Quantum Cohomology of G/P

Quantum Cohomology of $G / P$

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

Last update: 18 December 2013

Lecture 11: March 26, 1997

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

Fact: Since $G / P$ is projective and thus proper, we have $\begin{matrix} Mor (ℙ^{1}, G / P) & ≃ & Mor (ℙ^{1} \ {a finite set}, G / P) . \end{matrix}$ In particular $\begin{matrix} Mor (ℙ^{1}, G / P) & ≃ & Mor (ℂ^{\times}, G / P), (ℂ^{\times} = ℙ^{1} \ {0, \infty}) . \end{matrix}$

Lemma: Let $G'$ be a linear algebraic group. Then every principal $G' -bundle$ over $𝔸^{1} \supset ℂ$ is trivial, so it admits a section.

Proof.

W.L.O.G., assume that $G'$ is connected. Let $\begin{matrix} G' & ⟶ & E \\ ↓ \\ 𝔸^{1} \end{matrix}$ be a principal $G' -bundle.$ Let $B' \subset G'$ be a Borel subgroup of $G .$ Then have bundle $\begin{matrix} E / B' \\ ↓ \\ 𝔸^{1} \end{matrix}$ with fibre $G / B'$ which always admits a rational section. Since $G / B'$ is proper, we actually have a morphism $s : 𝔸^{1} \to E / B' .$ Now form the principal $B' -bundle$ $\begin{matrix} E_{new} & = & 𝔸^{1} \times_{E / B'} E \\ ↓ \\ 𝔸^{1} & ≃ & A^{1} \times_{E / B'} E / B' \end{matrix}$ It remains to show that $E_{new}$ has a section. Consider the normal series of $B' :$ $B' \supset B_{k} \supset B_{k - 1} \supset \dots \supset B_{0} = 0 dim B_{r} = r .$ $B_{r} / B_{r - 1}$ is abelian so is either $G_{a} = (ℂ, additive)$ or $G_{m} = (ℂ^{\times}, multiplicative) .$

Case 1 $-$ $G_{m} :$ Since the only line bundle over $𝔸^{1}$ is the trivial one, the associated line bundle over $𝔸^{1}$ is trivial.

Case 2 $-$ $G_{a} :$ Since $𝔸^{1}$ is affine, $H^{1} (𝔸^{1}, 𝒪_{𝔸^{1}}) = 0$ which is the obstruction for a $G_{a} -bundle$ to be trivial. $(H^{1} (𝔸^{1}, G_{a}) = H^{1} (𝔸^{1}, 𝒪_{𝔸^{1}}) = 0).$

$□$

Recall notation: for a variety $X$ over $ℂ,$ $\tilde{X} = Mor (ℂ^{\times}, X) .$

Theorem 1: The map $\begin{matrix} {\tilde{π}}_{P} : & \tilde{G} & ⟶ & \tilde{G / P} & = & Mor (ℙ^{1}, G / P) \\ g (t) & ⟼ & g (t) P \end{matrix}$ is surjective.

Proof.

Given $ϕ \in Mor (𝔸^{1}, G / P) ≃ Mor (ℙ^{1}, G / P),$ form the principal $P -bundle$ over $𝔸^{1} :$ $E = {(t, g) \in 𝔸^{1} \times G : ϕ (t) ≃ g P}$ with $P$ acting on the copy of $G$ from the right by right multiplications. By Lemma, $E$ admits a section, ie. $\exists$ $s : 𝔸^{1} \to E :$ $s (t) = (t, g (t)) \in E .$ Thus $g (t) \in Mor (𝔸^{1}, G) \in \tilde{G}$ is a lift of $ϕ .$ Similarly, can show that can also lift $ϕ$ to some $g' (t) \in Mor (ℙ^{1} \ {0}, G) .$

$□$

Next, we study the degree of the curve ${\tilde{π}}_{P} (g) \in Mor (ℙ^{1}, G / P)$ for $g \in \tilde{G} = Mor (ℂ^{\times}, G) .$

Recall notation

(1) For a variety $X$ over $ℂ,$ have $\begin{matrix} \tilde{X} & = & Mor (ℂ^{\times}, X) \end{matrix}$ and $\begin{matrix} {(\tilde{X})}_{0} & = & {ϕ \in \tilde{X} : ϕ |_{S^{'}} is trivial in π_{1} (X)} . \end{matrix}$ For example, for $s l (2, ℂ),$ $\begin{matrix} \tilde{B} & = & {(\begin{matrix} a (t) & b (t) \\ 0 & d (t) \end{matrix}) : a, b, d \in ℂ [t, t^{- 1}], a d = 1} . \end{matrix}$ Now $a, d \in ℂ [t, t^{- 1}]$ (Laurent polynomials) and $a d = 1 \Rightarrow a = λ t^{k}, d = \frac{1}{λ} t^{- k} .$ But must have $k = 0$ in order for $g (t) = (\begin{matrix} a (t) & b (t) \\ 0 & d (t) \end{matrix}) \in {(\tilde{B})}_{0} .$ Thus $\begin{matrix} {(\tilde{B})}_{0} & = & {(\begin{matrix} λ & b (t) \\ 0 & \frac{1}{λ} \end{matrix}) : λ \in ℂ^{\times}, b \in ℂ [t, t^{- 1}]} . \end{matrix}$ This is true in general: $\begin{matrix} {(\tilde{B})}_{0} & = & H ⋉ \tilde{U_{+}} . \end{matrix}$

Remark: Compare with $\begin{matrix} B_{af} & = & {(\begin{matrix} a (t) & b (t) \\ c (t) & d (t) \end{matrix}) : a, b, c, d \in ℂ [t], a d - b c = 1, c (0) = 0} . \end{matrix}$ Very different from ${(\tilde{B})}_{0} .$

(2) $\begin{matrix} π_{P} : & Q^{\lor} & ⟶ & H_{2} (G / P) \\ π_{P} (α_{i}^{\lor}) & = & {\begin{matrix} σ_{r_{i}}^{G / P} & if r_{i} \notin W_{P}, \\ 0 & if r_{i} \in W_{P} . \end{matrix} \end{matrix}$

Theorem 2:

(A)	Let $w_{1}, w_{2} \in W$ and $h_{1}, h_{2} \in Q^{\lor} .$ If $g \in B_{af}^{-} w_{1} t_{h_{1}} {(\tilde{B})}_{0} \cap B_{af} w_{2} t_{h_{2}} {(\tilde{B})}_{0},$ then $ϕ ≔ {\tilde{π}}_{B} (g) \in Mor (ℙ^{1}, G / B)$ satisfies $\begin{matrix} ϕ_{*} [ℙ^{1}] & = & π_{B} (h_{2} - h_{1}), \\ ϕ (\infty) & \in & B_{-} w_{1} \cdot B, \\ ϕ (0) & \in & B w_{2} \cdot B . \end{matrix}$
(B)	We have two disjoint unions: $\tilde{G} = ⨆_{x \in W_{af}} B_{af}^{-} x {(\tilde{B})}_{0} = ⨆_{y \in W_{af}} B_{af} y {(\tilde{B})}_{0} .$ Here, recall $\begin{matrix} B_{af} & = & {g \in Mor (ℙ^{1} \ {\infty}, G) : g (0) \in B}, \\ B_{af}^{-} & = & {g \in Mor (ℙ^{1} \ {0}, G) : g (0) \in B^{-}} . \end{matrix}$

Proof.

Proof of (A): Since $g \in B_{af}^{-} w_{1} t_{j_{1}} {(\tilde{B})}_{0},$ we can write $\begin{matrix} g (t) & = & b_{-} (t) n_{1} t^{- h_{1}} a_{1} u_{1} (t), t \in ℂ^{\times}, \end{matrix}$ where $b_{-} (t) \in B_{af}^{-},$ $u_{1} (t) \in {\tilde{U}}_{+},$ $a_{1} \in H$ and $n_{1}$ is a representative of $w_{1}$ in $G .$ Then by definition $\begin{matrix} ϕ (t) & = & g (t) \cdot B = b_{-} (t) w_{1} \cdot B, t \in ℂ^{\times} . \end{matrix}$ Since $b_{-} \in Mor (ℙ^{1} \ {0}, G)$ and $b_{-} (\infty) \in B_{-},$ we have $ϕ (\infty) \in B_{-} w_{1} \cdot B .$ Similarly, $ϕ (0) \in B w_{2} \cdot B .$ It remains to calculate $ϕ_{*} [ℙ^{1}] \in H_{2} (G / B) .$ We do this by calculating $⟨ ϕ_{*} [ℙ^{1}], λ ⟩$ for every dominant integral $λ \in {\underline{h}}^{*}$ considered as an element in $H^{2} (G / B) .$ 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 $v_{λ}^{+} \in V (λ) .$ Then we have the morphism $\begin{matrix} J : & G / B & ⟶ & ℙ (V (λ)) : & g \cdot B & ⟼ & ℂ g \cdot v_{λ}^{+} \end{matrix}$ and $λ \in H^{2} (G / B)$ is the pullback by $J$ of the standard generator of $𝔸^{2} (ℙ (V (λ))) .$ Thus $\begin{matrix} ⟨ ϕ_{*} [ℙ^{1}], λ ⟩ & = & the degree of J \circ ϕ : ℙ^{1} ⟶ ℙ (V (λ)) . \end{matrix}$ Using $\begin{matrix} g (t) & = & b_{-} (t) n_{1} t^{- h_{1}} a_{1} u_{1} (t), t \in ℂ^{\times}, \end{matrix}$ we have $\begin{matrix} g (t) \cdot v_{λ}^{+} & = & t^{- ⟨ λ, h_{1} ⟩} a_{1}^{λ} b_{-} (t) n_{1} \cdot v_{λ}^{+}, t \in ℂ^{\times}, \end{matrix}$ so in any chosen homogeneous coordinates, we can write $\begin{matrix} (J \circ ϕ) (t) & = & [V_{0} (t), V_{1} (t), \dots, V_{ℓ} (t)] \end{matrix}$ where each $V_{j} (t) \in ℂ [t, t^{- 1}]$ and has degree at most $- ⟨ λ, h_{1} ⟩$ and the degree $- ⟨ λ, h_{1} ⟩$ occurs. Similarly, using the fact that $g \in B_{af}^{+} w_{2} t_{h_{2}} {(\tilde{B})}_{0}$ we see that the minimal degree of the $V_{j} (t)'s$ is $- ⟨ λ, h_{2} ⟩ .$ Thus $\begin{matrix} degree of J \circ ϕ & = & max. deg - min. deg \\ = & ⟨ λ, h_{2} - h_{1} ⟩ . \end{matrix}$ Hence $\begin{matrix} ⟨ ϕ_{*} [ℙ^{1}], λ ⟩ & = & ⟨ λ, h_{2} - h_{1} ⟩ \\ \Rightarrow ϕ_{*} [ℙ^{1}] & = & h_{2} - h_{1} . \end{matrix}$ This finishes the proof of (A).

Proof of (B): First assume we have the unions, ie. $\begin{matrix} \tilde{G} = ⋃_{x \in W_{af}} B_{af}^{-} x {(\tilde{B})}_{0} = ⋃_{y \in W_{af}} B_{af} y {(\tilde{B})}_{0} . & (*) \end{matrix}$ We prove the disjointness. So assume $g \in (B_{af}^{-} x_{1} {(\tilde{B})}_{0}) \cap (B_{af}^{-} x_{1}^{'} {(\tilde{B})}_{0}) .$ Then also $g \in B_{af} y {(\tilde{B})}_{0}$ for some $y .$ Write $x_{1} = w_{1} t_{h_{1}}, x_{1}^{'} = w_{1}^{'} t_{h_{1}}^{'}, y = w_{2} t h_{2} .$ Then by (A), the curve ${\tilde{π}}_{B} (g) = φ$ satisfies $ϕ_{X} [ℙ^{1}] = h_{2} - h_{1} = h_{2}^{'} - h_{1}^{'}$ $\Rightarrow$ $h_{1} = h_{1}^{'} .$ Also $ϕ (\infty) \in w_{1} \cdot B \cap B_{-} w_{1}^{'} \cdot B .$ $\Rightarrow$ $w_{1} = w_{1}^{'} .$ Hence $x_{1} = x_{1}^{'} .$ This shows the first union in $(*)$ is disjoint. Similarly is the 2nd. Now we need to show $\tilde{G} \subset ⨆_{y \in W_{af}} B_{af} y {(\tilde{B})}_{0} .$ Since $[U_{\pm α_{i}}, i \in I_{af}]$ generate $\tilde{G},$ it suffices to show that $⨆_{y \in W_{af}} B_{af} y {(\tilde{B})}_{0}$ is stable under the left multiplication by $U_{\pm α_{i}}$ $\forall$ $i \in I_{af} .$ Clearly OK for $U_{α_{i}} \subset B_{af} .$ Only need to show $(U_{- α_{i}} \ {id}) ⨆_{y \in W_{af}} B_{af} y {(\tilde{B})}_{0} \subset ⨆_{y \in W_{af}} B_{af} y {(\tilde{B})}_{0} .$ Now we know: $U_{- α_{i}} \ {id} \subset B_{af} r_{i} U_{α_{i}} .$

Case 1: $\begin{matrix} \overline{y^{- 1} \cdot α_{i}} > 0 & \Rightarrow & U_{y^{- 1} \cdot α_{i}} \subset {(\tilde{B})}_{0} \\ \Rightarrow & U_{α_{i}} y {(\tilde{B})}_{0} \subset y {(\tilde{B})}_{0} \\ \Rightarrow & (U_{- α_{i}} \ {id}) B_{af} y {(\tilde{B})}_{0} \subset B_{af} y {(\tilde{B})}_{0} OK. \end{matrix}$

Case 2: $\overline{y^{- 1} \cdot α_{i}} < 0$ $\Rightarrow$ $U_{α_{i}} \ {id} \subset U_{- α_{i}} r_{i} H U_{- α_{i}}$ $\begin{matrix} \Rightarrow B_{af} r_{i} U_{α_{i}} \ {id} y {(\tilde{B})}_{0} & \subset & B_{af} r_{i} U_{- α_{i}} (r_{i} H) U_{- α_{i}} y {(\tilde{B})}_{0} \\ \subset & B_{af} r_{i} (r_{i} H) y {(\tilde{B})}_{0} \\ = & B_{af} y {(\tilde{B})}_{0} . \end{matrix}$

$□$

Definition: For $w_{1}, w_{2} \in W^{P},$ $τ \in H_{2} (G / P),$ set $\begin{matrix} \begin{matrix} M_{G / P, τ}^{w_{1}, w_{2}} & = & the variety of all ϕ \in Mor (ℙ^{1}, G / P) s.t. \end{matrix} \\ ϕ_{*} [ℙ^{1}] = τ, \\ ϕ (\infty) \in B_{-} w_{1} \cdot P, \\ ϕ (0) \in B w_{2} \cdot P . \end{matrix}$ It is a smooth irreducible variety of dimension $\begin{matrix} dim M_{G / P, τ}^{w_{1}, w_{2}} & = & ℓ (w_{2}) - ℓ (w_{1}) + ⟨ τ, c_{1} (T G / P) ⟩ . \end{matrix}$ Thus we have defined a map, for any $x_{1} = w_{1} t_{h_{1}},$ $x_{2} = w_{2} t_{h_{2}} \in U$ $\begin{matrix} {\tilde{π}}_{B} : & B_{af}^{-} x_{1} {(\tilde{B})}_{0} \cap B_{af} x_{2} {(\tilde{B})}_{0} & ⟶ & M_{G / B, π_{B} (h_{2} - h_{1})}^{w_{1}, w_{2}} . \end{matrix}$ Since ${\tilde{π}}_{B} (h {(\tilde{B})}_{0}) = {\tilde{π}}_{B} (g),$ we get a well-defined map, still denoted by ${\tilde{π}}_{B} :$ $\begin{matrix} {\tilde{π}}_{B} : & B_{af}^{-} x_{1} \cdot {(\tilde{B})}_{0} \cap B_{af} x_{2} \cdot {(\tilde{B})}_{0} & ⟶ & M_{G / B, π_{B} (h_{2} - h_{1})}^{w_{1}, w_{2}} . \end{matrix}$

Proposition: The map $\begin{matrix} {\tilde{π}}_{B} : & B_{af}^{-} x_{1} \cdot {(\tilde{B})}_{0} \cap B_{af} x_{2} \cdot {(\tilde{B})}_{0} & ⟶ & M_{G / B, π_{B} (h_{2} - h_{1})}^{w_{1}, w_{2}} \end{matrix}$ is bijective.

Proof.

We can in fact prove that $\begin{matrix} {\tilde{π}}_{B} |_{B_{af}^{-} x_{1} \cdot {(\tilde{B})}_{0}} : & B_{af}^{-} x_{1} \cdot {(\tilde{B})}_{0} & ⟶ & M_{G / B / π_{B} (h_{2} - h_{1})}^{w_{1}, w_{2}} \end{matrix}$ is injective. Indeed, if $g = b^{-} x_{1}$ and $g' = b^{-}' x_{1}$ where $b^{-}, b^{-}' \in B_{af}^{-}$ are such that ${\tilde{π}}_{B} (g \cdot {(\tilde{B})}_{0}) = {\tilde{π}}_{B} (g' \cdot ?????)$ ie. ${\tilde{π}}_{B} (g) = \tilde{π} (g'),$ then $b^{-} (t) x_{1} (t) \cdot B = b^{-}' (t) x_{1} (t) \cdot B .$ Here $x_{1} (t)$ is a representative of $x_{1} .$ Hence $\exists$ $b (t) \in \tilde{B}$ s.t. $b^{-} (t) x_{1} (t) = b^{-}' (t) x_{1} (t) b (t) .$ But $\tilde{B} = \tilde{H} ⋉ \tilde{U_{+}} = Γ \times H ⋉ \tilde{U_{+}} = Γ \times {(\tilde{B})}_{0} .$ So $\exists$ $h \in Γ,$ $b_{0} \in {(\tilde{B})}_{0}$ $\begin{matrix} b^{-} (t) x_{1} (t) & = & b^{-}' (t) x_{1} (t) t^{h} b_{0} (t) \end{matrix}$ or $b^{-} x_{1} \in b^{-}' x_{1} t_{h} {(\tilde{B})}_{0}$ or $b^{-} x_{1} \in B_{af}^{-} x_{1} {(\tilde{B})}_{0} \cap B_{af}^{-} x_{1} t_{h} {(\tilde{B})}_{0} .$ By the disjointness of the union $\begin{matrix} \tilde{G} & = & ⨆_{x \in W_{af}} B_{af}^{-} x {(\tilde{B})}_{0} \end{matrix}$ must have $t_{h} = id$ or $b (t) \in {(\tilde{B})}_{0} .$ Hence $g \cdot {(\tilde{B})}_{0} = g' \cdot {(\tilde{B})}_{0} .$ This shows that ${\tilde{π}}_{B}$ is injective. (Is this argument rigorous enough?) Now suppose $ϕ \in M_{G / B, π_{B} (h_{2} - h_{1})}^{w_{1}, w_{2}} .$ Let $g' \in \tilde{G}$ be any element such that ${\tilde{π}}_{B} (g') = ϕ .$ The by Theorem (B), there must exist $x_{1}^{'} = w_{1} t_{h_{1}^{'}}$ and $x_{2}^{'} = w_{2} t_{h_{2}^{'}} \in W_{af}$ s.t. $g' \in B_{af}^{-} x_{1}^{'} {(\tilde{B})}_{0} \cap B_{af} x_{2}^{'} {(\tilde{B})}_{0} .$ Let $g = g' t_{h - h_{1}^{'}} .$ Then $π_{B} (g) = π_{B} (g') = ϕ$ but now $g' \in B_{af}^{-} x_{1} {(\tilde{B})}_{0} \cap B_{af} x_{2}^{'} t_{h - h_{1}^{'}} {(\tilde{B})}_{0} .$ But since $\begin{matrix} ϕ_{*} [ℙ^{1}] & = & π_{B} (h_{2} - h_{1}) \end{matrix}$ we must have $x_{2}^{'} t_{h - h_{1}^{'}} = x_{2} .$ Hence $g' \in B_{af}^{-} x_{1} {(\tilde{B})}_{0} \cap B_{af} x_{2} {(\tilde{B})}_{0}$ or $g' \cdot {(\tilde{B})}_{0} \in (B_{af}^{-} x_{1} \cdot {(\tilde{B})}_{0}) \cap B_{af} x_{2} \cdot {(\tilde{B})}_{0} .$ This shows that $π_{B}$ is onto. Hence $π_{B}$ is bijective.

$□$

Remark: Note that in the definition of $M_{G / P, τ}^{w_{1}, w_{2}},$ we consider a reparametrization of a curve $ϕ$ or a shift of $ϕ$ by an element in $\tilde{H}$ as a new curve.

Connection of $M_{G / P, τ}^{w_{1}, w_{2}}$ to Schubert cells in $G_{af} / B_{af} :$

Introduce $\begin{matrix} W_{af}^{\pm} & = & {x \in W_{af} : β \in Δ_{+}^{re}, x \cdot β < 0 \Rightarrow \pm \overline{β} > 0} \end{matrix}$ so $W_{af}^{-}$ is as before the minimal coset representatives of $W_{af}^{-} / W .$ It is easy to see that $W_{af}^{-} w_{0} \subset W_{af}^{+}$ where $w_{0} \in W$ is the longest element of $W .$

Fact: For $x = w t_{h} \in W_{af}^{\pm},$ have $\begin{matrix} ℓ (x) & = & \pm ℓ_{s} (x) \end{matrix}$ where $ℓ_{s} (x),$ the stable length of $x,$ is defined to be $\begin{matrix} ℓ_{s} (w t_{h}) & = & ℓ (w) + ⟨ 2 ρ, h ⟩ . \end{matrix}$

Theorem 3: Let $x_{1} = w_{1} t_{h_{1}},$ $x_{2} = w_{2} t_{h_{2}}$ be in $W_{af}^{+} .$ Then we have a natural inverse isomorphism between smooth varieties: $\begin{matrix} B_{af}^{-} x_{1} \cdot B_{af} \cap B_{af} x_{2} \cdot B_{af} & ⇄_{π_{+}}^{π_{-}} & M_{G / B, π_{B} (h_{2} - h_{1})}^{w_{1}, w_{2}} \end{matrix}$ given by $\begin{matrix} π_{-} (g \cdot B_{af}) & = & {\tilde{π}}_{B} (g) if g \in B_{af}^{-} x_{1}, \\ π_{+} ({\tilde{π}}_{B} (g)) & = & g \cdot B_{af} if g \in B_{af} x_{2} . \end{matrix}$

Remark: Note that the intersection $B_{af}^{-} x_{1} \cdot B_{af} \cap B_{af} x_{2} \cdot B_{af}$ is smooth and has dimension $=$ $\begin{matrix} ℓ (x_{2}) - ℓ (x_{1}) & = & ℓ (w_{2}) + ⟨ 2 ρ, h_{2} ⟩ - ℓ (w_{1}) - ⟨ 2 ρ, h_{1} ⟩ \\ = & ℓ (w_{2}) - ℓ (w_{1}) + ⟨ 2 ρ, h_{2} - h_{1} ⟩ \\ = & dim M_{G / B, π_{B} (h_{2} - h_{1})}^{w_{1}, w_{2}} . \end{matrix}$

Lecture 12: April 8, 1997

Recall last lecture...

The fact $\tilde{G} = ⨆_{\underset{disjoint}{x \in W_{af}}} B_{af}^{-} x {(\tilde{B})}_{0} = ⨆_{\underset{disjoint}{y \in W_{af}}} B_{af} y {(\tilde{B})}_{0}$ is a special case of the following general fact:

Fact: If $V$ is a subgroup of $\tilde{G}$ such that for each $α \in Δ_{+}^{re},$ either $U_{α} \subset V$ or $U_{- α} \subset V,$ then we have two disjoint unions: $G_{af} = \tilde{G} = ⨆_{x \in W_{af}} B_{af}^{-} x V = ⨆_{y \in W_{af}} B_{af} y V .$

Two decompositions for any Kac-Moody group:

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

0)	$U_{-} = (U_{-} \cap (x^{- 1} B_{+} x)) (U_{-} \cap (x^{- 1} B_{-} x))$
1)	$(U_{+} \cap x B_{-} x^{- 1}) \times (B_{-} x \cdot B) \tilde{⟶} x B_{-} \cdot B$
2)	$(U_{+} \cap x B_{-} x^{- 1}) \times ((B_{-} x \cdot B) \cap B y \cdot B) \tilde{⟶} x B_{-} \cdot B \cap B y \cdot B$

\Rightarrow B_{-} x \cdot B \cap B y \cdot B \neq \emptyset ⟺ x \leq y,

and in this case,

B_{-} x \cdot B \cap B y \cdot B

is a non-singular irreducible affine variety of dimension

= ℓ (y) - ℓ (x) .

In order to prove Theorem 3 stated at the end of last Lecture, we need the following facts. Recall that $\begin{matrix} W_{af}^{+} & = & {x \in W_{af} : β \in Δ_{+}^{re}, x \cdot β < 0 \Rightarrow \overline{β} > 0} . \end{matrix}$

Proposition 1: The following are equivalent:

(0)	$x \in W_{af}^{+}$
(1)	$B_{af} \cap x^{- 1} B_{af}^{-} x \subset {(\tilde{B})}_{0}$
(2)	$x B_{af} x^{- 1} \cap B_{af}^{-} \subset x {(\tilde{B})}_{0} x^{- 1}$
(3)	$B_{af} x B_{af} \subset B_{af} x {(\tilde{B})}_{0}$
(1')	${(\tilde{B})}_{0} \cap x^{- 1} B_{af} x \subset B_{af}$
(2')	$x {(\tilde{B})}_{0} x^{- 1} \cap B_{af} \subset x B_{af} x^{- 1}$
(3')	$B_{af}^{-} x {(\tilde{B})}_{0} \subset B_{af}^{-} x B_{af}$

Proof.

The equivalence between (0) and (1) is clear because (2) says that if $β \in Δ_{+}^{re}$ and $x β < 0$ then $\overline{β} > 0 .$ It is also clear that (1) is equivalent to (2) because $x 1) x^{- 1} = 2).$ Now assume (1). We want to prove (3): It is enough to show that $x B_{af} \subset B_{af} x {(\tilde{B})}_{0} .$ Let $x b \in x B_{af} .$ Write $b = b_{1} b_{2}$ where $b_{1} \in B_{af} \cap x^{- 1} B_{af} x, b_{2} \in B_{af} \cap x^{- 1} B_{af}^{-} x .$ Then $x b = x b_{1} b_{2} = (x b_{1} x^{- 1}) x b_{2} .$ Now $x b_{1} x^{- 1} \in B_{af}$ and $b_{2} \in {(\tilde{B})}_{0}$ by 1). Hence $x b \in B_{af} x {(\tilde{B})}_{0} .$ This shows that $((0) \Leftrightarrow) (1) (\Leftrightarrow (2)) \Rightarrow (3).$ Now assume 3). We want to prove (1).

$□$

Proposition 2: For $x_{1}, x_{2} \in W_{af}^{+},$ the two maps $\begin{matrix} ϕ_{1} : & B_{af}^{-} x_{1} \cdot B_{af} & ⟶ & B_{af}^{-} x_{1} \cdot {(\tilde{B})}_{0} : & b^{-} x_{1} \cdot B_{af} & ⟼ & b^{-} x_{1} \cdot {(\tilde{B})}_{0} \\ ϕ_{2} : & B_{af} x_{2} \cdot {(\tilde{B})}_{0} & ⟶ & B_{af} x_{2} \cdot B_{af} : & b^{+} x_{2} \cdot {(\tilde{B})}_{0} & ⟼ & b^{+} x_{2} \cdot B_{af} \end{matrix}$ are both well-defined. Moreover, their restrictions to the following intersections give isomorphisms that are mutually inverses of each other $\begin{matrix} B_{af}^{-} x_{1} \cdot B_{af} \cap B_{af} x_{2} \cdot B_{af} & ⇄_{ϕ_{2}}^{ϕ_{1}} & B_{af}^{-} x_{1} \cdot {(\tilde{B})}_{0} \cap B_{af} x_{2} \cdot {(\tilde{B})}_{0} . \end{matrix}$

Proof.

$ϕ_{1}$ is well-defined because $B_{af}^{-} \cap x_{1} B_{af} x_{1}^{- 1} \subset x_{1} {(\tilde{B})}_{0} x_{1}^{- 1}$ ((2) in Prop. 1). $ϕ_{2}$ is well-defined because $B_{af} \cap x_{2} {(\tilde{B})}_{0} x_{2}^{- 1} \subset x_{2} B_{af} x_{2}^{- 1}$ ((2') in Prop. 1). Since $B_{af} x_{2} B_{af} \subset B_{af} x_{2} {(\tilde{B})}_{0} x_{2}^{- 1}$ ((3) in Prop. 1), we have $ϕ_{1} (B_{af}^{-} x_{1} \cdot B_{af} \cap B_{af} x_{2} \cdot B_{af}) \subset B_{af}^{-} x_{1} \cdot {(\tilde{B})}_{0} \cap B_{af} x_{2} \cdot {(\tilde{B})}_{0} .$ In more details, suppose that $m_{1} = b^{-} x_{1} \cdot B_{af} = b^{+} x_{2} \cdot B_{af} \in B_{af}^{-} x_{1} \cdot B_{af} \cap B_{af} x_{2} \cdot B_{af}$ where $b^{-} \in B_{af},$ $b^{+} \in B_{af} .$ Then $\exists$ $b \in B_{af}$ s.t. $b^{-} x_{1} = b^{+} x_{2} b .$ Write $b = b_{1} b_{2}$ where $b_{1} \in B_{af} \cap x_{2}^{- 1} B_{af} x_{2},$ $b_{2} \in B_{af} \cap x_{2}^{- 1} B_{af}^{-} x_{2} .$ Then $b^{-} x_{1} = b^{+} (x_{2} b_{1} x_{2}^{- 1}) x_{2} b_{2} .$ We know that $x_{2} b_{1} x_{2}^{- 1} \in B_{af}$ by definition of $b_{1}$ and that $b_{2} \in {(\tilde{B})}_{0}$ by (1) of Prop. 1. Then $ϕ (m_{1}) = b^{-} x_{1} \cdot {(\tilde{B})}_{0} = (b^{+} x_{2} b_{1} x_{2}^{- 1}) x_{2} \cdot {(\tilde{B})}_{0} \in B_{af} x_{2} \cdot {(\tilde{B})}_{0} .$ Moreover, by the definition of $ϕ_{2},$ we have $\begin{matrix} ϕ_{2} (ϕ_{1} (m_{1})) & = & (b^{+} x_{2} b_{1} x_{2}^{- 1}) x_{2} \cdot B_{af} \\ = & b^{+} x_{2} b_{1} \cdot B_{af} \\ = & b^{+} x_{2} \cdot B_{af} (since b_{1} \in B_{af}) \\ = & m_{1} . \end{matrix}$ This shows that $ϕ_{1}$ is injective and $ϕ_{2}$ is onto (when restricted to the intersections). Similarly we can show that $ϕ_{2} (B_{af}^{-} x_{1} \cdot {(\tilde{B})}_{0} \cap B_{af} x_{2} \cdot {(\tilde{B})}_{0}) \subset B_{af}^{-} x_{1} \cdot B_{af} \cap B_{af} x_{2} \cdot B_{af}$ and $ϕ_{1} (ϕ_{2} (m_{2})) = m_{2}$ for $m_{2} \in B_{af}^{-} x_{1} \cdot {(\tilde{B})}_{0} \cap B_{af} x' \cdot {(\tilde{B})}_{0} .$ Let's write out the details again: suppose that $m_{2} = b^{-} x_{1} \cdot {(\tilde{B})}_{0} \cap b^{+} x_{2} \cdot {(\tilde{B})}_{0} \in B_{af}^{-} x_{1} \cdot {(\tilde{B})}_{0} \cap B_{af} x_{2} \cdot {(\tilde{B})}_{0}$ where $b^{-} \in B_{af}^{-}$ and $b^{+} \in B_{af} .$ Then $\exists$ $b_{0} \in {(\tilde{B})}_{0}$ s.t. $b^{+} x_{2} = b^{-} x_{1} b_{0} .$ Write $b_{0} = b_{1} b_{2}$ where $b_{1} \in {(\tilde{B})}_{0} \cap x_{1}^{- 1} B_{af}^{-} x_{1}$ and $b_{2} \in {(\tilde{B})}_{0} \cap x_{1}^{- 1} B_{af} x_{1} .$ Then $b^{+} x_{1} = b^{-} (x_{1} b_{1} x_{1}^{- 1}) x_{1} b_{2} .$ Now $x_{1} b_{1} x_{1}^{- 1} \in B_{af}^{-}$ by definition and $b_{2} \in B_{af}$ by (1') of Prop. 1. Hence $b^{+} x_{2} \in B_{af}^{-} x_{1} B_{af},$ or $ϕ_{2} (m_{2}) = b^{+} x_{2} \cdot B_{af} \in B_{af}^{-} x_{1} \cdot B_{af} .$ In other words, $ϕ_{2} (B_{af}^{-} x_{1} \cdot B_{af} \cap B_{af} x_{2} \cdot B_{af}) \subset B_{af}^{-} x_{1} \cdot {(\tilde{B})}_{0} \cap B_{af} x_{1} \cdot {(\tilde{B})}_{0} .$ Moreover, by the definition of $ϕ_{1},$ we have $\begin{matrix} ϕ_{1} (ϕ_{2} (m_{2})) & = & b^{-} (x_{1} b_{1} x_{1}^{- 1}) x_{1} \cdot {(\tilde{B})}_{0} \\ = & b^{-} x_{1} b_{1} \cdot {(\tilde{B})}_{0} \\ = & b^{-} x_{1} \cdot {(\tilde{B})}_{0} (∴ b_{1} \in {(\tilde{B})}_{0}) \\ = & m_{2} . \end{matrix}$ This shows that when restricted to the intersections, both $ϕ_{1}$ and $ϕ_{2}$ 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 $x_{1} = w_{1} t_{h_{1}}$ and $x_{2} \in w_{2} t_{h_{2}}$ be in $W_{af}^{+} .$ Then we have mutually inverse isomorphisms $\begin{matrix} B_{af}^{-} x_{1} \cdot B_{af} \cap B_{af} x_{2} \cdot B_{af} & ⇄_{π_{+}}^{π_{-}} & M_{G / B, π_{B} (h_{2} - h_{1})}^{w_{1}, w_{2}} \end{matrix}$ defined by $\begin{matrix} π_{-} (g \cdot B_{af}) & = & {\tilde{π}}_{B} (g) if g \in B_{af}^{-} x_{1} s.t. g \cdot B_{af} \in B_{af}^{-} x_{1} \cdot B_{af} \cap B_{af} x_{2} \cdot B_{af}, \\ π_{+} ({\tilde{π}}_{B} (g)) & = & g \cdot B_{af} if g \in B_{af} x_{2} s.t. {\tilde{π}}_{B} (g) \in M_{G / B, π_{B} (h_{2} - h_{1})}^{w_{1}, w_{2}} . \end{matrix}$

Proof.

This is just Proposition 2 and the Proposition at the end of Lecture 11 (page 11-10.5(1)) combined, ie. $B_{af}^{-} x_{1} \cdot B_{af} \cap B_{af} x_{2} \cdot B_{af} ≃ B_{af}^{-} x_{1} \cdot {(\tilde{B})}_{0} \cap B_{af} x_{2} \cdot {(\tilde{B})}_{0} ≃ M_{G / B, π_{B} (h_{2} - h_{1})}^{w_{1}, w_{2}} .$

$□$

The Stable Bruhat order $\overset{s}{\leq}$ and the stable length $ℓ_{s}$

Say $h \in Q^{\lor}$ is sufficiently dominant if $⟨ ρ_{i}, h ⟩ ≫ 0$ for each $i \in I .$

Definition:

1)	For $x, y \in W_{af},$ write $" x \overset{s}{\leq} y "$ and say $" x$ is $\leq y$ under the stable Bruhat order" if $x t_{h} \leq y t_{h}$ for sufficiently dominant $h .$
2)	For $x = w t_{h} \in W_{af},$ define the stable length of $x$ to be $ℓ_{s} (x) = ℓ (w) + ⟨ 2 ρ, h ⟩ .$

Facts:

1)	For any $w \in W$ and $h$ dominant, have $x = w t_{h} \in W_{af}^{+} .$
2)	For any given $x \in W_{af},$ have $x t_{h} \in W_{af}^{+}$ for sufficiently dominant $h .$

Proof.

Clearly 2) follows from 1). We only prove 1). If $α < 0$ is a root for the finite $g,$ then for any $n > 0,$ $x \cdot (α + n δ) = w α + (n - ⟨ h, α ⟩) δ .$ Since $⟨ h, α ⟩ \leq 0,$ have $n - ⟨ h, α ⟩ \geq n > 0 .$ This always has $x \cdot (α + n δ) > 0 .$ This shows that if $β = α + n δ > 0$ is such that $x \cdot β < 0$ must have $α > 0 .$ Thus $x \in W_{af}^{+} .$

$□$

Proposition: $\begin{matrix} \underset{\underset{x_{1}}{| |}}{w_{1} t_{h_{1}}} \overset{s}{\leq} \underset{\underset{x_{2}}{| |}}{w_{2} t_{h_{2}}} & ⟺ & M_{G / B, π_{B} (h_{2} - h_{1})}^{w_{1}, w_{2}} \neq \emptyset \\ ⟺ & B_{af}^{-} x_{1} \cdot {(\tilde{B})}_{0} \cap B_{af} x_{2} \cdot {(\tilde{B})}_{0} \neq \emptyset . \end{matrix}$

Proof.

We have proved in Lecture 11 that $\begin{matrix} M_{G / B, π_{B} (h_{2} - h_{1})}^{w_{1}, w_{2}} & ≃ & B_{af}^{-} x_{1} \cdot {(\tilde{B})}_{0} \cap B_{af} x_{2} \cdot {(\tilde{B})}_{0} . \end{matrix}$ Now suppose $x_{1} \overset{s}{\leq} x_{2} .$ Then $\exists$ sufficiently dominant $h$ st. $x_{1} t_{h}, x_{2} t_{h} \in W_{af}^{+}$ and $x_{1} t_{h} \leq x_{2} t_{h} .$ This implies $B_{af}^{-} x_{1} t_{h} \cdot B_{af} \cap B_{af} x_{2} t_{h} \cdot B_{af} \neq \emptyset .$ But by Theorem 3, since $x_{1} t_{h}, x_{2} t_{h} \in W_{af}^{+},$ we have $\begin{matrix} M_{G / B, π_{B} (h_{2} - h_{1})}^{w_{1}, w_{2}} & ≃ & B_{af}^{-} x_{1} t_{h} \cdot B_{af} \cap B_{af} x_{2} t_{h} \cdot B_{af} \neq \emptyset . \end{matrix}$ Conversely, if $M_{G / B, π_{B} (h_{2} - h_{1})}^{w_{1}, w_{2}} \neq \emptyset,$ then for $h$ sufficiently dominant so that $x_{1} t_{h}, x_{2} t_{h} \in W_{af}^{+},$ we have $\begin{matrix} B_{af}^{-} x_{1} t_{h} \cdot B_{af} \cap B_{af} x_{2} t_{h} \cdot B_{af} & ≃ & M_{G / B, π_{B} (h_{2} - h_{1})}^{w_{1}, w_{2}} \neq \emptyset . \end{matrix}$ Thus $x_{1} t_{h} \leq x_{2} t_{h} .$ Hence $w_{1} t_{h_{1}} \overset{s}{\leq} w_{2} t_{h_{2}} .$

$□$

Proposition: For $x_{1}, x_{2} \in W_{af}^{+},$ have $x_{1} \overset{s}{\leq} x_{2} \Leftrightarrow x_{1} \leq x_{2} .$

Proof.

Suppose $x_{1}, x_{2} \in W_{af}^{+} .$ Then $\begin{matrix} x_{1} \overset{s}{\leq} x_{2} & ⟺ & M_{G / B, π_{B} (h_{2} - h_{1})}^{w_{1}, w_{2}} = B_{af}^{-} x_{1} \cdot B_{af} \cap B_{af} x_{2} \cdot B_{af} \neq \emptyset \\ ⟺ & x_{1} \leq x_{2} . \end{matrix}$

$□$

Proposition: For $w_{1}, w_{2} \in W,$ have $w_{1} \overset{s}{\leq} w_{2} \Leftrightarrow w_{1} \leq w_{2} .$

Proof 1.

Since $M_{G / B, π_{B} (0)}^{w_{1}, w_{2}} = B_{-} w_{1} \cdot B \cap B w_{2} \cdot B$ we have $\begin{matrix} w_{1} \overset{s}{\leq} w_{2} & ⟺ & B_{-} w_{1} \cdot B \cap B w_{2} \cdot B \neq \emptyset \\ ⟺ & w_{1} \leq w_{2} . \end{matrix}$

$□$

Proof 2.

We first prove that $W \subseteq W_{af}^{+} .$

Suppose $β = α + n δ > 0$ is s.t. $w \cdot β < 0 .$ Then we must have $α > 0 :$ for if $α < 0,$ then $n > 0,$ and thus $w \cdot β = w α + n δ > 0 .$ Contradiction. Hence $α > 0 .$ Hence $W \subset W_{af}^{+} .$

$□$

Question: Given $w \in W,$ for which $x \in W_{af}$ do we have $w \overset{s}{<} x$ and $ℓ_{s} (x) = ℓ_{s} (w) + 1 = ℓ (w) + 1 ?$

Answer: Iff $x$ is one of the following two forms: either $x = w r_{α}$ where $α \in {\overline{Δ}}_{+}$ and $ℓ (x) = ℓ (w) + 1$ (positive roots for the finite $𝔤)$ or $x = w r_{α} t_{α^{\lor}} = w r_{α + δ}$ where $α \in {\overline{Δ}}_{+}$ and $ℓ (w r_{α}) = ℓ (w) - ⟨ 2 ρ, α^{\lor} ⟩ + 1 .$

Proof.

Later.

$□$

Fact: If $w t_{h} \in W_{af}^{+}$ then $h \in Q^{\lor}$ is dominant.

Proof.

Proof by contradiction: Suppose $h$ is not dominant. Then $\exists$ $i$ s.t. $⟨ α_{i}, h ⟩ < 0 .$ Must have $⟨ α_{i}, h ⟩ \leq - 2 .$ Let $β = - α_{i} + δ \in Δ_{+}^{re} .$ Then $x \cdot β = - w α_{i} + (1 + ⟨ α_{i}, h ⟩) δ .$ But $\overline{β} = - α_{i} < 0 .$ Contradictory to $x \in W_{af}^{+}$ $\Rightarrow$ $h$ dominant.

$□$

Lecture 13: April 9, 1997

We first collect some facts about $ℓ_{s}$ and $\overset{s}{\leq} .$ Then talk about ${(W_{P})}_{af}$ and ${(W^{P})}_{af} .$

Proposition $ℓ_{s} :$ The following are true about $ℓ_{s} :$

(1)	$ℓ_{s} (w) = w$ $\forall$ $w \in W .$
(2)	$ℓ_{s} (x t_{h}) = ℓ_{s} (x) + ⟨ 2 ρ, h ⟩$ $\forall$ $x \in W_{af},$ $h \in Q^{\lor} .$
(3)	$ℓ_{s} (x w_{0}) = ℓ (w_{0}) - ℓ_{s} (x)$ $\forall$ $x \in W_{af},$ $w_{0} = longest$ in $W .$
(4)	$- ℓ (x) \leq ℓ_{s} (x) \leq ℓ (x)$ $\forall$ $x \in W_{af},$ $\begin{matrix} ℓ_{s} (x) = ℓ (x) & ⟺ & x \in W_{af}^{+}, \\ ℓ_{s} (x) = - ℓ (x) & ⟺ & x \in W_{af}^{-} . \end{matrix}$
(5)	For any $x, y \in W_{af}$ $\begin{matrix} ℓ_{s} (x y) & = & ℓ_{s} (y) + \sum_{\underset{x \cdot β < 0}{β \in Δ_{+}^{re}}} sign (\overline{y^{- 1} \cdot β}) \end{matrix}$ where $\begin{matrix} sign α & = & {\begin{matrix} 1 & if α \in {\overline{Δ}}_{+}, \\ - 1 & if α \in - {\overline{Δ}}_{+} . \end{matrix} \end{matrix}$

Recall $\begin{matrix} {\overline{Δ}}_{+} & = & the set of roots of the finite 𝔤 . \end{matrix}$

Proof.

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

(3): Write $x = w t_{h} .$ Then $\begin{matrix} ℓ_{s} (x w_{0}) & = & ℓ (w t_{h} w_{0}) = ℓ (w w_{0} t_{w_{0} h}) \\ = & ℓ (w w_{0}) + ⟨ 2 ρ, w_{0} h ⟩ \\ = & ℓ (w_{0}) - ℓ (w) - ⟨ 2 ρ, h ⟩ \\ = & ℓ (w_{0}) - ℓ_{s} (x) . \end{matrix}$

(4) We break the proof of (4) into a few parts: We first prove that $ℓ_{s} (x) = ℓ (x)$ for $x \in W_{af}^{+} :$ Assume $x = w t_{h} \in W_{af}^{+} .$ Then by the definition of $W_{af}^{+},$ if $α + n δ > 0$ is s.t. $x \cdot (α + n δ) = w α + (n - ⟨ α, h ⟩) δ < 0$ we must have $α > 0 .$ Thus if $w α < 0,$ then $n$ can only take values $0, 1, \dots, ⟨ α, h ⟩$ and if $w α > 0,$ then $n$ can only take values $0, 1, \dots, ⟨ α, h ⟩ - 1 .$ Thus the set $\begin{matrix} A & = & {α + n δ > 0 : x \cdot (α + n δ) < 0} \end{matrix}$ is contained in the set $\begin{matrix} B & = & {α + n δ : α > 0, w α < 0, n = 0, 1, \dots, ⟨ α, h ⟩} \cup {α + n δ : α > 0, w α > 0, n = 0, 1, \dots ⟨ α, h ⟩ - 1} . \end{matrix}$ Clearly $B \subset A .$ Thus $A = B .$ Hence $\begin{matrix} ℓ (x) = # B & = & \sum_{\underset{w α < 0}{α > 0}} (⟨ α, h ⟩ + 1) + \sum_{\underset{w α > 0}{α > 0}} ⟨ α, h ⟩ \\ = & \sum_{α > 0} ⟨ α, h ⟩ + \sum_{\underset{w α < 0}{α > 0}} \cdot 1 \\ = & ⟨ 2 ρ, h ⟩ + ℓ (w) \\ = & ℓ_{s} (x) . \end{matrix}$ This shows $\begin{matrix} ℓ_{s} (x) & = & ℓ (x) for x \in W_{af}^{+} . \end{matrix}$ We have proved (Lecture 8 that) $\begin{matrix} - ℓ_{s} (x) & = & ℓ (x) for x \in W_{af}^{-} . \end{matrix}$ To prove that $\begin{matrix} ℓ_{s} (x) & \leq & ℓ (x) for all x \in W_{af} \end{matrix}$ we need the following Lemma:

Lemma: Suppose that $h_{1} \in Q^{\lor}$ is dominant and regular. Then for all $x \in W_{af},$ we have $ℓ (x t_{h_{1}}) \leq ℓ (x) + ⟨ 2 ρ, h_{1} ⟩ .$ We will prove the Lemma later. Let's assume the Lemma for now. Let $x \in W_{af}$ be arbitrary. Let $h_{1}$ be sufficiently dominant so that $x t_{h_{1}} \in W_{af}^{+} .$ Then we have $\begin{matrix} ℓ_{s} (x) & = & ℓ_{s} (x t_{h_{1}}) - ⟨ 2 ρ, h_{1} ⟩ \\ = & ℓ (x t_{h_{1}}) - ⟨ 2 ρ, h_{1} ⟩ \\ \leq & ℓ (x) + ⟨ 2 ρ, h_{1} ⟩ - ⟨ 2 ρ, h_{1} ⟩ (Lemma) \\ = & ℓ (x) . \end{matrix}$ This shows that $ℓ_{s} (x) \leq ℓ (x)$ for all $x \in W_{af} .$ Now if $ℓ (x) = ℓ_{s} (x) = ℓ (w) + ⟨ 2 ρ, h ⟩$ for $x = w t_{h} \in W_{af},$ then since the set $\begin{matrix} B & = & {α + n δ : α > 0, w α < 0, n = 0, 1, \dots ⟨ α, h ⟩} \cup \\ {α + n δ : α > 0, w α < 0, n = 0, 1, \dots ⟨ α, h ⟩ - 1} \end{matrix}$ (if $⟨ α, h ⟩ < 0,$ then the first set in the union is taken to be $\emptyset .$ Similarly for the 2nd set) is obviously contained in the set $\begin{matrix} A & = & {α + n δ > 0 : x \cdot (α + n δ) < 0} . \end{matrix}$ But $# B = ℓ (w) + ⟨ 2 ρ, h ⟩$ $\Rightarrow$ $B = A .$ So for every $β \in Δ_{+}^{re} \in A$ have $\overline{β} > 0 .$ This shows that $x \in W_{af}^{+} .$ Similarly we can show $ℓ_{s} (x) \geq - ℓ (x)$ $\forall$ $x \in W_{af}$ and $ℓ_{s} (x) = - ℓ (x)$ $\Leftrightarrow$ $x \in W_{af}^{-} .$ This finishes the proof of (4) (except for the lemma). (Something is not right here).

We now prove (6): $\forall$ $x, y \in W_{af}$ (Do not trust this proof!) $\begin{matrix} ℓ_{s} (x y) & = & ℓ_{s} (y) + \sum_{\underset{x \cdot β < 0}{β \in Δ_{+}^{re}}} sign (\overline{y^{- 1} \cdot β}) . \end{matrix}$ Write $x = w_{1} t_{h_{1}},$ $y = w_{2} t_{h_{2}} .$ Then $\begin{matrix} ℓ_{s} (x y) & = & ℓ_{s} (w_{1} w_{2} t_{w_{2}^{- 1} h_{1} + h_{2}}) \\ = & ℓ (w_{1} w_{2}) + ⟨ 2 ρ, w_{2}^{- 1} h_{1} + h_{2} ⟩ \end{matrix}$ so $\begin{matrix} ℓ_{s} (x y) - ℓ_{s} (y) & = & ℓ (w_{1} w_{2}) - ℓ (w_{2}) + ⟨ 2 w_{1} ρ, h_{1} ⟩ \end{matrix}$ so need to show $\begin{matrix} ℓ (w_{1} w_{2}) - ℓ (w_{2}) + ⟨ 2 w_{1} ρ, h_{1} ⟩ & = & \sum_{\underset{x \cdot β < 0}{β \in Δ_{+}^{re}}} sign (\overline{y^{- 1} \cdot β}) . \end{matrix}$ Notice the special case: $x = w_{1},$ $y = w_{2},$ we are saying $\begin{matrix} ℓ (w_{1} w_{2}) - ℓ (w_{2}) & = & \sum_{\underset{w_{1} \cdot β < 0}{{\overline{Δ}}_{+} ∋ β > 0}} sign (w_{2}^{- 1} \cdot β) . \end{matrix}$ This is a statement about the finite Weyl group and can be proved by induction on $ℓ (w_{2}),$ for example. We assume this. Thus need to show $\begin{matrix} ⟨ 2 w_{2} ρ, h_{1} ⟩ & = & \sum_{\underset{x \cdot β < 0}{β \in Δ_{+}^{re}}} sign (\overline{y^{- 1} \cdot β}) - \sum_{\underset{w_{2} \cdot α < 0}{{\overline{Δ}}_{+} ∋ α > 0}} sign (w_{2}^{- 1} \cdot α) . \end{matrix}$ Let $\begin{matrix} A & = & {β = α + n δ > 0 : x \cdot β < 0} \\ = & {β = α + n δ > 0 : w_{1} α + (n - ⟨ α, h_{1} ⟩) δ < 0} . \end{matrix}$ For $β = α + n δ \in A,$ have $\begin{matrix} y^{- 1} \cdot β & = & w_{2}^{- 1} α + (n + ⟨ w_{2}^{- 1} α, h_{2} ⟩) δ \end{matrix}$ so $\begin{matrix} \overline{y^{- 1} \cdot β} & = & w_{2}^{- 1} α . \end{matrix}$ Break $A$ as a disjoint union $\begin{matrix} A & = & A_{1} \cup A_{2} \cup A_{3} \cup A_{4} \end{matrix}$ where $\begin{matrix} A_{1} & = & {β = α + n δ > 0 : α > 0, w_{1} α > 0, w_{1} α + (n - ⟨ α, h_{1} ⟩) δ < 0}, \\ A_{2} & = & {β = α + n δ > 0 : α > 0, w_{1} α < 0, w_{1} α + (n - ⟨ α, h_{1} ⟩) δ < 0}, \\ A_{3} & = & {β = α + n δ > 0 : α < 0, w_{1} α > 0, w_{1} α + (n - ⟨ α, h_{1} ⟩) δ < 0}, \\ A_{4} & = & {β = α + n δ > 0 : α < 0, w_{1} α < 0, w_{1} α + (n - ⟨ α, h_{1} ⟩) δ < 0}, \end{matrix}$ so $\begin{matrix} A_{1} & = & {β = α + n δ > 0 : α > 0, w_{1} α > 0, n = 0, 1, \dots, ⟨ α, h_{1} ⟩ - 1}, \\ A_{2} & = & {β = α + n δ > 0 : α > 0, w_{1} α < 0, n = 0, 1, \dots, ⟨ α, h_{1} ⟩}, \\ A_{3} & = & {β = α + n δ > 0 : α < 0, w_{1} α > 0, n = 0, 1, \dots, ⟨ α, h_{1} ⟩ - 1}, \\ A_{4} & = & {β = α + n δ > 0 : α < 0, w_{1} α < 0, n = 0, 1, \dots, ⟨ α, h_{1} ⟩} . \end{matrix}$ Note that $\sum_{α \in {\overline{Δ}}_{+}} sign (w_{2}^{- 1} α) = \sum_{\underset{w_{2}^{- 1} α > 0}{α > 0}} \cdot 1 + \sum_{\underset{w_{2}^{- 1} α < 0}{α > 0}} (- 1) = 2 ρ - 2 (ρ - w_{2} ρ) = 2 w_{2} ρ .$ Similarly, $\begin{matrix} \sum_{\underset{x \cdot β < 0}{β \in Δ_{+}^{re}}} sign (\overline{y^{- 1} \cdot β}) & = & \sum_{β \in A_{1} \cup A_{2} \cap A_{3} \cap A_{4}} sign (\overline{y^{- 1} \cdot β}) \end{matrix}$ so $\begin{matrix} \sum_{\underset{x \cdot β < 0}{β \in Δ_{+}^{re}}} sign (\overline{y^{- 1} \cdot β}) - \sum_{\underset{w_{1} \cdot α < 0}{α \in {\overline{Δ}}_{+}}} sign (w_{2}^{- 1} \cdot α) & = & \sum_{β \in A_{1}} sign (\overline{y^{- 1} \cdot β}) \\ = & ⟨ 2 w_{2} ρ, h_{1} ⟩ . \end{matrix}$ 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 $h_{1} \in Q^{\lor}$ is dominant and regular. Then for all $x \in W_{af},$ we have $ℓ (x t_{h_{1}}) \leq ℓ (x) + ⟨ 2 ρ, h_{1} ⟩ .$

Proof.

Set $\begin{matrix} A_{1} & = & {α + n δ > 0 : x t_{h_{1}} \cdot (α + n δ) < 0} \\ = & {α + n δ > 0 : x \cdot (α + (n - ⟨ h_{1}, α ⟩) δ) < 0} . \end{matrix}$ Write $A_{1}$ as $\begin{matrix} A_{1} & = & B_{1} \cup B_{2} \end{matrix}$ where: $\begin{matrix} B_{1} & = & A_{1} \cap {A_{1} \cap {α + n δ : α + (n - ⟨ α, h_{1} ⟩) δ > 0}}, \\ B_{2} & = & A_{1} \cap {A_{1} \cap {α + n δ : α + (n - ⟨ α, h_{1} ⟩) δ < 0}} . \end{matrix}$ The map $\begin{matrix} B_{1} & ⟶ & A : & α + n δ & ⟼ & α + (n - ⟨ α, h_{1} ⟩) δ \end{matrix}$ is injective: indeed, if $\begin{matrix} α + (n - ⟨ α, h_{1} ⟩) δ = α' + (n' - ⟨ α', h_{1} ⟩) δ \\ \Rightarrow α = α' and n - ⟨ α, h_{1} ⟩ = n' - ⟨ α', h_{1} ⟩ \\ \Rightarrow α = α', n = n' . Hence # B_{1} \leq # A = ℓ (x) . \end{matrix}$ Define the inclusion map $\begin{matrix} B_{2} ⟶ C & = & {α + n δ > 0 : α + n δ - ⟨ α, h_{1} ⟩ δ < 0} \\ = & {α + n δ > 0 : α > 0, n = 0, 1, \dots, ⟨ α, h_{1} ⟩ - 1} . \end{matrix}$ It is clear that $# c = \sum_{α} ⟨ α, h_{1} ⟩ = ⟨ 2 ρ, h_{1} ⟩$ $\Rightarrow # B_{2} \leq # C = ⟨ 2 ρ, h_{1} ⟩ .$ Hence $ℓ (x t_{h_{1}}) = # A = # B_{1} + # B_{2} \leq # A + # C = ℓ (x) + ⟨ 2 ρ, h_{1} ⟩ .$

$□$

In the next proposition, we collect some facts about $\overset{s}{\leq} :$

Proposition $\overset{s}{\leq} :$

(1)

For

x, y \in W_{af},

we have

\begin{matrix} x \overset{s}{\leq} y & ⟺ & x t_{h} \leq y t_{h} for sufficiently dominant h \\ ⟺ & y t_{- h} \leq x t_{- h} for sufficiently dominant h \\ ⟺ & x t_{h} \overset{s}{\leq} y t_{h} for all h \\ ⟺ & y w_{0} \overset{s}{\leq} x w_{0} where w_{0} = the longest in W . \end{matrix}

(2)

For

z \in W_{af}^{+},

we have

(2a)	$x \leq z$ $\Rightarrow$ $x \overset{s}{\leq} z .$
(2b)	$z \overset{s}{\leq} y$ $\Rightarrow$ $z \leq y .$

(3)

For

z \in W_{af}^{-},

we have

(3a)	$x \leq z$ $\Rightarrow$ $z \overset{s}{\leq} x .$
(3b)	$y \overset{s}{\leq} z$ $\Rightarrow$ $z \leq y .$

(4)

For

x, y \in W_{af}^{+},

x \overset{s}{\leq} y

\Leftrightarrow

x \leq y .

For

x, y \in W_{af}^{-},

x \overset{s}{\leq} y

\Leftrightarrow

y \leq x .

Proof.

(1) Only need to prove that $\begin{matrix} x \overset{s}{\leq} y & ⟺ & y t_{- h} \leq x t_{- h} for sufficiently dominant h \\ ⟺ & y 0 \overset{s}{\leq} x w_{0} \end{matrix}$

Lemma 1: If $x, y \in W_{af}^{-},$ then $x \leq y$ $\Leftrightarrow$ $x w_{0} \leq y w_{0} .$

Lemma 2: $x \overset{s}{\leq} y$ $\Leftrightarrow$ $w_{0} y \leq w_{0} x .$

Proof of Lemma 2.

If $ϕ \in M_{G / B, π_{B} (h_{2} - h_{1})}^{w_{1}, w_{2}},$ then $ϕ_{1}$ defined by $ϕ_{1} (t) ≔ ϕ (\frac{1}{t}) w_{0} \cdot B$ is in $M_{G / B, π_{B} (h_{2} - h_{1})}^{w_{0} w_{2}, w_{0} w_{1}} .$ This shows $x \overset{s}{\leq} y$ $\Leftrightarrow$ $w_{0} y \leq w_{0} x .$

$□$

I can not prove (1).

$□$

Proposition 1: Suppose that $h \in Q^{\lor} .$ Then $\begin{matrix} h \in Q_{+}^{\lor} ≔ \sum_{i \in ℤ} ℤ_{+} α_{i}^{\lor} & ⟺ & id \overset{s}{\leq} w t_{h} \forall w \in W \\ ⟺ & w \overset{s}{\leq} w_{0} t_{h} \forall w \in W \\ ⟺ & x \overset{s}{\leq} x t_{h} \forall w \in W_{af} . \end{matrix}$

Proposition 2: For $β \in Δ_{+}^{re}$ and $x \in W_{af}$ $\begin{matrix} r_{β} x \overset{s}{<} x & ⟺ & ℓ_{s} (r_{β} x) < ℓ_{s} (x) \\ ⟺ & \overline{x^{- 1} \cdot β} < 0, \\ x \overset{s}{<} r_{β} x & ⟺ & ℓ_{s} (x) ℓ_{s} (r_{β} x) \\ ⟺ & \overline{x^{- 1} \cdot β} > 0 . \end{matrix}$

Proposition 3: For $β \in Δ_{+}^{re}$ and $\overline{β} > 0,$ and $x \in W_{af}$ $\begin{matrix} x r_{β} \overset{s}{<} x & ⟺ & ℓ_{s} (x r_{β}) < ℓ_{s} (x) \\ ⟺ & x \cdot β < 0, \\ x \overset{s}{<} x r_{β} & ⟺ & ℓ_{s} (x) < ℓ_{s} (x r_{β}) \\ ⟺ & x \cdot β > 0 . \end{matrix}$

Proposition 4: $x \overset{s}{<} y$ $\Rightarrow$ $ℓ_{s} (x) < ℓ_{s} (y) .$

Proposition 5: If $x \overset{s}{\leq} y,$ then there exists a sequence of the form $x = x_{0} \overset{s}{<} x_{1} \overset{s}{<} x_{2} \overset{s}{<} \dots \overset{s}{<} x_{n} = y$ with $n \geq 0$ and $ℓ (x_{k}) = ℓ_{s} (x) + k$ for $0 \leq k \leq n .$

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

(a)

w \overset{s}{<} x

and

ℓ_{s} (x) = ℓ (w) + 1 .

(b)

x

is one of the following 2 cases:

(1)	$x = w r_{α},$ $α \in {\overline{Δ}}_{+}$ and $ℓ (x) = ℓ (w) + 1 .$
(2)	$x = w r_{α} t_{α^{\lor}} = w r_{α + δ},$ where $α \in {\overline{Δ}}_{+}$ and $ℓ (x) = ℓ (w) - ⟨ 2 ρ, α^{\lor} ⟩ + 1 .$

This is related to multiplication by

H^{2}

in the quantum cohomology.

We now turn to ${(W_{P})}_{af}$ and ${(W^{P})}_{af} :$

Fix a standard parabolic subgroup $P$ of $G .$ Let $\begin{matrix} Δ_{+} (P) & = & {α \in {\overline{Δ}}_{+} : 𝔤_{- α} \in P}, \\ Q_{P}^{\lor} & = & \sum_{α \in Δ_{+} (P)} ℤ α^{\lor} . \end{matrix}$ Set $\begin{matrix} {(W_{P})}_{af} & = & {w t_{h} : w \in W_{P}, h \in Q_{P}^{\lor}} . \end{matrix}$ this is the Weylf group of $\tilde{L_{P}},$ where $L_{P}$ is the Levi-factor of $P .$

Examples:

1)	$P = B,$ $W_{P} = id,$ ${(W_{P})}_{af} = id.$
2)	$P = G,$ $W_{P} = W,$ ${(W_{P})}_{af} = W_{af} .$
3)	$P = P_{i},$ $W_{P} = ⟨ 1, r_{i} ⟩, {(W_{P})}_{af} = ⟨ r_{α_{i}}, r_{δ - α_{i}} ⟩ .$

In general,

{(W_{P})}_{af}

is a Coxeter group; It is a subgroup of

W_{af},

but not a Coxeter subgroup, as seen in the example of

P = P_{i} .

The Length function $ℓ_{P} (y) :$

As a Coxeter group, ${(W_{P})}_{af}$ has a Length function $\begin{matrix} ℓ_{P} (y) & = & # {β > 0 : {\overline{β}}^{\lor} \in Q_{P}^{\lor}, y \cdot β < 0} . \end{matrix}$ Define ${(W^{P})}_{af} :$ $\begin{matrix} {(W^{P})}_{af} & = & {x \in W_{af} : β > 0, {\overline{β}}^{\lor} \in Q_{P}^{\lor} \Rightarrow x \cdot β > 0} . \end{matrix}$

Proposition: $\begin{matrix} W_{af} & = & {(W^{P})}_{af} {(W_{P})}_{af} \end{matrix}$ ie. each $z \in W_{af}$ can be uniquely written as a product $\begin{matrix} z & = & x y \end{matrix}$ where $x \in {(W^{P})}_{af}, y \in {(W_{P})}_{af} .$

Define $\begin{matrix} {\hat{π}}_{P} : & W_{af} & ⟶ & {(W^{P})}_{af} : & z & ⟼ & x . \end{matrix}$

The next proposition gives various properties of ${\hat{π}}_{P} :$

(Note: ????? ${\hat{π}}_{P}$ is what Peterson calls $π_{P}$ in class).

Proposition ${\hat{π}}_{P} :$

1)	${\hat{π}}_{P} (W) = W^{P} \subset {(W^{P})}_{af} \subset {(W_{af})}^{P}$ where ${(W_{af})}^{P}$ is the set of minimal representatives for $W_{af} / W_{P} .$
2)	${\hat{π}}_{P} (W_{af}^{\pm}) \subset W_{af}^{\pm} .$
3)	${\overset{\lor}{π}}_{P} (z) \leq z$ for all $z \in W_{af} .$
4)	For any $z, z' \in W_{af},$ $h \in Q^{\lor},$ have ${\hat{π}}_{P} (z t_{h}) = {\hat{π}}_{P} (z) {\hat{π}}_{P} (t_{h}) .$ $ℓ_{s} ({\hat{π}}_{P} (z t_{h})) = ℓ_{s} ({\hat{π}}_{P} (z)) + ⟨ C_{P}, h ⟩$ where $C_{P} = ρ + w_{P} ρ = \sum_{\underset{w_{0} w_{P} \cdot α < 0}{α \in {\overline{Δ}}_{+}}} α (w_{P} = longest in W).$ $z \overset{s}{\leq} z'$ $\Rightarrow$ ${\hat{π}}_{P} (z) \overset{s}{\leq} {\hat{π}}_{P} (z') .$ ${\hat{π}}_{P} (r_{β} z) < {\hat{π}}_{P} (z)$ $⟺$ $\overline{z^{- 1} \cdot β} \in Δ (g . p)$ $(\subset {\overline{Δ}}_{+}) .$ ${\hat{π}}_{P} (r_{β} z) = {\hat{π}}_{P} (z)$ $⟺$ $\overline{z^{- 1} \cdot β} \in Q_{P}^{+} .$ ${\hat{π}}_{P} (r_{β} z) > {\hat{π}}_{P} (z)$ $⟺$ $\overline{z^{- 1} \cdot β} \in - Δ (g / p)$ $(\subset {\overline{Δ}}_{+}) .$

Proposition: For $y \in {(W_{P})}_{af},$ $\begin{matrix} ℓ_{s, P} (y) & = & ℓ_{s} (y) \end{matrix}$ where $ℓ_{s, P}$ is the stable length function for ${(W_{P})}_{af} .$

Proposition: For $x \in {(W^{P})}_{af},$ $y \in {(W_{P})}_{af},$ $\begin{matrix} ℓ_{s} (x y) = ℓ_{s} (x) + ℓ_{s} (y), \\ ℓ (x) + ℓ_{s} (y) \leq ℓ (x y) . \end{matrix}$

Proposition: Any given $x \in {(W^{P})}_{af},$ can put $x t_{h} \in W_{af}^{+}, x t_{- h} \in W_{af}^{-}$ for sufficiently dominant $h \in {(Q^{\lor})}^{W_{P}},$ ie. $⟨ ρ_{i}, h ⟩ ≫ 0$ for all $i \in I$ such that $r_{i} \notin W_{P} .$

Notation: $\begin{matrix} \begin{matrix} {(\tilde{P})}_{0} & = & the identity component of \tilde{P}, \\ 𝔐_{P} & = & \tilde{G} / {(\tilde{P})}_{0}, \\ *_{P} & = & {(\tilde{P})}_{0} \in 𝔐_{P}, \end{matrix} \\ \begin{matrix} {\overset{\circ}{π}}_{P} : & 𝔐_{B} & ⟶ & 𝔐_{P} : & g *_{B} & ⟼ & g \cdot *_{P} . \end{matrix} \end{matrix}$ Have action of $Γ$ on $𝔐_{P} :$ $\begin{matrix} 𝔐_{P} \times Γ & ⟶ & 𝔐_{P} : & (g \cdot *_{P}) \cdot t & = & g t \cdot *_{P} . \end{matrix}$ This action is trivial if $t \in {t_{h} : h \in Q_{P}^{\lor}} .$ Set, for $z \in W_{af},$ $\begin{matrix} 𝔐_{P, z}^{\pm} & = & B_{af}^{\pm} z \cdot *_{P} . \end{matrix}$

Proposition: For $z \in W_{af}$ and $t \in Γ$ $\begin{matrix} 𝔐_{P, z}^{\pm} & = & 𝔐_{P, {\hat{π}}_{P} (z)}^{\pm}, \\ (𝔐_{P, z}^{\pm}) \cdot t & = & 𝔐_{P, z t}^{\pm}, \end{matrix}$ and for $x_{1}, x_{2} \in {(W^{P})}_{af},$ $𝔐_{P, x_{1}}^{-} \cap 𝔐_{P, x_{2}}^{+} \neq \emptyset ⟺ x_{1} \overset{s}{\leq} x_{2} .$

The moduli space $𝔐_{τ} = 𝔐_{τ, P} :$

Definition: Given a scheme $V / ℂ$ and a morphism $\begin{matrix} f : & V \times_{ℂ} ℙ^{1} & ⟶ & G / P, \end{matrix}$ we say that $f$ is of type $τ,$ for $τ \in H_{2} (G / P),$ if for any $ℂ -valued$ point $v$ of $V,$ the map $f_{V} : ℙ^{1} \to G / P$ defined by $\begin{matrix} ℙ^{1} & ≃ & ℂ \times_{ℂ} ℙ^{1} & \overset{v \times id}{⟶} & V \times_{ℂ} ℙ^{1} & \overset{f}{⟶} & G / P \end{matrix}$ satisfies $\begin{matrix} {(f_{V})}_{*} [ℙ^{1}] & = & τ . \end{matrix}$

The universal property of $(𝔐_{τ}, ev) :$

Proposition: Fix $τ \in H_{2} (G / P) .$ There exists a pair $(𝔐_{τ}, ev)$ where $𝔐_{τ}$ is a reduced scheme of finite type over $ℂ$ and $ev : 𝔐_{τ} \times_{ℂ} ℙ^{1} \to G / P$ is a morphism over $ℂ$ s.t.

1)	ev is of type $τ;$
2)	if $V$ is any reduced scheme of finite type over $ℂ$ and $f : V \times_{ℂ} ℙ^{1} \to G / P$ is a morphism over $ℂ,$ then $\exists!$ morphism $\hat{f} : V \to 𝔐_{τ}$ over $ℂ$ s.t. $\begin{matrix} f & = & ev \circ (\hat{f} \times id) . \end{matrix}$

Thus

(𝔐_{τ}, ev)

is unique up to a unique isomorphism. Moreover,

𝔐_{τ}

is quasi-projective, and it is either empty or else smooth and of dim

\begin{matrix} dim 𝔐_{τ} & = & dim G / P + ⟨ C_{1} T_{G / P}, τ ⟩ . \end{matrix}

Here we outline a proof of the fact that the Zariski tangent space to $𝔐_{τ}$ at $ϕ \in 𝔐_{τ}$ always has the above dimension: Suppose $\begin{matrix} ϕ : & ℙ^{1} & ⟶ & G / P \end{matrix}$ is s.t. $ϕ_{*} [ℙ^{1}] = τ .$ Then $\begin{matrix} T_{ϕ} 𝔐_{τ} & = & Γ (ℙ^{1}, ϕ^{*} T_{G / P}) . \end{matrix}$ Now as sheaves over $G / B,$ we have $\begin{matrix} 0 & ⟶ & 𝔞 & ⟶ & 𝔟 & ⟶ & T_{G / P} & ⟶ & 0 \end{matrix}$ 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 $ϕ,$ we have $\begin{matrix} 0 & ⟶ & ϕ^{*} 𝔞 & ⟶ & ϕ^{*} 𝔟 & ⟶ & ϕ^{*} T_{G / P} & ⟶ & 0 . \end{matrix}$ Thus we have the long exact sequence $\begin{matrix} 0 & ⟶ & H^{0} (ℙ^{1}, ϕ^{*} 𝔞) & ⟶ & H^{0} (ℙ^{1}, ϕ^{*} 𝔟) & ⟶ & H^{0} (ℙ^{1}, ϕ^{*} T_{G / P}) & ⟶ \\ ⟶ & H^{1} (ℙ^{1}, ϕ^{*} 𝔞) & ⟶ & H^{1} (ℙ^{1}, ϕ^{*} 𝔟) & ⟶ & H^{1} (ℙ^{1}, ϕ^{*} T_{G / P}) & ⟶ & 0 . \end{matrix}$ Since $𝔟$ is trivial as a vector bundle, have $\begin{matrix} H^{1} (ℙ^{1}, ϕ^{*} 𝔟) & = & 0 \\ \Rightarrow H^{1} (ℙ^{1}, ϕ^{*} T_{G / P}) & = & 0 \\ \Rightarrow dim Γ (ℙ^{1}, ϕ^{*} T_{G / P}) & = & dim H^{0} (ℙ^{1}, ϕ^{*} T_{G / P}) \\ = & χ (ϕ^{*} T_{G / P}) . \end{matrix}$ Using the general fact that for any vector bundle $E$ over $ℙ^{1},$ $\begin{matrix} χ (E) & = & dim E + ⟨ C_{1} (E), [ℙ^{1}] ⟩ . \end{matrix}$ We get $\begin{matrix} dim Γ (ℙ^{1}, ϕ^{*} T_{G / P}) & = & dim G / P + ⟨ C_{1} (ϕ^{*} T_{G / P}), [ℙ^{1}] ⟩ \\ = & dim G / P + ⟨ ϕ^{*} C_{1} (T_{G / P}), [ℙ^{1}] ⟩ \\ = & dim G / P + ⟨ C_{1} (T_{G / P}), ϕ^{*} [ℙ^{1}] ⟩ \\ = & dim G / P + ⟨ C_{1} (T_{G / P}), τ ⟩ . \end{matrix}$

Now for $τ \in H_{2} (G / P),$ $v, w \in W^{P},$ set $\begin{matrix} 𝔐_{τ}^{v, w} & = & B_{-} v \cdot P \times_{G / P} 𝔐_{τ} \times_{G / P} B w \cdot P . \end{matrix}$ By a theorem of Kleiman, we have

Proposition:

(1)	$𝔐_{τ}^{id, w_{0} w_{P}}$ is open and dense in $𝔐_{τ},$
(2)	$𝔐_{τ}^{v, w}$ is quasi-projective, and $\begin{matrix} dim 𝔐_{τ}^{v, w} & = & ⟨ C_{1} (T_{G / P}), τ ⟩ - ℓ (v) + ℓ (w) . \end{matrix}$

Kleiman's Theorem: Suppose $X$ is a homogeneous $G -space$ and $\begin{matrix} σ_{Y} : & Y & ⟶ & X \\ σ_{Z} : & Z & ⟶ & X \end{matrix}$ are smooth maps. Then for generic $g_{1}, g_{2} \in G,$ the set $\begin{matrix} g_{1} \cdot Y \times_{X} g_{2} \cdot Z & = & {(y, z) : g_{1} \cdot σ_{Y} (y) = g_{2} σ_{Z} (z)} \end{matrix}$ is a regular reduced variety of $dim = dim Y + dim Z - dim X .$

Lecture 14: April 15, 1997

Today we introduce two rings for each parabolic $P :$

1.	$R_{P}^{'} = q H^{T} {(G / P)}_{(q)} :$ $T -equivariant$ quantum cohomology of $G / P$ with the quantum parameter $q$ inverted,
2.	$R_{P} = q G_{T} (G / P) :$ $T -equivariant$ quantum cohomology of $G / P .$

Definition: $R_{P}^{'}$ is a free $S -module$ on symbols $σ_{P}^{(x)},$ $x \in {(W^{P})}_{af}$ with $z -grading$ $\begin{matrix} deg (s σ_{P}^{(x)}) & = & deg s + 2 ℓ_{s} (x) . \end{matrix}$

The ${\underline{A}}_{af}$ module structure on $R_{P}^{'}$

The $S -module$ structure on $R_{P}^{'}$ extends to an ${\underline{A}}_{af} -module$ structure on $R_{P}^{'}$ by $\begin{matrix} ν (A_{i}) \cdot σ_{P}^{(x)} & = & {\begin{matrix} - σ_{P}^{(r_{i} x)} & if \overline{x^{- 1} \cdot α_{i}} \in Δ (\underline{𝔤} / \underline{𝔭}), \\ 0 & otherwise \end{matrix} \end{matrix}$ where $ν$ is the automorphism of ${\underline{A}}_{af}$ defined at the end of Lecture 10 (page 10-13).

The map $ψ_{P} : H_{T} (Ω K) \to R_{P}^{'} :$

It is the $S -module$ map defined by $\begin{matrix} ψ_{P} (σ_{[x]}^{Ω}) & = & {\begin{matrix} σ_{P}^{(x)} & if x \in {(W^{P})}_{af}, \\ 0 & otherwise \end{matrix} \end{matrix}$ for all $x \in W_{af}^{-} .$

It should be easy to check that

1)	$ψ_{P} (σ) = j (σ) \cdot σ_{P}^{(id)} \in R_{P}^{'}$ $\forall$ $σ \in H_{T} (Ω K),$
2)	$ψ_{P}$ is an ${\underline{A}}_{af} -map.$

Theorem: There exists a unique commutative $S -algebra$ structure on $R_{P}^{'}$ such that

1)	$σ_{P}^{(id)} = 1,$
2)	$R_{P}^{'}$ is an $Ω -integrable$ ${\underline{A}}_{af} -module$ with the structure homomorphism $S \to R_{P}^{'} :$ $s \mapsto s σ_{P}^{(id)}$ and the ${\underline{A}}_{af} -module$ structure defined above.

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

1)	an $\underline{A} -module$ structure on $𝒪 (X),$
2)	an $S -map$ $f : H_{T} (Ω K) \to 𝒪 (X)$

such that

1)	$s \cdot p = π_{X} (s) p$ $\forall$ $s \in S,$ $p \in 𝒪 (X),$
2)	$π_{X}$ is an $\underline{A} -module$ map,
4)	$m : 𝒪_{(X)} \otimes_{S} 𝒪_{(X)} \to 𝒪 (X)$ is an $\underline{A} -module$ map,
5)	$f : H_{T} (Ω K) \to 𝒪 (X)$ is an $\underline{A} -module$ map.

Recall that we have used the notation

\begin{matrix} 𝒰 & = & Spec H^{T} (K / T), \\ 𝒜 & = & Spec H_{T} (Ω K) . \end{matrix}

Conditions 1)-4) say that

X = Spec 𝒪 (X)

is a

𝒰 -space,

where

𝒰

is a groupoid, and condition 5) says that

Spec f : X \to 𝒜

is a

𝒰 -space

morphism.

The geometrical models

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

$𝒰 = Spec H^{T} (K / T) :$ $G^{\lor}, 𝔤^{\lor}, \underline{h} = {({\underline{h}}^{\lor})}^{*} \subset {({\underline{𝔤}}^{\lor})}^{*} etc.$ For each $i \in I,$ let ${f_{i}^{\lor}}^{*} \subset {({\underline{𝔤}}^{\lor})}^{*}$ be such that $⟨ {f_{i}^{\lor}}^{*}, f_{j}^{\lor} ⟩ = δ_{?????}$ and with weight $α_{i}^{\lor} .$ Let $E = \sum_{i \in I} {f_{i}^{\lor}}^{*} \in {(𝔤^{\lor})}^{*} .$ Set $\begin{matrix} 𝒰 & = & {(E + h, u) \in (E + \underline{h}) \times U_{-}^{\lor} : u^{- 1} \cdot (E + h) \in {({\underline{n}}_{+}^{\lor})}^{⊥}} . \end{matrix}$ Notice that $𝒰$ can be identified with the following subset of $(E + \underline{h}) \times U_{-} \times (E + \underline{h}) :$ $\begin{matrix} 𝒰 & = & {(E + h_{1}, u, E + h_{2}) : u^{- 1} \cdot (E + h_{1}) = E + h_{2}} . \end{matrix}$ It thus has a groupoid structure as a subgroupoid of the direct product groupoid $(E + \underline{h}) \times U_{-} \times (E + \underline{h}) :$
$𝒜 = Spec H_{T} (Ω K) = {B^{\lor}}^{E + \underline{h}} = {(E + h, b) \in (E + \underline{h}) \times B^{\lor} : b \cdot (E + h) = E + h} .$
Action of $𝒰$ on ${B^{\lor}}^{E + \underline{h}} :$ $\begin{matrix} 𝒰 \times_{h} {B^{\lor}}^{E + \underline{h}} & ⟶ & {B^{\lor}}^{E + \underline{h}} : \\ (E + h, u, E + h') \cdot (E + h', b) & = & (E + h, u b u^{- 1}) . \end{matrix}$
The variety $Y^{E + \underline{h}} :$ $\begin{matrix} Y^{E + \underline{h}} & = & {(E + h, g \cdot B^{\lor}) \in (E + \underline{h}) \times G^{\lor} / B^{\lor} : g^{- 1} \cdot (E + h) ⊥ [{\underline{n}}_{+}^{\lor}, {\underline{n}}_{+}^{\lor}]} . \end{matrix}$ Have $\begin{matrix} {B^{\lor}}^{E + \underline{h}} & ⟶ & Y^{E + \underline{h}} : \\ (E + h, b) & ⟼ & (E + h, b w_{0} \cdot B^{\lor}) . \end{matrix}$ $𝒰$ acts on $Y^{E + \underline{h}} :$ $\begin{matrix} 𝒰 \times_{h} Y^{E + \underline{h}} & ⟶ & Y^{E + \underline{h}} : \\ (u + h, u, E + h') \cdot (E + h', g \cdot B \lor) & = & (E + h, u g \cdot B^{\lor}) . \end{matrix}$ The inclusion $\begin{matrix} {B^{\lor}}^{E + \underline{h}} & ⟶ & Y^{E + \underline{h}} \end{matrix}$ is $𝒰 -equivariant.$ $Spec R_{P}^{'} = {Y_{P}^{+}}^{E + \underline{h}} \cap {Y_{G}^{-}}^{E + \underline{h}} \subset Y^{E + h} (𝒰 -subset).$

The subring $Λ_{P}^{'} \subset R_{P}^{'} :$

For $h \in Q^{\lor},$ so $π_{P} (h) \in H_{2} (G / P),$ set $\begin{matrix} q_{π_{P} (h)} & = & σ_{P}^{({\hat{π}}_{P} (t_{h}))} \in R_{P}^{'}, \end{matrix}$ and $Λ_{P}^{'} = ℤ {q_{π_{P} (h)} : h \in Q^{\lor}} ≃ ℤ [H_{2} (G / P)] .$

Fact:

$Λ_{P}^{'} \subset R_{P}^{'}$ is a subring with $\begin{matrix} q_{τ} q_{τ'} & = & q_{τ + τ'} \\ deg q_{τ} & = & 2 ⟨ C_{1} T_{G / P}, τ ⟩ . \end{matrix}$
$q_{π_{P} (h)} \cdot σ_{P}^{(x)} = σ_{P}^{(x {\hat{π}}_{P} (t_{h}))} = σ_{P}^{({\hat{π}}_{P} (x t_{h}))} .$
The ${\underline{A}}_{af} -module$ structure on $R_{P}^{'}$ is $Λ_{P}^{'} -linear.$

Example $({\hat{π}}_{P} (t_{h})$ is not necessarily translational): ${s l}_{3}$ with extended Dynkin diagram $\begin{matrix} \end{matrix} .$ Let $W_{P} = ?????$ and $t = t_{θ} = r_{0} r_{θ} = \underset{\underset{\underset{{(W^{P})}_{af}}{⫙}}{⏟}}{r_{0} r_{2} r_{1}} \underset{\underset{\underset{{(W^{P})}_{af}}{⫙}}{⏟}}{r_{2}}$ $\Rightarrow$ ${\hat{π}}_{P} (t) = r_{0} r_{2} r_{1} .$

Fact: ${σ_{P}^{(w)} : w \in W^{P}}$ is a basis of $R_{P}^{'}$ over $S \times Λ_{P}^{'} .$

Proposition: Formulas for multiplications $*$ in $R_{P}^{'}$ and ${\underline{A}}_{af}$ on $R_{P}^{'} :$ $\begin{matrix} A_{i} \cdot (s σ) & = & (A_{i} \cdot s) σ + (r_{i} \cdot s) (A_{i} \cdot σ) \\ A_{i} \cdot σ_{P}^{(w)} & = & {\begin{matrix} - σ_{P}^{(r_{i} w)} & if w^{- 1} \cdot α_{i} \in Δ (\underline{𝔤} / \underline{𝔭}), \\ 0 & otherwise, \end{matrix} \\ (A_{i} \cdot σ) * σ' & = & A_{i} \cdot [σ * (r_{i} \cdot σ')] + σ * (A_{i} \cdot σ') . \end{matrix}$

The operator $A_{0}^{'} :$

Assume that $G$ is simple. $α_{0} = δ - θ,$ $Π_{af} = Π \cup {0},$ $A_{0}^{'} = ν (A_{0}) = - w_{0} A_{0} w_{0}$ where $w_{0} \in W$ is the longest element.

Proposition: $\begin{matrix} A_{0}^{'} \cdot (s σ) & = & - (A_{θ^{\lor}} \cdot s) σ + (r_{θ} \cdot s) (A_{0}^{'} \cdot σ), \\ A_{0}^{'} \cdot σ_{P}^{(w)} & = & {\begin{matrix} - q_{π_{P} (w^{- 1} \cdot θ^{\lor})}^{- 1} σ_{P}^{({\hat{π}}_{P} (r_{θ} w))} & if w^{- 1} \cdot θ \in Δ (\underline{𝔤} / \underline{𝔭}), \\ 0 & otherwise, \end{matrix} \\ (A_{0}^{'} \cdot σ) * σ' & = & A_{0}^{'} \cdot (σ * (r_{θ} \cdot σ')) - σ * (A_{θ^{\lor}} \cdot σ') . \end{matrix}$

Theorem: $ψ_{P} : H_{T} (Ω K) \to R_{P}^{'}$ is a homomorphism of $S -algebras,$ and $\begin{matrix} ψ_{P} (σ) * σ' & = & j (σ) \cdot σ' \end{matrix}$ for $σ \in H_{T} (Ω K)$ and $σ' \in R_{P}^{'} .$

Proof.

This is a direct consequence of $R_{P}^{'}$ being $Ω -integrable.$

$□$

The structure constants $J_{P, z}^{x, y},$ $x, y, z \in {(W^{P})}_{af} :$

For $x, y, z \in {(W^{P})}_{af},$ define structure constants $J_{P, z}^{x, y} \in S$ by $\begin{matrix} σ_{P}^{(x)} \times σ_{P}^{(y)} & = & \sum_{z \in {(W^{P})}_{af}} J_{P, z}^{x, y} σ_{P}^{(z)} . \end{matrix}$

Facts:

(1)	$deg J_{P, z}^{x, y} = 2 (ℓ_{s} (x) + ℓ_{s} (y) - ℓ_{s} (z)),$
(2)	$J_{P, z}^{x, y} = J_{B, z}^{x, y},$
(3)	$J_{P, z π_{P} (t t')}^{x {\hat{π}}_{P} (t), y {\hat{π}}_{P} (t')} = J_{P, z}^{x, y},$
(4)	For $x, y, z \in {(W^{P})}_{af}$ with $x \in W_{af}^{-},$ $\begin{matrix} J_{P, z}^{x, y} & = & {\begin{matrix} ϵ (x y z) j_{x}^{z y^{- 1}} & if ℓ (y z^{- 1}) + ℓ_{s} (z) = ℓ_{s} (y), \\ 0 & otherwise. \end{matrix} \end{matrix}$

Multiplication by $H^{2}$ in $R_{B}^{'} :$

Theorem: For $i \in I$ and $w \in W$ $\begin{matrix} σ_{B}^{(r_{i})} * σ_{B}^{(w)} & = & \sum_{\underset{ℓ (w r_{α}) = ℓ (w) + 1}{α \in {\overline{Δ}}_{+}}} ⟨ ρ_{i}, α^{\lor} ⟩ σ_{B}^{(w r_{α})} \\ + \sum_{\underset{ℓ (w r_{α}) = ℓ (w) + 1 - ⟨ 2 ρ, α^{\lor} ⟩}{α \in {\overline{Δ}}_{+}}} ⟨ ρ_{i}, α^{\lor} ⟩ q_{π_{B} (α^{\lor})} σ_{B}^{(w r_{α})} \\ - (ρ_{i} - w \cdot ρ_{i}) σ_{B}^{(w)} . \end{matrix}$

Remark: One way of looking at the above formula is $\begin{matrix} σ_{B}^{(r_{i})} + ρ_{i} & = & {(ρ_{i})}_{R} + \sum_{\underset{ℓ (r_{α}) = ⟨ 2 ρ, α^{\lor} ⟩ - 1}{α \in {\overline{Δ}}_{+}}} ⟨ ρ_{i}, α^{\lor} ⟩ q_{π_{B} (α^{\lor})} A_{r_{α}}, \end{matrix}$ where the left hand side is a multiplication operator on $R_{B}^{'}$ and the right hand side is an element in ${\underline{A}}_{af}$ considered as an operator on $R_{B}^{'} .$ The right hand side is a commuting family of elements in ${\underline{A}}_{af} .$

A fact with no classical analogy:

$\begin{matrix} {\underline{A}}_{af} \otimes_{Λ_{B}^{-}} Λ_{B}^{'} & ≅ & {End}_{{[R_{B}^{'}]}^{W}} R_{B}^{'} where Λ_{B}^{-} = \sum_{\underset{h dominant}{h \in Q^{\lor}}} ℤ q ????? \\ {[R_{B}^{'}]}^{W} & ≅ & {End}_{{\underline{A}}_{af} \otimes_{Λ_{B}^{-}} Λ_{B}^{'}} R_{B}^{'} \\ {End}_{\underline{A} \otimes Λ_{B}^{'}} R_{B}^{'} & ≅ & {\underline{A}}_{R} \otimes Λ_{B}^{'} \end{matrix}$

The ring $R_{P}$

Define $\begin{matrix} R_{P} & = & \sum_{\underset{x \overset{s}{\geq} id}{x \in {(W^{P})}_{af}}} s σ_{P}^{(x)} . \end{matrix}$ (Recall that $x = w t_{h} \overset{s}{\geq} id$ $\Leftrightarrow$ $h \in Q_{+}^{\lor}).$ It is clear from the way $\underline{A}$ acts that $R_{P}$ is an $\underline{A} -stable$ submodule of $R_{P}^{'} .$

Fact: For $z \in W_{af},$ $\sum_{\underset{x \overset{s}{\geq} z}{x \in {(W^{P})}_{af}}} s σ_{P}^{(x)}$ is an $R_{P} -submodule$ of $R_{P}^{'} .$

Let $Λ_{P} = Λ_{P}^{'} \cap R_{P} = \sum_{τ \in π_{P} (Q_{+}^{\lor})} z q_{τ} .$ Then $\begin{matrix} R_{P} \otimes_{Λ_{P}} Λ_{P}^{'} & ≅ & R_{P}^{'} \end{matrix}$ and ${σ_{P}^{(w)} : w \in W^{P}}$ is an $S \otimes Λ_{P} -basis$ of $R_{P} .$ The augumentation homomorphism is defined to be $\begin{matrix} ε : & Λ_{P} & ⟶ & ℤ : & ε (q_{τ}) & = & δ_{τ, 0} . \end{matrix}$

Fact: The map $\begin{matrix} R_{P} \otimes_{Λ_{P}} ℤ & \tilde{⟶} & H^{T} (G / P) \\ σ_{P}^{(w)} \otimes 1 & ⟼ & σ_{P}^{(w)} \end{matrix}$ is an isomorphism as $\underline{A} -modules$ and $S -algebras.$

Thus it is reasonable to call $R_{P}$ the $T -equivariant$ quantum cohomology of $G / P .$ It specializes to the $T -equivariant$ cohomology of $G / P$ when the quantum parameters $Λ_{P}$ go to $0 .$

Poincare Duality

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

Define the $S \otimes Λ_{P} -linear$ map $\begin{matrix} \int : & R_{P} & ⟶ & S \otimes Λ_{P} \end{matrix}$ by $\begin{matrix} \int σ_{P}^{(w)} & = & δ_{w, w_{0} w_{P}} for w \in W^{P} . \end{matrix}$

Theorem: $\int σ_{P}^{(v)} * (w_{0} \cdot σ_{P}^{(w_{0} w w_{P})}) = δ_{v, w} .$

Corollary: Have an isomorphism $\begin{matrix} P D : & R_{P} & ≃ & {Hom}_{S \otimes Λ_{P}} (R_{P}, S \otimes Λ_{P}) \end{matrix}$ defined by $\begin{matrix} P D (φ) (φ') & = & \int φ \times φ', \end{matrix}$ or concretely $\begin{matrix} P D (σ_{P}^{(w)}) & = & w_{0} \cdot σ_{P}^{(w_{0} w w_{P})} . \end{matrix}$

The Euler Class $χ_{G / P} :$

$\begin{matrix} χ_{G / P} & \overset{def}{=} & {P D}^{- 1} ({tr}_{R_{P} / S \otimes Λ_{P}}) \end{matrix}$ where ${tr}_{R_{P} / S \otimes Λ_{P}} \in {Hom}_{S \otimes Λ_{P}} (R_{P}, S \otimes Λ_{P})$ is defined by $\begin{matrix} {tr}_{R_{P} / S \otimes Λ_{P}} (ϕ) & = & trace over S \otimes Λ_{P} of (ℓ_{φ} : ϕ' \mapsto φ * φ') . \end{matrix}$ In other words, $\begin{matrix} {tr}_{R_{P} / S \otimes Λ_{P}} (ϕ) & = & \int ϕ * χ_{G / P} . \end{matrix}$ Write $\begin{matrix} σ_{P}^{(v)} * σ_{P}^{(w)} & = & \sum_{u \in W^{P}} b_{u}^{v, w} σ_{P}^{(u)} . \end{matrix}$ Then $\begin{matrix} {tr}_{R_{P} (S \otimes Λ_{P})} (σ_{P}^{(v)}) & = & \sum_{w \in W^{P}} b_{w}^{v, w} \\ = & \sum_{w \in W^{P}} \int σ_{P}^{(v)} * σ_{P}^{(w)} * (w_{0} \cdot σ_{P}^{(w_{0} w w_{P})}) \end{matrix}$ $\begin{matrix} \Rightarrow χ_{G / P} & = & \sum_{w \in W^{P}} σ_{P}^{(w)} * (w_{0} \cdot σ_{P}^{(w_{0} w w_{P})}) . \end{matrix}$

Facts:

1)	$ϕ * χ_{G / P} = 0$ $\Leftrightarrow$ $ϕ$ is nilpotent,
2)	$χ_{G / P}$ annihilates $Ω_{R_{P} / S \otimes Λ_{P}} .$

Example: For $s l (3)$ and $P = B,$ $χ_{G / B}$ is invertible $\Leftrightarrow$ $q_{1} q_{2} (q_{1} + q_{2})$ is invertible.

Lecture 15: April 16, 1997

More facts on $R_{B} :$

Fact 1: For $w \in W,$ $\begin{matrix} \sum_{\underset{u v = w [red]}{u, v \in W}} ϵ (u) σ_{B}^{(u^{- 1})} * σ_{B}^{(v)} & = & δ_{w, 1}, \\ \sum_{\underset{u v = w [red]}{u, v \in W}} σ_{B}^{(u)} * ϵ (v) σ_{B}^{(v^{- 1})} & = & δ_{w, 1} . \end{matrix}$

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

Fact 2: For $σ \in R_{B},$ $\begin{matrix} σ & = & \sum_{w \in W} [A_{w_{0}} \cdot (σ \cdot (w_{0} \cdot σ_{B}^{(w_{0} w)}))] * ϵ (w) σ_{B}^{(w)} . \end{matrix}$ What does this mean? This is not expressing $σ$ in the basis ${ϵ (w) σ_{B}^{(w)} : w \in W}$ of $R_{B}$ as an $S \otimes Λ_{B} -module.$

Fact 3: $R_{B}$ is a free ${(R_{B})}^{\underline{A}} -module$ with basis ${σ_{B}^{(w)} : w \in W} .$

Fact 4: ${(R_{B})}^{\underline{A}}$ is a polynomial ring on the $q_{π_{B} (α_{i}^{\lor})}'s$ and the $σ_{B}^{(r_{i})} + ρ_{i}$ for $i \in I .$

Fact 5: ${(R_{B})}^{\underline{A}} \to ℤ \otimes_{S} R_{B}$ is onto over $ℚ .$ (?)

The $S -subalgebra$ $R_{P}^{-}$ of $R_{P}^{'} :$

Define $R_{P}^{-} = Im ψ_{P} = \sum_{x \in {(W^{P})}_{af} \cap W_{af}^{-}} S σ_{P}^{(x)} .$ Then $\begin{matrix} R_{B}^{-} & ≃ & H_{T} (Ω K) \end{matrix}$ but in general $\begin{matrix} H_{T} (Ω K) & \begin{matrix} \end{matrix} & R_{P}^{-} . \end{matrix}$ We have:

$R_{P}^{-} = {\underline{A}}_{af} \cdot σ_{P}^{(id)} .$
Every ${\underline{A}}_{af} -submodule$ of $R_{P}^{'}$ is an $R_{P}^{-} -submodule$ of ?????.
$\begin{matrix} R_{P}^{-} \otimes_{Λ_{P}^{-}} Λ_{P}^{'} & ≅ & R_{P}^{'} \end{matrix}$ where $Λ_{P}^{-} = Λ_{P}^{'} \cap R_{P}^{-} = \sum_{\underset{h is dominant}{h \in Q^{\lor}}} ℤ q_{π_{P} (h)} .$

Remark: Working with the case when $G$ is simple, connected but not necessarily simply connected so $Ω K$ is no longer connected, we get the following fact: Assume that $a_{i} = 1$ for all $i \in I$ in $θ = \sum_{i \in I} a_{i} α_{i} .$ Let $P = P_{ρ_{i}}$ so $W_{P} = {⟨ r_{j} ⟩}_{j \in I, j \neq i} .$ Let $w = w_{0} w_{P} .$ Let $Q$ be a standard parabolic Then $\begin{matrix} σ_{Q}^{{\hat{π}}_{Q} (w)} * (w σ_{Q}^{(v)}) & = & q_{π_{Q} (ρ_{i}^{\lor} - v^{- 1} ρ_{i}^{\lor})} σ_{Q}^{({\hat{π}}_{Q} (w v))} \end{matrix}$ for all $v \in W^{Q} .$ Consequently $σ^{({\hat{π}}_{Q} (w))}$ is invertible in $R_{Q}^{'}$ (no clue! How is $Q$ related to $P = P_{ρ_{i}} ?)$

Example: $G = {S L}_{3},$ (?) $W_{P} = r_{2},$ $G / P = ℙ^{2} .$ $σ^{21} \times σ^{21} \times σ^{21} = q^{2}$ (?).

A Filtration:

For $h \in Q^{\lor},$ define an $\underline{A} -submodule$ $F_{P, h}^{-}$ of $R_{P}^{-}$ (depends only on $h$ mod $Q_{P}^{\lor})$ by $F_{P, h}^{-} = R_{P}^{-} \cap q_{- π_{P} (h)} R_{P} = \sum_{\underset{x \overset{s}{\geq} {\hat{π}}_{P} (t_{- h})}{x \in {(W^{P})}_{af} \cap W_{af}^{-}}} S σ_{P}^{(x)}$ (a finite sum). Then $F_{P, h}^{-} * F_{P, h'}^{-} \subset F_{P, h + h'}^{-} .$

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

Fact: When $h \in Q^{\lor}$ is dominant, $\begin{matrix} F_{P, h}^{-} & = & ψ_{P} (T_{t_{- ω (h)}}) \end{matrix}$ where $F_{t_{- ω (h)}}$ is the Bruhat-Filtration in $H_{T} (Ω K)$ in Lecture 9 and $ω (h) = - w_{0} \cdot h$ is the diagram automorphism. Have

$F_{P, h}^{-} * F_{P, h'}^{-} = F_{P, h + h'}^{-} .$
$σ * F_{P, h}^{-} \subset F_{P, h'}^{-}$ $\Leftrightarrow$ $σ \in F_{P, h' - h}^{-} .$

More on $G / B$ and $G / P :$

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

An analogous statement is true for quantum cohomology:

Consider the $S -algebra$ $\begin{matrix} R^{P, Q} & = & \sum_{\underset{\underset{x \overset{s}{\geq} id}{y \in {(W_{Q})}_{af}}}{x \in {(W^{P})}_{af}}} s σ_{Q}^{(x y)} . \end{matrix}$ Let $\begin{matrix} R_{\leq n}^{P, Q} & = & \sum_{\underset{\underset{ℓ_{s} (y) \leq n}{x \overset{s}{\geq} id}}{\underset{y \in {(W_{Q})}_{af}}{x \in {(W^{P})}_{af}}}} s σ_{Q}^{(x y)} . \end{matrix}$

Fact: $R_{\leq m}^{P, Q} R_{\leq n}^{P, Q} \subset R_{\leq m + n}^{P, Q} .$

Define ${\overline{R}}^{P, Q} = gr R^{P, Q} = \sum_{n \in ℤ} {\overline{R}}_{n}^{P, Q}$ where $\begin{matrix} {\overline{R}}_{n}^{P, Q} & = & R_{\leq n}^{P, Q} / R_{\leq (n - 1)}^{P, Q} . \end{matrix}$

Fact: $\begin{matrix} \underset{\underset{= R_{P}}{⏟}}{R_{G / P}} \otimes_{ℤ} (ℤ \otimes_{S} R_{P Q}^{'}) & ≃ & {\overline{R}}^{P, Q} . \end{matrix}$

Define $\begin{matrix} R_{-}^{P, Q} & = & \sum_{\underset{y \in {(W_{Q})}_{af} \cap {(W_{P})}_{af}^{-}}{\underset{x \overset{s}{\geq} id}{x \in {(W^{P})}_{af}}}} s σ_{Q}^{(x y)}, \\ R_{- n}^{P, Q} & = & \sum_{\underset{\underset{ℓ_{s} (y) \leq n}{y \in {(W_{Q})}_{af} \cap {(W_{P})}_{af}^{-}}}{\underset{x \overset{s}{\geq} id}{x \in {(W^{P})}_{af}}}} s σ_{Q}^{(x y)} . \end{matrix}$

Fact: $\begin{matrix} gr R_{-}^{P, Q} & ≃ & \underset{\underset{= R_{P}}{⏟}}{R_{G / P}} \otimes [Im (H_{*} Ω_{0} (K \cap P) ⟶ ℤ \otimes_{S} R_{Q}^{'})] . \end{matrix}$

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

Fact:

For $G = S L (n, ℂ)$ 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 $G \supset P \supset B .$ The fact $\begin{matrix} gr R_{-}^{P, B} ≃ R_{P} \otimes (Im (H_{*} Ω_{0} (K \cap P) ⟶ ℤ \otimes_{S} R_{B}^{'})) & (*) \end{matrix}$ has the following meaning in terms of the geometric models: Recall the (Peterson) variety $Y \subset G^{\lor} / B^{\lor} .$ It contains $2^{ℓ} - T -fixed$ points ${w_{P} : P parabolic} .$ Label them by $y_{P} .$ Set $\begin{matrix} Y_{P}^{+} & = & Y \cap B_{-}^{\lor} w_{P} \cdot B^{\lor}, \\ Y_{P}^{-} & = & Y \cap B_{+}^{\lor} w_{P} \cdot B^{\lor} (B_{+}^{\lor} = B^{\lor}) . \end{matrix}$ Then $\begin{matrix} R_{P} & ≃ & 𝒪 (Y_{P}^{+}), \\ H_{*} (Ω_{0} (K \cap P)) & ≃ & 𝒪 (Y_{P}^{-}) . \end{matrix}$ Can think of $gr R_{-}^{P, B}$ as the subring of $𝒪 (Y_{G}^{-} \cap Y_{B}^{+})$ that are regular at $y_{P}$ (not quite sure this is true) so $(*)$ says that near $y_{P},$ the variety $P$ looks like $Y_{P}^{+} \times Y_{P}^{-} .$

The quantum cohomology $q H^{•} (G / P) :$

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

For $n \geq 3,$ consider the open subscheme $V_{n} (ℂ)$ of ${(ℙ_{ℂ}^{1})}^{n} :$ $\begin{matrix} V_{n} (ℂ) & = & {(z_{1}, \dots, z_{n}) \in {(ℙ_{ℂ}^{1})}^{n} : z_{i} \neq z_{j}, i \neq j, z_{1} = \infty, z_{2} = 0, z_{3} = 1} . \end{matrix}$ For $τ \in H_{2} (G / P),$ let $\begin{matrix} 𝔐_{τ} & = & {ϕ : ℙ^{1} \to G / P : ϕ_{*} [ℙ^{1}] = τ}, \\ 𝔐_{n, τ} & = & 𝔐_{τ} \times V_{n} (ℂ) \end{matrix}$ so $\begin{matrix} dim 𝔐_{n, τ} & = & ⟨ C_{1} (T_{G / P}), τ ⟩ + dim G / P + n - 3 . \end{matrix}$ Set $\begin{matrix} ev : & 𝔐_{n, τ} & ⟶ & {(G / P)}^{n} : \\ ev (ϕ, z_{1}, \dots, z_{n}) & = & (ϕ_{(z_{1})}, ϕ (z_{2}), \dots, ϕ (z_{n})) . \end{matrix}$ Roughly speaking, $𝔐_{n, τ}$ admits a compactification $\overline{𝔐_{n, τ}}$ which admits a fundamental class $[{\overline{M}}_{n, τ}] .$ (Manin-Konstevich).

Now for $ϕ_{1} \otimes \dots \otimes ϕ_{n} \in H^{•} ({(G / P)}^{n}) ≃ H {(G / P)}^{\otimes n},$ have $\int_{\overline{𝔐_{n, z}}} {ev}^{*} (ϕ_{1} \otimes \dots \otimes ϕ_{n}) \in ℤ (or ℂ ?)$ Using Poincare duality, can regard above as giving a $ℤ -linear$ map $\begin{matrix} J_{n, τ} : & \otimes^{n - 1} H^{•} (G / P) & ⟶ & H^{•} (G / P) \end{matrix}$ of degree $= - 2 [⟨ C_{1} (T_{G / P}), τ ⟩ + (n - 3)] .$ In other words, for any $n$ subvariety $X_{1}, \dots, X_{n}$ of $G / P$ with $\begin{matrix} \sum_{i = 1}^{n} codim X_{i} & = & 2 {dim}_{ℂ} 𝔐_{n, τ} \end{matrix}$ we have $\begin{matrix} ⟨ J_{n, τ} ({P D}^{- 1} [X_{1}] \otimes \dots \otimes {P D}^{- 1} [X_{n - 1}]), [X_{n}] ⟩ & = & # (\overline{𝔐_{n, τ}} \times_{{(G / P)}^{n}} (g_{1} X_{1} \times \dots \times g_{n} X_{n})) (ℂ) \end{matrix}$ for all $(g_{1}, \dots, g_{n})$ in a dense open subset of ${(G (ℂ))}^{n} .$ These numbers are the Gromov-Witten invariants.

Fact: For $ϕ \in H^{2} (G / P)$ and $n \geq 4$ $\begin{matrix} J_{n, τ} (ϕ_{1} \otimes \dots \otimes ϕ_{n - 2} \otimes ϕ) & = & ⟨ ϕ, τ ⟩ J_{n, τ} (ϕ_{1} \otimes \dots \otimes ϕ_{n - 2}) . \end{matrix}$

Now let $D = ℚ [[ε]]$ with indeterminant $ε .$ Given $ν \in ε (H^{*} (G / P) \otimes_{ℤ} D),$ can make $H^{*} (G / P) \otimes_{ℤ} D$ into a commutative associative $D -algebra$ with unit $σ_{P}^{id}$ with quantum product $*_{ν}$ by $\begin{matrix} σ *_{ν} σ' & = & \sum_{n, τ} J_{n, τ} (σ \otimes σ' \otimes \frac{ν^{n - 3}}{(n - 3)!}) \end{matrix}$ where $ν^{n - 3} = ν \otimes \dots \otimes ν$ $((n - 3) -times).$ In particular, for $ϕ \in H^{2} (G / P),$ define $\begin{matrix} σ *_{ε ϕ} σ' & = & \sum_{τ} J_{3, τ} (σ \otimes σ') exp ε ⟨ ϕ, τ ⟩ . \end{matrix}$ The "potential" function for $J_{3, τ} "$ satisfy WDVV-equation.

The small quantum cohomology:

Make $H^{*} (G / P) \otimes_{ℤ} Λ_{P}$ into a $Λ_{P} -algebra$ $q H^{*} (G / P)$ by $\begin{matrix} σ * σ' & = & \sum_{τ \in π_{P} (Q_{+}^{\lor})} q_{τ} J_{3, τ} (σ \otimes σ') . \end{matrix}$

Theorem:

(1)	$*$ is associative.
(2)	$q H^{*} (G / P)$ is $ℤ -graded.$
(3)	For $i \in I,$ $w \in W$ $\begin{matrix} σ_{B}^{r_{i}} * σ_{B}^{w} & = & \sum_{\underset{ℓ (w r_{α}) = ℓ (w) + 1}{α \in {\overline{Δ}}_{+}}} ⟨ ρ_{i}, α^{\lor} ⟩ σ_{B}^{w r_{α}} \\ + \sum_{\underset{ℓ (w r_{α}) = ℓ (w) + 1 - ⟨ 2 ρ, α^{\lor} ⟩}{α \in {\overline{Δ}}_{+}}} ⟨ ρ_{i}, α^{\lor} ⟩ q_{π_{B} (α^{\lor})} σ_{B}^{w r_{α}} . \end{matrix}$

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 $q H^{•} (G / P)$ and $q H^{•} (G / B) :$

Let $τ \in H_{2} (G / P) .$ Then there exists a unique $h \in Q^{\lor}$ s.t. $\begin{matrix} π_{P} (h) & = & τ \end{matrix}$ and $- 1 \leq ⟨ α, h ⟩ \leq 0$ for all $α \in - Δ (\underline{𝔭} / \underline{𝔟}) .$

Define a standard parabolic $P_{1} \subset P$ by $\begin{matrix} Δ (p_{1} / b) & = & {α \in Δ (\underline{𝔭} / \underline{𝔟}) : ⟨ α, h ⟩ = 0} . \end{matrix}$ There have birational morphisms $\begin{matrix} 𝔐_{π_{B} (h), G / B} & ⟶ & 𝔐_{π_{P} (h), G / P_{1}} \times_{G / P_{1}} G / B \\ 𝔐_{π_{P_{1}} (h), G / P_{1}} & ⟶ & 𝔐_{τ, G / P} . \end{matrix}$ This gives a commutative diagram: $\begin{matrix} \otimes^{n - 1} H^{*} (G / P) & \overset{can}{⟶} & \otimes^{n - 1} H^{*} (G / P_{1}) & \overset{can}{⟶} & \otimes^{n - 1} H^{*} (G / B) \\ ↓ J_{n, τ} & ↓ J_{n, π_{P_{1}} (h)} & ↓ J_{n, π_{B} (h)} \\ H^{*} (G / P) & ⟵_{\underset{P / P_{1}}{over fibre}}^{integration} & H^{*} (G / P_{1}) & \overset{can}{↪} & H^{*} (G / B) \end{matrix}$ This will be used in Lecture 16 to prove $ℤ \otimes_{S} R_{P} ≃ q H^{*} (G / P) .$

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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