semisimple lie algebras math 649, 2013
TRANSCRIPT
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Semisimple Lie AlgebrasMath 649, 2013
Dan Barbasch
March 7
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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Theorem (4, Weyl’s normal form)
Let h ⊆ g be a CSA. There is a basis Xα ∈ gα, α ∈ ∆ such that
[Xα,X−α] = Hα, [H,Xα] = α(H)Xα, [Xα,Xβ] = Nα,βXα+β
satisfyingNα,β = 0 if α + β /∈ ∆
Nα,β = −N−α,−β.
Furthermore,
N2α,β =
q(1− p)
2α(Hα)
where β + nα, p ≤ n ≤ q is the α-string through β.
A proof can be found in the texts of Helgason, Jacobson, orSamelson.
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Some Proofs
Let β + pα, . . . , β + qα be the string of α through β. This is anirreducible representation of sl(2)α. Note that p ≤ 0 ≤ q.
1 [X−α, [Xα,Xβ]] =q(1− p)
2α(Hα)B(Xα,X−α)Xβ. Exercise.
2 If α, β, γ ∈ ∆ and α + β + γ = 0, then Nα,β = Nβ,γ = Nγ,α.
3 If α, β, α + β ∈ ∆, then Nα,β · N−α,α+β =q(1− p)
2α(Hα).
Combining these facts we find, Nα,β · N−α,−β = −q(1− p)
2α(Hα).
The number on the right is nonpositive, which is consistent withNα,β = −N−α,−β.
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Real Forms
A real vector space V is said to have a complex structure if thereis an R-linear J : V −→ V such that J2 = −Id . Then V is acomplex vector space with scalar multiplication
(a + ib)v = av + bJv .
Conversely if E is a complex vector space, it is also a real vectorspace with complex structure via J = i Id .
Definition
A real algebra g is said to have a complex structure if there is acomplex structure J such that ad X ◦ J = J ◦ ad X .For a complex algebra, a real form is a subalgebra gR ⊂ gsatisfying gR ∩ igR = (0) and g = gR + igR.
Recall also that if g is a real Lie algebra, its complexification isdefined as gc := g⊗R C with the obvious bracket structure. Theng is a real form of gc .
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Real Forms
Exercise
Show that a real Lie algebra g is semisimple, solvable, nilpotent ifand only if gc is semisimple, solvable, nilpotent.
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Compact Forms
Definition
A real Lie algebra g is called compact if the Cartan-Killing form Bis negative definite.
Let g be complex semisimple and ∆ be the roots. A subset ∆+ iscalled a positive system if
1 α, β ∈ ∆+, then α + β ∈ ∆+ or else is not a root,
2 ∆+ ∪ (−∆+) = ∆.
Such systems always exist; choose an H0 ∈ hR such thatα(H0) 6= 0 for any α ∈ ∆.
∆+ := {α : α(H0) > 0} is a positive system.
Conversely any positive system is given by this procedure, but thiswill be proved later.
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Compact Forms
Theorem (5)
Every semisimple Lie algebra g has a a compact real form.
Proof.
Choose a positive system ∆+ and a basis in Weyl’s normal form.The subspace
gk = ihR +∑α∈∆+
R(Xα − X−α) +∑α∈∆+
R(Xα + iX−α)
is a real Lie subalgebra, and it is an exercise to verify that B isnegative definite when restricted to this subspace; the vectors Xαhave to satisfy B(Xα,X−α) = 1 since [Xα,X−α] = Hα.
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Adjoint Group
Recallad : g→ Der(g). (1)
For g semisimple this is an isomorphism. Recall
Aut(g) := {A ∈ GL(g) | A([x , y ]) = [Ax ,Ay ]}. (2)
Int(g) := the closure of the subgroup of Aut(g) generated by ead x
with x ∈ g.The Lie algebras of these groups are Der(g) and ad g ⊂ Der(g).Int(g) is also called the adjoint group of g. It is connected.
When g is semisimple, g ∼= Der(g), so Int(g) is the connectedcomponent of the identity of Aut(g).Note that for A ∈ Aut(g), ad Ax = A ◦ ad x ◦ A−1 becausead(Ax)(y) = [Ax , y ] = A([x ,A−1y ]) = A ◦ ad x ◦ A−1(y).Then Aut(g) ⊂ O(B), the orthogonal group of B :
B(Ax ,Ay) = Tr(ad Ax ◦ ad Ay) = Tr(A ◦ ad x ◦ A−1 ◦ A ◦ ad y ◦ A−1) =
= Tr(ad x ◦ ad y).(3)Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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Theorem (6)
If B is negative definite, Aut(g) and Int(g) are closed subgroups ofan orthogonal group, therefore compact.
Proof.
From the above discussion, Aut(g) ⊂ O(B), the group leaving Binvariant. When B is negative definite, this group is compact.
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Theorem (7, Haar measure)
Let G be a compact Hausdorff topological group. Then G has aunique biinvariant (Borel) measure dµ
f ∈ Cc(G ) 7→∫G
f (x)dµ(x) (4)
such that∫G
f (gx)dµ(x) =
∫G
f (xg)dµ(x) =
∫G
f (x)dµ(x) (5)
for all g ∈ G and f ∈ Cc(G ).
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Proposition
Suppose (π,V ) is a representation of a compact group G . V afinite dimensional complex vector space. Then V has aG−invariant inner product.
Proof.
Let 〈 , 〉 be any inner product. Define
(v ,w) =
∫G〈π(x)v , π(x)w〉dµ(x). (6)
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Exercise
Complete the proof.
Corollary
Any finite dimensional representation of G is completely reducible.
Proof.
Let 〈 , 〉 be a G –invariant inner product. If W ⊆ V is an invariantsubspace, then W⊥ is also G –invariant.
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Complete Reducibility
Theorem
Let (π,V ) be a finite dimensional representation of a complexsemisimple Lie algebra. Then (π,V ) is completely reducible.
Proof.
Let gR be a compact real form. A nontrivial result asserts thatthere is a simply connected Lie group GR with Lie algebra gR.Standard properties of Lie groups imply that (π,V ) exponentiatesto a representation of GR. Let W be a g–invariant subspace. It isGR–invariant, so it has a GR–invariant complement, W ′. Then W ′
is gR invariant. Since g = gR ⊗R C, and W ′ is complex, W ′ is ginvariant.
References: F. Warner, Hausner-Schwartz.
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An algebraic proof
We would like to give an algebraic proof that works for other fieldsas well. Let us consider the case of sl(2) first. Recall that finitedimensional irreducible modules F (n) are parametrized by n ∈ N.F (n) has the following properties:
1 h acts semisimply, the eigenvalues are n, n − 2, . . . ,−n withmultiplicity 1.
2 e and f act nilpotently.
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Denote
h :=Kh, n := Ke, b := Kh + Ke,
n := Kh + Kf , b := Kh + Kf .(7)
Let λ ∈ K. Define the Verma module
M(λ) := U(g)⊗U(b) Kλ (8)
where Kλ is the 1−dimensional module of b
π(h)11λ = λ11λ, π(e)11λ = 0. (9)
Then M(λ) is a representation of g:
π(x)(y ⊗ 11λ) := xy ⊗ 11λ. (10)
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Proposition
1 M(λ) is semisimple as an h module. The eigenvalues areλ− 2n, n ∈ N occurring with multiplicity 1.
2 e acts locally nilpotently and f acts freely on M(λ).
3 M(λ) is irreducible except when λ ∈ N.
In this case, the Jordan–Holder series is
0→ M(−λ− 2)→ M(λ)→ F (λ)→ 0. (11)
M(−λ− 2) is the largest proper submodule. F (λ) is the uniqueirreducible quotient.
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Proof.
M(λ) has a basis {f n ⊗ 11λ}.
f : f n ⊗ 11λ 7→ f n+1 ⊗ 11λ (12)
h · f n ⊗ 11λ = f n(h − 2n)⊗ 11λ = (λ− 2n)f n ⊗ 11λ (13)
ef n = ef · f n−1 = (h + fe)f n−1 = (λ− 2n + 2)f n−1+
+ (λ− 2n + 4)f n−1 + · · ·+ [nλ− n(n − 1)]f n−1 =
= n(λ− n + 1)f n−1.
(14)
all terms tensored with 11λ.
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Lemma
Suppose (π,V ) is a module generated by a vector v such thate · vλ = 0, h · vλ = λvλ. Then there is a unique g−homomorphism
M(λ)→ V → 0 (15)
that sends 11λ to vλ.
Proof.
The map f n ⊗ 11λ 7→ π(f )nvλ is well defined.
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Proposition (1)
Assume h acts semisimply on the finite dimensional representationV . Then V is completely reducible.
Proof.
Write V = ⊕Ni=−NVi as a decomposition with respect to π(h).
Then Homg(F (N),V ) ' Homg(M(N),V ),because any map from F (N) to V can be composed with thecanonical one M(N) −→ F (N), and any nonzero mapM(N) −→ V must have a nontrivial kernel which must beM(−N − 2) because V is finite dimensional.There is an inclusion
0→ ⊕mF (N)→ V (16)
where m = dim VN ; choose a basis v1, . . . , vm of VN , and map thehighest weight of the i−th copy of F (N) into vi . The quotient ofV by the image contains no copies of F (N).
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Proof of Proposition 1
proof continued.
To show that there is an invariant complement, we have to showthat ⊕F (N) is also a quotient of V . We have
Homg(V ,F (N)) ' Homg(F (N)∗,V ∗) ' Homg(F (N),V ∗) (17)
because F (N) ∼= F (N)∗. But V ∗ ' ⊕Ni=−NV ∗i and
dim Vi = dim V ∗i
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It is not so easy to show that h always acts semisimply. Assume(π,V ) is a representation of a semisimple Lie algebra, and letW ⊆ V be an invariant subspace.A complement W ′ is associated with a projection e ∈ End(V )such that e2 = e, and V is the 1-eigenspace of e. Two suchprojections e, e ′ have the property that a = e − e ′ is zero on Wand maps V to W . Let
X := {a ∈ End(V ) | a(V ) ⊆W , a|W = 0}. (18)
We note also that
π(x)W ⊆W ∀x ∈ g⇔ π(x)e − eπ(x) ∈ X ∀x . (19)
In fact in this case X is a representation of g,
x · a := π(x)a− a · π(x). (20)
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Proposition
W has a g-invariant complement iff for any e, there is a ∈ X suchthat
π(x)e − eπ(x) = x · a. (21)
Exercise
Show that e − a is the projection which gives the invariantcomplement.
We can phrase this as follows. e defines a map
f : g→ X , f (x) := x · e = π(x) ◦ e − e ◦ π(x). (22)
This map satisfies
f ([x , y ]) = x · f (y)− y · f (x) (Jacobi identity). (23)
We would like to show that for such a map, there is a ∈ Xsatisfying
f (x) = x · a. (24)
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This fits into a more general framework. Let (π,V ) be arepresentation of g. Define
C i (g,V ) := Hom(Λig∗,V ). (25)
Then there is a mapd : Cn → Cn+1 (26)
given by
dω(x0, . . . , xn) =∑
(−1)iπ(xi ) · ω(x0 · · · xi · · · xn) (27)
−∑
(−1)i+jω([xi , xj ], x0 · · · xi · · · xj · · · xn).
Exercise
Show that d2 = 0.
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Define H i (g,V ) := ker dn/ im dn−1. Then
C 0(g,V ) = Hom(C,V ) ' V , C 1(g,V ) ' Hom(g,V ), (28)
ϕa ∈ C 0, dϕa(x) = π(x)a, (29)
f ∈ C 1, df (x , y) = −f ([x , y ]) + x · f (y) + y · f (x). (30)
We are trying to show that H1(g,V ) = 0 for any V .Note also that
H0(g,V ) = V g = {v ∈ V | π(x)v = 0 ∀x}.
If x ∈ g, define ιx : Cn → Cn−1 by
(ιxω)(x1, . . . , xn−1) := ω(x , x1, . . . , xn−1). (31)
Also define
Ox : Cn → Cn (32)
Oxω(x1, . . . , xn) = −π(x) · ω(x1, . . . , xn) (33)
+∑
ω(x1, . . . , [x , xi ], . . . , xn). (34)
This is a representation of g.Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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In general if (π,V ) and (ρ,W ) are representations of g,HomC(V ,W ) is a representation as well:
(x · f )(v) = −f (π(x) · v) + ρ(x)(f (v)). (35)
In fact it is a g× g representation which we have restricted to thediagonal.
Proposition
The following formulas hold:
ix · d + d · ix = Ox [Ox ,Oy ] = O[x ,y ] Ox ◦ d = d ◦ Ox . (36)
Proof.
Exercise.
Dan Barbasch Semisimple Lie Algebras Math 649, 2013