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: 10 December 2013

Lecture 1: February 7 1997

Course Outline:

Let $\begin{matrix} G : semisimple algebraic group over ℂ \\ B \subset G : a Borel \\ P \supset B : a parabolic \\ K \subset G : maximal compact \\ T = K \cap B : maximal torus in K \\ W = N_{K} (T) / T : Weyl group \end{matrix}$ Then $G / B = K / T$ and $W$ acts on the de Rham cohomology space $H^{*} (K / T) .$ Moreover, since $K / T$ maps to the classifying space $B_{T},$ we have a morphism $H^{*} (B_{T}) \to H^{*} (K / T)$ of algebras. The map $G / B \to G / P : g B \mapsto g P$ gives inclusion $\begin{matrix} H^{*} (G / P) & ↪ & H^{*} (G / B) & = & H^{*} (K / T) . \end{matrix}$ $G / P$ is a smooth projective variety.

The de Rham cohomology $H^{*} (G / P)$ can be used to answer the following question: suppose that three subvarieties $x_{1}, x_{2}$ and $x_{3}$ of $G / P$ are in general position, and that $\sum_{k = 1}^{3} dim x_{k} = dim G / P .$ What is the number of points in the intersection $x_{1} \cap x_{2} \cap x_{3} ?$

The quantum cohomology $q H^{*} (G / P)$ answers a more general question: what is the number of holomorphic maps $ϕ : ℙ^{1} \to G / P$ with a fixed degree such that $\begin{matrix} ϕ (0) & \in & X_{1} \\ ϕ (1) & \in & X_{2} \\ ϕ (\infty) & \in & X_{3} ? \end{matrix}$ Some features of $q H^{*} (G / P) :$

There is no natural homomorphism $q H^{*} (G / P) \to q H^{*} (G / B);$
If $P \subset Q \subset G$ is a filtration, $\exists$ sth. similar to $gr H^{*} (G / P) = H^{*} (G / Q) \otimes H^{*} (Q / P);$
$W$ does not act on $q H^{*} (G / B) .$

So take equivariant cohomology $H^{T} (G / P),$ where $T$ acts on $G / P$ from the left by left translations.

Can define $T -equivariant$ quantum cohomology $q H^{T} (G / P) .$ Then

$W$ acts on $q H^{T} (G / P);$
The affine Weyl group $W_{af}$ acts on ${q H^{T} (G / P)}_{(q z)}$ (the parameter $q$ inverted);
Have creation and annihilation operators;
Have Schubert basis for $q H^{T} (G / P);$
Have "stable" Bruhat order on $W_{af};$
Formula for multiplication by $H^{2};$
Special for symmetric spaces of the form $G / P;$
Borel presentation;
Pieri formula.

Geometrical models - the variety $Y .$

For each parabolic, have $y_{p} \in Y$ and $\begin{matrix} Y_{p}^{\pm} & ≔ & {y \in Y : lim_{t \to \infty} ????? = y_{p}} \\ Y & = & ⨆_{p} Y_{p}^{\pm} over ℂ \end{matrix}$ and $\begin{matrix} 𝒪 (Y_{p}^{+}) & ≅ & q H^{*} (G / P) over ℤ \\ 𝒪 (Y_{p}^{-}) & ≅ & H_{*} (Ω (H \cap P)) \end{matrix}$ where $Ω (K \cap P)$ is the group of loops in $K \cap P .$ Moreover, $Y_{p}^{-} ≃ ℂ^{n}$ for some $n,$ and $Y = \overline{Y_{G}^{-}} = \overline{Y_{B}^{+}}$ $\Rightarrow$ $\begin{matrix} 𝒪 (Y_{G}^{-}) & ⟶ & 𝒪 (Y_{G}^{-} \cap Y_{B}^{+}) & ⟵ & 𝒪 (Y_{B}^{+}) \\ | | ≀ & | | ≀ & | | \\ H_{*} (Ω K) & {q H^{*} (G / B)}_{(q)} & q H^{*} (G / B) \end{matrix}$ Will express the Schubert basis elements as matrix entries of some representations. The variety $Y$ lies in $G^{\lor} / B^{\lor},$ where $G^{\lor}$ is the Langland dual of $G .$

Lecture 2: February 11, 1997

Kac-Moody root datum

Definition: A generalized Cartan matrix is a matrix $A = {(a_{i j})}_{i, j \in I}$ with integer entries for some finite set $I$ such that

(1)	$a_{i i} = 2$ $\forall i \in I$
(2)	$a_{i j} \leq 0$ $\forall i \neq j$
(3)	$a_{i j} = 0 \Leftrightarrow a_{j i} = 0$

A Kac-Moody root datum consists of

a generalized Cartan matrix $A = {(a_{i j})}_{i, j \in I}$
two finitely generated free $ℤ -modules$ $h_{ℤ}^{\lor}$ and $h_{ℤ}$ end????? with a perfect pairing $⟨ ⟩$ between them
two maps $\begin{matrix} I & ⟶ & h_{ℤ}^{\lor} : & i & ⟼ & α_{i} \\ I & ⟶ & h_{ℤ} : & i & ⟼ & α_{i}^{\lor} \end{matrix}$ such that $⟨ α_{j}, α_{i}^{\lor} ⟩ = a_{i j} (backwards)$

?????

Can form direct sums of root data
Can form the "dual" root data: $\begin{matrix} (A, h_{ℤ}^{\lor}, h_{ℤ}) & ⟼ & (A^{t}, h_{ℤ}, h_{ℤ}^{\lor}) \\ α_{i} & ⟷ & α_{i}^{\lor} \end{matrix}$

?????ition: $\begin{matrix} Simple roots: & π = {α_{i}}_{i \in I} \subset h_{ℤ}^{\lor} \\ Simple coroots: & π^{\lor} = {α_{i}^{\lor}}_{i \in I} \subset h_{ℤ} \\ Weight lattice: & h_{ℤ}^{\lor} \\ Coweight lattice: & h_{ℤ} \\ Root lattice: & Q = ⨁_{i \in I} ℤ α_{i} \\ Co-root lattice: & Q^{\lor} = ⨁_{i \in I} ℤ α_{i}^{\lor} \end{matrix}$ where $\begin{matrix} Q & ⟶ & h_{ℤ}^{\lor} & isom. ⟺ of adjoint type \\ Q^{\lor} & ⟶ & h_{ℤ} & isom. ⟺ of simply connected type. \end{matrix}$

Remark: In the classical case, root datum comes from connected reductive algebraic groups over $ℂ .$

Definition: We say that $A = {(A_{i j})}_{i, j \in I}$ is symmetrizable if $A = (diagonal) \cdot (symmetric) .$

Assumption: Will assume that $A$ is symmetrizable.

The numbers $m_{i j} :$ Define, for $i \neq j,$ $i, j \in I$ $m_{i j} = {\begin{matrix} 2 & if a_{i j} a_{j i} = 0 \\ 3 & if a_{i j} a_{j i} = 1 \\ 4 & if a_{i j} a_{j i} = 2 \\ 6 & if a_{i j} a_{j i} = 3 \\ \infty & if a_{i j} a_{j i} \geq 4 \end{matrix}$ The Weyl group $W$ is the group with generators $r_{i},$ $i \in I$ with relations $\begin{matrix} r_{i}^{2} & = & 1, i \in I, \\ {(r_{i} r_{j})}^{m_{i j}} & = & 1, i, j \in I, i \neq j . \end{matrix}$ The $r_{i}'s$ are called the simple reflections.

Notation:

$w = r_{i_{1}} r_{i_{2}} \dots r_{i_{n}}$ [red] means that this is a reduced expression, i.e., $n$ is the minimum number such that $w$ is a product of $n$ simple reflections. Also write $n = ℓ (w) .$
If $W$ is finite, use $w_{0}$ to denote the longest element.
From now on, write $h_{ℤ}^{\lor} = h_{ℤ}^{*} .$ Using $⟨ ⟩ .$

?????tions of $W$ on $h_{ℤ}^{*},$ $h_{ℤ}, Q,$ $Q^{\lor},$ $S = S (h_{ℤ}^{*})$

$W$ acts on $h_{ℤ}^{*}$ by $r_{i} λ = λ - ⟨ λ, α_{i}^{\lor} ⟩ α_{i}$ $W (Q) \subset Q .$
$W$ acts on $h_{ℤ}$ by $r_{i} h = h - ⟨ α_{i}, h ⟩ α_{i}^{\lor}$ $W (Q^{\lor}) \subset Q^{\lor}$
The $W$ actions on $h_{ℤ}^{*}$ and on $h_{ℤ}$ preserve the pairing ?????
The $W$ actions on $Q$ and on $Q^{\lor}$ are faithful
$W$ acts on $S = S (h_{ℤ}^{*}),$ the symmetric algebra of $h_{ℤ}^{*}$ (????? the action on $h_{ℤ}^{*}).$ $W$ acts by algebra automorphisms. The action of $W$ on $S$ will be denoted by $s \overset{w}{⟼} w \cdot s$

The Nil-Hecke ring $A$

Definition: the Nil-Hecke ring $\underline{A}$ associated to the root datum $(A = {(a_{i j})}_{i, j \in I}, h_{ℤ}^{*}, h_{ℤ}, {α_{i}}_{i \in I}, {α_{i}^{\lor}}_{i \in I})$ is the associated ring with $1$ with generators $\hat{λ}, A_{i}, \hat{λ} \in h_{ℤ}^{*}, i \in I$ and relations: $\begin{matrix} \hat{λ} + \hat{μ} & = & \hat{λ + μ}, \\ \hat{λ} \hat{μ} & = & \hat{μ} \hat{λ}, λ, μ \in h_{ℤ}^{*}, \\ A_{i} \hat{λ} & = & \overset{\lor}{r_{i} λ} A_{i} + ⟨ λ, α_{i}^{\lor} ⟩, λ \in h_{ℤ}^{*}, i \in I \\ A_{i} A_{i} & = & 0, i \in I, \\ \underset{\underset{m_{i j}}{⏟}}{A_{i} A_{j} A_{i}} \dots & = & \underset{\underset{m_{i j}}{⏟}}{A_{j} A_{i} A_{j}} \dots, (i \neq j, i, j \in I) . \end{matrix}$ The grading on $\underline{A}$ is defined to be $\begin{matrix} deg \hat{λ} & = & 2, \\ deg A_{i} & = & - 2 . \end{matrix}$ For $w \in W$ and for any $w = r_{i_{1}} r_{i_{2}} \dots r_{i_{n}} [red],$ set $\begin{matrix} A_{w} & = & A_{i_{1}} A_{i_{2}} \dots A_{i_{n}} \\ (A_{id} & = & 1) . \end{matrix}$ Then it is clear that

(1)	$A_{w}$ is independent of the reduced expression
(2)	$A_{v} A_{w} = {\begin{matrix} A_{v w} & if ℓ (v) + ℓ (w) = ℓ (v w), \\ 0 & otherwise. \end{matrix}$

Clearly

S \subseteq A

as a subring.

Proposition: ${A_{w} : w \in W}$ is an $S -basis$ for $\underline{A} .$

(Does this need a proof?)

Proposition: The map $\begin{matrix} Z W & ⟶ & \underline{A} : & r_{i} & ⟼ & 1 - \hat{α_{i}} A_{i} & = & A_{i} \hat{α_{i}} - 1, i \in I \end{matrix}$ defines an injective ring homomorphism.

Proof.

Only need to check $r_{i}^{2} = 1,$ $i \in I$ and ${(r_{i} r_{j})}^{m_{i j}} = 1$ for $i \neq j .$ Injectivity is clear (?)

$□$

Proposition: The following defines an $A -module$ structure on $S :$ $\begin{matrix} s' \cdot s & = & s' s \\ A_{i} \cdot s & = & \frac{1}{α_{i}} (s - r_{i} \cdot s) \end{matrix}$

Proof.

The induced $r_{i}$ action on $S$ is $s \mapsto r_{i} \cdot s,$ as the usual one.

$□$

Remark: Suppose we need to check certain specified operators for $s \in S$ and $A_{i}$ on some space $M$ is an action. We first check $1 - \hat{α_{i}} A_{i} = A_{i} \hat{α_{i}} - 1 .$ Then this is how $r_{i}$ acts. If this gives a $W -action,$ we are done.

Proposition: For $s \in S,$ $i \in I$ and $w \in W$ $\begin{matrix} w s & = & (w \cdot s) w \\ A_{i} s & = & r_{i} (s) A_{i} + A_{i} \cdot s \\ A_{i} s & = & s A_{i} + (A_{i} \cdot s) r_{i} \end{matrix}$ in $\underline{A} .$

Proof.

Note: All of this proof was crossed out in the scanned notes.

Just need to check that $r_{i} = 1 - \hat{α_{i}} A_{i} = A_{i} \hat{α_{i}} - 1$

$□$

The anti-automorphism $*$ on $A$

$\begin{matrix} * (S) & = & S \\ * A_{w} & = & A_{w^{- 1}} \end{matrix}$ To check that this is an anti-automorphism, need to check only $\hat{λ} A_{i} = A_{i} \hat{r_{i} λ} + ⟨ λ, α_{i}^{\lor} ⟩ 1 λ \in h_{ℤ}^{*}, i \in I .$ This is easy. Now since $r_{i} = 1 - \hat{α_{i}} A_{i} = A_{i} \hat{α_{i}} - 1$ we get $* r_{i} = \hat{α_{i}} A_{i} - 1 = - r_{i} .$ $(A_{i} r_{i} = - A_{i}, r_{i} A_{i} = A_{i})$ Consequently, $* w = {(- 1)}^{ℓ (w)} w^{- 1}$

Definitions of $\underline{A}$ on $M \otimes_{S} N$ and ${Hom}_{S} (M, N)$

Assume that $M$ and $N$ are $\underline{A} -module$ and are thus ?????odules. Form $\begin{matrix} M \otimes_{S} N & = & M \otimes N / {s m \otimes n - m \otimes s n} \\ {Hom}_{S} (M, N) & = & {f : M \to N : f (s m) = s f (m)} \end{matrix}$ want to define $\underline{A} -module$ structures on $M \otimes_{S} N$ and ${Hom}_{S} (M, N) .$

????? $M \otimes_{S} N :$ $\begin{matrix} s \cdot (m \otimes n) & = & s m \otimes n \\ A_{i} \cdot (m \otimes n) & = & A_{i} \cdot m \otimes n + r_{i} \cdot m \otimes A_{i} \cdot n \\ = & m \otimes A_{i} \cdot n + A_{i} \cdot m \otimes r_{i} \cdot n \end{matrix}$ ????? check that this is an action, we first need to show that the above operators are well-defined. The $s \cdot$ operator is clearly OK. ????? $s \in S,$ $i \in I,$ we have, by definition $\begin{matrix} A_{i} \cdot (s m \otimes n - m \otimes s n) & = & (A_{i} s) \cdot m \otimes n + (r_{i} s) \cdot m \otimes A_{i} \cdot n \\ - A_{i} \cdot m \otimes s n - r_{i} \cdot m \otimes (A_{i} s) \cdot n \end{matrix}$ Using $\begin{matrix} \begin{matrix} A_{i} s & = & (r_{i} \cdot s) A_{i} + A_{i} \cdot s \\ r_{i} s & = & (r_{i} \cdot s) r_{i} \end{matrix} & \begin{matrix} r_{i} & = & 1 - \hat{α_{i}} A_{i} \\ r_{i} \cdot s & = & s - α_{i} A_{i} \cdot s \end{matrix} \end{matrix}$ we get $\begin{matrix} A_{i} \cdot (s m \otimes n - m \otimes s n) & = & (r_{i} \cdot s) A_{i} \cdot m \otimes n + (A_{i} \cdot s) m \otimes n \\ + (r_{i} \cdot s) r_{i} \cdot m \otimes A_{i} \cdot n - A_{i} \cdot m \otimes s n \\ - r_{i} \cdot m \otimes (r_{i} \cdot s) A_{i} \cdot n - r_{i} \cdot m \otimes (A_{i} \cdot s) n \\ = & (r_{i} \cdot s) r_{i} \cdot m \otimes A_{i} \cdot n - r_{i} \cdot m \otimes (r_{i} \cdot s) A_{i} \cdot n \\ + (s - \hat{α_{i}} A_{i} \cdot s) α_{i} \cdot m \otimes n - A_{i} \cdot m \otimes s n \\ + (A_{i} \cdot s) m \otimes n - (m - \hat{α_{i}} A_{i} \cdot m) \otimes (A_{i} \cdot s) n \\ = & (r_{i} \cdot s) r_{i} \cdot m \otimes A_{i} \cdot n - r_{i} \cdot m \otimes (r_{i} \cdot s) A_{i} \cdot n \\ + s A_{i} \cdot m \otimes n - A_{i} m \otimes s n \\ - ((A_{i} \cdot s) \hat{α_{i}} A_{i} \cdot m \otimes n - \hat{α_{i}} A_{i} \cdot m \otimes (A_{i} \cdot s) n) \\ + (A_{i} \cdot s) m \otimes n - m \otimes (A_{i} \cdot s) n \\ \in & ⟨ S' m \otimes n - m \otimes s' n : m, n \in M, N ⟩ . \end{matrix}$ Hence $A_{i}$ is well-defined.

Next, since $r_{i} = 1 - \hat{α_{i}} A_{i},$ we have $\begin{matrix} A_{i} \cdot m \otimes n + r_{i} \cdot m \otimes A_{i} \cdot n & = & A_{i} \cdot m \otimes n + m \otimes A_{i} \cdot n - \hat{α_{i}} A_{i} \cdot m \otimes A_{i} \cdot n \\ = & A_{i} \cdot m \otimes n - A_{i} \cdot m \otimes \hat{α_{i}} A_{i} \cdot n + m \otimes A_{i} \cdot n \\ = & A_{i} \cdot m \otimes r_{i} \cdot n + m \otimes A_{i} \cdot n \\ = & m \otimes α_{i} \cdot n + A_{i} \cdot m \otimes r_{i} \cdot n . \end{matrix}$ This gives the 2nd expression for $A_{i} \cdot (m \otimes n) .$

Now for $s \in S$ and $i \in I,$ we need to show $A_{i} \cdot (s \cdot (m \otimes n)) = (r_{i} \cdot s) \cdot (A_{i} \cdot (m \otimes n)) + (A_{i} \cdot s) \cdot (m \otimes n)$ $\begin{matrix} l.h.s. & = & (A_{i} s) \cdot m \otimes n + (r_{i} s) \cdot m \otimes A_{i} \cdot n \\ r.h.s. & = & (r_{i} \cdot s) A_{i} \cdot m \otimes n + (r_{i} \cdot s) r_{i} \cdot m \otimes A_{i} \cdot n + (A_{i} \cdot s) m \otimes n \\ = & (A_{i} s) \cdot m \otimes n + (r_{i} s) \cdot m \otimes α_{i} \cdot n \\ = & l.h.s. \end{matrix}$ From this, we see that $r_{i} = 1 - \hat{α_{i}} A_{i} = A_{i} \hat{α_{i}} - 1$ acts by $\begin{matrix} r_{i} \cdot (m \otimes n) & = & m \otimes n - \hat{α_{i}} A_{i} \cdot m \otimes n - \hat{α_{i}} r_{i} \cdot m \otimes A_{i} \cdot n \\ = & r_{i} \cdot m \otimes n - r_{i} \cdot m \otimes \hat{α_{i}} A_{i} \cdot n \\ = & r_{i} \cdot m \otimes r_{i} \cdot n \end{matrix}$ This clearly induces an action of $W$ on $M \otimes_{S} N .$ Thus we have proved that we indeed have an action of $\underline{A}$ on $M \otimes_{S} N .$

On ${Hom}_{S} (M, N),$ define: $\begin{matrix} (s \cdot f) (m) & = & s f (m) \\ (A_{i} \cdot f) (m) & = & f (A_{i} \cdot m) + A_{i} \cdot f (r_{i} \cdot m) \\ = & A_{i} \cdot f (m) - r_{i} \cdot f (A_{i} \cdot m) \end{matrix}$ Need to check that this is indeed an action. Clearly $s \cdot$ is o????? First, since $r_{i} = 1 - \hat{α_{i}} A_{i}$ and since $f$ is $S -linear,$ we have $\begin{matrix} f (A_{i} \cdot m) + A_{i} \cdot f (r_{i} \cdot m) & = & f (A_{i} \cdot m) + A_{i} \cdot (f (m) - \hat{α_{i}} f (A_{i} \cdot m)) \\ = & A_{i} \cdot f (m) + f (A_{i} \cdot m) - (A_{i} \hat{α_{i}}) \cdot f (A_{i} \cdot m) \\ = & A_{i} \cdot f (m) - r_{i} \cdot f (A_{i} \cdot m) \end{matrix}$ This shows that the two expressions for $A_{i} \cdot f$ are equal. Now we show that $\begin{matrix} ????? (s m) & = & s (A_{i} f) (m) . \end{matrix}$ $\begin{matrix} ????? h.s. & = & f ((A_{i} s) \cdot m) + A_{i} \cdot f ((r_{i} s) \cdot m) \\ ????? h.s. & = & s f (A_{i} \cdot m) + s A_{i} \cdot f (r_{i} \cdot m) \\ = & f (s A_{i} \cdot m) + (A_{i} r_{i} (s) + A_{i} \cdot s) f (r_{i} \cdot m) \\ = & f (s A_{i} \cdot m) + A_{i} \cdot f (r_{i} (s) r_{i} \cdot m) + f ((A_{i} \cdot s) r_{i} \cdot m) \end{matrix}$ ????? $\begin{matrix} r_{i} s & = & r_{i} (s) r_{i} \\ A_{i} s & = & s A_{i} + (A_{i} \cdot s) r_{i} \end{matrix}$ ????? see that $l.h.s. = r.h.s.$

????? shows that $A_{i} \cdot f \in {Hom}_{S} (M, N) .$

????? need to check that $A_{i} \cdot (s \cdot f) = (r_{i} \cdot s) \cdot (A_{i} \cdot f) + (A_{i} \cdot s) \cdot f$ $\begin{matrix} ????? m & ⟼ & s f (A_{i} \cdot m) + A_{i} s \cdot f (r_{i} \cdot m) \\ ????? m & ⟼ & (r_{i} \cdot s) f (A_{i} \cdot m) + (r_{i} \cdot s) A_{i} \cdot f (r_{i} \cdot m) + (A_{i} \cdot s) f (m) \\ = (r_{i} \cdot s) f (A_{i} \cdot m) + A_{i} s \cdot f (r_{i} \cdot m) - f ((A_{i} \cdot s) r_{i} \cdot m) + f ((A_{i} \cdot s) m) \end{matrix}$ ????? $A_{i} s = (r_{i} \cdot s) A_{i} + A_{i} \cdot s and s A_{i} = A_{i} s - (A_{i} \cdot s) r_{i}$ ????? see $l.h.s = r.h.s.$

Finally, for $r_{i} = 1 - \hat{α_{i}} A_{i} = A_{i} \hat{α_{i}} - 1,$ we see that $\begin{matrix} (r_{i} \cdot f) (m) & = & f (m) - \hat{α_{i}} f (A_{i} \cdot m) - \hat{α_{i}} A_{i} \cdot f (r_{i} \cdot m) \\ = & f (r_{i} \cdot m) - \hat{α_{i}} A_{i} \cdot f (r_{i} \cdot m) \\ = & r_{i} \cdot f (r_{i} \cdot m) . \end{matrix}$ Consequently, $(w \cdot f) (m) = w \cdot f (w^{- 1} \cdot m)$ This is certainly an action of $W$ on ${Hom}_{S} (M, N) .$ Hence we have an action of $\underline{A}$ on ${Hom}_{S} (M, N) .$

All these proofs seem to be longer than necessary.

But anyway, we have showed that $\begin{matrix} s \cdot (m \otimes n) & = & s m \otimes n \\ A_{i} \cdot (m \otimes n) & = & A_{i} \cdot m \otimes n + r_{i} \cdot m \otimes A_{i} = m \otimes A_{i} \cdot n + A_{i} \cdot m \otimes r_{i} \\ w \cdot (m \otimes n) & = & w \cdot m \otimes w \cdot n \\ (s \cdot f) (m) & = & s f (m) \\ (A_{i} \cdot f) (m) & = & f (A_{i} \cdot m) + A_{i} \cdot f (r_{i} \cdot m) = A_{i} \cdot f (m) - r_{i} \cdot f (A_{i} \cdot m) \\ (w \cdot f) (m) & = & w \cdot f (w^{- 1} \cdot m) \end{matrix}$ make $M \otimes_{S} N$ and ${Hom}_{S} (M, N)$ $\underline{A} -modules$ again.

?????ition Given $A$ modules $M,$ $N$ and $P$ (they are therefore also $S -modules),$ the following canonical $S -module$ maps are also $A -module$ maps:

1.	${Hom}_{S} (S, M) ≅ M,$ $S \otimes_{S} M ≃ M ≃ M \otimes_{S} S$
2.	$M \otimes_{S} N ≃ N \otimes_{S} M$
3.	$M \otimes_{S} (N \otimes_{S} P) ≃ (M \otimes_{S} N) \otimes_{S} P$
4.	${Hom}_{S} (M \otimes_{S} N, P) ≃ {Hom}_{S} ({Hom}_{S} (M, N), P)$
5.	$M \otimes_{S} {Hom}_{S} (N, P) \to {Hom}_{S} ({Hom}_{S} (M, N), P)$
6.	${Hom}_{S} (M, N) \otimes_{S} P \to {Hom}_{S} (M, N \otimes_{S} P)$

Definition: For an $\underline{A} -module$ $P,$ set $P^{A} = {p \in P : A_{i} \cdot p = 0 \forall i \in I} .$

Proposition: For $\underline{A} -modules$ $M$ and $N,$ ${Hom}_{\underline{A}} (M, N) ≃ {({Hom}_{S} (M, N))}^{A} .$

Example: Regard $\underline{A}$ as an $\underline{A}$ module by left multiplications. Then our previous constructions define an $\underline{A} -module$ structure on $\underline{A} \otimes_{S} \underline{A} .$ Define: $\begin{matrix} Δ : & \underline{A} & ⟶ & \underline{A} \otimes_{S} \underline{A} \end{matrix}$ by $\begin{matrix} Δ a & = & a \cdot (1 \otimes 1) \end{matrix}$ Thus $\begin{matrix} Δ w & = & w \otimes w \\ Δ s & = & s \otimes 1 = 1 \otimes s \\ Δ A_{i} & = & A_{i} \otimes 1 + r_{i} \otimes A_{i} = 1 \otimes A_{i} + A_{i} \otimes r_{i} \end{matrix}$ For any two $\underline{A}$ modules $M$ and $N,$ since we have $a \cdot (m \otimes n) = a_{(1)} \cdot m \otimes a_{(2)} \cdot n$ for $a \in S$ or $a = A_{i},$ $i \in I,$ where $Δ a = a_{(1)} \otimes a_{(2)},$ we have $a \cdot (m \otimes n) = a_{(1)} \cdot m \otimes a_{(2)} \cdot n \forall a \in A .$

Proposition: In the finite case, $\begin{matrix} Δ A_{w_{0}} & = & \sum_{w \in W} A_{w_{0} w} \otimes w_{0} A_{w} \\ = & \sum_{w \in W} A_{w} \otimes w_{0} A_{w_{0} w} \end{matrix}$

Proof.

It is easy to show by induction on $ℓ (w)$ that for any $w \in W$ $Δ A_{w} = A_{w} \otimes w + \sum_{v < w} A_{v} \otimes a_{v}$ for some $a_{v} \in A .$ So $Δ A_{w_{0}} = \sum_{w \in W} A_{w} \otimes a_{w}$ with $a_{w_{0}} = w_{0} .$ Now for any $i \in I,$ $\begin{matrix} A_{i} A_{w_{0}} & = & 0 \\ \Rightarrow & 0 & = & Δ (A_{i}) Δ (A_{w_{0}}) \\ \Rightarrow & 0 & = & (A_{i} \otimes 1 + r_{i} \otimes A_{i}) \sum_{w \in W} A_{w} \otimes a_{w} \\ = & \sum_{w \in W} (A_{i} A_{w} \otimes a_{w} + r_{i} A_{w} \otimes A_{i} a_{w}) \\ = & \sum_{w \in W} (A_{i} A_{w} \otimes a_{w} + (1 - \hat{α_{i}} A_{i}) A_{w} \otimes A_{i} a_{w}) \\ = & \sum_{w \in W} A_{i} A_{w} \otimes r_{i} a_{w} - A_{w} \otimes A_{i} a_{w} \\ \Rightarrow & \sum_{w \in W} A_{i} A_{w} \otimes r_{i} a_{w} & = & \sum_{w \in W} A_{w} \otimes A_{i} a_{w} \end{matrix}$ Now $\begin{matrix} l.h.s. & = & \sum_{r_{i} w > w} A_{r_{i} w} \otimes r_{i} a_{w} \\ \Rightarrow & a_{r_{i} w} & = & - A_{i} a_{w} if r_{i} w < w \\ \Rightarrow & a_{w_{0}} & = & w_{0} A_{w_{0} w} . ? \end{matrix}$

$□$

Lecture 3: February 12, 1997

Recall that a Kac-Moody root datum consists of

a generalized Cartan matrix $A = {(a_{i j})}_{i, j \in I};$
two finitely generated free $ℤ -modules$ $h_{ℤ}^{\lor} = h_{ℤ}^{*}$ and $h_{ℤ}$ with a perfect pairing $⟨ ⟩$ between then;
two maps $\begin{matrix} I & ⟶ & h_{ℤ}^{\lor} : & i & ⟼ & α_{i} \\ I & ⟶ & h_{ℤ} : & i & ⟼ & α_{i}^{\lor} \end{matrix}$ such that $⟨ α_{i}, α_{j}^{\lor} ⟩ = a_{j i} .$

The weight lattice is

h_{ℤ}^{\lor}

The coweight lattice is

h_{ℤ}

The root lattice is

Q \overset{def}{=} ⨁_{i \in I} ℤ α_{i}

The co-root lattice is

Q^{\lor} \overset{def}{=} ⨁_{i \in I} ℤ α_{i}^{\lor}

Say the datum is of the adjoint type if

Q \to h_{ℤ}^{\lor} : α_{i} \mapsto α_{i}

is an isomorphism.
Say the datum is of the simply connected type if

Q^{\lor} \to h_{ℤ} : α_{i}^{\lor} \mapsto α_{i}^{\lor}

is an isomorphism.

For $s l (2, ℂ),$ use $e, f, h$ for the standard generators such that $\begin{matrix} [h, e] & = & 2 e \\ [h, f] & = & - 2 f \\ [e, f] & = & h \end{matrix}$ Given a Kac-Moody root datum $(A, I, h_{ℤ}^{\lor}, h_{ℤ}, ⟨ ⟩),$ set $\underline{h} = ℂ \otimes_{ℤ} h_{ℤ}$ and regard it as a commutative Lie algebra. Set $h_{i} = α_{i}^{\lor}$

Theorem (see Kac?): For any Kac-Moody root datum, there exists a Lie algebra $\underline{g}$ over $ℂ$ (of Kac-Moody type) and Lie algebra homomorphisms $\begin{matrix} ϕ : & \underline{h} & ⟶ & \underline{g} \\ ϕ_{i} : & {s l}_{2} (ℂ) & ⟶ & \underline{g} & \forall i \in I \end{matrix}$ such that

(1)	$\begin{matrix} ϕ_{i} (h) & = & ϕ (h_{i}) \\ [ϕ (h), ϕ_{i} (e)] & = & ⟨ α_{i}, h ⟩ ϕ_{i} (e) h \in \underline{h} \\ [ϕ (h), ϕ_{i} (f)] & = & - ⟨ α_{i}, h ⟩ ϕ_{i} (f) i \in I \\ [ϕ_{i} (e), ϕ_{j} (f)] & = & 0 (i \neq j) \end{matrix}$
(2)	For each $i \in I,$ $\underline{g}$ as an ${s l}_{2} (ℂ)$ module via $ϕ_{i}$ (using the adj. rep) is a direct sum of finite-dimensional ${s l}_{2} (ℂ) -module.$
(3)	If $\underline{g}', ϕ',$ or $ϕ_{i}^{'}$ are another such system, then there exists a unique $ψ : g \to g'$ such that $ϕ' = ψ \circ ϕ$ and $ϕ_{i}^{'} = ψ \circ ϕ_{i} .$ Thus $(g, ϕ, ϕ_{i}, i \in I)$ is unique.

Definition:

(1)	An ${s l}_{2} (ℂ) -module$ $V$ over $ℂ$ is integrable if it is a direct sum of finite-dim. modules.
(2)	An $\underline{h} -module$ $V$ over $ℂ$ is integrable if $V = ⨁_{μ \in h_{ℤ}^{*}} V_{μ}$ where $V_{μ} = {v \in V : h v = μ (v) v for all h \in \underline{h}}$
(3)	A $\underline{g} -module$ $V$ over $ℂ$ is integrable if it is ${s l}_{2} (ℂ) -integrable$ (via $ϕ_{i},$ for each $i \in I)$ and $\underline{h} -integrable$ via $ϕ .$

So the adjoint representation of

\underline{g}

\underline{g}

is integrable.

Definition:

$ϕ : \underline{h} \to \underline{g}$ is injective, so call $\underline{h} \subset g$ the Cartan subalgebra
$ϕ_{i} : {s l}_{2} (ℂ) \to \underline{g}$ is injective for each $i \in I .$ Set $\begin{matrix} e_{i} & = & ϕ_{i} (e) \\ f_{i} & = & ϕ_{i} (f) \\ h_{i} & = & ϕ_{i} (h) = α_{i}^{\lor} \end{matrix}$
$Z (g),$ the center of $g,$ is contained in $\underline{h} .$
Set $\begin{matrix} n_{+} & = & {⟨ e_{i} ⟩}_{i \in I} = Lie subalgebra generated by {e_{i}, i \in I} \\ n_{-} & = & {⟨ f_{i} ⟩}_{i \in I} \end{matrix}$ Then $\underline{g} = n_{-} + \underline{h} + n_{+}$ triangular decomposition

Every ideal of

\underline{g}

contained totally in

n_{-}

n_{+}

0 .

\begin{matrix} b_{+} & \overset{def}{=} & \underline{h} + n_{+} = b (Borel subalgebra) \\ b_{-} & \overset{def}{=} & \underline{h} + n_{-} \end{matrix}

Fact: $\underline{g}$ is the Lie algebra over $ℂ$ with generators $h \in \underline{h}, e_{i}, f_{i}, i \in I$ with relations $\begin{matrix} [h, h'] & = & 0 \\ [h, e_{i}] & = & ⟨ α_{i}, h ⟩ e_{i} \\ [h, f_{i}] & = & - ⟨ α_{i}, h ⟩ f_{i} \\ [e_{i}, f_{j}] & = & δ_{i j} h_{i} \\ {(ad e_{i})}^{1 - a_{i j}} e_{j} & = & 0 (i \neq j) \\ {(ad f_{i})}^{1 - a_{i j}} f_{j} & = & 0 (i \neq j) \end{matrix}$

Warning: $\underline{h} ⫋ centralizer of \underline{h} in \underline{g}$

The $Q -grading$ of $\underline{g} :$

For $β \in Q,$ the root lattice, set $g_{β} = {x \in g : [h, x] = ⟨ β, h ⟩ x \forall h \in \underline{h}} .$ Then $\underline{g} = ⨁_{β \in Q} g_{β}$ and $[g_{β_{1}}, g_{β_{2}}] \subset g_{β_{1} + β_{2}}$

?????ice that $g_{0} \underline{h} g_{α_{i}} = ℂ e_{i} g_{- α_{i}} = ℂ f_{i} i \in I .$ $\begin{matrix} ℤ_{+} & = & {0, 1, 2, \dots} \\ Q_{+} & = & ⨁_{i \in I} ℤ_{+} α_{i} \subset Q sub-semigroup \end{matrix}$ ????? $β, ν \in Q,$ say $β \geq ν$ if $β \cdot ν \in Q_{+} .$

????? $n_{\pm} = ⨁_{β \in Q_{\pm}} g_{β}$ $\begin{matrix} Δ & = & {β \in Q : g_{β} \neq 0, β \neq 0}, set of roots \\ Δ_{+} & = & Δ \cap Q_{+}, set of positive roots \\ Π & = & {α_{i} : i \in I}, set of simple roots \end{matrix}$

????? $\begin{matrix} Δ_{-} & = & - Δ_{+} \end{matrix}$

????? $\begin{matrix} Δ_{+} \cup Δ_{-} & = & Δ \\ Δ_{+} \cap Δ_{-} & = & \emptyset \\ n_{\pm} & = & ⨁_{β \in Δ_{\pm}} g_{β} \end{matrix}$

The principle $ℤ -grading$ of $\underline{g} :$

Let $ρ^{\lor} \in Q^{\lor}$ be the unique(?) element such that $⟨ α_{i}, ρ^{\lor} ⟩ \equiv 1 \forall i \in I .$ For $β \in Q,$ the integer $ht (β) = ⟨ β, ρ^{\lor} ⟩$ is called the height of $β .$ For $n \in ℤ,$ set ${\underline{g}}_{n} = ⨁_{\underset{ht (β) = n}{β \in Q}} g_{β}$ Thus is a $ℤ -grading$ for $\underline{g} .$

The set of real roots:

Need to define the Weyl group first. To define the Weyl group, need to define the Kac-Moody group.

Compact involution of $\underline{g} :$

This is the conjugation-linear automorphism of $\underline{g}$ such that $\begin{matrix} e_{i} & ⟷ & - f_{i} i \in ????? \\ h & ⟷ & - h h \in {\underline{h}}_{ℝ} \overset{det}{=} ℝ \otimes_{ℤ} h_{ℤ} \subset \underline{h} . \end{matrix}$

Kac-Moody group

????? $(ℂ) :$

For $u \in ℂ,$ set $t \in ℂ^{\times},$ set $\begin{matrix} x (u) & = & (\begin{matrix} 1 & u \\ 0 & 1 \end{matrix}) \\ y (u) & = & (\begin{matrix} 1 & 0 \\ u & 1 \end{matrix}) \\ h (t) & = & (\begin{matrix} t & 0 \\ 0 & t^{- 1} \end{matrix}) \end{matrix} \in {S L}_{2} (ℂ) .$ ?????l finite-dimensional representation of ${S L}_{2} (ℂ)$ is said to be ?????ational if its matrix entries are regular functions on ${S L}_{2} (ℂ) .$ ????? representation of ${S L}_{2} (ℂ)$ on a vector space $V$ over $ℂ$ is said to be differentiable if it is a direct sum of finitely many finite dimensional rational representations.

?????t: Integrable representations of ${s l}_{2} (ℂ) \leftrightarrow differentiable$ rep. of ${S L}_{2} (ℂ) .$ (This is because ${S L}_{2} (ℂ)$ is an algebraic group).

The complex torus $H :$

Define $H = Hom (h_{ℤ}^{\lor}, ℂ^{\times}) .$ For $h \in h_{ℤ}$ and $t \in ℂ^{\times},$ define $t^{h} \in H$ by $t^{h} (λ) = t^{⟨ λ, h ⟩}, λ \in h_{ℤ}^{\lor} .$ Thus, for each such $h \in h_{ℤ},$ the map $\begin{matrix} ℂ^{\times} & ⟶ & H : & t & ⟼ & t^{h} \end{matrix}$ is a homomorphism. Moreover $t^{h + h'} = t^{h} \cdot t^{h^{'}} .$ A representation of $H$ on $V / ℂ$ is said to be differentiable if it is a direct sum of $1 -dimensional$ rational representations of $H .$

Fact: Differentiable representations of $H$ $⟷$ integrable representations of $\underline{h} .$

Next: The Kac-Moody group $G$ corresponding to the Kac-Moody root datum we started with at the beginning.

The Kac-Moody group $G :$

Given the Kac-Moody root datum, there is a group $G$ with homomorphisms $\begin{matrix} ϕ : & H & ⟶ & G \\ ϕ_{i} : & {S L}_{2} (ℂ) & ⟶ & G & i \in I \end{matrix}$ $\begin{matrix} ϕ_{i} (h (t)) & = & ϕ (t^{h_{i}}) \\ ϕ (t^{h}) ϕ_{i} (x (u)) ϕ (t^{- h}) & = & ϕ_{i} (x (t^{⟨ α_{i}, h ⟩} u)) \\ ϕ (t^{h}) ϕ_{i} (y (u)) ϕ (t^{- h}) & = & ϕ_{i} (y (t^{- ⟨ α_{i}, h ⟩} u)) \\ ϕ_{i} (x (u)) ϕ_{j} (y (v)) & = & ϕ_{j} (y (v)) ϕ_{i} (x (u)), i \neq j \end{matrix}$ There exists a representation Ad of $G$ on $\underline{g}$ such that under $ϕ$ and $ϕ_{i},$ $i \in I,$ the corresponding representations of $H$ and ${S L}_{2} (ℂ)$ on $\underline{g}$ differentiate to the representations of $\underline{h}$ and ${s l}_{2} (ℂ)$ on $\underline{g}$ defined by ad.

If $(G',$ $ϕ'$ and $ϕ_{i}^{'})$ is another system with above properties, then there exists a unique $ψ : G \to G'$ such that $ϕ' = ψ \circ ϕ ϕ_{i}^{'} = ψ \circ ϕ_{i}$

$G$ is generated by the images of $ϕ$ and $ϕ_{i},$ $i \in I .$
A $G -module$ is said to be differentiable if it is differentiable as $H$ and ${S L}_{2} (ℂ)$ modules under $ϕ$ and each $ϕ_{i}^{'},$ $i \in I .$ Thus $differentiable G -module ⟷ integrable \underline{g} -module.$
There exists a faithful differentiable $G -module$ (probably not Ad).
$ϕ : H \to G$ is injective. So we call $H \subset G$ the Cartan subgroup. Have $\begin{matrix} Z (G) \overset{def}{=} ker Ad \subset H \\ ker ϕ_{i} \subset Z ({S L}_{2} (ℂ)) \end{matrix}$

The Weyl group $W :$

For each $i \in I,$ $u \in ℂ,$ set $\begin{matrix} x_{i} (u) & = & ϕ_{i} (x (u)) = ϕ_{i} (\begin{matrix} 1 & u \\ 0 & 1 \end{matrix}) \in G \\ y_{i} (u) & = & ϕ_{i} (y (u)) = ϕ_{i} (\begin{matrix} 1 & 0 \\ u & 1 \end{matrix}) \in G \\ n_{i} & = & y_{i} (1) x_{i} (- 1) y_{i} (1) = ϕ_{i} (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}) \in G \end{matrix}$ ????? $\begin{matrix} \overset{\overset{m_{i j}}{⏞}}{n_{i} n_{j} n_{i}} \dots & = & \overset{\overset{m_{i j}}{⏞}}{n_{j} n_{i} n_{j}} \dots \end{matrix}$ ????? for $w = r_{i_{1}} \dots r_{i_{n}}$ [red], let $n_{w} = n_{i_{1}} n_{i_{2}} \dots n_{i_{n}}$ ????? $n_{w} \cdot g_{β} = g_{w \cdot β}$ ????? in general $n_{w} n_{w^{- 1}} \neq id .$ ????? $N = {⟨ n_{i}, H ⟩}_{i \in I} \subset G$ ????? the subgroup of $G$ generated by ${n_{i}, i \in I}$ and $H .$ ????? $\begin{matrix} N / H & ≃ & W \\ n_{i} / H & ⟶ & r_{i} \end{matrix}$ Warning: can have $H ⊊ Z_{G} (H) .$

The real roots:

Note that $W \cdot Δ = Δ$ so $W$ permutes the root system.

Set $Δ^{re} = ⋃_{i \in I} W \cdot α_{i}$ and call elements in $Δ^{re}$ the real roots.

If $β = w \cdot α_{i} \in Δ^{re}$ for some $i \in I,$ then $g_{β} = n_{w} \cdot g_{α_{i}}$ so ${dim}_{ℂ} g_{β} = 1$ and $g_{m β} = 0 for | m | > 1 .$ Also, define $r_{β} = w r_{i} w^{- 1} \in W .$ Then

(1)	$r_{w_{1} β} = w_{1} r_{β} w_{1}$ for any $w_{1} \in W$
(2)	$\begin{matrix} r_{β} \cdot λ & = & λ - ⟨ λ, β^{\lor} ⟩ β λ \in h_{ℤ}^{*} \\ r_{β} \cdot h & = & h - ⟨ β, h ⟩ β^{\lor} h \in h_{ℤ} . \end{matrix}$

Lecture 4: February 19, 1997

I am moving the part on Bruhat decomposition of $G / P$ to the end of lecture 3. The main part of this lecture is on

Equivariant Cohomology (due to Borel):

Let $L$ be a topological group. $(L = K$ or $T$ in our applications). A principal $L -bundle$ is a topological space $E$ equipped with $c$ continuous free right action of $L$ $\begin{matrix} E \times L & ⟶ & L \end{matrix}$ and a projection $E \to B$ such that locally $E = B' \times L$ where with $L$ acts on $B' \times L$ by $(b, ℓ) \overset{ℓ_{1}}{⟼} (b, ℓ ℓ_{1}),$ $B' \underset{open}{\subset} B .$ Have $B ≃ E / L$ with the quotient topology.
The universal principal $L -bundle$ $E_{L}$ is a principal $L -bundle$ such that $E_{L}$ is contractible. Set $B_{L} = E_{L} / L .$ It is called the Classifying space of $L .$

Example: $\begin{matrix} B_{S^{1}} & = & ℂ P^{\infty} \\ E_{T} & = & E_{K} (because T \subset K) \end{matrix}$

Definition: An $L -space$ is a topological space $X$ endowed with a continuous left $L -action:$ $\begin{matrix} L \times X & ⟶ & X : & (ℓ, x) & ⟼ & ℓ \cdot x \cap ℓ x . \end{matrix}$

Definition: Given an $L -space$ $X,$ form the space $E_{L} \times^{L} X = (E_{L} \times X) / L$ where $(e, x) \cdot ℓ = (e ℓ^{- 1}, ℓ x)$ is a free left $L -action.$ The $L -equivariant$ cohomology of $X$ is by definition the singular homology of $E_{L} \times^{L} X :$ $H^{L} (X) = H^{*} (E_{L} \times^{L} X)$

Structures on $H^{L} (X) :$

It is a graded ring, where the grading is nothing but the grading on $H^{*} (E_{L} \times^{L} X) .$ (And so is the ring structure ?????
The fibration $E_{L} \times^{L} X \to E_{L} / L = B_{L}$ gives a graded ring homomorphism $\begin{matrix} H^{*} (B_{L}) & ⟶ & H^{*} (E_{L} \times^{L} X) \end{matrix}$ i.e. $\begin{matrix} H^{L} (pt) & ⟶ & H^{L} (X) \end{matrix}$ Thus $H^{L} (X)$ has a natural $H^{L} (pt) -module$ structure

Functoriality:

Given an $L -map$ of $L -spaces$ $\begin{matrix} f : & X & ⟶ & Y, \end{matrix}$ form the map $\begin{matrix} E_{L} \times^{L} X & ⟶ & E_{L} \times^{L} Y \\ [e, x] & ⟼ & [e, f (x)] . \end{matrix}$ Have commutative diagram $\begin{matrix} \begin{matrix} E_{L} \times^{L} X & ⟶ & E_{L} \times^{L} Y \\ π_{x} ↓ & ↓ π_{y} \\ B_{L} & \overset{id}{⟶} & B_{L} \end{matrix} & \begin{matrix} [e, x] & ⟼ & [e, f (x)] \\ ↧ & ↧ \\ [e] & = & [e] \end{matrix} \end{matrix}$ have a graded ring homomorphism $\begin{matrix} f^{*} : & H^{L} (X) & ⟵ & H^{L} (Y) \end{matrix}$ which is also a $H^{*} (B_{L}) = H^{L} (pt) -module$ map.

????? special case of $Y = pt$ with $\begin{matrix} f : & X & ⟶ & pt, \end{matrix}$ ?????es $\begin{matrix} π_{x} : & E_{L} \times^{L} X & ⟶ & E_{L} \times^{L} pt & = & B_{L}, \\ f^{*} : & H^{L} (pt) & ⟶ & H^{L} (X) \end{matrix}$ ????? just the one considered before.

$L -equivariant$ homology

This is the space of ${Hom}_{H^{L} (pt)} (H^{L} (X), H^{L} (pt)) .$ Than any $L -space$ map $\begin{matrix} f : & X & ⟶ & Y \end{matrix}$ induces $\begin{matrix} f_{*} : & {Hom}_{H^{L} (pt)} (H^{L} (X), H^{L} (pt)) & ⟶ & {Hom}_{H^{L} (pt)} (H^{L} (Y), H^{L} (pt)) . \end{matrix}$

The restriction homomorphism (or the evaluation at 0):

This is the map $\begin{matrix} v_{0} (X) : & H^{L} (X) & ⟶ & H^{*} (X) \end{matrix}$ induced by the map $\begin{matrix} E_{L} \times X & ⟶ & E_{L} \times^{L} X . \end{matrix}$ This is a graded $ℤ -ring$ homomorphism.

For any $L -space$ map $f : X \to Y,$ have commutative diagram $\begin{matrix} H^{L} (X) & \overset{v_{0} (X)}{⟶} & H^{*} (X) \\ f^{*} ↑ & ↑ f^{*} \\ H^{L} (Y) & \overset{v_{0} (Y)}{⟶} & H^{*} (Y) \end{matrix}$

Examples:

L

acts freely on

X .

Then

H^{L} (X) ≃ H^{*} (X / L)

Proof.

Have the following fibre bundle with contractible fibre $E_{L} .$ $\begin{matrix} E_{L} \times^{L} X & ⟵ & E_{L} \\ ↓ \\ X / L \end{matrix}$ Thus $H^{L} (X) = H^{*} (E_{L} \times^{L} X) ≃ H^{*} (X / L) .$

$□$

L

acts trivially on

X .

Then

H^{L} (X) ≃ H^{L} (pt) \otimes H^{*} (X)

Proof.

Have $E_{L} \times^{L} X ≃ B_{L} \times X .$

$□$

Proposition: Have $H^{*} (B_{T}) ≃ S (h_{ℤ}^{*}) .$

Proof.

For $λ \in h_{ℤ}^{*},$ define $\begin{matrix} e^{λ} : & T & ⟶ & ℂ^{\times} : & e^{λ} (e^{h}) & = & e^{⟨ λ, h ⟩}, h \in h_{ℤ} . \end{matrix}$ If $E$ is a principal $T -bundle,$ form the complex line bundle $ℒ_{- λ} = E \times_{T} ℂ$ by $[e t, c] = [e, e^{- λ} (t) c] t \in T, e \in E, c \in ℂ .$ Then $λ ⟼ C_{1} (E \times_{T} ℂ), the first Chern class$ gives a homomorphism $\begin{matrix} S (h_{ℤ}^{*}) & ⟶ & H^{*} (E / T) . \end{matrix}$ In particular, take $E = E_{T} = E_{K} .$ Then get $\begin{matrix} S (h_{ℤ}^{*}) & ⟶ & H^{*} (E_{T} / T) & = & H^{T} (pt) . \end{matrix}$ One can then show that this is an isomorphism of graded rings if $λ \in h_{ℤ}^{*}$ is given $deg = 2 .$

$□$

The second $S -module$ structure on $H^{T} (K / T)$

Set $E_{u} = E_{T} = E_{K} .$ The map $E_{u} \times^{T} (K / T) ⟶ E_{u} / T : [e, k T] ⟼ e k T$ is another ring homomorphism, which we will denote by $π_{R}$ for reasons that will be clear next time; $\begin{matrix} π_{R} : & S & ⟶ & H^{T} (K / T) . \end{matrix}$

Remarks:

$π_{R},$ together with the map $\begin{matrix} π_{L} : & S & ⟶ & H^{T} (K / T) \end{matrix}$ incuded by $K / T \to pt,$ will be the source and target maps for the Hopf algebroid structure on $H^{T} (K / T)$ that will be discussed next lecture.
Set $\begin{matrix} E_{u}^{(2)} & = & E_{u} \times_{E_{u} / K} E_{u} = {(e_{1}, e_{2}) \in E \times E : e_{1} K = e_{2} K} \\ \subset & E_{u} \times E_{u} \end{matrix}$ It is a $(K \times K) -inv.$ subset of $E_{u} \times E_{u} .$ Since $K$ acts on $E_{u}$ freely, we have the identification $E_{u}^{(2)} ≃ E_{u} \times K : (e_{1}, e_{2}) ⟼ (e_{1}, k) if e_{2} = e_{1} .$ Under this identification, the $T \times T$ action on $E_{u}^{(2)}$ becomes the action $(e, k) \overset{(t_{1}, t_{2})}{⟼} (e_{1} t_{1}, t_{1}^{- 1} k t_{2})$ of $T \times T$ on $E_{u} \times K .$ (easy to check this: $(e, k) ⟼ (e_{1}, e_{1} k) \overset{(t_{1}, t_{2})}{⟼} (e_{1} t_{1}, e_{1} k t_{2}) ⟼ (e_{1} t_{1}, e_{1} t_{1} t_{1}^{- 1} k t_{2}) ⟼ (e_{1} t_{1}, t_{1}^{- 1} k t_{2})).$ Thus we have $E_{u}^{(2)} / T \times T ≃ E_{u} \times^{T} K / T$ The map $E_{u} \times^{T} (K / T) ⟶ E_{u} / T : [e, k T] ⟼ e k T$ now is just the projection from $E_{u}^{(2)} / T \times T$ to the 2nd factor $E_{u} ?????$ This will be used in the next lecture.

Proposition: For any $T -space$ $Y,$ we have $H^{T} (K \times^{T} Y) ≃ H^{T} (K / T) \otimes_{S} H^{T} (Y)$ where the $S -module$ structure on $H^{T} (K / T)$ is via the second ring homomorphism $π_{K} : S ⟶ H^{T} (K / T) .$ (The $S -module$ structure on $H^{T} (Y)$ is the usual one).

Proof.

Consider the following commutative square: $\begin{matrix} \begin{matrix} (????? K \times^{T} Y) & \overset{p_{1}}{⟶} & E_{u} \times^{K} (K \times^{T} Y) & ≃ & E_{u} \times^{T} Y \\ p_{2} ↓ & ↓ q_{1} \\ (????? K \times^{T} pt) & \overset{q_{2}}{⟶} & E_{u} \times^{K} (K \times^{T} pt) \\ | ≀ & | ≀ \\ ????? (K / T) & E_{u} / T \end{matrix} & = & \begin{matrix} {[e, [k, y]]}_{T} & \overset{p_{1}}{⟼} & {[e, [k, y]]}_{K} & \tilde{⟼} & [e k, y] \\ p_{2} ↓ & ↓ q_{1} \\ {[e, [k, y]]}_{T} & \overset{q_{2}}{⟼} & {[e, [k, p]]}_{K} \\ ↧ ≀ & ↧ ≀ \\ [e, k] & [e k] & = & [e k, pt] \end{matrix} \end{matrix}$ notice that $q_{1}^{*} : S \to H^{T} (Y)$ is the usual homo. (induced from $Y \to pt).$ ????? $q_{2}^{*} = π_{R} : S \to H^{T} (K / T)$ is the second homomorphism. Now since the square is commutative, ie. $q_{2} \circ p_{2} = q_{1} \circ p_{1},$ we get a ring homomorphism $\begin{matrix} H^{T} (K / T) \otimes_{S} H^{T} (Y) & ⟶ & H^{T} (K \times^{T} Y) . \\ x \otimes y & ⟼ & p_{2}^{*} (x) p_{1}^{*} (y) \end{matrix} assuming even col ?????$ To show that this is an isomorphism, we first notice that the fibration $p_{1}$ has fibre $K / T$ which is a $C W -complex$ if only even dimensions. Thus Leray-Hitsch theorem tells us that $H^{T} (K \times^{T} Y)$ is a free module over $H^{T} (Y)$ with basis coming from $H^{*} (K / T) .$ The special case of $Y = pt$ says that $H^{T} (K / T)$ is a free $S = H^{T} (pt) -module$ with basis coming from $H^{*} (K / T) .$ Using a basis of $H^{*} (K / T),$ we see that the map $H^{T} (K / T) \otimes_{S} H^{T} (Y) ⟶ H^{T} (K \times^{T} Y)$ is an isomorphism.

$□$

Definition: The morphism $\begin{matrix} ε : & H^{T} (K / T) & ⟶ & S \end{matrix}$ induced by $\begin{matrix} T / T & ↪ & K / T \end{matrix}$ is called the co-unit map.

Definition: For any $K -space$ $X,$ the map $\begin{matrix} Δ_{X} : & H^{T} (X) & ⟶ & H^{T} (K / T) \otimes_{S} H^{T} (X) \end{matrix}$ induced by the $T -map$ $\begin{matrix} μ_{K} : & K \times^{T} X & ⟶ & X : & [k, x] & ⟼ & k x, \end{matrix}$ ie. $\begin{matrix} Δ_{x} : & H^{T} (X) & \overset{μ_{K}^{*}}{⟶} & H^{T} (K \times^{T} X) & ≃ & H^{T} (K / T) \otimes_{S} H^{T} (X) \end{matrix}$ is called the co-module map.

Proposition: For any $K -space$ $X,$ we have $(ε \otimes id) \circ Δ_{X} = id |_{H^{T} (X)}$ and $\begin{matrix} (Δ_{K / T} \otimes id) \circ Δ_{X} & = & (id \otimes Δ_{X}) \circ Δ_{X} : & H^{T} (X) & ⟶ & H^{T} (K / T) \otimes_{S} H^{T} (K / T) \otimes_{S} H^{T} (X) . \end{matrix}$

Definition: A groupoid scheme $(𝒴, 𝒮)$ consists of two schemes $𝒴$ and $𝒮$ and five morphisms: $\begin{matrix} P_{L}, P_{R} : & 𝒴 & ⟶ & 𝒮 \\ ℓ : & 𝒮 & ⟶ & 𝒴 \\ i : & 𝒴 & ⟶ & 𝒴 \\ μ : & 𝒴 \times_{S} 𝒴 & ⟶ & 𝒴 \end{matrix}$ (fibre produ????? $- \times_{S}$ refers to $P_{L},$ and $\times_{S} -$ refers to $P_{R})$ They must satisfy: $\begin{matrix} P_{L} \circ ℓ = {id}_{𝒴} = P_{R} \circ ℓ \\ P_{L} \circ i = P_{R} P_{R} \circ i = P_{L} \\ P_{L} \circ μ = p_{2} \circ p_{1} P_{R} \circ μ = P_{R} \circ p_{2} \\ μ \circ ({id}_{𝒴}, ℓ \circ P_{R}) = {id}_{𝒴} \\ μ \circ (ℓ \circ P_{L}, {id}_{𝒴}) = {id}_{𝒴} \\ μ \circ ({id}_{𝒴}, i) = i \circ P_{R} μ \circ (i, {id}_{𝒴}) = ℓ \circ P_{R} \\ μ \circ ({id}_{𝒴} \times μ) = μ \circ (μ \times {id}_{𝒴}) \end{matrix}$ These imply $i \circ i = {id}_{𝒴} .$

If $𝒴 = Spec R$ and $𝒮 = Spec S,$ then $𝒴 \times_{S} 𝒴 = Spec (R \times_{S} R) .$

Lecture 5: February 25, 1997

Recall the concept of a groupoid:

A groupoid is a small category with every morphism invertible

Example: Let $G$ be a group acting on a space $X .$ Then we can form a groupoid $(𝒴, 𝒮),$ where $𝒮 = X,$ $𝒴 = {(x, g, y) : x, y \in X, x = g \cdot y}$ Multiplication is given by $(x, g, y) (x', g', y') = (x, g g', y') if y = x'$ Source map: $𝒴 ⟶ S : (x, g, y) ⟼ y$ Target map: $𝒴 ⟶ S : (x, g, y) ⟼ x$ Inverse map: $𝒴 ⟶ 𝒴 : (x, g, y) ⟼ (y, g^{- 1}, x)$ Units: $S ⟶ 𝒴 : x ⟼ (x, e, x) .$

An action $ϕ : 𝒴 \times_{S} X \to X$ of a groupoid scheme $(𝒴, S)$ on a scheme $X$ ????? $S$ with structure morphism $P_{X} : X \to S$ is one such that

(1)	$ϕ \circ (μ \times {id}_{X}) = ϕ \circ ({id}_{𝒴} \times ϕ)$
(2)	$P_{X} \circ ϕ = P_{L} \circ P_{1}$ where $P_{1} : 𝒴 \times X \to 𝒴,$ $(y, x) \mapsto ?????$
(3)	$ϕ \circ ((e \circ P_{X}) \times {id}_{X}) = {id}_{X} .$

The groupoid scheme $𝒰 = Spec H^{T} (K / T)$

Let $E_{u}$ be the principal $K$ (and thus also $T)-bundle.$ For $n \geq 1,$ let $\begin{matrix} E_{u}^{n} & = & E_{u} \times \dots \times E_{u} & n times \\ K^{n} & = & K \times \dots \times K & n times \\ T^{n} & = & T \times \dots \times T & n times \end{matrix}$ Set $E_{u}^{(n)} = {(e_{1}, \dots, e_{n}) \in E_{u}^{n} : e_{1} L = \dots = e_{n} K} \subseteq E_{u}^{n} .$ As a subset of $E_{u}^{n},$ the set $E_{u}^{(n)}$ is invariant under the $K^{n} -action,$ so $E_{u}^{(n)}$ is a principal $K^{n} -bundle.$

Set $B^{(n)} = E_{u}^{(n)} / T^{n}$ Then it is easy to check that $B^{(2)}$ is a groupoid over $B^{(1)} = E / T = B_{T}$ with the following structure maps: (This is a subquotient of the coarse groupoid $E \times E$ over $E):$

Source and target maps: $\begin{matrix} p_{1} : & B^{(1)} & = & E_{u}^{(2)} / T^{2} & ⟶ & E / T : & [e_{1}, e_{2}] & ⟼ & [e_{1}] \\ p_{2} : & B^{(2)} & = & E_{u}^{(2)} / T^{2} & ⟶ & E / T : & [e_{1}, e_{2}] & ⟼ & [e_{2}] \end{matrix}$
identities: $\begin{matrix} d (= diagonal) : & B^{(1)} & = & E / T & ⟶ & B^{(2)} & : & [e] & ⟼ & [e, e] \end{matrix}$
inverse: $\begin{matrix} t (= transposition) : & B^{(2)} & ⟶ & B^{(2)} & : & [e_{1}, e_{2}] & ⟼ & [e_{2}, e_{1}] \end{matrix}$
multiplication: $\begin{matrix} μ : & B^{(2)} \times_{B^{(1)}} B^{(2)} & ≃ & B^{(3)} & ⟶ & B^{(2)} & : & ([e_{1}, e_{2}], [e_{2}, e_{3}]) & ⟼ & [e_{1}, ?????] \end{matrix}$

We now pull back all the above structure maps on cohomology:

First note that

E_{u}^{(2)} ≃ E_{u} \times K

\begin{matrix} (e_{1}, e_{2}) & ⟶ & (e_{1}, k) \end{matrix} if e_{2} = e_{1} k

Under this identification, the

T^{2}

action on

E_{u}^{(2)}

becomes

\begin{matrix} (e_{1}, k) & ⟼ & (e_{1}, e_{1} k) & ⟼ & (e_{1} t_{1}, e_{1} k t_{2}) \\ ⟼ & (e_{1} t_{1}, e_{1} t_{1} t_{1}^{- 1} k t_{2}) \\ ⟼ & (e_{1} t_{1}, t_{1}^{- 1} k t_{2}) \end{matrix}

Thus we get an induced identification

\begin{matrix} E_{u}^{(2)} / T^{2} & ≃ & E_{u} \times^{T} K / T \\ [e_{1}, e_{2}] & ⟼ & [e_{1}, k T] & if e_{2} = e_{1} k \end{matrix}

Similarly, we have

\begin{matrix} E_{u}^{(3)} & ≃ & E_{u} \times K \times K & : & (e_{1}, e_{1} k_{1}, e_{1} k_{1} k_{2}) & ⟼ & (e_{1}, k_{1}, k_{2}) \end{matrix}

and

\begin{matrix} (e_{1} t_{1}, e_{1} k_{1} t_{2}, e_{1} k_{1} k_{2} t_{3}) & = & (e_{1} t_{1}, e_{1} t_{1} t_{1}^{- 1} k_{1} t_{2}, e_{1} k_{1} t_{2} t_{2}^{- 1} k_{2} t_{3}) \\ ⟼ & (e_{1} t_{1}, t_{1}^{- 1} k_{1} t_{2}, t_{2}^{- 1} k_{2} t_{3}) \end{matrix}

E_{u}^{(3)} / T^{3} ≃ (E_{u} \times K \times K) / T^{3}

where the

T^{3}

action on

E_{u} \times K \times K

(e_{1}, k_{1}, k_{2}) \cdot (t_{1}, t_{2}, t_{3}) = (e_{1} t_{1}, t_{1}^{- 1} k_{1} t_{2}, t_{2}^{- 1} k_{2} t_{3})

But

(E_{u} \times K \times K) / T^{3} ≃ E_{u} \times^{T} (K \times^{T} K / T)

so we have the identifications

\begin{matrix} B^{(2)} & ≃ & E_{u} \times^{T} K / T \\ B^{(3)} & ≃ & E_{u} \times^{T} (K \times^{T} K / T) \end{matrix}

Hence

\begin{matrix} H^{•} (B^{(2)}) & ≃ & H^{T} (K / T) \\ H^{•} (B^{(3)}) & ≃ & H^{T} (K \times^{T} K / T) \\ ≃ & H^{T} (K / T) \otimes_{S} H^{T} (K / T) (from last time) \end{matrix}

where the last identification is due to the general fact we proved last time that for any

K -space

Y,

\begin{matrix} H^{T} (K \times^{T} Y) & ≃ & H^{T} (K / T) \otimes_{S} H^{T} (Y) . \end{matrix}

We also have

H^{•} (B^{(1)}) = H^{•} (E / T) = S

Therefore, the pull-backs on cohomology of all the structure maps for the groupoid

B^{(2)}

over

B^{(1)}

give the groupoid structure on

𝒰 = Spec H^{T} (K / T) .

Summary

Set $R = H^{T} (K / T),$ $S = H^{T} (pt) = H (B_{T}) = H^{•} (B^{(1)}) .$ Then from: $\begin{matrix} p_{1} : & B^{(2)} & ⟶ & B^{(1)} : & [e_{1}, e_{2}] & ⟼ & [e_{1}] \\ p_{2} : & B^{(2)} & ⟶ & B^{(1)} : & [e_{1}, e_{2}] & ⟼ & [e_{2}] \\ d : & B^{(1)} & ⟶ & B^{(2)} : & [e] & ⟼ & [e, e] \\ t : & B^{(2)} & ⟶ & B^{(2)} : & [e_{1}, e_{2}] & ⟼ & [e_{2}, e_{1}] \\ μ : & B^{(3)} & ⟶ & B^{(2)} : & [e_{1}, e_{2}, e_{3}] & ⟼ & [e_{1}, e_{3}] \end{matrix}$ we get: $\begin{matrix} π_{L} & = & p_{1}^{*} : & S & ⟶ & R \\ π_{R} & = & p_{2}^{*} : & S & ⟶ & R \\ ε & = & d^{*} : & R & ⟶ & S \\ c & = & t^{*} : & R & ⟶ & R \\ Δ & = & μ^{*} : & R & ⟶ & R \otimes_{S} R \end{matrix}$

Theorem: The above maps $π_{L},$ $π_{R},$ $ε,$ $c$ and $Δ$ make $(𝒰 = Spec R, \underline{h} = Spec S)$ into a groupoid scheme. Moreover, if $X$ is any $K -space,$ the map $\begin{matrix} Δ_{X} & = & {K \times^{T} X ⟶ X : [k, x] ⟼ k x}^{*} : & H^{T} (X) & ⟶ & H^{T} (K / T) \otimes H^{T} (X) \end{matrix}$ is the composition of an action of $(𝒰, \underline{h})$ on $Spec H^{T} (X),$ (assuming that $H^{*} (X)$ is even)

Characteristic operators

Definition: A characteristic operator for $(K, T)$ is a rule that assigns to each $K -space$ $X$ an $H^{K} (X) -linear$ endomorphism $ϕ_{X} : H^{T} (X) \to H^{T} (X)$ such that if $F : X \to Y$ is a $K -map$ then $F^{*} \circ ϕ_{Y} = ϕ_{X} \circ F^{*} .$

Remark: When $K = T,$ any $H^{T} (X) -linear$ endomorphism of $H^{T} (X)$ must be a multiplication operator by characters. This is why the name characteristic operators.

Fact: The set $\underline{\hat{A}}$ of all characteristic operators is an $S -algebra.$

Definition: We say that a characteristic operator is of compact support if there exists a compact subset $K_{0} \subset K$ which is $T -stable$ such that given any $K -space$ $X,$ a $T -stable$ subset $X_{0}$ of $X$ and an element $z \in H^{T} (X)$ vanishing in $H^{T} (K_{0} X_{0}),$ the element $ϕ_{X} (z_{0})$ must vanish in $H^{T} (X_{0}) .$

Remark: In the finite case, can take $K_{0} = K$ and every characteristic operator is compact.

Definition-Notation: $\begin{matrix} {\underline{\hat{A}}}_{c} & = & the S -subalgebra of \underline{\hat{A}} of all characteristic operators of compact support. \end{matrix}$

Proposition: For any characteristic operator $a$ and any $K -space$ $X,$ we have $\begin{matrix} Δ_{X} \circ a & = & (a \otimes id) \circ Δ_{X} : & H^{T} (X) & ⟶ & H^{T} (K / T) \otimes_{S} H^{T} (X) \end{matrix}$

Corollary 1: For a characteristic operator $a,$ we have $\begin{matrix} a = 0 & ⟺ & a = 0 on H^{T} (K / T) \\ ⟺ & ε \circ a = 0 \in {Hom}_{S} (H^{T} (K / T), S) . \end{matrix}$

Proof.

If $ε \circ a = 0 : H^{T} (K / T) \to S,$ then for any $K -space$ $X,$ $\begin{matrix} a on H^{T} (X) & = & (ε \otimes id) \circ Δ_{X} \circ a (because (ε \otimes id) \circ Δ_{X} = {id}_{H^{T} (X)}) \\ = & (ε \otimes id) \circ (a \otimes id) \otimes Δ_{X} (by Proposition) \\ = & (ε \circ a \otimes id) \circ Δ_{X} \\ = & 0 . \end{matrix}$

$□$

Corollary 2: $\underline{\hat{A}}$ has no $S -torsion.$

Proof.

If $s \in S$ and $a \in \underline{\hat{A}}$ are such that $s a = 0 and a \neq 0$ then for any $z \in H^{T} (K / T)$ $\begin{matrix} 0 = (ε \circ s a) (z) & = & ε (s (a \cdot z)) \\ = & s ε (a \cdot z) \end{matrix}$ But since $a \neq 0,$ we know by Corollary 1 that $ε \circ a \neq ?????$ so $\exists z \neq 0$ s.t. $ε (a \cdot z) \neq 0 \in S .$ Since $S$ is a polynomial algebra, it has no $S -torsion.$ Thus $S = 0 .$ This shows that $\underline{\hat{A}}$ has no $S -torsion.$

$□$

Corollary 3 (added by me) (of Corollary 1): The action of $a \in \underline{\hat{A}}$ on $H^{T} (X)$ is expressed using $\begin{matrix} Δ_{X} : & H^{T} (X) & ⟶ & H^{T} (K / T) \otimes_{S} H^{T} (X) \end{matrix}$ and the map $ε \circ a : H^{T} (K / T) \to S$ by $\begin{matrix} a on H^{T} (X) & = & (ε \circ a \otimes id) \circ Δ_{X} . \end{matrix}$

Remark: Should think of $\underline{\hat{A}}$ as the dual of $H^{T} (K / T)$ by $a \mapsto ε \circ a \in Hom (H^{T} (K / T), S) .$

Integration over the fibre

Assume that $P : E \to B$ is a fibration over a pathwise connected base $B$ with $b_{0} \in B .$ Let $F = P^{- 1} (b_{0}) .$ Assume that this fibration is orientable. This means that the holonomy around $b_{0}$ acts trivially on $H^{*} (F) .$ Since $B$ is pathwise connected, the weak homotopy type of $F$ is independent of the choice of $b_{0} .$ Then we have, assuming $H^{γ} (F) = 0$ for $γ > n$ $\begin{matrix} {Hom}_{ℤ} (H^{n} (F), ℤ) & ⟶ & ({Hom}_{H^{*} (B)} (H^{*} (E), H^{*} (B)) degree - n) \end{matrix}$ denoted by $\begin{matrix} τ & ⟼ & \int_{τ} \end{matrix}$ obtained as follows by using the Serre spectral sequence: $\begin{matrix} H^{m + n} (E) & ⟶ & E_{\infty}^{m, n} & ≃ & E_{2}^{m, n} & ≃ & H^{m} (B, H^{n} (F)) & \overset{τ}{⟶} & H^{m} (B, ℤ) . \end{matrix}$

Remark

(1)	This is just the identity map when $B = pt.$
(2)	It is functorial over pullbacks.
(3)	It preserves certain Mayer-Vietoris sequences
(4)	Can do this for relative cohomology as well.

The $\underline{A} -action$ on $H^{*} (E / T)$ for any principal $K -bundle$ $E$

If $E$ is a principal $T -bundle,$ then we have a ring homomorphism $\begin{matrix} ch : & S & ⟶ & H^{even} (E / T) & : & λ & ⟼ & c_{1} (ℒ_{- λ} = E \times_{T} ℂ_{e^{- λ}}) & \in & H^{2} (E / T) . \end{matrix}$ We call it the characteristic homomorphism. Using the characteristic homomorphism, we get an $S -module$ structure on $H^{*} (E / T) :$ $\begin{matrix} s \cdot z & = & ch (s) z . \end{matrix}$

Now assume that $E$ is also a principal $K -bundle,$ so thus also a $T -bundle.$ Then we can use the $K -action$ to define the following $W -action$ on $H^{*} (E / T) :$ for $w \in W,$ $\begin{matrix} w \cdot z & = & w^{*} z \end{matrix}$ where $w : E / T \to E / T :$ $w \cdot e T = e w T .$ Because of the following basic property of the characteristic map, $\begin{matrix} w^{*} c_{1} (ℒ_{- λ}) & = & c_{1} (w^{*} ℒ_{- λ}) = c_{1} (ℒ_{- w \cdot λ}) \\ ie. w^{*} ch (λ) & = & ch (w \cdot λ) \end{matrix}$ we have, for any $w \in W$ and $s \in S$ $\begin{matrix} w s & = & (w \cdot s) w \end{matrix}$ as operators on $H^{*} (E / T) .$ Therefore we have an action of the smashed product algebra $ℂ W ⋉ S$ on $H^{*} (E / T) .$

Now for each $i \in I,$ consider the fibre bundle $\begin{matrix} E / T \\ ↓ π_{i} \\ E / K_{i} \end{matrix}$ which has fibre $K_{i} / T ≃ P_{i} / B ≃ ℂ P^{1}$ so it has a preferred orientation $σ_{i} \in {Hom}_{ℤ} (H^{*} (K_{i} / T, ℤ))$ namely the fundamental cycle. Integration over the fibre gives $\begin{matrix} H^{*} (E / T) & ⟶ & H^{* - 2} (E / K_{i}) & : & z & ⟼ & \int_{σ_{i}} z \end{matrix}$ Now define $\begin{matrix} A_{i} : & H^{*} (E / T) & ⟶ & H^{* - 2} (E / T) & : & A_{i} \cdot z & = & π_{i}^{*} \int_{σ_{i}} z . \end{matrix}$

Proposition: For any $z \in H^{*} (E / T),$ $\begin{matrix} \begin{matrix} α_{i} \cdot (A_{i} \cdot z) & = & z - r_{i} \cdot z \end{matrix} & (*) \end{matrix}$

Proof.

We will check this over $ℚ$ (Why?). The fibration $π_{i} : E / T \to E / K_{i}$ gives a $H^{*} (E / K_{i}) -module$ structure on $H^{*} (E / T) .$ Since the fibre is $≃ ℂ P^{1},$ this is in fact a free $H^{*} (E / K_{i}) -module,$ a basis of which is given by $1$ and $\frac{1}{2} ch (α_{i}) \in H^{*} (E / T) .$ For $z_{0} \in H^{*} (E / T)$ we use the same letter to denote the pull back $π_{i}^{*} ????? \in H^{*} (E / T) .$ We will check $(*)$ for $z = z_{0}$ and $z = \frac{1}{2} ch (α_{i}) z_{0} .$ Clearly $A_{i} \cdot z_{0} = 0$ and $r_{i} \cdot z_{0} = z_{0} .$ Thus $(*)$ holds for $z = z_{0} .$ Now for $z = \frac{1}{2} ch (α_{i}) z_{0},$ $\begin{matrix} α_{i} \cdot (A_{i} \cdot z) & = & α_{i} \cdot (A_{i} \cdot (\frac{ch (α_{i})}{2}) z_{0}) . \end{matrix}$

Lemma: $A_{i} \cdot ch (α_{i}) = 2 .$ (a calculation over $ℂ P^{1})$

Assume Lemma. Then $α_{i} \cdot (A_{i} \cdot z) = α_{i} \cdot z_{0} = ch (α_{i}) z_{0} .$ On the other hand, $\begin{matrix} z - r_{i} \cdot z_{0} & = & \frac{1}{2} ch (α_{i}) z_{0} - r_{i} \cdot (\frac{1}{2} ch (α_{i}) z_{0}) \\ = & \frac{1}{2} ch (α_{i}) z_{0} - r_{i} \cdot (\frac{1}{2} ch (α_{i})) r_{i} \cdot z \\ = & \frac{1}{2} ch (α_{i}) z_{0} + \frac{1}{2} ch (α_{i}) z_{0} \\ = & ch (α_{i}) z_{0} . \end{matrix}$ Hence $(*)$ holds for $z = \frac{1}{2} ch (α_{i}) z_{0} .$

It is strange to carry the $\frac{1}{2}$ around. Why necessary?

$□$

Therefore we have

Theorem: For any principal $K -bundle$ $E,$ the following define an $\underline{A} -action$ on $H^{*} (E / T) :$ $\begin{matrix} s \cdot z & = & ch (s) z \\ w \cdot z & = & w^{*} z \\ A_{i} \cdot z & = & π_{i}^{*} \int_{σ_{i}} z \end{matrix}$ Moreover, the characteristic morphism $\begin{matrix} ch : & S & ⟶ & H^{even} (E / T) : & λ & ⟼ & c_{1} (ℒ_{- λ}) \end{matrix}$ is an $\underline{A} -map,$ where $A_{i}$ acts on $s \in S$ by $\begin{matrix} A_{i} \cdot s & = & \frac{s - r_{i} \cdot s}{α_{i}} \end{matrix}$ as before (see Lecture 2).

Example: $E = K$ with right action of $K$ by right multiplications. Then the $A_{i}'s$ on $H^{*} (K / T)$ are the BGG-operators.

Example: If $E_{1} \overset{f}{\to} E_{2}$ is a $K -map,$ then $f^{*} : H^{*} (E_{2} / T) \to H^{*} (E_{1} / T)$ is clearly an $\underline{A} -map.$

$\underline{A} -action$ on $H^{T} (X)$ for $K -space$ $X :$

Example: Let $X$ be a $K -space$ and let $E_{u} = E_{K}$ be the universal principal bundle of $K .$ Let $\begin{matrix} E & = & E_{u} \times X \end{matrix}$ with the $K -action$ given by $\begin{matrix} (e, x) \cdot k & = & (e k^{- 1}, k x) . \end{matrix}$ Then $\begin{matrix} E / T & = & E_{u} \times^{T} X \end{matrix}$ so get an action of $\underline{A}$ on $H^{T} (X) .$ If $f : X \to Y$ is a $K -map,$ then $\begin{matrix} E_{u} \times X & ⟶ & E_{u} \times Y & : & (e, x) & ⟼ & (e, f (x)) \end{matrix}$ is a $K -map,$ so $\begin{matrix} f^{*} : & H^{T} (Y) & ⟶ & H^{T} (X) \end{matrix}$ is an $\underline{A} -map.$ Finally, the $\underline{A} -action$ on $H^{T} (X)$ is clearly $H^{K} (X) -linear.$ Thus we can think of elements of $\underline{A}$ as characteristic operators.

Property: For any $K -space$ $X,$ the morphism $\begin{matrix} S & ⟶ & H^{T} (X) & (= {(X \to pt)}^{*}) \end{matrix}$ is an $\underline{A} -map.$

Proof.

This is the same as the characteristic morphism. ?????

$□$

Proposition: For any $K -space$ $X,$ the multiplication map $\begin{matrix} H^{T} (X) \otimes_{S} H^{T} (X) & ⟶ & H^{T} (X) \end{matrix}$ is an $\underline{A} -map.$

$T -equivariant$ homology

For a $T -space$ $X,$ the $T -equivariant$ homology of $X$ is defined to be ${Hom}_{S} (H^{T} (X), S) .$
Suppose that $X$ is a $K -space.$ Then $H^{T} (X)$ is an $\underline{A} -module.$ Since $S$ is also an $\underline{A} -module,$ we know that ${Hom}_{S} (H^{T} (X), S)$ is then also a $\underline{A} -module$ (see Lecture 2): $\begin{matrix} (s \cdot f) (z) & = & s f (z) \\ (A_{i} \cdot f) (z) & = & f (A_{i} \cdot z) + A_{i} \cdot f (r_{i} \cdot z) = A_{i} \cdot f (z) - r_{i} \cdot f (α_{i} \cdot z) \\ (w \cdot f) (z) & = & w \cdot f (w^{- 1} \cdot z) \end{matrix}$
If $F : X \to Y$ is a $K -map,$ then we have shown that $F^{*} : H^{T} (Y) \to H^{T} (X)$ is an $\underline{A} -map.$ Define $\begin{matrix} F_{*} : & {Hom}_{S} (H^{T} (X), S) & ⟶ & {Hom}_{S} (H^{T} (Y), S) \end{matrix}$ by $(F_{*} f) (z_{Y}) = f (F^{*} z_{Y}) .$ Then $F_{*}$ is an $\underline{A} -map$ as well. Let's check $F_{*} (A_{i} \cdot f) = A_{i} \cdot (F_{*} f) .$ So let $t \in H^{T} (Y),$ need to show $\begin{matrix} (A_{i} \cdot f) (F^{*} z) & = & f (α_{i} \cdot (F_{*} f)) (z) . \end{matrix}$ Now $\begin{matrix} lhs & = & f (A_{i} \cdot F^{*} z) + A_{i} \cdot f (r_{i} \cdot F^{*} z) \\ rhs & = & (F_{*} f) (A_{i} \cdot z) + A_{i} \cdot F_{*} f (r_{i} \cdot z) \\ = & f (F^{*} (A_{i} \cdot z)) + A_{i} \cdot f (F^{*} (r_{i} \cdot z)) . \end{matrix}$ Since $F^{*}$ is an $\underline{A} -map,$ we indeed have $lhs = rhs.$

Example: Suppose $Y$ is a $T -space$ such that $H^{r} (Y) = 0$ for $r > n .$ Then integration over the fibre for $\begin{matrix} Y & ⟶ & E_{u} \times^{T} Y \\ ↓ \\ E_{u} / T \end{matrix}$ gives a map $\begin{matrix} {Hom}_{ℤ} (H^{n} (Y), ℤ) & ⟶ & {Hom}_{S} (H^{T} (Y), S) \\ \begin{matrix} \end{matrix} & \begin{matrix} \end{matrix} \\ τ & ⟼ & \int_{τ} \end{matrix}$ For each $i \in I,$ we have a map $\begin{matrix} {Hom}_{ℤ} (H^{n} (Y), ℤ) & ⟶ & {Hom}_{ℤ} (H^{n + 2} (K_{i} \times^{T} Y), ℤ) & : & τ & ⟼ & σ_{i} * τ \end{matrix}$ where $σ_{i} * τ \in {Hom}_{ℤ} (H^{n + 2} (K_{i} \times^{T} Y), ℤ)$ is the composition $\begin{matrix} H^{n + 2} (K_{i} \times^{T} Y) & \underset{\int_{τ}}{⟶} & H^{2} (K_{i} / T) & \underset{σ_{i}}{⟶} & ℤ \end{matrix}$ using integration over the fibre first for the bundle $\begin{matrix} Y & ⟶ & K_{i} \times^{T} Y \\ ↓ \\ K_{i} / T \end{matrix} .$ Consequently we have a map $\begin{matrix} {Hom}_{ℤ} (H^{n} (Y), ℤ) & ⟶ & {Hom}_{ℤ} (H^{n + 2} (K_{i} \times^{T} Y), Z) & ⟶ & {Hom}_{S} (H^{T} (K_{i} \times^{T} Y), S) \\ τ & ⟼ & σ_{i} * τ & ⟼ & \int_{σ_{i} * τ} . \end{matrix}$ Now suppose that $X$ is a $K -space$ with $K -action$ $\begin{matrix} μ : & K \times X & ⟶ & X . \end{matrix}$ Assume that $F : Y \to X$ is a $T -equivariant$ map. Then for $τ \in {Hom}_{ℤ} (H^{n} (Y), Z),$ we have $\int_{τ} \in {Hom}_{S} (H^{T} (Y), S),$ so $F_{*} \int_{τ} \in {Hom}_{S} (H^{T} (X), S)$ and thus $A_{i} \cdot F_{*} \int_{τ} \in {Hom}_{S} (H^{T} (X), S) .$ On the other hand, we have $\begin{matrix} K_{i} \times^{T} Y & \overset{F_{i}}{⟶} & K_{i} \times^{T} X & \overset{μ}{⟶} & X \\ [k_{i}, y] & ⟼ & [k_{i}, f (y)] & ⟼ & k_{i} \cdot f (y) \end{matrix}$ and $\int_{σ_{i} * τ} \in {Hom}_{S} (H^{T} (K_{i} \times^{T} Y), S) .$

Fact: $\begin{matrix} A_{i} \cdot F_{*} \int_{τ} & = & μ_{*} {F_{i}}_{*} \int_{σ_{i} * τ} \in {Hom}_{S} (H^{T} (X), S) . \end{matrix}$

Proof.

$□$

This fact will be used in the next lecture for $Y = X_{w}^{P},$ a Schubert variety in ?????

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.

page history