A ne Lie Algebras(Under Construction) · A ne Lie Algebras(Under Construction) Shuai Wang July 2016...

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Affine Lie Algebras(Under Construction) Shuai Wang July 2016 1 Introduction Preparation for my oral qualification, minor topic. Basic knowledge about affine Lie algebras. 2 concrete examples Example 2.1 ( ˆ sl 3 (C)). We can realize it as a loop algebra. g = sl 3 . E 1 := E 12 ,E 2 := E 23 , [E 1 ,E 2 ]= E 13 . ˆ sl 3 . ˆ sl 3 (C)=(C[t, t -1 ] sl 3 ) Cc Cd. [t i x + λc + μd, t j y + λ 0 c + μ 0 d]= t i+j [x, y]+ μjt j y - μ 0 it i x + δ i,-j < x, y > c. <, >. x, y sl 3 are matrices < x, y >= tr(xy). e i ,f i . e 1 =1 E 1 ,e 2 =1 E 2 ; f 1 =1 F 1 ,f 2 =1 F 2 . e 0 ,f 0 ,h 0 . e 0 = t [F 1 ,F 2 ]= t E 31 . f 0 = t -1 [E 1 ,E 2 ]= t -1 E 13 . h 0 =[e 0 ,f 0 ]= -1 (H 1 + H 2 )+ c = -1 (E 11 + E 33 )+ c H. H =1 H 0 Cc Cd. Note that c is just the central element c = h 0 + h 1 + h 2 . Π. α 1 = 1 = 2 1 (c)= α 1 (d)=0 α 2 = 1 = 2 1 (c)= α 1 (d)=0 θ = α 1 + α 2 = 1 - 3 (c)= θ(d)=0 δ : δ(1 H 0 )= δ(c)=0(d)=1. α 0 = -θ + δ. 1

Transcript of A ne Lie Algebras(Under Construction) · A ne Lie Algebras(Under Construction) Shuai Wang July 2016...

Page 1: A ne Lie Algebras(Under Construction) · A ne Lie Algebras(Under Construction) Shuai Wang July 2016 1 Introduction Preparation for my oral quali cation, minor topic. Basic knowledge

Affine Lie Algebras(Under Construction)

Shuai Wang

July 2016

1 Introduction

Preparation for my oral qualification, minor topic. Basic knowledge about affine Lie algebras.

2 concrete examples

Example 2.1 (sl3(C)). We can realize it as a loop algebra.

• g = sl3.E1 := E12, E2 := E23, [E1, E2] = E13.

• sl3.sl3(C) = (C[t, t−1]⊗ sl3)⊕ Cc⊕ Cd.

[ti ⊗ x+ λc+ µd, tj ⊗ y + λ′c+ µ′d] =

ti+j ⊗ [x, y] + µjtj ⊗ y − µ′iti ⊗ x+ δi,−j < x, y > c.

• <,>. x, y ∈ sl3 are matrices< x, y >= tr(xy).

• ei, fi.e1 = 1⊗ E1, e2 = 1⊗ E2; f1 = 1⊗ F1, f2 = 1⊗ F2.

• e0, f0, h0.e0 = t⊗ [F1, F2] = t⊗ E31.

f0 = t−1 ⊗ [E1, E2] = t−1 ⊗ E13.

h0 = [e0, f0] = −1⊗ (H1 +H2) + c = −1⊗ (E11 + E33) + c

• H.H = 1⊗H0 ⊕ Cc⊕ Cd.

Note that c is just the central element c = h0 + h1 + h2.

• Π.α1 = ε1 = ε2, α1(c) = α1(d) = 0

α2 = ε1 = ε2, α1(c) = α1(d) = 0

θ = α1 + α2 = ε1 − ε3, θ(c) = θ(d) = 0

δ : δ(1⊗H0) = δ(c) = 0, δ(d) = 1.

α0 = −θ + δ.

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• H∗ = H∨.H∗ = span{α0, α1, α2, γ}

whereγ(h0) = 1, γ(h1) = γ(h2) = γ(d) = 0, γ(c) = 1.

• Weight decomposition.sl3 = H ⊕(k,α)6=(0,0) (tk ⊗ gα)

= (1⊗H0 ⊕ Cc⊕ Cd)H ⊕(k,α)6=(0,0) tk ⊗ gα

[1⊗ x+ λc+ µd, tk ⊗ gα] = α(x)tk ⊗ gα + kµtk ⊗ gα = (α+ kδ)(x)tk ⊗ gα.

Thus we know tk ⊗ gα is exactly the weight space of α+ kδ, and the multiplicity is given by

– if α 6= 0, dim(Lα+kδ) = 1.

– if α = 0, k 6= 0, dim(Lkδ) = dim(tk ⊗ H0) = 2, in general for a untwisted affine Kac-Moodyalgebra, the dimension is l = rank(A) = dim(H0) = dim(H)− 2, it’s one less than the size of theCartan matrix.

– ρ, ρ0.ρ0 = α1 + α2.

ρ = ρ0 + h∨γ = α1 + α2 + 3γ.

– Coxeter number and dual Coxeter number.

h = a0 + a1 + a2 = 3

h∨ = c0 + c1 + c2 = 3.

– Weyl group= t(M∗)W 0

– the lattice M∗

M∗ = Zα1 ⊕ Zα2.

Example 2.2 (Macdonald’s identity for sl2(C)). First recall Kac’s denominator formula

Πα∈Φ+(1− e−α)mα =∑w∈W

ε(w)ew(ρ)−ρ

– root system Φ.Φ = {α+ nδ|α ∈ Φ0, n ∈ Z} ∪ {nδ|n ∈ Z}

– positive roots.Φ+ = {α+ nδ|n > 0, α ∈ Φ0} ∪ {(Φ0)+} ∩ {nδ|n > 0}

– LHS.Πα∈(Φ0)+(1− e−α)Πn>0(1− e−nδ)lΠα∈Φ0(1− e−α−nδ).

– RHS.

(Πα∈(Φ0)+(1− e−α))(∑α∈M∗

χ0(h∨α)e−c(h∨α)

2h∨ δ)

wherec(h∨α) =< h∨ + ρ0, h∨α+ ρ0 > − < ρ0 + α0 > .

– Let q = e−δ, we have

Πn>0{(1− qn)lΠα∈Φ0(1− qne−α)} =∑α∈M∗

χ0(h∨α)qc(h∨α)

2h∨

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Now for sl2 we have

– the Cartan matrix 2 −2

−2 2

– mα+nδ = 1,mnδ = l = 1.

– Φ0 = {α1,−α1}.– M∗ = Zα1

– ρ0 = 12α1.

– h∨ = 2, let α = nα1, we have

c(h∨α) =< 2nα1 +1

2α1, 2nα1 +

1

2α1 > − <

1

2α1,

1

2α1 >

= 2(4n2 + 2n)⇒ c(h∨α)

2h∨= 2n2 + n.

χ0(h∨α) =e2nα1 − e−(2n+1)α1

1− e−α1

let z = e−α1 , we have

χ0(h∨α) =z−2n − z2n+1

1− z.

– So we have

Πn>0(1− qn)(1− qnz)(1− qnz−1) =∑n∈Z

z−2n − z2n+1

1− zq2n2+n

this can be written as

Πn>0(1− qn)(1− qnz)(1− qnz−1) =1

1− z∑n∈Z

(−1)nznqn(n−1)

2 .

It’s the classical Jacobi’s triple identity. But if you only want to prove this identity, it’s elementary.

3 E6

We can realize the Weyl group of E6 as the configuration group of 27 lines on a smooth cubic surface in P3.

Proposition 3.1. If l(wsα) > l(w), then w(α) is positive.

Proof. We prove by induction on l(w). If l(w) = 0, then nothing to prove. If l(w) > 0, then pick an s suchthat l(w) > l(sw),

l(swsα) ≥ l(wsα)− 1 > l(w)− 1 = l(sw)

Then by induction for sw, we have sw(α) is positive. If w(α) is negative, since s only change thepositivity of exactly one root, say β, then we must have w(α) = −β ⇒ sw(α) = β. From this we knowswsα(sw)−1 = s(sα is a simple reflection, then so is its conjugate, we know all reflections are of the form si,and it’s obvious β is the new root with eigenvalue −1), in other words wsα = sw, this contradict the factthat l(wsα) > l(w) > l(sw).

Simlilarly, if we have l(wsα) < l(w), then w(α) is negative. Simply let u = wsα, then l(wsα) > l(u), thusu(α) = wsα(α) = −w(α) is positive, that is w(α) is negative.

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Proposition 3.2 (Length and eigenvector). Assume l(wsα) < l(w), then

wsα = sαw ⇔ ρw(α) = −α

Proof. From last proposition, we know that w(α) is negative. If w(α) = −α, then wsαw−1 = sα by similar

argument as above. That is wsα = sαw.On the other hand, if wsαw

−1 = sα, since we know if w(α) = β, then wsαw−1 = sβ , thus we must have

w(α) = α or −α, we already know w(α) is negative, thus w(α) = −α.

Proposition 3.3 (Divisors on G/B). The U -orbit of w ∈ W in G/B has codimension 1 if and only ifw = w0sα for some simple root α.

Proof. The stablizer of [wB] isU ∩ wUw−1.

Since this is a T -stable set, we know it’s a union of some one dimensional unipotent subgroups Uλ. Let E(w) ={γ ∈ Φ−|w−1(w) ∈ Φ−}. Then we only need to check when this set has cardinality 1. E(w0sα) = w0(α), onthe other hand if E(w) = w0(α), then w−1w0sα(β) ∈ Φ+ for any β ∈ Φ+, which means w−1w0sα = id.

This result is useful, we can prove the Picard group of the flag variety G/B is generated by Uw0sαB, it’sfinitely generated and has the same rank as the group G itself. Let Dα = Uw0sαB, we have the followingproposition

Proposition 3.4 (Pic(G/B) ∼= Zrank(G) = Z|Φ+|).

{Dα|αsimple root}

is a basis of Pic(G/B). The divisor class D =∑nαDα is effective(i.e contains a positive divisor) if and only

if nα ≥ 0 for all α.

Proof. The first few statements are quite obvious form the following sequence

⊕αZ[Dα]→ Pic(G/B)→ Pic(Uw0B)→ 0.

The last term is 0 since Uw0B is affine. On the otherhand, if∑nαDα ≡ 0, then we can find a f ∈ K(G/B),

f has no poles and zeros in Uw0B, but this is the top cell which is dense in G/B, then we know f is aconstant. Then for any divisor class D if D is invariant under the U action, then it must be of the form∑nαDα, the reason is

D = D′ + div(f);D′ =∑

nαDα.

Both D and D′ are U invariant, then so is div(f), and since U is unipotent, we know f is actually invariantunder the U -action. f is thus a constant on the top cell, f is constant. Then U acts on the linear system ofall positive divisors in the same divisor class as D(it’s nonempty by our assumption), which is a projectivespace, thus by Borel fixed point theorem, we must have a U -invariant divisor in it, {Dα} is a basis, so thisinvariant must be

∑α nαDα itself, then we know nα ≥ 0.

Remark (Pic(SL(2)) = 0). The same decomposition still works, but this time the left hand side are severalcopies of B, they’re affine(isomorphic to (A1 − {0})× A1), then Pic(SL(2))− 0, same argument works forall SL(n).

Proposition 3.5 (Construction of involutions). Any involution w ∈ W can be obtained starting from theinvolution id by a sequence of length-increasing operations that are

• either multiplication of an involution by a simple reflection sα with which it commutes

• or conjugation by a simple reflection sα with which it does not commute.

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Proof. We prove it by induction on the length l(w).

• If l(w) = 0, then w = id, nothing to prove.

• If l(w) > 0, then we can find a simple reflection sα(i.e α is a simple root), such that l(wsα) < l(w)

– If sα commutes with w, then wsαwsα = w2 = id, thus wsα is an involution with smaller length.w = wsαsα.

– If sα does not commute with w, then we have l(wsα) = l(w)− 2. This is because l(wsα) < l(w),thus we can find a reduced expression of w that ends with sα(This is simply because consider areduced expression of wsα = s1 . . . sr, r < l(w), then w = s1 . . . srsα since l(w) ≥ r+ 1, this mustbe a reduced expression of w). Thus w = w−1 = sαsr . . . s1. Since l(wsα) < l(w), we can usethe Strong exchanging condition, which says wsα = sα . . . si . . . sr, but in our situation wsα 6=sαs1 . . . sr = s1 . . . sr, otherwise sαw = wsα, reduced to our first situation. Thus l(sαwsα) <1 + l(w)− 2 < l(w), and sαwsα is an involution. We’re done.

Corollary 3.5.1. Any involution element in a Weyl group W can be written as a product of commutingreflections in W

Proof. First find an sα such that l(wsα) < l(w),

• If we have w(α) = −α, by our previous lemma, we have sαw = wsα, so wsα is also an involutionwith smaller length, by induction, it can be written into a product of commuting reflections sβ1

. . . sβk .Consider the inner product

< α, βi >= − < α,wsαβi >= − < sαα,wβi >= − < −α,−βi >= − < α, βi >

So we have < α, βi >= 0,∀i, thus w = sβ1 . . . sβksα is also a product of commuting reflections.

• If wsα 6= sαw, then by the construction of involutions, l(sαwsα) < l(w) and sαwsα is also an involution,thus sαwsα = sβ1

. . . sβk for orthogonal {βi}, then so are {s′i = sαsβisα} and we have w = s′1 . . . s′k,

we’re done.

Proposition 3.6 (Conjugate orthogonal roots in E6). In the root system of E6, we can find at most 4pairwisely orthogonal roots. Actually, we can prove a nicer result: there’re 5 conjugacy classes of involutionsin the Weyl group of E6

1;R1;R1R2;R1R2R3;R1R2R3R4

Where Ri = sβi , and {β1, β2, β3, β4} are 4 orthogonal roots.

Proof. Simply because we have a filtration of the root system of E6:

Φ(E6) ⊃ Φ(A5) ⊃ Φ(A3) ⊃ Φ(A1)

One way to see this is to compute the number of roots and the order of the group.

The interesting thing here is that when we want to consider real algebraic geometry in some situations,then we always have to consider the Gal(C/R) action, which is an involution. Then we can find a way toconsider real solutions to our questions.

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4 A classification of generalized cartan matrices

We shall prove that there’re only three types of generalized Cartan matrices:

• Finite type:Al, Bl, Cl, Dl, G2, F4, E6, E7, E8

• Affine type:

–A1

1, A(1)l (l ≥ 2), B

(1)l (l ≥ 3), C

(1)l (l ≥ 2), D

(1)l (l ≥ 4)

G(1)2 , F

(1)4 , E

(1)6 , E

(1)7 , E

(1)8

–A

(2)2 , A

(2)2l (l ≥ 2), A

(2)2l+1(l ≥ 2), D

(2)l+1(l ≥ 2), E

(2)6

–D

(3)4

5 Basic knowledge about Fourier analysis

Theorem 5.1. Suppose f is an integrable function on S1, and ∀n ∈ Zf(n) = 0, then f = 0 at points wheref is continuous.

Proof. Without losing of generality, we may assume f is a real function defined on [−π, π] bounded by M ,

and f(0) > 0, we may choose α > 0 such that f(θ) > f(0)2 ∀|θ| < α. We need a sequence of trigonometric

polynomials pk(θ) such that pk → δ in some sense, we also like them to have similar properties as compactsupported functions actually we only need |pk(θ)| < (1 − ε

2 ), for our purpose we also like it to be greaterthan 1 near 0. Let’s consider

pk(θ) = (ε+ cosθ)k

We may choose ε small enough and a positive number β < α such that p1 ≥ 1+ ε2 ,∀θ < β and p1 < 1− ε

2 .Then

∀n ∈ Z, f(n) = 0→∫f(θ)pk(θ)dθ = 0

However we also have

|∫|θ|>α

fpk| ≤ 2πM(1− ε

2)k → 0

∫β<|θ|<α

f(θ)pk(θ)dθ > 0

∫|θ|<α

f(θ)pk(θ)dθ ≥ 2αf(0)

2(1 +

ε

2)k → +∞

Which is a contradiction.

Corollary 5.1.1 (Uniqueness of Fourier series). If f is a continuous on the circle and f(n) = 0 for alln ∈ Z, then f = 0

Corollary 5.1.2. If f is a continuous function on S1. Suppose that∑n∈Z |f(n)| < +∞. Then the Fourier

series converges to f uniformly on S1.

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Proof. Since∑n∈Z |f(n)| < +∞,

∑n∈Z f(n)einθ univormli converges to a continuous function g. And by

the same reason,∑

commutes with∫

, thus g(n) = f(n). The result follows from the previous corollary.

Proposition 5.2. Suppose that f is a twice continuous differentiable function on S1, then

f(n) = O(1

|n|2)

Proof.

2π =

∫ 2π

0

f(θ)e−inθdθ

= f(θ)e−inθ|2πθ=0 − (−1

in)

∫ 2π

0

e−inθf ′(θ)

=1

−n2

∫ 2π

0

f ′′(θ)e−inθ

Since f ′′e−inθ is bounded, the result follows.

Let Hn denote the Hilbert space L2(S1,Cn) and GL(H), B(H) denote all invertible bounded operators inH and all bounded operators in H respectively.

6 Basic knowledge about Hilbert space and Banach space

Proposition 6.1. If |A| < 1, then I −A ∈ GL(H).

Proof. First we claim ∀y ∈ H,∃!x ∈ H s.t (I − A)x = y, which means we can define (I − A)−1. This isbecause the operator A+ yI for a fixed y has a unique fixed point by the contraction mapping theorem.

And it’s bounded simply because |x| ≤ |y|+ |A||x|, that is |x| ≤ 11−|A| |y|. We conclude that (I −A)−1 ≤

11−|A| .

Corollary 6.1.1. GL(H) is open in B(H).

Proof. If A ∈ GL(H), B = A + (B − A) = A−1(I + A−1(B − A)), thus we know if |B − A| < 1 − |A|,I +A−1(B −A) ∈ GL(H)⇒ B ∈ GL(H).

7 Compact operators

8 Hilbert-Schmidt Operators

Let H be a separable Hilbert space with a orthogonal basis {ei}∞i=0.

Definition 8.1. T is called a Hilbert-Schmidt operator if∑∞i=0 |Tei|2 < +∞. And the Hilbert-Schmidt

norm of such T is defined to be |T |2 := (∑|Tei|2)

12 .

Since the Hilbert-Schmidt operators form a two-sided ideal in B(H), then by [J,A−1] = −A−1[J,A]A−1,we conclude that GLres(H) is a group. Let H = H+ ⊕H− be a Hilbert-Schmidt operator with a polariza-tion(i.e a decomposition as the orthogonal sum of two closed subspaces).

Definition 8.2 (The restricted general linear group of Hilbert space). GLres(H) is the subgroup of GL(H)consisting of operators such that [J,A] is a Hilbert-Schmidt operator.

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Actually, we can compute [J,A], we can write A as a blocked matrix w.r.t the polarization

A =

a b

c d

, [J,A] =

0 2b

−2c 0

That is A is a Hilbert-Schmidt operator ⇔ b, c are Hilbert-Schmidt operators.There’s also another formulation of GLres(H), first define BJ(H) which is the Banach algebra of all

bounded operators A : H → H such that [J,A] is a Hilbert-Schmidt. And the norm on it is defined to be

|A|J := |A|+ |[J,A]|2

Then GLres(H) can be viewed as the group of units in BJ(H).

9 Fredholm Operators

Finite dimensional kernels and cokernels.

10 Induced representations

Example 10.1 (V ect(S1)⊗C). Actually V ect(S1) can be identified with Maps(S1,R), where we view R asthe lie algebra of S1. Then we can apply the Peter-Weyl theorem which tells us

V ect(S1)⊗ C ∼= ⊕k∈ZVkWhere Vk denotes the one dimensional representation S1 → GL(1,C), eiθ 7→ eikθ

The Peter-Weyl theorem tells us what does C(G/H) looks like, we can also consider a closely relatedspace V ect(G/H) the tangent vector field on X = G/H. Let T be the tangent space to X at its base-point[H].(If we have a Riemannian structure on G, then T can be identified with the orthogonal complement ofTeH in TeG)

Proposition 10.2. Tangent vector field ξ on X = G/H can be identified with H-equivariant maps ξ : G→T , in the sense that

ξ(gh−1) = hξ(g)

In other words, we haveV ect(X) ∼= MapsH(G;T )

And the corresonding G-action on MapsH(G,T ) is given by g • ξ(x) = ξ(xg−1)

Proof. Since ξ(gH) ∈ TgH(G/H)⇒ g−1ξ(gH) ∈ T , we can define

ξ : G→ T

g 7→ g−1ξ(gH)

It’s obvious that ξ(gh−1) = hg−1ξ(gh−1H) = hg−1ξ(gH) = hξ(g).

Conversely, if we have ξ ∈MapsH(G;T ), we can define ξ(gH) = gξ(g).

Then consider the G-action on V ect(X), we have (gξ)(xH) = gξ(g−1xH), then we know (gξ)(x) =

x−1gξ(g−1xH) = ξ(g−1x).

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Remark. Here we want to say a little more,

(gξ)(x) = ξ(g−1x)

is indeed a G-action, since ((g1g2)ξ)(x) = ξ(g−12 g−1

1 x) = (g2(g1ξ))(x). But also note in many cases, whenwe want to consider left actions, we always use left multiplication directly, and right multiplication by theinverse if we want to consider the right action.

It’s quite interesting that many theories are not so hard to understand, but we have to spend quite a lotof time to understand the exact meaning of each terminology.

In general, we haveV ect(X)⊗ C = MapsH(G;TC)

Now, we can apply the Peter-Weyl theorem

MapsH(G;TC) = {C(G)⊗ TC}H = ⊕P (P ⊗ (P ⊗ TC)H)

Where the direct sum is over all irreducible representations of G.The first identity simply says that Maps(G;TC) is n = dim(TC) copies of C(G)

Remark. To understand the several identities above, we have to look closer to different types of actions.

• right H-action on Ghy g = gh−1

• H-action on C(G) induced by the right H-actio on G.

(hy f)(g) = f(h−1 y g) = f(gh)

• H-action on C(G)⊗ TC induced by the H-action on G/H:

hy gH = hgH

• H ↪→ G-action on MapH(G;TC)

(g y f)(x) = f(g−1x)

Then it’s a natural action form MapH(G;TC) to MapH(G;TC).

• Defining property of MapH(G;TC)

f(gh−1) = hy f(g)

More precisely, MapH(G;TC) are those H-equivariant maps from G to TC, since f(gh−1) = f(hy g)

• right/left G-action on G

left actionxy g = xg

right actionxy g = xg−1

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• right/left G-action on C(G) induced by the right/left G-action on G.

left translation(xy f)(g) = f(x−1 y g) = f(x−1g)

right translation(xy f)(g) = f(x−1 y g) = f(gx)

• G×G action on Calg(G). Where the first copy of G acts as left translation and the second copy actsas right translation.

• G ↪→ G×G-action on Calg(G), similar.

• G×G-action on P ⊗ P , where P is an irreducible representation of G, natural action.

• right H-action on Calg(G), which just means the right translation on functions.

• right H-action on P ⊗ P , natural action.

Then we want to understand the following

• There’s an isomorphism of representations of G×G given by

⊕P ⊗ P → Calg(G)

η ⊗ ξ → fP,ξ,η

Here I only want to show that the right translation of the second copy of G is related to the G-representation on P . The right translation by x is given by

(xy f)(g) = f(gx) =< η, gxξ >= fP,xξ,η(g)

• C(G/H) is just the part of C(G) which is invariant under the right translation of H, and

C(G/H) ∼= ⊕PP ⊗ P

The first statement just says that f ∈ C(G/H) ⇔ ∀g, f(gH) = f(g), that means f is invariant underthe right translation by H. We already know that the right translation is represented by the H actionon P , thus we have

Cfin(G/H) = ⊕PP ⊗ PH

Therefore

C(G/H) ∼= ⊕PP ⊗ PH

•MapH(G;T ) ∼= (C(G)⊗ T )H

Here H acts on C(G) by right translation. We have a natural correspondence ξ ∈ Map(G,T ) ⇔∑fi ⊗ ei, where fi is just the ith coordinates of ξ.∑

fi ⊗ ei ∈ (C(G)⊗ T )H ⇔∑fi(xh)⊗ h(ei) =

∑fi(x)⊗ ei ⇔

∑fi(x)⊗ h(ei) =

∑fi(xh

−1)⊗ ei ⇔

h(ξ(x)) = ξ(xh−1)⇔ ξ ∈MapH(G;T )

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•(C(G)⊗ T )H = ⊕PP ⊗ (P ⊗ T )H

This is straightforward from the fact that H acts on C(G) by right translation, and right translationsis represented by the representation on P .

•(P ⊗ T )H ∼= HomH(P, T )

For unitary representations we have P = P ∗, since At = A−1 ⇒ A = (At)−1. Thus we have

P ⊗ T ∼= P ∗ ⊗ T ∼= Hom(P, T )

Recall that the H-action on f ∈ P ∗ is given by

< hy f, x >=< f, h−1 y x >

Then we knowf =

∑fi ⊗ ei ∈ (P ∗ ⊗ T )H ⇔∑

(hy fi ⊗ h(ei))(x) = (∑

fi ⊗ ei)(x)⇔∑fi(h

−1x)⊗ h(ei) =∑

fi(x)⊗ ei ⇔∑

fi(x)⊗ h(ei) =∑

fi(hx)⊗ ei ⇔

h(f(x)) = h(∑

fi(x)ei) =∑

fi(hx)⊗ ei = f(hx)

That is f is an intertwining operator between P and T as H-modules.

Example 10.3 (Representations of real orthogonal groups).

Example 10.4 (V ect(S2)⊗C). We can view S2 as O3/O2, and then T ∼= R2, we already know all irreduciblerepresentations fo O3 are given by spherical harmonic polynomials Hk. So we only need to consider (Hk ⊗C2)O2 .

Actually, we havedim(Hk ⊗ C2)O2 = dimO2(C, Hk ⊗ C2)

= 2dimO2(C2, Hk) = 2, k ≥ 1

The unique copy of C2 is given by xk−1y ⊕ xk−1z, this is meaningful only for k ≥ 1.To get more feeling about this argument, we can compute (P1⊗C2)O2 which turns out to be a 2 dimensional

subspace generated by(something like, check it!) a(y ⊗ e1 + z ⊗ e2) + b(y ⊗ e2 + z ⊗ e1).

V ect(S2)⊗ C ∼= ⊕k≥1(Hk ⊗Hk)

We can also interpret the disappearance of H0 by the fact that it’s O3 invariant, however, there’s no O3

invariant vector field on S2.

Remark (Characters). For a representation ρ : G → GL(V ), we define χV : G → C, χV (g) = tr(ρ(g)), wealso define x : T → C, actually in general χV is a polynomial in the representation ring R(U(G))of G. Justconsider the trivial example SO(2) ↪→ GL(2,C), then χV = x+ x−1, where x : SO(2)→ C; eiθ 7→ eiθ.

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11 Loop groups as groups of operators in Hilbert space

11.1 LGL(n,C)→ GLres(H(n))

Proposition 11.1. If γ : S1 → GL(b,C) is smooth, then the multiplication operator Mγ ∈ GLres(H(n)).

Proof. We only need to consider the case n = 1.Then to prove H(1)+ → H

(1)− , H

(1)− → H

(1)+ are Hilbert-

Schmidt, we only need to check that∑

(|k| + 1)|γk|2 < +∞, this follows from the basic fact that f(n) =O(|n|2) for a twice differentiable function f , not to say a smooth function.

12 Birkhoff factorization

Theorem 12.1 (Birkhoff factorization). ∀γ ∈ LGL(n,C) can be factorized

γ = γ−λγ+

Where γ− ∈ L−GL(n,C), γ+ ∈ L+GL(n,C), and λ is a loop which is a homomorphism from S1 to thediagonal matrices in GL(n,C). i.e λ is of the form

z 7→

zk1

· · ·

zkn

We have an interesting application of the Birkhoff factorization, we know in algebraic geometry that any

vector bundle on CP 1 is a direct sum of line bundles. This is also true for holomorphic vector bundles overS2.This was first pointed out by Grothendieck.

Theorem 12.2. Any holomorphic vector bundle E on S2 is isomorphic to a su Lk1 ⊕ Lk2 · · · ⊕ Lkn , whereki ∈ Z+ and L is the canonical line bundle on S2, which is defined by the cluthing function S1 → GL(1,C) =C× : z 7→ z.

Proof. Since U0 = S2 − {∞} and U∞ = S2 − {0} both are separable by holomorphic functions, they’reStein manifolds, by Cartan’s theorem B holomorphic bundles on them are all trivial. So E is obtained byattaching U0 × Cn and U∞ × Cn via the clunthing function

γ : U0 ∩ U∞ → GL(n,C)

Then we can apply the Birkhoff factorization, we can factorize γ as γ−λγ+, since γ−, γ+ can be viewed asinvertible bundle maps on U∞×Cn and U0×Cn respectively. Thus E is isomorphic to a holomorphic bundlewhich is defined by λ. That is

E ∼= Lk1 ⊕ · · · ⊕ Lkn

13 Borel-Weil Theory

One reason why we want to consider several homogeneous spaces of compact groups ”arises from their rolein constructing the irreducible representations”. Actually in Borel-Weil theory we only need G/T ∼= GC/B

+.

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• Every character (root) α : T → T extends uniquely to a holomorphic homomorphism α : B+ → C× =GL(1,C)

• Then on X = GC/B+ we can construct a line bundle Lα = GC ×B+ Cα. i.e GC × C/ < (gb, ξ) ∼

(g, α(b)ξ) >

• GC acts on Lα (by left multiplication?) and thus acts on H0(Lα) = Γ(X,Lα)

Theorem 13.1 (Borel-Weil). If α is not an antidominant weight then H0(X,Lα) = 0. If α is an antidom-inant weight then H0(X,Lα) is an irreducible representation of lowest weight α

We first discuss the relation between global sections on holomorphic line bundles on X and holomorphicmaps from X to a projective space.

• If we have a holomorphic map f : X → P (V ), then the pull-back of the tautological line bundle onP (V ) is a holomorphic line bundle on X. We denote it by Lf . And it can be viewed as a subspace ofX × V . φ : V → C defines a glbal section of L∗f , i.e we have a map V ∗ → Γ(L∗f )

• If L is base-point free, i.e for every x ∈ X, we can find a global section of L which does not vanish atx.Let Γ = Γ(X,L) Then we can construct a map fL : X → P (Γ∗), for x ∈ X Γ∗ 3 fL(x) : σ 7→ σ(x).

Then given an irreducible representation V of GC, to view it as Γ(X,L) for some L, we only need tofind a holomorphic map f : GC/B

+ → P (V ∗), i.e if we can find a B+ invariant ray in V ∗, we can just takeout the highest weight vector in V ∗. By our previous argument, V is just the dual of V ∗, we have a mapV → Γ(L∗f )

We know Λk(Cn) is an irreducible representation of Un. First consider the tautological bundleS onGr(k, V ), by the splitting principal c1(S) = c1(Det) = −σ1, thus the determinat bundle Det does not haveany nonzero global section. Instead,let’s consider Det∗.

Example 13.2 (Γ(Gr(k, V ), Det∗) ∼= Λk(V ∗)).

α : Λk(V ∗)→ Γ(Det∗)

α1 ∧ · · · ∧ αk → {v1 ∧ · · · ∧ vk 7→ det(< αi, vj >)}

It’s quite straightforward that α is injective. To prove it’s surjective, we first prove that (Λ(V ))∗ →Γ(Det∗) is surjective. (Before this, we actually need some holomorphic argument to make sure ∀s ∈ Γ(Det∗)is anticommutative and multilinear)

U ⊂ V k consistig of k-tuples of linear independent vectors in V , then the composition of the natural mapU → Λk(V ) with s. First fix v2, ,vk then the resulting function is linear and defined on the complement of W =span < v2, . . . , vk >, we may assume codim(W ) ≥ 2(otherwise Gr(k, V ) is trivial), by the Hartogs’s theoremit can be extends to be a linear function on V ( Here we still need the holomorphic property of the map). Dothe same thing for other variables, then we get a multilinear map V k → C, since it’s anticommutative on asubset which is not discret, we conclude it actually comes from a linear map Λk(V )→ C.

Finally Λk(V ∗) ∼= (Λk(V ))∗ is given by det(< αi, vj >).

14 Spin representations of so(Q)

so(Q) ∼= Λ2V → Cliffeven(Q)→ End(ΛevenW )⊕ End(ΛoddW )

Actually, we could pretend that we don’t know anything about the Clifford algebra, we only need toremember, the spin representations come from some natural actions of Λ2V on two isotropic spaces w.r.t thequadratic form Q.

Λ2V → so(V,Q)

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a ∧ b→ (φa∧b : v 7→ 2(Q(b, v)a−Q(a, v)b))

Λ2V → Cliffeven(Q)

φa∧b 7→1

2(a • b− b • a) = a • b−Q(a, b)

For dimV = 2kl : V ∼= W ⊕W ′ → End(Λ•W )

W 3 w 7→ (Lw : α 7→ w ∧ α)

W ′ 3 w′ 7→ (D2w′∗ : α1 ∧ · · · ∧ αn 7→∑i

(−1)i+1(2Q(w′, αi)α1 ∧ · · · ∧ αi · · · ∧ αn))

Then we can check thatl2(w) = l2(w′) = 0

l(w) ◦ l(w′) + l(w′) ◦ l(w) = 2Q(w,w′)id

Then l can be naturally extended to be a morphism l : Cliff(Q) → End(∧•W ) by counting dimensions.Since Cliffeven(Q) preserves the decomposition wedgeW ∼= ∧evenW ⊕ ∧oddW , we also have

Cliffeven(Q) ∼= End(∧evenW )⊕ End(∧oddW )

Then it’s straightforward to check that(remember we’re now working under the standard antidiagonalquadratic form instead of the one given by the identity matrix)

h 3 Hi = Ei,i − En+i,n+i 7→1

2ei ∧ en+i 7→

1

2(ei • en+i − 1) 7→ 1

2(Lei ◦D2e∗n+i

− id)

Then we get1

2(Lei ◦D2e∗n+i

− id)(eI) = eI , i ∈ I

1

2(Lei ◦D2e∗n+i

− id)(eI) = 0, i /∈ I

Thus eI is a weight space with weight 12 (∑i∈I Li−

∑i/∈I Li), since the Weyl group of so(2k) is (Z/2Z)koSk,

we know all even I’s are conjugate,similar for odd ones, then ∧evenW and ∧oddW are two irreduciblerepresentations of so(2k) with highest weights

α =1

2(L1 + · · ·+ Ln), β =

1

2(L1 + · · ·+ Ln−1 − Ln)

And to be more precise, we have

n = 2k,∧evenW ∼= Γα,∧oddW ∼= Γβ

n = 2k + 1,∧evenW ∼= Γβ ,∧oddW ∼= Γα

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15 Borel-Weil-Bott Theory

15.1 Bruhat decomposition

G/B = tw∈WBwB

Note that the decompositions seems quite symmetric, but actually, every orbit is different from the others.To understand this decomposition more, we adapt the following description.

Theorem 15.1 (Bruhat decomposition for GL(n,C)). For any g ∈ GL(n,C), we have

g = nπb

where n is upper-trianglular and with 1’s on the diagonal, π is a permutation matrix, b is a upper triangularmatrix.

Remark. This decomposition is not unique, for example

1 ∗ ∗

1 ∗

1

=

1 ∗

1 ∗

1

1 ∗

1

1

This is the reason why we want to consider reduced echelon form. That is a matrix w looks like

∗ ∗ 1

∗ 1

1

1

which is characterized by the following two properties

• the jth column ends in a 1 in the πthi row.

• wik = 0, k > πj

Then given any g ∈ GL(n,C), we have a unique algorithm to get an echelon form by multiplying anupper triangular matrix on the right(quite similar as the Gram-Schmidt process). Then the sequence π =(π1, . . . , πn) is necessarliy a permutation of {1, 2, . . . , n}. We thus get the decomposition g = nπb, and thetwo properties can be reformulated as

• wπ−1 ∈ U , U denotes the subgroup of lower triangular subgroup with 1’s on the diagonal.

• π−1w ∈ U−, U− denotes the subgroup of lower triangular matrices with 1’s on the diagonal.

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In other words, our algorithm produce a unique n = wπ−1 which belongs to N ∩ πU−π−1. We can alsoview this decomposition as the orbit decomposition of the U -action on the homogeneous space GL(n,C)/B,not that π1B and π2B are in different orbit if π1 6= π2, simply because if n1π1b1 = n2π2b2, then we haveA = π1b = nπ2, actually, π1, π2 are determined by the positions of the last 1′s in every column in A, wethus have π1 = π2. This decomposition also tells us in every orbit, we have an element like πb. We concludeU acts on GL(n,C)/B by left multiplication, we can get |Sn| orbits, the orbit correponding to π ∈ Sn ishomeomorphic to Uπ = U ∩ πU−π−1. Note that the number of entries in the matrix w is the number ofpairs πi > πj , i < j. We define this number to be lπ, the number of crossing of π. For example, forπ = (n, n− 1, . . . , 1)

1 2 . . . . . . n

1 2 . . . . . . 1

lπ = n(n−1)2 , by dimension argument, we know almost all elements are in this orbit, and in this case, if

g = nπb, we naturally have nπ ∈ πU−, so it’s fair to say, for almost all g ∈ GL(n,C), the decompositiong = nπb is unique.

Example 15.2 (GLn(Fq)). One the one hand, we know

|GL(GLn(Fq))| = (qn − 1)(qn − q) . . . (qn − qn−1)

|B| = (q − 1)nqn(n−1)

2

We get

|GLn(Fq)| = Πnk=1

qk − 1

q − 1

On the other hand, we have the Bruhat decomposition, thus, we get an identity

Πnk=1

qk − 1

q − 1=

∑π∈Sn

qlπ

Remark. Given the Brihat decomposition of GC/B ∼= G/T , for a compact connected Lie group G, we caneasily prove the fact that any element g ∈ G is conjugate to an elment in T . Consider

fg : G/T → G/T

xT → gxT

The assertion above amounts to prove that we can always find a fixed point, since G is connected, fg ishomotopic to the identity map, and apply the Lefschetz fixed-point theorem, we only need to prove the Eulercharacteristic

χ(G/T ) =∑i

(−1)iHi(G/T,Z)

However the Bruhat decomposition tells us that we only have cells in even dimension, χ(G/T ) > 0, we’redone. Actually, we can prove the following two assertion by choosing a regular element in T .

• any element of G is conjugate to an element in T

• any connected abelian(this cannot be omitted) subgroup of G is conjugate to a subgroup of T .

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15.2 Induced representation and Borel-Weil theorem

Consider G = Un and is complexification GC = GL(n,C). Let V be an irreducible unitary representationof G. Then V is naturally a irreducible representation of GL(n,C). The construction of the Borel-Weiltheorem is the following

• Decompose V under the action of the maximal torus T ⊂ GL(n,C). Then we can find a basis{v1, . . . , vn} of V has the property that

u • v = λ(u)v = uk11 uk22 . . . uknn v,∀v ∈ {v1, . . . , vn}

• Observe that uEiju−1 = uiu

−1j Eij , we know Eijv is either a weight vector or 0. If v is of weight

k = (k1, . . . , kn), and Eijv is a nonzero weight vector, then we have the weight of Eijv is k′ =(k1, . . . , ki + 1, . . . , kj − 1, . . . , kn).

• By ordering the weights lexicographically, we can find ”some” highest weight vector, pick ”one ofthem”, still denote by v. Then necessarily we have Eijv = 0,∀i < j, thus v is actually an eigenvectorof the B-action, where B is the subgroup of upper triangular matrices. In other words, we have

b11 ∗ ∗

∗ ∗

bnn

• v = b • v = λ(b)v = bk111 . . . b

knnnv

• Fix this highest weight vector v, we define

φ : V ∗ →M−k = MapholB (GC,C)

f 7→ {φ(f) : g 7→ f(gv)}

Then we can check φ(f)(gb) = f(gbv) = λ(b)f(gv) = λ(b)φ(f)(g). In other words, M−k is inducedfrom the 1-dimensional representation λ−1 of B.

• (Borel-Weil)V ∗ →M−k

is an isomorphism of representations. φ above is not a zero map, since it’s an GC equivariant map,we only need to check that φ is surjective, by Shur’s lemma again, we only need to check that M−k isirreducible. For this pupose, we only need to prove that M−k contains at most a unique U−-invariantsubspace, which means, it has a unique lowest weight vector.

• To finish the proof of the Borel-Weil theorem, recall that from the Bruhat decomposition of GC, weknow we have an open dense subset of GC(which is the top dimensional cell of GC), and any element gin this cell has a decomposition g = nb for some n ∈ U−, b ∈ B. Now if h ∈M−k is fixed by U−, thenwe have f(nb) = nh(b) = h(b) = λ(b)h(1), in other words h|U−B is uniquely determined by h(1), sinceh is continuous and the cell U−B is open dense in GC, so h itself is determined by h(1). Together withthe fact that we can always find at least 1 U− invariant vector in M−k.We get

dim((M−k)U−

= 1

• As a corollary of the Borel-Weil theorem, we know that any irreducible representation of G = Uncontains a unique highest weight vector.

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Remark (λ or λ−1, and left or right?). Let H ⊂ G be groups. Note that our definition of H invariantfunctions on G is that

f(gh) = h−1f(g)

ThusM−k = MapholB (GC,C)

f(gb) = λ(b)f(g) = bk111 . . . bknnnf(g)

is by definition induced from λ−1, that’s the reason we call it M−k. And M−k is an irreducible representationwith lowest weight −k, but NOT highest weight −k. To be more precise, we have

• V ∼= M∗−k: highest weight (k1, . . . , kn), lowest weight (kn, . . . , k1)

• V ∗ ∼= M−k: highest weight (−kn, . . . ,−k1), lowest weight (−k1, . . . , kn)

• Mk: highest weight (kn, . . . , k1), lowest weight (k1, . . . , kn).

• the subscribe k in Mk denotes the lowest weight of this representation. the subscribe k in M∗k denotethe ”transpose-inverse” of the lowest weight, in other words, the lowest weight of M∗k is (−kn, . . . ,−k1)

• if k is not dominant, then M∗−k = 0;M−k = 0, that is Borel-Weil theorem is not for every weight, it’sjust for some special weights.

H acts on the right, and remember to take the inverse, G acts on the left.

Remark (Induced representation and H0(G/B,Lα)).

Example 15.3 ((SkCn)∗). (SkCn)∗ is induced from a 1-dimensional representation of B:

λ :

b11 ∗ ∗

∗ ∗

bnn

7→ b−k11

We can also see it’s induced from the 1-dimensional representation of H = GL1,n−1 defined by the sameproperty. We prove it for the GL1,n−1 case. Then if f ∈ MapholH (G,C), then f(gh) = hk11f(g), we get fis independent of the last n − 1 columns. So we may view f as a holomorphic function on C − {0}, whichhas the property that f(cz) = ckf(z), by analytic continuation, we know f is a homogeneous polynomial ofdegree k w.r.t entries in the first column.

Example 15.4 ((∧kCn)∗). (∧kCn)∗ is induced from the 1-dimensional representation of B defined by

λ :

b11 ∗ ∗

∗ ∗

bnn

7→ b−1

11 . . . b−1kk

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Or we can also view it as induced from the 1-dimensional representation of H = GLk,n−k defined bya b

0 c

7→ det(a)−1

By similar reasons as above, we know ∀f ∈ MapholH (G,C) is independent of the last n − k columns. Andsince

f(g

a b

0 c

) = det(a)f(g)

we know f only depends on the wedge product of the first k-columns g1 ∧ g2 · · · ∧ gk and on any given 1-dimensional space ∧kW , f is naturally linear, in other words, f is a global section of the kth wedge powerof the dual of the tautological bundle on the Grassmannian Gr(k, n),i.e at every point [W ] ∈ Gr(k, n), thefibre is isomorphic to (∧kW )∗. In short, we have

MapholH (GC,C) ∼= H0(G/B,∧kS∗)

And more over {pi1...ik = det(gi1...ik,12...k)|{i1, . . . , ik ⊂ {1, 2 . . . , n}}} is a set of basis point free globalsections, which means this bundle is very ample, and we get the Plucker embedding:

Gr(k, n)→ P(∧kCn)

[W ]→ [w1 ∧ · · · ∧ wk]

Remark (⊗detm). If V is an irreducible representation of G with highest weight k = (k1, . . . , kn), thenV ⊗ detk is an irreducible representation of k′ = (k1 +m, . . . , kn +m)

Theorem 15.5 (M−k and (Sk1Cn)∗ ⊗ · · · ⊗ (Skn−1Cn)∗). M−k is a subrepresentation of (Sk1−knCn)∗ ⊗· · · ⊗ (Skn−1−knCn)∗ ⊗ detkn

Proof. By multiplying a certain power of the determinant, we reduce to the case kn = 0. Then ∀f ∈MapholB (GC,C) can be viewed as a function of n vectors. Since kn = 0, by similar argument as above, weknow f is independent of the last column. So f is actually a function of n − 1 vectors. And if we fix thecolumns g2, . . . , gn−1, we get a well-defined holomorphic function on Cn − span{g2, . . . , gn−1}. By Hartog’stheorem, we know a holomorphic function on Cn cannot have singular locus of codimension greater than 1.In other words, f can be extended to be a hlomorphic function on Cn. And it still has the homogeneousproperty, that means f depends polynomially on g1 of degree k1, similar for all other k′is.

Example 15.6 (The spin representation of O2n). Note that we have a natrual embedding

Un → O2n

A+ iB 7→

A B

−B A

Pin(2n)/O2n

∼= Un/Un, where Un is the 2-folds covering of Un in Pin(2n). Then we have Un ∼= {(u, c)|u ∈Un, c

2 = det(u)} Then we can construct a representation from the 1-dimensional representation λ : (u, c) 7→c−1 of Un, which turns out to ve the spin representation. We can restrict the spin representation on Unagain,and get a representation of Un which is isomorphic to ∧Cn ⊗ λ.

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Now we can consider the vector bundle construction in the Borel-Weil theory, take root λ ∈ X(T ) as anexample, we can defined the following line bundle on G/B

L−λ = G× C/((gb, c) ∼ (g, λ(b)c))

G/B G/Pα

p

π

First we write down the explicit isomorphim

V ∗ ∼= M−α = MapholB (G,C)→ H0(G/B,Lα)

f 7→ {gb 7→ (gb, f(gb))} ∈ {G/B → Lα}It’s well-defined since

(gb, f(gb)) = (gb, α−1(b)f(gb)) = (g, f(g))

conversely, given any holomorphic section σ : G/B → Lα, σ(gb) = (gb, f(gb)), we can construct a homomor-phic function on G/B, that is

H0(G/B,Lα)→MapholB (G,C) = M−α = V ∗

σ 7→ {gb→ f(gb)}Since σ is a well-defined global section of Lα, we have

(gb, f(gb)) ∼ (g, α−1(b)f(gb)) = (g, f(g))⇒ f(gb) = α(b)f(g)

In short, we get another version of the Borel-Weil theorem

Theorem 15.7 (Borel-Weil). When α is a dominant weight, H0(G/B,Lα) is dual to the irreducible highest-weight representation of G with highest weight α.

H0(G/B,Lα) ∼= M−α = MapholB (G,C) ∼= V ∗

Example 15.8 (V ∗ ∼= MapholB (G,C) and H0(G/B,L−k)). We want to compute an explicit example G =SL(3,C). First, we collect all the information we need about the structure of SL(3,C).

• The maximal torus

T = {

a

b

c

|abc = 1} ∼= Gm ×Gm

• The character lattice X(T ) ∼= Z⊕ Z is the same as the character of the Borel subgroup

a ∗ ∗

b ∗

c

7→ ak1bk2

Note that we have many different ways to write down a character, for example ak1bk2 = bk2−k1c−k1 .

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Page 21: A ne Lie Algebras(Under Construction) · A ne Lie Algebras(Under Construction) Shuai Wang July 2016 1 Introduction Preparation for my oral quali cation, minor topic. Basic knowledge

• Positive roots

α1 :

a

b

c

7→ ab−1;α2 :

a

b

c

7→ bc−1;α1 + α2 :

a

b

c

7→ ac−1

Note again that ac−1 = a2b

• The root subgroup and the parabolic subgroup corresponding to α = α1

Uα1 =

1

∗ 1

1

;Pα1 =

∗ ∗ ∗

∗ ∗ ∗

• SL(2,C) ∼=< Uα, U−α > is obvious. And we see that

α∨ : t 7→

t

t−1

1

• The bruhat decomposition Pα = B

⊔BsαB = B

⊔UαsαB. To be more precise, we have

Pα =

0 1

−1 0

1

B⊔

1

∗ 1

1

B

Note that the second cell is just all the matrices of the form

{

a b e

c d f

g

∈ Pα|a 6= 0}

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Page 22: A ne Lie Algebras(Under Construction) · A ne Lie Algebras(Under Construction) Shuai Wang July 2016 1 Introduction Preparation for my oral quali cation, minor topic. Basic knowledge

And Pα/B ∼= CP 1, we have many ways to see this, either from the Bruhat decomposition or from thefollowing free action of Pα on CP 1

a b e

c d f

g

• [x0, x1] = [ax0 + bx1, cx0 + dx1]

The stablizer of [1, 0] is exactly the Borel subgroup B.

• The charts of Pα/B and the Bruhat decomposition.

Pα/B → CP 1

gB 7→ g • [1, 0]

Then we easily find D+(x0) are precisely those matrices in the second cell in the Bruhat decompositionabove. D+(x1) are all the matrices of the form

{

a b e

c d f

g

|c 6= 0}

which is corresponding to another Bruhat decomposition of Pα

Pα =

1 0

0 1

1

B⊔

∗ 1

−1 0

1

B

And the coordinates [x0, x1] is precisely [a, c].

We want to consider the natural projection π : G/B → G/Pα, because every fibre above y ∈ G/Pα, ishomeomorphic to a copy of Pα/B, which we already know is P1. Then Lα|π−1y is a line bundle, so it’s mustbe of the form O(n), we can determine this n with the information we already have. After that, it’s easy tosee that Lα has global section if and only if α is dominant. By the first Bruhat decomposition, we know thatany element (g, v) ∈ p−1(D+(x0)) can be uniquely written in a form

(g, v) ∼ (

1 0 0

x1 1 0

0 0 1

, c)

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Page 23: A ne Lie Algebras(Under Construction) · A ne Lie Algebras(Under Construction) Shuai Wang July 2016 1 Introduction Preparation for my oral quali cation, minor topic. Basic knowledge

We know p−1(D+(x0)) must be trivial, for further purpose, we given an explicit trivilization

φ0 : A× C→ p−1(D+(x0))

([1, x1], c) 7→ (

1 0 0

x1 1 0

0 0 1

, c)

Since we have

(

a b e

c d f

g

, v) = (

1 0 0

ca 1 0

1

a b e

ad−bca = 1

agaf−cea

g

, v)

= (

1 0 0

ca 1 0

1

, λ(

a b e

1ag

af−cea

g

)v) = (

1 0 0

ca 1 0

1

, ak1(ag)−k2v)

then the inverse is given byφ−1

0 : p−1(D+(x0))→ A× C

(

a b e

c d f

g

, v) 7→ ([1,

c

a], ak1−k2g−k2v)

Similarly, on p−1(D+(x1)), we have

φ1 : A× C→ p−1(D+(x1))

([x0, 1], v) 7→ (

x0 1 0

−1 0 0

1

, v)

And the inverseφ−1i : p−1(D+(x1))→ A× C

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Page 24: A ne Lie Algebras(Under Construction) · A ne Lie Algebras(Under Construction) Shuai Wang July 2016 1 Introduction Preparation for my oral quali cation, minor topic. Basic knowledge

(

a b e

c d f

g

, v) 7→ ([

a

c, 1], ck1−k2g−k2v)

To compute the degree of the restriction L−λ, we can either compute the transition function or find ameromorphic section and compute its poles and zeros. Since I always got confused by the transition functions,I prefer to use the second method, before doing that we give the transition function on D+(x0x1)

L−λ|D+(x0x1) = {h0 : D+(x0)→ C, h1 : D+(x1)→ C|h0 =xk1−k21

xk1−k20

h1}

L−λ|π−1y∼= O(−(k1 − k2))

We extend the constant section σ on D+(x0) be be a global meromorphic section, on D+(x0), it’s given by

σ|D+(x0) = (

1 0 0

x1 1 0

0 0 1

, 1)

by φ−11 , we know we can extend it on D+(x1)

(

1 0 0

x1 1 0

0 0 1

, 1) 7→ [

1

x1, xk1−k21 ] = ([x0, 1], x

−(k1−k2)0 )

Then we know σ has a pole at [0, 1] of order k1 − k2, we get once more

L−λ|π−1y∼= O(−(k1 − k2))

Actually, we have the following proposition

Proposition 15.9 (L−λ|π−1y∼= O(− < α∨, λ >)). This is obvious for SL(3,C) from our computations

above, λ ◦ α∨(t) = tk1(t−1)k2 = tk1−k2 , thus we have < α∨, λ >= k1 − k2

15.3 Bott’s theorem

We first state the Bott’s theorem. Let C be the closed dominant Weyl chamber corresponding to α. i.e{λ ∈X(T )| < α∨, λ >≥ 0|∀α ∈ R+} = {λ ∈ X(T )| < α∨, λ >≥ 0|∀α ∈ Π}.

Theorem 15.10 (Bott). Let G be a split reductive simply-connected algebraic group over an algebraicallyclosed field, and fix a split maximal torus T and a Borel subgroup B containing T . Let W be the Weyl groupacting via the shifted action on the integral weights. Let λ be an integral weight of G, and Lλ be the bundleconstructed above, then

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Page 25: A ne Lie Algebras(Under Construction) · A ne Lie Algebras(Under Construction) Shuai Wang July 2016 1 Introduction Preparation for my oral quali cation, minor topic. Basic knowledge

• If λ ∈ C,Hi(G/B,Lλ) = 0,∀i ≥ 1;

• If λ+ ρ lies on some hyperplanes defined by the root α, then Hi(G/B,Lλ) = 0,∀i ≥ 0;

• Otherwise, ∃!w ∈W ,such that w−1 • λ = µ ∈ C, then

Hi(G/B,Lλ) = H0(G/B,Lµ), i = l(w)

Hi(G/B,Lλ) = 0, i 6= l(w)

Remark (Relations between G/B, Lλ, Π, R+, and C). In some suitable sense, they’re actually differentnames for the same thing. Let’s take SL(3,C) as an example again. The standard choice of fundamentalroots is {α1, α2}, the corresponding Borel subgroup B+ is the group of upper-triangular matrices. And letλ = α1 +α2, which is obviously a dominant weight( Note that α1 and α2 themselves are not dominant,sinceα1 = 2w1−w2;α2 = −w1+2w2). Then if we choose another system of fundamental weights Π′ = {−α1,−α2},the corresponding Borel is the group of lower triangular matrices B−, and −λ = −α1−α2 is now a dominantweight w.r.t to this root system, our question is

H0(G/B+,Lλ) ∼= H0(G/B−,L−λ)?

There’re some other unimportant but misleading things, say, we can also talk about the root system ofGL(3,C), which is the complexification of Un, then all the irreducible representations are classified by highestweights, which are naturally dominant weights, but the dominant weights of GL(3,C) are given by

a

b

c

7→ ak1bk2ck3 ; k1 ≥ k2 ≥ k3

But if a weight λ :

a

b

c

→ ak1bk2 of SL(3,C) is dominant if and only if k1 ≥ k2 ≥ 0. To be more

precise, the two fundamental weights are given by

w1 :

a

b

c

7→ a;w2 :

a

b

c

7→ ab

This is computed from the relation between the fundamental roots and fundamental weightsα1

α2

=

2 −1

−1 2

w1

w2

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Page 26: A ne Lie Algebras(Under Construction) · A ne Lie Algebras(Under Construction) Shuai Wang July 2016 1 Introduction Preparation for my oral quali cation, minor topic. Basic knowledge

Remark. Before proving the Bott’s theorem, we want to review several terminologies,

• X(T ), X∗(T )

• Y (T ), X∗(T )

• roots

• coroots

• weights, integral weights

• Killing form on X(T ) and the natural paring

X∗(T )×X∗(T )→ Z

And how to compute them in practice?

• Weyl group action on weights and shifted Weyl group action and how to compute them in practice?

• simply-connected v.s not simply-connected

• semi-simple v.s reductive

Remark (Lα and H0(G/B,Lα)).

Figure 1: The Universe

16 Conclusion

“I always thought something was fundamentally wrong with the universe” [1]

References

[1] D. Adams. The Hitchhiker’s Guide to the Galaxy. San Val, 1995.

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