## Quantum Cohomology of $G/P$

Last update: 10 December 2013

## Lecture 1: February 7 1997

### Course Outline:

Let $G:semisimple algebraic group over ℂ B⊂G:a Borel P⊃B:a parabolic K⊂G:maximal compact T=K∩B:maximal torus in K W=NK(T)/T:Weyl group$ Then $G/B=K/T$ and $W$ acts on the de Rham cohomology space ${H}^{*}\left(K/T\right)\text{.}$ Moreover, since $K/T$ maps to the classifying space ${B}_{T},$ we have a morphism ${H}^{*}\left({B}_{T}\right)\to {H}^{*}\left(K/T\right)$ of algebras. The map $G/B\to G/P:gB↦gP$ gives inclusion $H*(G/P) ↪ H*(G/B) = H*(K/T).$ $G/P$ is a smooth projective variety.

The de Rham cohomology ${H}^{*}\left(G/P\right)$ 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 ${\sum }_{k=1}^{3}\text{dim} {x}_{k}=\text{dim} G/P\text{.}$ What is the number of points in the intersection ${x}_{1}\cap {x}_{2}\cap {x}_{3}\text{?}$

The quantum cohomology $q{H}^{*}\left(G/P\right)$ answers a more general question: what is the number of holomorphic maps $\varphi :{ℙ}^{1}\to G/P$ with a fixed degree such that $ϕ(0) ∈ X1 ϕ(1) ∈ X2 ϕ(∞) ∈ X3?$ Some features of $q{H}^{*}\left(G/P\right):$

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

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

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

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

### Geometrical models - the variety $Y\text{.}$

For each parabolic, have ${y}_{p}\in Y$ and $Yp± ≔ { y∈Y:limt→∞ ?????=yp } Y = ⨆pYp± over ℂ$ and $𝒪(Yp+) ≅ qH*(G/P) over ℤ 𝒪(Yp-) ≅ H*(Ω(H∩P))$ where $\Omega \left(K\cap P\right)$ is the group of loops in $K\cap P\text{.}$ Moreover, ${Y}_{p}^{-}\simeq {ℂ}^{n}$ for some $n,$ and $Y=YG-‾= YB+‾$ $⇒$ $𝒪(YG-) ⟶ 𝒪(YG-∩YB+) ⟵ 𝒪(YB+) ||≀ ||≀ || H*(ΩK) qH*(G/B)(q) qH*(G/B)$ Will express the Schubert basis elements as matrix entries of some representations. The variety $Y$ lies in ${G}^{\vee }/{B}^{\vee },$ where ${G}^{\vee }$ is the Langland dual of $G\text{.}$

## Lecture 2: February 11, 1997

### Kac-Moody root datum

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

 (1) ${a}_{ii}=2$ $\forall i\in I$ (2) ${a}_{ij}\le 0$ $\forall i\ne j$ (3) ${a}_{ij}=0⇔{a}_{ji}=0$

A Kac-Moody root datum consists of

• a generalized Cartan matrix $A={\left({a}_{ij}\right)}_{i,j\in I}$
• two finitely generated free $ℤ\text{-modules}$ ${h}_{ℤ}^{\vee }$ and ${h}_{ℤ}$ end????? with a perfect pairing $⟨ ⟩$ between them
• two maps $I ⟶ hℤ∨: i ⟼ αi I ⟶ hℤ: i ⟼ αi∨$ such that $⟨αj,αi∨⟩ =aij(backwards)$

?????

• Can form direct sums of root data
• Can form the "dual" root data: $(A,hℤ∨,hℤ) ⟼ (At,hℤ,hℤ∨) αi ⟷ αi∨$

?????ition: $\begin{array}{cc}\text{Simple roots:}& \pi ={\left\{{\alpha }_{i}\right\}}_{i\in I}\subset {h}_{ℤ}^{\vee }\\ \text{Simple coroots:}& {\pi }^{\vee }={\left\{{\alpha }_{i}^{\vee }\right\}}_{i\in I}\subset {h}_{ℤ}\\ \text{Weight lattice:}& {h}_{ℤ}^{\vee }\\ \text{Coweight lattice:}& {h}_{ℤ}\\ \text{Root lattice:}& Q=\underset{i\in I}{⨁}ℤ{\alpha }_{i}\\ \text{Co-root lattice:}& {Q}^{\vee }=\underset{i\in I}{⨁}ℤ{\alpha }_{i}^{\vee }\end{array}$ where $Q ⟶ hℤ∨ isom. ⟺ of adjoint type Q∨ ⟶ hℤ isom. ⟺ of simply connected type.$

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

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

Assumption: Will assume that $A$ is symmetrizable.

The numbers ${m}_{ij}\text{:}$ Define, for $i\ne j,$ $i,j\in I$ $mij= { 2 if aijaji=0 3 if aijaji=1 4 if aijaji=2 6 if aijaji=3 ∞ if aijaji≥4$ The Weyl group $W$ is the group with generators ${r}_{i},$ $i\in I$ with relations $ri2 = 1,i∈I, (rirj)mij = 1,i,j∈I, i≠j.$ The ${r}_{i}\text{'s}$ are called the simple reflections.

Notation:

• $w={r}_{{i}_{1}}{r}_{{i}_{2}}\cdots {r}_{{i}_{n}}$ [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 $n=\ell \left(w\right)\text{.}$
• If $W$ is finite, use ${w}_{0}$ to denote the longest element.
• From now on, write ${h}_{ℤ}^{\vee }={h}_{ℤ}^{*}\text{.}$ Using $⟨ ⟩\text{.}$

?????tions of $W$ on ${h}_{ℤ}^{*},$ ${h}_{ℤ},Q,$ ${Q}^{\vee },$ $S=S\left({h}_{ℤ}^{*}\right)$

• $W$ acts on ${h}_{ℤ}^{*}$ by $riλ=λ- ⟨λ,αi∨⟩ αi$ $W\left(Q\right)\subset Q\text{.}$
• $W$ acts on ${h}_{ℤ}$ by $rih=h- ⟨αi,h⟩ αi∨$ $W\left({Q}^{\vee }\right)\subset {Q}^{\vee }$
• The $W$ actions on ${h}_{ℤ}^{*}$ and on ${h}_{ℤ}$ preserve the pairing ?????
• The $W$ actions on $Q$ and on ${Q}^{\vee }$ are faithful
• $W$ acts on $S=S\left({h}_{ℤ}^{*}\right),$ the symmetric algebra of ${h}_{ℤ}^{*}$ (????? the action on ${h}_{ℤ}^{*}\text{).}$ $W$ acts by algebra automorphisms. The action of $W$ on $S$ will be denoted by $s⟼ww·s$

### The Nil-Hecke ring $A$

Definition: the Nil-Hecke ring $\underset{_}{A}$ associated to the root datum $\left(A={\left({a}_{ij}\right)}_{i,j\in I},{h}_{ℤ}^{*},{h}_{ℤ},{\left\{{\alpha }_{i}\right\}}_{i\in I},{\left\{{\alpha }_{i}^{\vee }\right\}}_{i\in I}\right)$ is the associated ring with $1$ with generators $λˆ,Ai, λˆ∈hℤ*, i∈I$ and relations: $λˆ+μˆ = λ+μˆ, λˆμˆ = μˆλˆ, λ,μ∈hℤ*, Aiλˆ = riλ∨Ai+ ⟨λ,αi∨⟩, λ∈hℤ*,i∈I AiAi = 0,i∈I, AiAjAi ⏟mij ⋯ = AjAiAj ⏟mij ⋯, (i≠j,i,j∈I).$ The grading on $\underset{_}{A}$ is defined to be $deg λˆ = 2, deg Ai = -2.$ For $w\in W$ and for any $w=ri1 ri2⋯ rin [red],$ set $Aw = Ai1Ai2 ⋯Ain (Aid = 1).$ Then it is clear that

 (1) ${A}_{w}$ is independent of the reduced expression (2) ${A}_{v}{A}_{w}=\left\{\begin{array}{cc}{A}_{vw}& \text{if} \ell \left(v\right)+\ell \left(w\right)=\ell \left(vw\right),\\ 0& \text{otherwise.}\end{array}$
Clearly $S\subseteq A$ as a subring.

Proposition: $\left\{{A}_{w}:w\in W\right\}$ is an $S\text{-basis}$ for $\underset{_}{A}\text{.}$

(Does this need a proof?)

Proposition: The map $ZW ⟶ A_: ri ⟼ 1-αiˆAi = Aiαiˆ-1,i∈I$ defines an injective ring homomorphism. Proof. Only need to check ${r}_{i}^{2}=1,$ $i\in I$ and ${\left({r}_{i}{r}_{j}\right)}^{{m}_{ij}}=1$ for $i\ne j\text{.}$ Injectivity is clear (?) $\square$

Proposition: The following defines an $A\text{-module}$ structure on $S\text{:}$ $s′·s = s′s Ai·s = 1αi (s-ri·s)$ Proof. The induced ${r}_{i}$ action on $S$ is $s↦{r}_{i}·s,$ as the usual one. $\square$

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 $1-\stackrel{ˆ}{{\alpha }_{i}}{A}_{i}={A}_{i}\stackrel{ˆ}{{\alpha }_{i}}-1\text{.}$ Then this is how ${r}_{i}$ acts. If this gives a $W\text{-action,}$ we are done.

Proposition: For $s\in S,$ $i\in I$ and $w\in W$ $ws = (w·s)w Ais = ri(s)Ai+ Ai·s Ais = sAi+(Ai·s) ri$ in $\underset{_}{A}\text{.}$ Proof. Note: All of this proof was crossed out in the scanned notes. Just need to check that $ri=1-αiˆ Ai=Aiαiˆ-1$ $\square$

### The anti-automorphism $*$ on $A$

$*(S) = S *Aw = Aw-1$ To check that this is an anti-automorphism, need to check only $λˆAi=Ai riλˆ+ ⟨λ,αi∨⟩ 1λ∈hℤ*,i∈I.$ This is easy. Now since $ri=1-αiˆAi =Aiαiˆ-1$ we get $*ri=αiˆAi -1=-ri.$ $\text{(}{A}_{i}{r}_{i}=-{A}_{i},{r}_{i}{A}_{i}={A}_{i}\text{)}$ Consequently, $*w=(-1)ℓ(w) w-1$

### Definitions of $\underset{_}{A}$ on $M{\otimes }_{S}N$ and ${\text{Hom}}_{S}\left(M,N\right)$

Assume that $M$ and $N$ are $\underset{_}{A}\text{-module}$ and are thus ?????odules. Form $M⊗SN = M⊗N/ {sm⊗n-m⊗sn} HomS(M,N) = { f:M→N:f(sm) =sf(m) }$ want to define $\underset{_}{A}\text{-module}$ structures on $M{\otimes }_{S}N$ and ${\text{Hom}}_{S}\left(M,N\right)\text{.}$

????? $M{\otimes }_{S}N\text{:}$ $s·(m⊗n) = sm⊗n Ai·(m⊗n) = Ai·m⊗n+ ri·m⊗Ai·n = m⊗Ai·n+Ai·m ⊗ri·n$ ????? check that this is an action, we first need to show that the above operators are well-defined. The $s·$ operator is clearly OK. ????? $s\in S,$ $i\in I,$ we have, by definition $Ai· ( sm⊗n-m⊗sn ) = (Ais)·m⊗n+ (ris)·m⊗Ai ·n -Ai·m⊗sn- ri·m⊗(Ais) ·n$ Using $Ais = (ri·s)Ai+ Ai·s ris = (ri·s)ri ri = 1-αiˆAi ri·s = s-αiAi·s$ we get $Ai· ( sm⊗n- m⊗sn ) = (ri·s)Ai ·m⊗n+(Ai·s) m⊗n +(ri·s)ri· m⊗Ai·n-Ai·m ⊗sn -ri·m⊗(ri·s) Ai·n-ri·m⊗ (Ai·s)n = (ri·s)ri·m ⊗Ai·n-ri·m ⊗(ri·s)Ai·n +(s-αiˆAi·s) αi·m⊗n-Ai·m ⊗sn +(Ai·s)m⊗n- ( m-αiˆAi·m ) ⊗(Ai·s)n = (ri·s)ri·m ⊗Ai·n-ri·m ⊗(ri·s)Ai·n +sAi·m⊗n-Ai m⊗sn - ( (Ai·s)αiˆAi ·m⊗n-αiˆAi ·m⊗(Ai·s)n ) +(Ai·s)m⊗n- m⊗(Ai·s)n ∈ ⟨ S′m⊗n-m⊗ s′n:m,n∈M,N ⟩ .$ Hence ${A}_{i}$ is well-defined.

Next, since ${r}_{i}=1-\stackrel{ˆ}{{\alpha }_{i}}{A}_{i},$ we have $Ai·m⊗n+ri·m ⊗Ai·n = Ai·m⊗n+ m⊗Ai·n- αiˆAi· m⊗Ai·n = Ai·m⊗n- Ai·m⊗αiˆ Ai·n+m⊗ Ai·n = Ai·m⊗ri·n +m⊗Ai·n = m⊗αi·n+Ai· m⊗ri·n.$ This gives the 2nd expression for ${A}_{i}·\left(m\otimes n\right)\text{.}$

Now for $s\in S$ and $i\in I,$ we need to show $Ai·(s·(m⊗n))= (ri·s)· (Ai·(m⊗n))+ (Ai·s)·(m⊗n)$ $l.h.s. = (Ais)·m⊗n+ (ris)·m⊗Ai ·n r.h.s. = (ri·s)Ai·m ⊗n+(ri·s)ri ·m⊗Ai·n+ (Ai·s)m⊗n = (Ais)·m⊗n+ (ris)·m⊗αi·n = l.h.s.$ From this, we see that ${r}_{i}=1-\stackrel{ˆ}{{\alpha }_{i}}{A}_{i}={A}_{i}\stackrel{ˆ}{{\alpha }_{i}}-1$ acts by $ri·(m⊗n) = m⊗n-αiˆAi ·m⊗n-αiˆri ·m⊗Ai·n = ri·m⊗n-ri ·m⊗αiˆAi ·n = ri·m⊗ri·n$ This clearly induces an action of $W$ on $M{\otimes }_{S}N\text{.}$ Thus we have proved that we indeed have an action of $\underset{_}{A}$ on $M{\otimes }_{S}N\text{.}$

On ${\text{Hom}}_{S}\left(M,N\right),$ define: $(s·f)(m) = sf(m) (Ai·f)(m) = f(Ai·m)+Ai ·f(ri·m) = Ai·f(m)-ri ·f(Ai·m)$ Need to check that this is indeed an action. Clearly $s·$ is o????? First, since $ri=1-αiˆAi$ and since $f$ is $S\text{-linear,}$ we have $f(Ai·m)+Ai ·f(ri·m) = f(Ai·m)+Ai· (f(m)-αiˆf(Ai·m)) = Ai·f(m)+f(Ai·m) -(Aiαiˆ)· f(Ai·m) = Ai·f(m)-ri ·f(Ai·m)$ This shows that the two expressions for ${A}_{i}·f$ are equal. Now we show that $?????(sm) = s(Aif)(m).$ $?????h.s. = f((Ais)·m)+ Ai·f((ris)·m) ?????h.s. = sf(Ai·m)+sAi ·f(ri·m) = f(sAi·m)+ ( Airi(s)+ Ai·s ) f(ri·m) = f(sAi·m)+ Ai·f(ri(s)ri·m) +f((Ai·s)ri·m)$ ????? $ris = ri(s)ri Ais = sAi+(Ai·s)ri$ ????? see that $\text{l.h.s.}=\text{r.h.s.}$

????? shows that $Ai·f∈HomS(M,N).$

????? need to check that $Ai·(s·f)= (ri·s)· (Ai·f)+ (Ai·s)·f$ $????? m ⟼ sf(Ai·m)+ Ais·f(ri·m) ????? m ⟼ (ri·s)f(Ai·m) +(ri·s)Ai·f (ri·m)+(Ai·s) f(m) =(ri·s)f (Ai·m)+Ais· f(ri·m)-f ( (Ai·s)ri·m ) +f((Ai·s)m)$ ????? $Ais=(ri·s) Ai+Ai·sand sAi=Ais-(Ai·s) ri$ ????? see $\text{l.h.s}=\text{r.h.s.}$

Finally, for ${r}_{i}=1-\stackrel{ˆ}{{\alpha }_{i}}{A}_{i}={A}_{i}\stackrel{ˆ}{{\alpha }_{i}}-1,$ we see that $(ri·f)(m) = f(m)-αiˆf (Ai·m)- αiˆAi·f (ri·m) = f(ri·m)- αiˆAi·f (ri·m) = ri·f(ri·m).$ Consequently, $(w·f)(m)= w·f(w-1·m)$ This is certainly an action of $W$ on ${\text{Hom}}_{S}\left(M,N\right)\text{.}$ Hence we have an action of $\underset{_}{A}$ on ${\text{Hom}}_{S}\left(M,N\right)\text{.}$

All these proofs seem to be longer than necessary.

But anyway, we have showed that $s·(m⊗n) = sm⊗n Ai·(m⊗n) = Ai·m⊗n+ri· m⊗Ai=m⊗Ai ·n+Ai·m⊗ri w·(m⊗n) = w·m⊗w·n (s·f)(m) = sf(m) (Ai·f)(m) = f(Ai·m)+Ai ·f(ri·m)= Ai·f(m)-ri ·f(Ai·m) (w·f)(m) = w·f(w-1·m)$ make $M{\otimes }_{S}N$ and ${\text{Hom}}_{S}\left(M,N\right)$ $\underset{_}{A}\text{-modules}$ again.

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

 1 ${\text{Hom}}_{S}\left(S,M\right)\cong M,$ $S{\otimes }_{S}M\simeq M\simeq M{\otimes }_{S}S$ 2 $M{\otimes }_{S}N\simeq N{\otimes }_{S}M$ 3 $M{\otimes }_{S}\left(N{\otimes }_{S}P\right)\simeq \left(M{\otimes }_{S}N\right){\otimes }_{S}P$ 4 ${\text{Hom}}_{S}\left(M{\otimes }_{S}N,P\right)\simeq {\text{Hom}}_{S}\left({\text{Hom}}_{S}\left(M,N\right),P\right)$ 5 $M{\otimes }_{S}{\text{Hom}}_{S}\left(N,P\right)\to {\text{Hom}}_{S}\left({\text{Hom}}_{S}\left(M,N\right),P\right)$ 6 ${\text{Hom}}_{S}\left(M,N\right){\otimes }_{S}P\to {\text{Hom}}_{S}\left(M,N{\otimes }_{S}P\right)$

Definition: For an $\underset{_}{A}\text{-module}$ $P,$ set $PA= { p∈P:Ai·p=0 ∀i∈I } .$

Proposition: For $\underset{_}{A}\text{-modules}$ $M$ and $N,$ $HomA_(M,N)≃ (HomS(M,N))A.$

Example: Regard $\underset{_}{A}$ as an $\underset{_}{A}$ module by left multiplications. Then our previous constructions define an $\underset{_}{A}\text{-module}$ structure on $\underset{_}{A}{\otimes }_{S}\underset{_}{A}\text{.}$ Define: $Δ: A_ ⟶ A_⊗SA_$ by $Δa = a·(1⊗1)$ Thus $Δw = w⊗w Δs = s⊗1=1⊗s ΔAi = Ai⊗1+ri⊗Ai =1⊗Ai+Ai⊗ri$ For any two $\underset{_}{A}$ modules $M$ and $N,$ since we have $a·(m⊗n)= a(1)·m⊗ a(2)·n$ for $a\in S$ or $a={A}_{i},$ $i\in I,$ where $\Delta a={a}_{\left(1\right)}\otimes {a}_{\left(2\right)},$ we have $a·(m⊗n)= a(1)·m⊗ a(2)·n ∀a∈A.$

Proposition: In the finite case, $ΔAw0 = ∑w∈WAw0w ⊗w0Aw = ∑w∈WAw⊗ w0Aw0w$ Proof. It is easy to show by induction on $\ell \left(w\right)$ that for any $w\in W$ $ΔAw=Aw⊗w+ ∑v for some ${a}_{v}\in A\text{.}$ So $ΔAw0=∑w∈W Aw⊗aw$ with ${a}_{{w}_{0}}={w}_{0}\text{.}$ Now for any $i\in I,$ $AiAw0 = 0 ⇒ 0 = Δ(Ai) Δ(Aw0) ⇒ 0 = ( Ai⊗1+ ri⊗Ai ) ∑w∈W Aw⊗aw = ∑w∈W ( AiAw⊗ aw+ri Aw⊗Aiaw ) = ∑w∈W ( AiAw⊗aw+ (1-αiˆAi) Aw⊗Aiaw ) = ∑w∈WAi Aw⊗riaw- Aw⊗Aiaw ⇒ ∑w∈WAi Aw⊗riaw = ∑w∈WAw ⊗Aiaw$ Now $l.h.s. = ∑riw>w Ariw⊗ riaw ⇒ ariw = -Aiawif riw $\square$

## Lecture 3: February 12, 1997

Recall that a Kac-Moody root datum consists of

• a generalized Cartan matrix $A={\left({a}_{ij}\right)}_{i,j\in I};$
• two finitely generated free $ℤ\text{-modules}$ ${h}_{ℤ}^{\vee }={h}_{ℤ}^{*}$ and ${h}_{ℤ}$ with a perfect pairing $⟨ ⟩$ between then;
• two maps $I ⟶ hℤ∨: i ⟼ αi I ⟶ hℤ: i ⟼ αi∨$ such that $⟨αi,αj∨⟩ =aji.$
The weight lattice is ${h}_{ℤ}^{\vee }$
The coweight lattice is ${h}_{ℤ}$
The root lattice is $Q\stackrel{\text{def}}{=}\underset{i\in I}{⨁}ℤ{\alpha }_{i}$
The co-root lattice is ${Q}^{\vee }\stackrel{\text{def}}{=}\underset{i\in I}{⨁}ℤ{\alpha }_{i}^{\vee }$
Say the datum is of the adjoint type if $Q\to {h}_{ℤ}^{\vee }:{\alpha }_{i}↦{\alpha }_{i}$ is an isomorphism.
Say the datum is of the simply connected type if ${Q}^{\vee }\to {h}_{ℤ}:{\alpha }_{i}^{\vee }↦{\alpha }_{i}^{\vee }$ is an isomorphism.

For $sl\left(2,ℂ\right),$ use $e,f,h$ for the standard generators such that $[h,e] = 2e [h,f] = -2f [e,f] = h$ Given a Kac-Moody root datum $\left(A,I,{h}_{ℤ}^{\vee },{h}_{ℤ},⟨ ⟩\right),$ set $h_=ℂ⊗ℤhℤ$ and regard it as a commutative Lie algebra. Set ${h}_{i}={\alpha }_{i}^{\vee }$

Theorem (see Kac?): For any Kac-Moody root datum, there exists a Lie algebra $\underset{_}{g}$ over $ℂ$ (of Kac-Moody type) and Lie algebra homomorphisms $ϕ: h_ ⟶ g_ ϕi: sl2(ℂ) ⟶ g_ ∀i∈I$ such that

 (1) $ϕi(h) = ϕ(hi) [ ϕ(h), ϕi(e) ] = ⟨αi,h⟩ ϕi(e)h∈h_ [ ϕ(h), ϕi(f) ] = -⟨αi,h⟩ ϕi(f)i∈I [ ϕi(e), ϕj(f) ] = 0(i≠j)$ (2) For each $i\in I,$ $\underset{_}{g}$ as an ${sl}_{2}\left(ℂ\right)$ module via ${\varphi }_{i}$ (using the adj. rep) is a direct sum of finite-dimensional ${sl}_{2}\left(ℂ\right)\text{-module.}$ (3) If $\underset{_}{g}\prime ,\varphi \prime ,$ or ${\varphi }_{i}^{\prime }$ are another such system, then there exists a unique $\psi :g\to g\prime$ such that $\varphi \prime =\psi \circ \varphi$ and ${\varphi }_{i}^{\prime }=\psi \circ {\varphi }_{i}\text{.}$ Thus $\left(g,\varphi ,{\varphi }_{i},i\in I\right)$ is unique.

Definition:

 (1) An ${sl}_{2}\left(ℂ\right)\text{-module}$ $V$ over $ℂ$ is integrable if it is a direct sum of finite-dim. modules. (2) An $\underset{_}{h}\text{-module}$ $V$ over $ℂ$ is integrable if $V=⨁μ∈hℤ* Vμ$ where $Vμ= { v∈V:hv=μ(v)v for all h∈h_ }$ (3) A $\underset{_}{g}\text{-module}$ $V$ over $ℂ$ is integrable if it is ${sl}_{2}\left(ℂ\right)\text{-integrable}$ (via ${\varphi }_{i},$ for each $i\in I\text{)}$ and $\underset{_}{h}\text{-integrable}$ via $\varphi \text{.}$
So the adjoint representation of $\underset{_}{g}$ on $\underset{_}{g}$ is integrable.

Definition:

• $\varphi :\underset{_}{h}\to \underset{_}{g}$ is injective, so call $\underset{_}{h}\subset g$ the Cartan subalgebra
• ${\varphi }_{i}:{sl}_{2}\left(ℂ\right)\to \underset{_}{g}$ is injective for each $i\in I\text{.}$ Set $ei = ϕi(e) fi = ϕi(f) hi = ϕi(h)= αi∨$
• $Z\left(g\right),$ the center of $g,$ is contained in $\underset{_}{h}\text{.}$
• Set $n+ = ⟨ei⟩i∈I =Lie subalgebra generated by {ei,i∈I} n- = ⟨fi⟩i∈I$ Then $g_=n-+h_ +n+$ triangular decomposition
Every ideal of $\underset{_}{g}$ contained totally in ${n}_{-}$ or ${n}_{+}$ is $0\text{.}$ $b+ =def h_+n+=b (Borel subalgebra) b- =def h_+n-$

Fact: $\underset{_}{g}$ is the Lie algebra over $ℂ$ with generators $h∈h_,ei, fi,i∈I$ with relations $[h,h′] = 0 [h,ei] = ⟨αi,h⟩ei [h,fi] = -⟨αi,h⟩fi [ei,fj] = δijhi (ad ei)1-aij ej = 0(i≠j) (ad fi)1-aij fj = 0(i≠j)$

Warning: $h_⫋ centralizer of h_ in g_$

### The $Q\text{-grading}$ of $\underset{_}{g}\text{:}$

For $\beta \in Q,$ the root lattice, set $gβ= { x∈g:[h,x]= ⟨β,h⟩x ∀h∈h_ } .$ Then $g_=⨁β∈Q gβ$ and $\left[{g}_{{\beta }_{1}},{g}_{{\beta }_{2}}\right]\subset {g}_{{\beta }_{1}+{\beta }_{2}}$

?????ice that $g0h_ gαi=ℂei g-αi=ℂfi i∈I.$ $ℤ+ = {0,1,2,…} Q+ = ⨁i∈Iℤ+αi ⊂Qsub-semigroup$ ????? $\beta ,\nu \in Q,$ say $\beta \ge \nu$ if $\beta ·\nu \in {Q}_{+}\text{.}$

????? $n±= ⨁β∈Q± gβ$ $Δ = { β∈Q:gβ≠0, β≠0 } ,set of roots Δ+ = Δ∩Q+,set of positive roots Π = {αi:i∈I}, set of simple roots$

????? $Δ- = -Δ+$

????? $Δ+∪ Δ- = Δ Δ+∩ Δ- = ∅ n± = ⨁β∈Δ± gβ$

### The principle $ℤ\text{-grading}$ of $\underset{_}{g}\text{:}$

Let ${\rho }^{\vee }\in {Q}^{\vee }$ be the unique(?) element such that $⟨αi,ρ∨⟩ ≡1∀i∈I.$ For $\beta \in Q,$ the integer $ht(β)=⟨β,ρ∨⟩$ is called the height of $\beta \text{.}$ For $n\in ℤ,$ set $g_n= ⨁β∈Qht(β)=n gβ$ Thus is a $ℤ\text{-grading}$ for $\underset{_}{g}\text{.}$

### 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 $\underset{_}{g}\text{:}$

This is the conjugation-linear automorphism of $\underset{_}{g}$ such that $ei ⟷ -fi i∈????? h ⟷ -h h∈h_ℝ =detℝ⊗ℤ hℤ⊂h_.$

## Kac-Moody group

### ?????$\left(ℂ\right)\text{:}$

For $u\in ℂ,$ set $t\in {ℂ}^{×},$ set $x(u) = (1u01) y(u) = (10u1) h(t) = (t00t-1) ∈SL2(ℂ).$ ?????l finite-dimensional representation of ${SL}_{2}\left(ℂ\right)$ is said to be ?????ational if its matrix entries are regular functions on ${SL}_{2}\left(ℂ\right)\text{.}$ ????? representation of ${SL}_{2}\left(ℂ\right)$ 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 ${sl}_{2}\left(ℂ\right)↔\text{differentiable}$ rep. of ${SL}_{2}\left(ℂ\right)\text{.}$ (This is because ${SL}_{2}\left(ℂ\right)$ is an algebraic group).

### The complex torus $H\text{:}$

Define $H=Hom(hℤ∨,ℂ×).$ For $h\in {h}_{ℤ}$ and $t\in {ℂ}^{×},$ define ${t}^{h}\in H$ by $th(λ)= t⟨λ,h⟩, λ∈hℤ∨.$ Thus, for each such $h\in {h}_{ℤ},$ the map $ℂ× ⟶ H: t ⟼ th$ is a homomorphism. Moreover $th+h′= th·th′.$ A representation of $H$ on $V/ℂ$ is said to be differentiable if it is a direct sum of $1\text{-dimensional}$ rational representations of $H\text{.}$

Fact: Differentiable representations of $H$ $⟷$ integrable representations of $\underset{_}{h}\text{.}$

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 $ϕ: H ⟶ G ϕi: SL2(ℂ) ⟶ G i∈I$ $ϕi(h(t)) = ϕ(thi) ϕ(th)ϕi (x(u))ϕ (t-h) = ϕi (x(t⟨αi,h⟩u)) ϕ(th)ϕi (y(u))ϕ (t-h) = ϕi (y(t-⟨αi,h⟩u)) ϕi(x(u)) ϕj(y(v)) = ϕj(y(v)) ϕi(x(u)), i≠j$ There exists a representation Ad of $G$ on $\underset{_}{g}$ such that under $\varphi$ and ${\varphi }_{i},$ $i\in I,$ the corresponding representations of $H$ and ${SL}_{2}\left(ℂ\right)$ on $\underset{_}{g}$ differentiate to the representations of $\underset{_}{h}$ and ${sl}_{2}\left(ℂ\right)$ on $\underset{_}{g}$ defined by ad.

If $\text{(}G\prime ,$ $\varphi \prime$ and ${\varphi }_{i}^{\prime }\text{)}$ is another system with above properties, then there exists a unique $\psi :G\to G\prime$ such that $ϕ′=ψ∘ϕ ϕi′=ψ∘ϕi$

• $G$ is generated by the images of $\varphi$ and ${\varphi }_{i},$ $i\in I\text{.}$
• A $G\text{-module}$ is said to be differentiable if it is differentiable as $H$ and ${SL}_{2}\left(ℂ\right)$ modules under $\varphi$ and each ${\varphi }_{i}^{\prime },$ $i\in I\text{.}$ Thus $differentiable G-module⟷ integrable g_-module.$
• There exists a faithful differentiable $G\text{-module}$ (probably not Ad).
• $\varphi :H\to G$ is injective. So we call $H\subset G$ the Cartan subgroup. Have $Z(G)=defker Ad⊂H ker ϕi⊂Z (SL2(ℂ))$

### The Weyl group $W\text{:}$

For each $i\in I,$ $u\in ℂ,$ set $xi(u) = ϕi(x(u))= ϕi (1u01) ∈G yi(u) = ϕi(y(u))= ϕi (10u1) ∈G ni = yi(1)xi(-1) yi(1)=ϕi (0-110) ∈G$ ????? $ninjni ⏞mij ⋯ = njninj ⏞mij ⋯$ ????? for $w={r}_{{i}_{1}}\cdots {r}_{{i}_{n}}$ [red], let $nw=ni1ni2 ⋯nin$ ????? $nw·gβ= gw·β$ ????? in general $nwnw-1≠ id.$ ????? $N=⟨ni,H⟩i∈I ⊂G$ ????? the subgroup of $G$ generated by $\left\{{n}_{i},i\in I\right\}$ and $H\text{.}$ ????? $N/H ≃ W ni/H ⟶ ri$ Warning: can have $H⊊{Z}_{G}\left(H\right)\text{.}$

### The real roots:

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

Set $Δre= ⋃i∈IW ·αi$ and call elements in ${\Delta }^{\text{re}}$ the real roots.

If $\beta =w·{\alpha }_{i}\in {\Delta }^{\text{re}}$ for some $i\in I,$ then $gβ=nw·gαi$ so $dimℂ gβ=1$ and $gmβ=0for |m|>1.$ Also, define $rβ=wriw-1 ∈W.$ Then

 (1) ${r}_{{w}_{1}\beta }={w}_{1}{r}_{\beta }{w}_{1}$ for any ${w}_{1}\in W$ (2) $rβ·λ = λ- ⟨λ,β∨⟩ β λ∈hℤ* rβ·h = h-⟨β,h⟩ β∨h∈hℤ.$

## 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. $\text{(}L=K$ or $T$ in our applications). A principal $L\text{-bundle}$ is a topological space $E$ equipped with $c$ continuous free right action of $L$ $E×L ⟶ L$ and a projection $E\to B$ such that locally $E=B\prime ×L$ where with $L$ acts on $B\prime ×L$ by $\left(b,\ell \right)\stackrel{{\ell }_{1}}{⟼}\left(b,\ell {\ell }_{1}\right),$ $B\prime \underset{\text{open}}{\subset }B\text{.}$ Have $B\simeq E/L$ with the quotient topology.
• The universal principal $L\text{-bundle}$ ${E}_{L}$ is a principal $L\text{-bundle}$ such that ${E}_{L}$ is contractible. Set ${B}_{L}={E}_{L}/L\text{.}$ It is called the Classifying space of $L\text{.}$

Example: $BS1 = ℂP∞ ET = EK(because T⊂K)$

Definition: An $L\text{-space}$ is a topological space $X$ endowed with a continuous left $L\text{-action:}$ $L×X ⟶ X: (ℓ,x) ⟼ ℓ·x∩ℓx.$

Definition: Given an $L\text{-space}$ $X,$ form the space $EL×LX= (EL×X)/L$ where $\left(e,x\right)·\ell =\left(e{\ell }^{-1},\ell x\right)$ is a free left $L\text{-action.}$ The $L\text{-equivariant}$ cohomology of $X$ is by definition the singular homology of ${E}_{L}{×}^{L}X\text{:}$ $HL(X)= H*(EL×LX)$

### Structures on ${H}^{L}\left(X\right)\text{:}$

1. It is a graded ring, where the grading is nothing but the grading on ${H}^{*}\left({E}_{L}{×}^{L}X\right)\text{.}$ (And so is the ring structure ?????
2. The fibration ${E}_{L}{×}^{L}X\to {E}_{L}/L={B}_{L}$ gives a graded ring homomorphism $H*(BL) ⟶ H*(EL×LX)$ i.e. $HL(pt) ⟶ HL(X)$ Thus ${H}^{L}\left(X\right)$ has a natural ${H}^{L}\left(\text{pt}\right)\text{-module}$ structure

### Functoriality:

Given an $L\text{-map}$ of $L\text{-spaces}$ $f: X ⟶ Y,$ form the map $EL×LX ⟶ EL×LY [e,x] ⟼ [e,f(x)].$ Have commutative diagram $EL×LX ⟶ EL×LY πx ↓ ↓ πy BL ⟶id BL [e,x] ⟼ [e,f(x)] ↧ ↧ [e] = [e]$ have a graded ring homomorphism $f*: HL(X) ⟵ HL(Y)$ which is also a ${H}^{*}\left({B}_{L}\right)={H}^{L}\left(\text{pt}\right)\text{-module}$ map.

????? special case of $Y=\text{pt}$ with $f: X ⟶ pt,$ ?????es $πx: EL×LX ⟶ EL×Lpt = BL, f*: HL(pt) ⟶ HL(X)$ ????? just the one considered before.

### $L\text{-equivariant}$ homology

This is the space of $HomHL(pt) (HL(X),HL(pt)).$ Than any $L\text{-space}$ map $f: X ⟶ Y$ induces $f*: HomHL(pt) (HL(X),HL(pt)) ⟶ HomHL(pt) (HL(Y),HL(pt)).$

### The restriction homomorphism (or the evaluation at 0):

This is the map $v0(X): HL(X) ⟶ H*(X)$ induced by the map $EL×X ⟶ EL×LX.$ This is a graded $ℤ\text{-ring}$ homomorphism.

For any $L\text{-space}$ map $f:X\to Y,$ have commutative diagram $HL(X) ⟶v0(X) H*(X) f* ↑ ↑ f* HL(Y) ⟶v0(Y) H*(Y)$

Examples:

1. $L$ acts freely on $X\text{.}$ Then $HL(X)≃H*(X/L)$ Proof. Have the following fibre bundle with contractible fibre ${E}_{L}\text{.}$ $EL×LX ⟵ EL ↓ X/L$ Thus ${H}^{L}\left(X\right)={H}^{*}\left({E}_{L}{×}^{L}X\right)\simeq {H}^{*}\left(X/L\right)\text{.}$ $\square$

2. $L$ acts trivially on $X\text{.}$ Then $HL(X)≃ HL(pt)⊗ H*(X)$ Proof. Have $EL×LX ≃BL×X.$ $\square$

Proposition: Have ${H}^{*}\left({B}_{T}\right)\simeq S\left({h}_{ℤ}^{*}\right)\text{.}$ Proof. For $\lambda \in {h}_{ℤ}^{*},$ define $eλ: T ⟶ ℂ×: eλ(eh) = e⟨λ,h⟩,h∈hℤ.$ If $E$ is a principal $T\text{-bundle,}$ form the complex line bundle ${ℒ}_{-\lambda }=E{×}_{T}ℂ$ by $[et,c]= [e,e-λ(t)c] t∈T, e∈E, c∈ℂ.$ Then $λ⟼C1(E×Tℂ), the first Chern class$ gives a homomorphism $S(hℤ*) ⟶ H*(E/T).$ In particular, take $E={E}_{T}={E}_{K}\text{.}$ Then get $S(hℤ*) ⟶H*(ET/T) =HT(pt).$ One can then show that this is an isomorphism of graded rings if $\lambda \in {h}_{ℤ}^{*}$ is given $\text{deg}=2\text{.}$ $\square$

### The second $S\text{-module}$ structure on ${H}^{T}\left(K/T\right)$

Set ${E}_{u}={E}_{T}={E}_{K}\text{.}$ The map $Eu×T(K/T)⟶ Eu/T: [e,kT]⟼ekT$ is another ring homomorphism, which we will denote by ${\pi }_{R}$ for reasons that will be clear next time; $πR: S ⟶ HT(K/T).$

Remarks:

1. ${\pi }_{R},$ together with the map $πL: S ⟶ HT(K/T)$ incuded by $K/T\to \text{pt},$ will be the source and target maps for the Hopf algebroid structure on ${H}^{T}\left(K/T\right)$ that will be discussed next lecture.
2. Set $Eu(2) = Eu×Eu/KEu= { (e1,e2)∈ E×E:e1K=e2 K } ⊂ Eu×Eu$ It is a $\left(K×K\right)\text{-inv.}$ subset of ${E}_{u}×{E}_{u}\text{.}$ Since $K$ acts on ${E}_{u}$ freely, we have the identification $Eu(2)≃ Eu×K: (e1,e2)⟼ (e1,k) if e2=e1.$ Under this identification, the $T×T$ action on ${E}_{u}^{\left(2\right)}$ becomes the action $(e,k)⟼(t1,t2) (e1t1,t1-1kt2)$ of $T×T$ on ${E}_{u}×K\text{.}$ (easy to check this: $(e,k)⟼ (e1,e1k) ⟼(t1,t2) (e1t1,e1kt2) ⟼(e1t1,e1t1t1-1kt2) ⟼(e1t1,t1-1kt2)).$ Thus we have $Eu(2)/T×T ≃Eu×TK/T$ The map $Eu×T(K/T) ⟶Eu/T: [e,kT]⟼ ekT$ now is just the projection from ${E}_{u}^{\left(2\right)}/T×T$ to the 2nd factor ${E}_{u}\text{?????}$ This will be used in the next lecture.

Proposition: For any $T\text{-space}$ $Y,$ we have $HT(K×TY)≃ HT(K/T)⊗S HT(Y)$ where the $S\text{-module}$ structure on ${H}^{T}\left(K/T\right)$ is via the second ring homomorphism $πK:S⟶HT(K/T).$ (The $S\text{-module}$ structure on ${H}^{T}\left(Y\right)$ is the usual one). Proof. Consider the following commutative square: $(?????K×TY) ⟶p1 Eu×K(K×TY) ≃ Eu×TY p2 ↓ ↓ q1 (?????K×Tpt) ⟶q2 Eu×K(K×Tpt) |≀ |≀ ?????(K/T) Eu/T = [e,[k,y]]T ⟼p1 [e,[k,y]]K ⟼∼ [ek,y] p2 ↓ ↓ q1 [e,[k,y]]T ⟼q2 [e,[k,p]]K ↧≀ ↧≀ [e,k] [ek] = [ek,pt]$ notice that ${q}_{1}^{*}:S\to {H}^{T}\left(Y\right)$ is the usual homo. (induced from $Y\to \text{pt).}$ ????? ${q}_{2}^{*}={\pi }_{R}:S\to {H}^{T}\left(K/T\right)$ 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 $HT(K/T)⊗SHT(Y) ⟶ HT(K×TY). x⊗y ⟼ p2*(x)p1*(y) assuming even col?????$ To show that this is an isomorphism, we first notice that the fibration ${p}_{1}$ has fibre $K/T$ which is a $CW\text{-complex}$ if only even dimensions. Thus Leray-Hitsch theorem tells us that ${H}^{T}\left(K{×}^{T}Y\right)$ is a free module over ${H}^{T}\left(Y\right)$ with basis coming from ${H}^{*}\left(K/T\right)\text{.}$ The special case of $Y=\text{pt}$ says that ${H}^{T}\left(K/T\right)$ is a free $S={H}^{T}\left(\text{pt}\right)\text{-module}$ with basis coming from ${H}^{*}\left(K/T\right)\text{.}$ Using a basis of ${H}^{*}\left(K/T\right),$ we see that the map $HT(K/T)⊗SHT(Y) ⟶HT(K×TY)$ is an isomorphism. $\square$

Definition: The morphism $ε: HT(K/T) ⟶ S$ induced by $T/T ↪ K/T$ is called the co-unit map.

Definition: For any $K\text{-space}$ $X,$ the map $ΔX: HT(X) ⟶ HT(K/T)⊗SHT(X)$ induced by the $T\text{-map}$ $μK: K×TX ⟶ X: [k,x] ⟼ kx,$ ie. $Δx: HT(X) ⟶μK* HT(K×TX) ≃ HT(K/T)⊗SHT(X)$ is called the co-module map.

Proposition: For any $K\text{-space}$ $X,$ we have $(ε⊗id)∘ΔX= id|HT(X)$ and $(ΔK/T⊗id)∘ΔX = (id⊗ΔX)∘ΔX: HT(X)⟶ HT(K/T)⊗SHT(K/T)⊗SHT(X).$

Definition: A groupoid scheme $\left(𝒴,𝒮\right)$ consists of two schemes $𝒴$ and $𝒮$ and five morphisms: $PL,PR: 𝒴 ⟶ 𝒮 ℓ: 𝒮 ⟶ 𝒴 i: 𝒴 ⟶ 𝒴 μ: 𝒴×S𝒴 ⟶ 𝒴$ (fibre produ????? $-{×}_{S}$ refers to ${P}_{L},$ and ${×}_{S}-$ refers to ${P}_{R}\text{)}$ They must satisfy: $PL∘ℓ=id𝒴 =PR∘ℓ PL∘i=PR PR∘i=PL PL∘μ=p2∘p1 PR∘μ=PR∘ p2 μ∘(id𝒴,ℓ∘PR) =id𝒴 μ∘(ℓ∘PL,id𝒴) =id𝒴 μ∘(id𝒴,i)= i∘PRμ∘ (i,id𝒴)=ℓ∘ PR μ∘(id𝒴×μ)= μ∘(μ×id𝒴)$ These imply $i\circ i={\text{id}}_{𝒴}\text{.}$

If $𝒴=\text{Spec} R$ and $𝒮=\text{Spec} S,$ then $𝒴×S𝒴=Spec (R×SR).$

## 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 $X\text{.}$ Then we can form a groupoid $\left(𝒴,𝒮\right),$ where $𝒮=X,$ $𝒴= { (x,g,y): x,y∈X, x=g·y }$ Multiplication is given by $(x,g,y) (x′,g′,y′)= (x,gg′,y′) if y=x′$ 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 $\varphi :𝒴{×}_{S}X\to X$ of a groupoid scheme $\left(𝒴,S\right)$ on a scheme $X$ ????? $S$ with structure morphism ${P}_{X}:X\to S$ is one such that

 (1) $\varphi \circ \left(\mu ×{\text{id}}_{X}\right)=\varphi \circ \left({\text{id}}_{𝒴}×\varphi \right)$ (2) ${P}_{X}\circ \varphi ={P}_{L}\circ {P}_{1}$ where ${P}_{1}:𝒴×X\to 𝒴,$ $\left(y,x\right)↦\text{?????}$ (3) $\varphi \circ \left(\left(e\circ {P}_{X}\right)×{\text{id}}_{X}\right)={\text{id}}_{X}\text{.}$

### The groupoid scheme $𝒰=\text{Spec} {H}^{T}\left(K/T\right)$

Let ${E}_{u}$ be the principal $K$ (and thus also $T\text{)-bundle.}$ For $n\ge 1,$ let $Eun = Eu×⋯×Eu n times Kn = K×⋯×K n times Tn = T×⋯×T n times$ Set $Eu(n)= { (e1,…,en) ∈Eun:e1L= ⋯=enK } ⊆Eun.$ As a subset of ${E}_{u}^{n},$ the set ${E}_{u}^{\left(n\right)}$ is invariant under the ${K}^{n}\text{-action,}$ so ${E}_{u}^{\left(n\right)}$ is a principal ${K}^{n}\text{-bundle.}$

Set $B(n)= Eu(n)/Tn$ Then it is easy to check that ${B}^{\left(2\right)}$ is a groupoid over ${B}^{\left(1\right)}=E/T={B}_{T}$ with the following structure maps: (This is a subquotient of the coarse groupoid $E×E$ over $E\text{):}$

• Source and target maps: $p1: B(1) = Eu(2)/T2 ⟶ E/T: [e1,e2] ⟼ [e1] p2: B(2) = Eu(2)/T2 ⟶ E/T: [e1,e2] ⟼ [e2]$
• identities: $d(=diagonal): B(1) = E/T ⟶ B(2) : [e] ⟼ [e,e]$
• inverse: $t(=transposition): B(2) ⟶ B(2) : [e1,e2] ⟼ [e2,e1]$
• multiplication: $μ: B(2)×B(1)B(2) ≃ B(3) ⟶ B(2) : ([e1,e2],[e2,e3]) ⟼ [e1,?????]$
We now pull back all the above structure maps on cohomology:

First note that $Eu(2)≃Eu×K$ by $(e1,e2) ⟶ (e1,k) if e2=e1k$ Under this identification, the ${T}^{2}$ action on ${E}_{u}^{\left(2\right)}$ becomes $(e1,k) ⟼ (e1,e1k) ⟼ (e1t1,e1kt2) ⟼ (e1t1,e1t1t1-1kt2) ⟼ (e1t1,t1-1kt2)$ Thus we get an induced identification $Eu(2)/T2 ≃ Eu×TK/T [e1,e2] ⟼ [e1,kT] if e2=e1k$ Similarly, we have $Eu(3) ≃ Eu×K×K : (e1,e1k1,e1k1k2) ⟼ (e1,k1,k2)$ and $(e1t1,e1k1t2,e1k1k2t3) = ( e1t1, e1t1t1-1 k1t2, e1k1t2 t2-1k2t3 ) ⟼ ( e1t1, t1-1k1 t2,t2-1 k2t3 )$ so $Eu(3)/T3≃ (Eu×K×K)/T3$ where the ${T}^{3}$ action on ${E}_{u}×K×K$ is $(e1,k1,k2)· (t1,t2,t3)= ( e1t1, t1-1k1t2, t2-1k2t3 )$ But $(Eu×K×K)/T3 ≃Eu×T (K×TK/T)$ so we have the identifications $B(2) ≃ Eu×T K/T B(3) ≃ Eu×T (K×TK/T)$ Hence $H•(B(2)) ≃ HT(K/T) H•(B(3)) ≃ HT(K×TK/T) ≃ HT(K/T)⊗S HT(K/T) (from last time)$ where the last identification is due to the general fact we proved last time that for any $K\text{-space}$ $Y,$ $HT(K×TY) ≃ HT(K/T)⊗S HT(Y).$ We also have $H•(B(1))= H•(E/T)=S$ Therefore, the pull-backs on cohomology of all the structure maps for the groupoid ${B}^{\left(2\right)}$ over ${B}^{\left(1\right)}$ give the groupoid structure on $𝒰=\text{Spec} {H}^{T}\left(K/T\right)\text{.}$

### Summary

Set $R={H}^{T}\left(K/T\right),$ $S={H}^{T}\left(\text{pt}\right)=H\left({B}_{T}\right)={H}^{•}\left({B}^{\left(1\right)}\right)\text{.}$ Then from: $p1: B(2) ⟶ B(1): [e1,e2] ⟼ [e1] p2: B(2) ⟶ B(1): [e1,e2] ⟼ [e2] d: B(1) ⟶ B(2): [e] ⟼ [e,e] t: B(2) ⟶ B(2): [e1,e2] ⟼ [e2,e1] μ: B(3) ⟶ B(2): [e1,e2,e3] ⟼ [e1,e3]$ we get: $πL = p1*: S ⟶ R πR = p2*: S ⟶ R ε = d*: R ⟶ S c = t*: R ⟶ R Δ = μ*: R ⟶ R⊗SR$

Theorem: The above maps ${\pi }_{L},$ ${\pi }_{R},$ $\epsilon ,$ $c$ and $\Delta$ make $\left(𝒰=\text{Spec} R,\underset{_}{h}=\text{Spec} S\right)$ into a groupoid scheme. Moreover, if $X$ is any $K\text{-space,}$ the map $ΔX = { K×TX⟶X: [k,x]⟼kx } * : HT(X) ⟶ HT(K/T)⊗HT(X)$ is the composition of an action of $\left(𝒰,\underset{_}{h}\right)$ on $\text{Spec} {H}^{T}\left(X\right),$ (assuming that ${H}^{*}\left(X\right)$ is even)

### Characteristic operators

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

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

Fact: The set $\underset{_}{\stackrel{ˆ}{A}}$ of all characteristic operators is an $S\text{-algebra.}$

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

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

Definition-Notation: $Aˆ_c = the S-subalgebra of Aˆ_ of all characteristic operators of compact support.$

Proposition: For any characteristic operator $a$ and any $K\text{-space}$ $X,$ we have $ΔX∘a = (a⊗id)∘ΔX: HT(X) ⟶ HT(K/T)⊗SHT(X)$

Corollary 1: For a characteristic operator $a,$ we have $a=0 ⟺ a=0on HT(K/T) ⟺ ε∘a=0∈ HomS(HT(K/T),S).$ Proof. If $\epsilon \circ a=0:{H}^{T}\left(K/T\right)\to S,$ then for any $K\text{-space}$ $X,$ $aonHT(X) = (ε⊗id)∘ ΔX∘a (because (ε⊗id)∘ ΔX= idHT(X)) = (ε⊗id)∘ (a⊗id)⊗ΔX (by Proposition) = (ε∘a⊗id)∘ ΔX = 0.$ $\square$

Corollary 2: $\underset{_}{\stackrel{ˆ}{A}}$ has no $S\text{-torsion.}$ Proof. If $s\in S$ and $a\in \underset{_}{\stackrel{ˆ}{A}}$ are such that $sa=0anda≠0$ then for any $z\in {H}^{T}\left(K/T\right)$ $0=(ε∘sa)(z) = ε(s(a·z)) = sε(a·z)$ But since $a\ne 0,$ we know by Corollary 1 that $\epsilon \circ a\ne \text{?????}$ so $\exists z\ne 0$ s.t. $\epsilon \left(a·z\right)\ne 0\in S\text{.}$ Since $S$ is a polynomial algebra, it has no $S\text{-torsion.}$ Thus $S=0\text{.}$ This shows that $\underset{_}{\stackrel{ˆ}{A}}$ has no $S\text{-torsion.}$ $\square$

Corollary 3 (added by me) (of Corollary 1): The action of $a\in \underset{_}{\stackrel{ˆ}{A}}$ on ${H}^{T}\left(X\right)$ is expressed using $ΔX: HT(X) ⟶ HT(K/T)⊗SHT(X)$ and the map $\epsilon \circ a:{H}^{T}\left(K/T\right)\to S$ by $aonHT(X) = (ε∘a⊗id) ∘ΔX.$

Remark: Should think of $\underset{_}{\stackrel{ˆ}{A}}$ as the dual of ${H}^{T}\left(K/T\right)$ by $a↦\epsilon \circ a\in \text{Hom}\left({H}^{T}\left(K/T\right),S\right)\text{.}$

### Integration over the fibre

Assume that $P:E\to B$ is a fibration over a pathwise connected base $B$ with ${b}_{0}\in B\text{.}$ Let $F={P}^{-1}\left({b}_{0}\right)\text{.}$ Assume that this fibration is orientable. This means that the holonomy around ${b}_{0}$ acts trivially on ${H}^{*}\left(F\right)\text{.}$ Since $B$ is pathwise connected, the weak homotopy type of $F$ is independent of the choice of ${b}_{0}\text{.}$ Then we have, assuming ${H}^{\gamma }\left(F\right)=0$ for $\gamma >n$ $Homℤ(Hn(F),ℤ) ⟶ ( HomH*(B) (H*(E),H*(B)) degree -n )$ denoted by $τ ⟼ ∫τ$ obtained as follows by using the Serre spectral sequence: $Hm+n(E) ⟶ E∞m,n ≃ E2m,n ≃ Hm(B,Hn(F)) ⟶τ Hm(B,ℤ).$

Remark

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

### The $\underset{_}{A}\text{-action}$ on ${H}^{*}\left(E/T\right)$ for any principal $K\text{-bundle}$$E$

If $E$ is a principal $T\text{-bundle,}$ then we have a ring homomorphism $ch: S ⟶ Heven(E/T) : λ ⟼ c1(ℒ-λ=E×Tℂe-λ) ∈ H2(E/T).$ We call it the characteristic homomorphism. Using the characteristic homomorphism, we get an $S\text{-module}$ structure on ${H}^{*}\left(E/T\right):$ $s·z = ch(s)z.$

Now assume that $E$ is also a principal $K\text{-bundle,}$ so thus also a $T\text{-bundle.}$ Then we can use the $K\text{-action}$ to define the following $W\text{-action}$ on ${H}^{*}\left(E/T\right):$ for $w\in W,$ $w·z = w*z$ where $w:E/T\to E/T:$ $w·eT=ewT\text{.}$ Because of the following basic property of the characteristic map, $w*c1(ℒ-λ) = c1(w*ℒ-λ)= c1(ℒ-w·λ) ie.w*ch(λ) = ch(w·λ)$ we have, for any $w\in W$ and $s\in S$ $ws = (w·s) w$ as operators on ${H}^{*}\left(E/T\right)\text{.}$ Therefore we have an action of the smashed product algebra $ℂW⋉S$ on ${H}^{*}\left(E/T\right)\text{.}$

Now for each $i\in I,$ consider the fibre bundle $E/T ↓ πi E/Ki$ which has fibre ${K}_{i}/T\simeq {P}_{i}/B\simeq ℂ{P}^{1}$ so it has a preferred orientation ${\sigma }_{i}\in {\text{Hom}}_{ℤ}\left({H}^{*}\left({K}_{i}/T,ℤ\right)\right)$ namely the fundamental cycle. Integration over the fibre gives $H*(E/T) ⟶ H*-2(E/Ki) : z ⟼ ∫σiz$ Now define $Ai: H*(E/T) ⟶ H*-2(E/T) : Ai·z = πi*∫σiz.$

Proposition: For any $z\in {H}^{*}\left(E/T\right),$ $αi·(Ai·z) = z-ri·z (*)$ Proof. We will check this over $ℚ$ (Why?). The fibration ${\pi }_{i}:E/T\to E/{K}_{i}$ gives a ${H}^{*}\left(E/{K}_{i}\right)\text{-module}$ structure on ${H}^{*}\left(E/T\right)\text{.}$ Since the fibre is $\simeq ℂ{P}^{1},$ this is in fact a free ${H}^{*}\left(E/{K}_{i}\right)\text{-module,}$ a basis of which is given by $1$ and $\frac{1}{2}\text{ch}\left({\alpha }_{i}\right)\in {H}^{*}\left(E/T\right)\text{.}$ For ${z}_{0}\in {H}^{*}\left(E/T\right)$ we use the same letter to denote the pull back ${\pi }_{i}^{*}\text{?????}\in {H}^{*}\left(E/T\right)\text{.}$ We will check $\left(*\right)$ for $z={z}_{0}$ and $z=\frac{1}{2}\text{ch}\left({\alpha }_{i}\right){z}_{0}\text{.}$ Clearly ${A}_{i}·{z}_{0}=0$ and ${r}_{i}·{z}_{0}={z}_{0}\text{.}$ Thus $\left(*\right)$ holds for $z={z}_{0}\text{.}$ Now for $z=\frac{1}{2}\text{ch}\left({\alpha }_{i}\right){z}_{0},$ $αi·(Ai·z) = αi· ( Ai·(ch(αi)2) z0 ) .$ Lemma: ${A}_{i}·\text{ch}\left({\alpha }_{i}\right)=2\text{.}$ (a calculation over $ℂ{P}^{1}\text{)}$ Assume Lemma. Then $αi·(Ai·z)= αi·z0=ch (αi)z0.$ On the other hand, $z-ri·z0 = 12ch(αi)z0- ri·(12ch(αi) z0) = 12ch(αi)z0- ri·(12ch(αi)) ri·z = 12ch(αi)z0+ 12ch(αi) z0 = ch(αi)z0.$ Hence $\left(*\right)$ holds for $z=\frac{1}{2}\text{ch}\left({\alpha }_{i}\right){z}_{0}\text{.}$ It is strange to carry the $\frac{1}{2}$ around. Why necessary? $\square$

Therefore we have

Theorem: For any principal $K\text{-bundle}$ $E,$ the following define an $\underset{_}{A}\text{-action}$ on ${H}^{*}\left(E/T\right)\text{:}$ $s·z = ch(s)z w·z = w*z Ai·z = πi* ∫σiz$ Moreover, the characteristic morphism $ch: S ⟶ Heven(E/T): λ ⟼ c1(ℒ-λ)$ is an $\underset{_}{A}\text{-map,}$ where ${A}_{i}$ acts on $s\in S$ by $Ai·s = s-ri·s αi$ as before (see Lecture 2).

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

Example: If ${E}_{1}\stackrel{f}{\to }{E}_{2}$ is a $K\text{-map,}$ then ${f}^{*}:{H}^{*}\left({E}_{2}/T\right)\to {H}^{*}\left({E}_{1}/T\right)$ is clearly an $\underset{_}{A}\text{-map.}$

### $\underset{_}{A}\text{-action}$ on ${H}^{T}\left(X\right)$ for $K\text{-space}$$X\text{:}$

Example: Let $X$ be a $K\text{-space}$ and let ${E}_{u}={E}_{K}$ be the universal principal bundle of $K\text{.}$ Let $E = Eu×X$ with the $K\text{-action}$ given by $(e,x)·k = (ek-1,kx).$ Then $E/T = Eu×TX$ so get an action of $\underset{_}{A}$ on ${H}^{T}\left(X\right)\text{.}$ If $f:X\to Y$ is a $K\text{-map,}$ then $Eu×X ⟶ Eu×Y : (e,x) ⟼ (e,f(x))$ is a $K\text{-map,}$ so $f*: HT(Y) ⟶ HT(X)$ is an $\underset{_}{A}\text{-map.}$ Finally, the $\underset{_}{A}\text{-action}$ on ${H}^{T}\left(X\right)$ is clearly ${H}^{K}\left(X\right)\text{-linear.}$ Thus we can think of elements of $\underset{_}{A}$ as characteristic operators.

Property: For any $K\text{-space}$ $X,$ the morphism $S ⟶ HT(X) (=(X→pt)*)$ is an $\underset{_}{A}\text{-map.}$ Proof. This is the same as the characteristic morphism. ????? $\square$

Proposition: For any $K\text{-space}$ $X,$ the multiplication map $HT(X)⊗SHT(X) ⟶ HT(X)$ is an $\underset{_}{A}\text{-map.}$

### $T\text{-equivariant}$ homology

• For a $T\text{-space}$ $X,$ the $T\text{-equivariant}$ homology of $X$ is defined to be ${\text{Hom}}_{S}\left({H}^{T}\left(X\right),S\right)\text{.}$
• Suppose that $X$ is a $K\text{-space.}$ Then ${H}^{T}\left(X\right)$ is an $\underset{_}{A}\text{-module.}$ Since $S$ is also an $\underset{_}{A}\text{-module,}$ we know that ${\text{Hom}}_{S}\left({H}^{T}\left(X\right),S\right)$ is then also a $\underset{_}{A}\text{-module}$ (see Lecture 2): $(s·f)(z) = sf(z) (Ai·f)(z) = f(Ai·z)+Ai ·f(ri·z)= Ai·f(z)-ri ·f(αi·z) (w·f)(z) = w·f(w-1·z)$
• If $F:X\to Y$ is a $K\text{-map,}$ then we have shown that ${F}^{*}:{H}^{T}\left(Y\right)\to {H}^{T}\left(X\right)$ is an $\underset{_}{A}\text{-map.}$ Define $F*: HomS(HT(X),S) ⟶ HomS(HT(Y),S)$ by $\left({F}_{*}f\right)\left({z}_{Y}\right)=f\left({F}^{*}{z}_{Y}\right)\text{.}$ Then ${F}_{*}$ is an $\underset{_}{A}\text{-map}$ as well. Let's check ${F}_{*}\left({A}_{i}·f\right)={A}_{i}·\left({F}_{*}f\right)\text{.}$ So let $t\in {H}^{T}\left(Y\right),$ need to show $(Ai·f) (F*z) = f(αi·(F*f)) (z).$ Now $lhs = f(Ai·F*z)+ Ai·f(ri·F*z) rhs = (F*f)(Ai·z) +Ai·F*f(ri·z) = f(F*(Ai·z))+ Ai·f(F*(ri·z)).$ Since ${F}^{*}$ is an $\underset{_}{A}\text{-map,}$ we indeed have $\text{lhs}=rhs.$

Example: Suppose $Y$ is a $T\text{-space}$ such that ${H}^{r}\left(Y\right)=0$ for $r>n\text{.}$ Then integration over the fibre for $Y ⟶ Eu×TY ↓ Eu/T$ gives a map $Homℤ(Hn(Y),ℤ) ⟶ HomS(HT(Y),S) ⫙ ⫙ τ ⟼ ∫τ$ For each $i\in I,$ we have a map $Homℤ(Hn(Y),ℤ) ⟶ Homℤ(Hn+2(Ki×TY),ℤ) : τ ⟼ σi*τ$ where ${\sigma }_{i}*\tau \in {\text{Hom}}_{ℤ}\left({H}^{n+2}\left({K}_{i}{×}^{T}Y\right),ℤ\right)$ is the composition $Hn+2(Ki×TY) ⟶∫τ H2(Ki/T) ⟶σi ℤ$ using integration over the fibre first for the bundle $Y ⟶ Ki×TY ↓ Ki/T .$ Consequently we have a map $Homℤ(Hn(Y),ℤ) ⟶ Homℤ(Hn+2(Ki×TY),Z) ⟶ HomS(HT(Ki×TY),S) τ ⟼ σi*τ ⟼ ∫σi*τ.$ Now suppose that $X$ is a $K\text{-space}$ with $K\text{-action}$ $μ: K×X ⟶ X.$ Assume that $F:Y\to X$ is a $T\text{-equivariant}$ map. Then for $\tau \in {\text{Hom}}_{ℤ}\left({H}^{n}\left(Y\right),Z\right),$ we have ${\int }_{\tau }\in {\text{Hom}}_{S}\left({H}^{T}\left(Y\right),S\right),$ so $F*∫τ∈ HomS(HT(X),S)$ and thus $Ai·F*∫τ∈ HomS(HT(X),S).$ On the other hand, we have $Ki×TY ⟶Fi Ki×TX ⟶μ X [ki,y] ⟼ [ki,f(y)] ⟼ ki·f(y)$ and $∫σi*τ∈ HomS(HT(Ki×TY),S).$

Fact: $Ai·F*∫τ = μ*Fi* ∫σi*τ ∈HomS (HT(X),S).$ Proof. ? $\square$

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.