}}\text{:}$ $𝒰×hB∨E+h_ ⟶ B∨E+h_: (E+h,u,E+h′)· (E+h′,b) = (E+h,ubu-1).$
• The variety ${Y}^{E+\underset{_}{h}}\text{:}$ $YE+h_ = { (E+h,g·B∨)∈ (E+h_)× G∨/B∨:g-1 ·(E+h)⊥ [n_+∨,n_+∨] } .$ Have $B∨E+h_ ⟶ YE+h_: (E+h,b) ⟼ (E+h,bw0·B∨).$ $𝒰$ acts on ${Y}^{E+\underset{_}{h}}\text{:}$ $𝒰×hYE+h_ ⟶ YE+h_: (u+h,u,E+h′)· (E+h′,g·B∨) = (E+h,ug·B∨).$ The inclusion $B∨E+h_ ⟶ YE+h_$ is $𝒰\text{-equivariant.}$ $Spec RP′= YP+E+h_∩ YG-E+h_⊂ YE+h (𝒰-subset).$

### The subring ${\Lambda }_{P}^{\prime }\subset {R}_{P}^{\prime }\text{:}$

For $h\in {Q}^{\vee },$ so ${\pi }_{P}\left(h\right)\in {H}_{2}\left(G/P\right),$ set $qπP(h) = σP(πˆP(th)) ∈RP′,$ and $ΛP′=ℤ {qπP(h):h∈Q∨} ≃ℤ[H2(G/P)].$

Fact:

• ${\Lambda }_{P}^{\prime }\subset {R}_{P}^{\prime }$ is a subring with $qτqτ′ = qτ+τ′ deg qτ = 2⟨C1TG/P,τ⟩.$
• ${q}_{{\pi }_{P}\left(h\right)}·{\sigma }_{P}^{\left(x\right)}={\sigma }_{P}^{\left(x{\stackrel{ˆ}{\pi }}_{P}\left({t}_{h}\right)\right)}={\sigma }_{P}^{\left({\stackrel{ˆ}{\pi }}_{P}\left(x{t}_{h}\right)\right)}\text{.}$
• The ${\underset{_}{A}}_{\text{af}}\text{-module}$ structure on ${R}_{P}^{\prime }$ is ${\Lambda }_{P}^{\prime }\text{-linear.}$

Example $\text{(}{\stackrel{ˆ}{\pi }}_{P}\left({t}_{h}\right)$ is not necessarily translational): ${sl}_{3}$ with extended Dynkin diagram $\begin{array}{c} 0 1 2 \end{array}\text{.}$ Let ${W}_{P}=\text{?????}$ and $t=tθ=r0rθ= r0r2r1⏟⫙(WP)af r2⏟⫙(WP)af$ $⇒$ ${\stackrel{ˆ}{\pi }}_{P}\left(t\right)={r}_{0}{r}_{2}{r}_{1}\text{.}$

Fact: $\left\{{\sigma }_{P}^{\left(w\right)}:w\in {W}^{P}\right\}$ is a basis of ${R}_{P}^{\prime }$ over $S×{\Lambda }_{P}^{\prime }\text{.}$

Proposition: Formulas for multiplications $*$ in ${R}_{P}^{\prime }$ and ${\underset{_}{A}}_{\text{af}}$ on ${R}_{P}^{\prime }\text{:}$ $Ai·(sσ) = (Ai·s)σ+ (ri·s) (Ai·σ) Ai·σP(w) = { -σP(riw) if w-1·αi ∈Δ(𝔤_/𝔭_), 0 otherwise, (Ai·σ)*σ′ = Ai· [σ*(ri·σ′)] +σ*(Ai·σ′).$

### The operator ${A}_{0}^{\prime }\text{:}$

Assume that $G$ is simple. ${\alpha }_{0}=\delta -\theta ,$ ${\Pi }_{\text{af}}=\Pi \cup \left\{0\right\},$ $A0′=ν(A0) =-w0A0w0$ where ${w}_{0}\in W$ is the longest element.

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

Theorem: ${\psi }_{P}:{H}_{T}\left(\Omega K\right)\to {R}_{P}^{\prime }$ is a homomorphism of $S\text{-algebras,}$ and $ψP(σ)*σ′ = j(σ)·σ′$ for $\sigma \in {H}_{T}\left(\Omega K\right)$ and $\sigma \prime \in {R}_{P}^{\prime }\text{.}$ Proof. This is a direct consequence of ${R}_{P}^{\prime }$ being $\Omega \text{-integrable.}$ $\square$

### The structure constants ${J}_{P,z}^{x,y},$$x,y,z\in {\left({W}^{P}\right)}_{\text{af}}\text{:}$

For $x,y,z\in {\left({W}^{P}\right)}_{\text{af}},$ define structure constants ${J}_{P,z}^{x,y}\in S$ by $σP(x)× σP(y) = ∑z∈(WP)af JP,zx,y σP(z).$

Facts:

 (1) $\text{deg} {J}_{P,z}^{x,y}=2\left({\ell }_{s}\left(x\right)+{\ell }_{s}\left(y\right)-{\ell }_{s}\left(z\right)\right),$ (2) ${J}_{P,z}^{x,y}={J}_{B,z}^{x,y},$ (3) ${J}_{P,z{\pi }_{P}\left(tt\prime \right)}^{x{\stackrel{ˆ}{\pi }}_{P}\left(t\right),y{\stackrel{ˆ}{\pi }}_{P}\left(t\prime \right)}={J}_{P,z}^{x,y},$ (4) For $x,y,z\in {\left({W}^{P}\right)}_{\text{af}}$ with $x\in {W}_{\text{af}}^{-},$ $JP,zx,y = { ϵ(xyz) jxzy-1 if ℓ(yz-1) +ℓs(z)= ℓs(y), 0 otherwise.$

### Multiplication by ${H}^{2}$ in ${R}_{B}^{\prime }\text{:}$

Theorem: For $i\in I$ and $w\in W$ $σB(ri)* σB(w) = ∑ α∈Δ‾+ ℓ(wrα)=ℓ(w)+1 ⟨ρi,α∨⟩ σB(wrα) +∑α∈Δ‾+ℓ(wrα)=ℓ(w)+1-⟨2ρ,α∨⟩ ⟨ρi,α∨⟩ qπB(α∨) σB(wrα) -(ρi-w·ρi) σB(w).$

Remark: One way of looking at the above formula is $σB(ri)+ρi = (ρi)R+ ∑α∈Δ‾+ℓ(rα)=⟨2ρ,α∨⟩-1 ⟨ρi,α∨⟩ qπB(α∨) Arα,$ where the left hand side is a multiplication operator on ${R}_{B}^{\prime }$ and the right hand side is an element in ${\underset{_}{A}}_{\text{af}}$ considered as an operator on ${R}_{B}^{\prime }\text{.}$ The right hand side is a commuting family of elements in ${\underset{_}{A}}_{\text{af}}\text{.}$

### A fact with no classical analogy:

$A_af ⊗ΛB- ΛB′ ≅ End[RB′]W RB′where ΛB-= ∑h∈Q∨h dominant ℤq????? [RB′]W ≅ EndA_af⊗ΛB-ΛB′ RB′ EndA_⊗ΛB′ RB′ ≅ A_R⊗ ΛB′$

### The ring ${R}_{P}$

Define $RP = ∑x∈(WP)afx≥sid sσP(x).$ (Recall that $x=w{t}_{h}\stackrel{s}{\ge }\text{id}$ $⇔$ $h\in {Q}_{+}^{\vee }\text{).}$ It is clear from the way $\underset{_}{A}$ acts that ${R}_{P}$ is an $\underset{_}{A}\text{-stable}$ submodule of ${R}_{P}^{\prime }\text{.}$

Fact: For $z\in {W}_{\text{af}},$ $∑x∈(WP)afx≥sz sσP(x)$ is an ${R}_{P}\text{-submodule}$ of ${R}_{P}^{\prime }\text{.}$

Let $ΛP=ΛP′ ∩RP=∑τ∈πP(Q+∨) zqτ.$ Then $RP⊗ΛP ΛP′ ≅ RP′$ and $\left\{{\sigma }_{P}^{\left(w\right)}:w\in {W}^{P}\right\}$ is an $S\otimes {\Lambda }_{P}\text{-basis}$ of ${R}_{P}\text{.}$ The augumentation homomorphism is defined to be $ε: ΛP ⟶ ℤ: ε(qτ) = δτ,0.$

Fact: The map $RP⊗ΛPℤ ⟶∼ HT(G/P) σP(w)⊗1 ⟼ σP(w)$ is an isomorphism as $\underset{_}{A}\text{-modules}$ and $S\text{-algebras.}$

Thus it is reasonable to call ${R}_{P}$ the $T\text{-equivariant}$ quantum cohomology of $G/P\text{.}$ It specializes to the $T\text{-equivariant}$ cohomology of $G/P$ when the quantum parameters ${\Lambda }_{P}$ go to $0\text{.}$

### Poincare Duality

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

Define the $S\otimes {\Lambda }_{P}\text{-linear}$ map $∫: RP ⟶ S⊗ΛP$ by $∫σP(w) = δw,w0wP for w∈WP.$

Theorem: $\int {\sigma }_{P}^{\left(v\right)}*\left({w}_{0}·{\sigma }_{P}^{\left({w}_{0}w{w}_{P}\right)}\right)={\delta }_{v,w}\text{.}$

Corollary: Have an isomorphism $PD: RP ≃ HomS⊗ΛP(RP,S⊗ΛP)$ defined by $PD(φ)(φ′) = ∫φ×φ′,$ or concretely $PD(σP(w)) = w0· σP(w0wwP).$

### The Euler Class ${\chi }_{G/P}\text{:}$

$χG/P =def PD-1(trRP/S⊗ΛP)$ where $trRP/S⊗ΛP ∈HomS⊗ΛP (RP,S⊗ΛP)$ is defined by $trRP/S⊗ΛP (ϕ) = trace over S⊗ΛP of (ℓφ:ϕ′↦φ*φ′).$ In other words, $trRP/S⊗ΛP (ϕ) = ∫ϕ*χG/P.$ Write $σP(v)* σP(w) = ∑u∈WP buv,w σP(u).$ Then $trRP(S⊗ΛP) (σP(v)) = ∑w∈WP bwv,w = ∑w∈WP∫ σP(v)* σP(w)* (w0·σP(w0wwP))$ $⇒χG/P = ∑w∈WP σP(w)* (w0·σP(w0wwP)).$

Facts:

 1) $\varphi *{\chi }_{G/P}=0$ $⇔$ $\varphi$ is nilpotent, 2) ${\chi }_{G/P}$ annihilates ${\Omega }_{{R}_{P}/S\otimes {\Lambda }_{P}}\text{.}$

Example: For $sl\left(3\right)$ and $P=B,$ ${\chi }_{G/B}$ is invertible $⇔$ ${q}_{1}{q}_{2}\left({q}_{1}+{q}_{2}\right)$ is invertible.

## Lecture 15: April 16, 1997

More facts on ${R}_{B}\text{:}$

Fact 1: For $w\in W,$ $∑u,v∈Wuv=w [red] ϵ(u)σB(u-1) *σB(v) = δw,1, ∑u,v∈Wuv=w [red] σB(u)* ϵ(v)σB(v-1) = δw,1.$

Remark: Recall from Lecture 7 that similar identities hold for ${H}^{T}\left(K/T\right)\text{.}$ They can now be considered as a corollary of this fact here about $q{H}^{T}\left(K/T\right)\text{.}$ Does this follow from any Hopf algebroid structure on $q{H}^{T}\left(K/T\right)\text{?}$

Fact 2: For $\sigma \in {R}_{B},$ $σ = ∑w∈W [ Aw0· (σ·(w0·σB(w0w))) ] *ϵ(w) σB(w).$ What does this mean? This is not expressing $\sigma$ in the basis $\left\{ϵ\left(w\right){\sigma }_{B}^{\left(w\right)}:w\in W\right\}$ of ${R}_{B}$ as an $S\otimes {\Lambda }_{B}\text{-module.}$

Fact 3: ${R}_{B}$ is a free ${\left({R}_{B}\right)}^{\underset{_}{A}}\text{-module}$ with basis $\left\{{\sigma }_{B}^{\left(w\right)}:w\in W\right\}\text{.}$

Fact 4: ${\left({R}_{B}\right)}^{\underset{_}{A}}$ is a polynomial ring on the ${q}_{{\pi }_{B}\left({\alpha }_{i}^{\vee }\right)}\text{'s}$ and the ${\sigma }_{B}^{\left({r}_{i}\right)}+{\rho }_{i}$ for $i\in I\text{.}$

Fact 5: ${\left({R}_{B}\right)}^{\underset{_}{A}}\to ℤ{\otimes }_{S}{R}_{B}$ is onto over $ℚ\text{.}$ (?)

### The $S\text{-subalgebra}$${R}_{P}^{-}$ of ${R}_{P}^{\prime }\text{:}$

Define $RP-=Im ψP= ∑x∈(WP)af∩Waf- SσP(x).$ Then $RB- ≃ HT(ΩK)$ but in general $HT(ΩK) ⇒ ⇒ RP-.$ We have:

• ${R}_{P}^{-}={\underset{_}{A}}_{\text{af}}·{\sigma }_{P}^{\left(\text{id}\right)}\text{.}$
• Every ${\underset{_}{A}}_{\text{af}}\text{-submodule}$ of ${R}_{P}^{\prime }$ is an ${R}_{P}^{-}\text{-submodule}$ of ?????.
• $RP-⊗ΛP- ΛP′ ≅ RP′$ where $ΛP-=ΛP′ ∩RP-= ∑h∈Q∨h is dominant ℤqπP(h).$

Remark: Working with the case when $G$ is simple, connected but not necessarily simply connected so $\Omega K$ is no longer connected, we get the following fact: Assume that ${a}_{i}=1$ for all $i\in I$ in $\theta =\sum _{i\in I}{a}_{i}{\alpha }_{i}\text{.}$ Let $P={P}_{{\rho }_{i}}$ so ${W}_{P}={⟨{r}_{j}⟩}_{j\in I,j\ne i}\text{.}$ Let $w={w}_{0}{w}_{P}\text{.}$ Let $Q$ be a standard parabolic Then $σQπˆQ(w) *(wσQ(v)) = qπQ(ρi∨-v-1ρi∨) σQ(πˆQ(wv))$ for all $v\in {W}^{Q}\text{.}$ Consequently ${\sigma }^{\left({\stackrel{ˆ}{\pi }}_{Q}\left(w\right)\right)}$ is invertible in ${R}_{Q}^{\prime }$ (no clue! How is $Q$ related to $P={P}_{{\rho }_{i}}\text{?)}$

Example: $G={SL}_{3},$ (?) ${W}_{P}={r}_{2},$ $G/P={ℙ}^{2}\text{.}$ ${\sigma }^{21}×{\sigma }^{21}×{\sigma }^{21}={q}^{2}$ (?).

### A Filtration:

For $h\in {Q}^{\vee },$ define an $\underset{_}{A}\text{-submodule}$ ${F}_{P,h}^{-}$ of ${R}_{P}^{-}$ (depends only on $h$ mod ${Q}_{P}^{\vee }\text{)}$ by $FP,h-=RP-∩ q-πP(h) RP= ∑x∈(WP)af∩Waf-x≥sπˆP(t-h) SσP(x)$ (a finite sum). Then $FP,h-* FP,h′-⊂ FP,h+h′-.$

Remark: In the geometric models to be given later, elements of ${F}_{P,h}^{-}$ correspond to trivializing certain line bundles on the (Peterson) variety $Y\text{.}$ (a $𝒴\text{).}$

Fact: When $h\in {Q}^{\vee }$ is dominant, $FP,h- = ψP(Tt-ω(h))$ where ${F}_{{t}_{-\omega \left(h\right)}}$ is the Bruhat-Filtration in ${H}_{T}\left(\Omega K\right)$ in Lecture 9 and $\omega \left(h\right)=-{w}_{0}·h$ is the diagram automorphism. Have

• ${F}_{P,h}^{-}*{F}_{P,h\prime }^{-}={F}_{P,h+h\prime }^{-}\text{.}$
• $\sigma *{F}_{P,h}^{-}\subset {F}_{P,h\prime }^{-}$ $⇔$ $\sigma \in {F}_{P,h\prime -h}^{-}\text{.}$

### More on $G/B$ and $G/P\text{:}$

Fix parabolic $P$ and $Q$ s.t. $G\supset P\supset Q\text{.}$ Recall a (classical) fact on ${H}^{•}\left(G/Q\right)\text{:}$ the fibration $P/Q ⟶ G/Q ↓ G/P$ gives rise to a filtration on ${H}^{•}\left(G/Q\right)$ such that $Gr H*(G/Q) ≃ H*(P/Q)⊗ H*(G/P).$

An analogous statement is true for quantum cohomology:

Consider the $S\text{-algebra}$ $RP,Q = ∑x∈(WP)afy∈(WQ)afx≥sid sσQ(xy).$ Let $R≤nP,Q = ∑ x∈(WP)af y∈(WQ)af x≥sid ℓs(y)≤n sσQ(xy).$

Fact: $R≤mP,Q R≤nP,Q⊂ R≤m+nP,Q.$

Define $R‾P,Q= gr RP,Q= ∑n∈ℤ R‾nP,Q$ where $R‾nP,Q = R≤nP,Q/ R≤(n-1)P,Q.$

Fact: $RG/P⏟=RP ⊗ℤ (ℤ⊗SRPQ′) ≃ R‾P,Q.$

Define $R-P,Q = ∑ x∈(WP)af x≥sid y∈(WQ)af∩(WP)af- sσQ(xy), R-nP,Q = ∑ x∈(WP)af x≥sid y∈(WQ)af∩(WP)af- ℓs(y)≤n sσQ(xy).$

Fact: $gr R-P,Q ≃ RG/P⏟=RP⊗ [ Im ( H*Ω0 (K∩P)⟶ ℤ⊗SRQ′ ) ] .$

Corollary: If $ℤ{\otimes }_{S}{R}_{P/Q}$ and $ℤ{\otimes }_{S}{R}_{\text{?????}/Q}$ are reduced, then $ℤ\otimes {R}_{G/Q}$ is reduced.

Fact:

• For $G=SL\left(n,ℂ\right)$ every ${R}_{G/P}={R}_{P}$ is reduced.
• Other cases where very ${R}_{P}$ is reduced are: ${G}_{2},$ ${B}_{2},$ ?????

Remark (from informal lecture in the common room after the lecture): Look at the case $G\supset P\supset B\text{.}$ The fact $gr R-P,B≃ RP⊗ (Im(H*Ω0(K∩P)⟶ℤ⊗SRB′)) (*)$ has the following meaning in terms of the geometric models: Recall the (Peterson) variety $Y\subset {G}^{\vee }/{B}^{\vee }\text{.}$ It contains ${2}^{\ell }-T\text{-fixed}$ points $\left\{{w}_{P}:P \text{parabolic}\right\}\text{.}$ Label them by ${y}_{P}\text{.}$ Set $YP+ = Y∩B-∨wP· B∨, YP- = Y∩B+∨wP·B∨ (B+∨=B∨).$ Then $RP ≃ 𝒪(YP+), H*(Ω0(K∩P)) ≃ 𝒪(YP-).$ Can think of $\text{gr} {R}_{-}^{P,B}$ as the subring of $𝒪\left({Y}_{G}^{-}\cap {Y}_{B}^{+}\right)$ that are regular at ${y}_{P}$ (not quite sure this is true) so $\left(*\right)$ says that near ${y}_{P},$ the variety $P$ looks like ${Y}_{P}^{+}×{Y}_{P}^{-}\text{.}$

### The quantum cohomology $q{H}^{•}\left(G/P\right)\text{:}$

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

For $n\ge 3,$ consider the open subscheme ${V}_{n}\left(ℂ\right)$ of ${\left({ℙ}_{ℂ}^{1}\right)}^{n}\text{:}$ $Vn(ℂ) = { (z1,…,zn)∈ (ℙℂ1)n: zi≠zj, i≠j , z1=∞, z2 =0, z3=1 } .$ For $\tau \in {H}_{2}\left(G/P\right),$ let $𝔐τ = {ϕ:ℙ1→G/P:ϕ*[ℙ1]=τ} , 𝔐n,τ = 𝔐τ×Vn(ℂ)$ so $dim 𝔐n,τ = ⟨C1(TG/P),τ⟩ +dim G/P+n-3.$ Set $ev: 𝔐n,τ ⟶ (G/P)n: ev(ϕ,z1,…,zn) = (ϕ(z1),ϕ(z2),⋯,ϕ(zn)).$ Roughly speaking, ${𝔐}_{n,\tau }$ admits a compactification $\stackrel{‾}{{𝔐}_{n,\tau }}$ which admits a fundamental class $\left[{\stackrel{‾}{M}}_{n,\tau }\right]\text{.}$ (Manin-Konstevich).

Now for ${\varphi }_{1}\otimes \cdots \otimes {\varphi }_{n}\in {H}^{•}\left({\left(G/P\right)}^{n}\right)\simeq H{\left(G/P\right)}^{\otimes n},$ have $∫𝔐n,z‾ ev*(ϕ1⊗⋯⊗ϕn) ∈ℤ(or ℂ?)$ Using Poincare duality, can regard above as giving a $ℤ\text{-linear}$ map $Jn,τ: ⊗n-1H•(G/P) ⟶ H•(G/P)$ of degree $=-2\left[⟨{C}_{1}\left({T}_{G/P}\right),\tau ⟩+\left(n-3\right)\right]\text{.}$ In other words, for any $n$ subvariety ${X}_{1},\dots ,{X}_{n}$ of $G/P$ with $∑i=1n codim Xi = 2dimℂ𝔐n,τ$ we have $⟨ Jn,τ ( PD-1[X1] ⊗⋯⊗ PD-1[Xn-1] ) ,[Xn] ⟩ = # ( 𝔐n,τ‾ ×(G/P)n ( g1X1×⋯× gnXn ) ) (ℂ)$ for all $\left({g}_{1},\dots ,{g}_{n}\right)$ in a dense open subset of ${\left(G\left(ℂ\right)\right)}^{n}\text{.}$ These numbers are the Gromov-Witten invariants.

Fact: For $\varphi \in {H}^{2}\left(G/P\right)$ and $n\ge 4$ $Jn,τ ( ϕ1⊗⋯⊗ ϕn-2⊗ϕ ) = ⟨ϕ,τ⟩ Jn,τ ( ϕ1⊗⋯⊗ ϕn-2 ) .$

Now let $D=ℚ\left[\left[\epsilon \right]\right]$ with indeterminant $\epsilon \text{.}$ Given $\nu \in \epsilon \left({H}^{*}\left(G/P\right){\otimes }_{ℤ}D\right),$ can make ${H}^{*}\left(G/P\right){\otimes }_{ℤ}D$ into a commutative associative $D\text{-algebra}$ with unit ${\sigma }_{P}^{\text{id}}$ with quantum product ${*}_{\nu }$ by $σ*νσ′ = ∑n,τ Jn,τ ( σ⊗σ′⊗ νn-3(n-3)! )$ where ${\nu }^{n-3}=\nu \otimes \cdots \otimes \nu$ $\text{(}\left(n-3\right)\text{-times).}$ In particular, for $\varphi \in {H}^{2}\left(G/P\right),$ define $σ*εϕσ′ = ∑τJ3,τ (σ⊗σ′) exp ε⟨ϕ,τ⟩.$ The "potential" function for ${J}_{3,\tau }\text{"}$ satisfy WDVV-equation.

### The small quantum cohomology:

Make ${H}^{*}\left(G/P\right){\otimes }_{ℤ}{\Lambda }_{P}$ into a ${\Lambda }_{P}\text{-algebra}$ $q{H}^{*}\left(G/P\right)$ by $σ*σ′ = ∑τ∈πP(Q+∨) qτJ3,τ (σ⊗σ′).$

Theorem:

 (1) $*$ is associative. (2) $q{H}^{*}\left(G/P\right)$ is $ℤ\text{-graded.}$ (3) For $i\in I,$ $w\in W$ $σBri* σBw = ∑α∈Δ‾+ℓ(wrα)=ℓ(w)+1 ⟨ρi,α∨⟩ σBwrα +∑α∈Δ‾+ℓ(wrα)=ℓ(w)+1-⟨2ρ,α∨⟩ ⟨ρi,α∨⟩ qπB(α∨) σBwrα.$

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 $q{H}^{•}\left(G/P\right)$ and $q{H}^{•}\left(G/B\right)\text{:}$

Let $\tau \in {H}_{2}\left(G/P\right)\text{.}$ Then there exists a unique $h\in {Q}^{\vee }$ s.t. $πP(h) = τ$ and $-1\le ⟨\alpha ,h⟩\le 0$ for all $\alpha \in -\Delta \left(\underset{_}{𝔭}/\underset{_}{𝔟}\right)\text{.}$

Define a standard parabolic ${P}_{1}\subset P$ by $Δ(p1/b) = { α∈Δ(𝔭_/𝔟_) :⟨α,h⟩=0 } .$ 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 $ℤ{\otimes }_{S}{R}_{P}\simeq q{H}^{*}\left(G/P\right)\text{.}$

## 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.