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Semisimple Lie AlgebrasMath 649, 2013
Dan Barbasch
March 7
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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.
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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.
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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 .
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
Real Forms
Exercise
Show that a real Lie algebra g is semisimple, solvable, nilpotent ifand only if gc is semisimple, solvable, nilpotent.
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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.
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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.
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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
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.
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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 ).
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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)
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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.
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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.
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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.
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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)
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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.
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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.
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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.
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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).
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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)
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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)
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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.
Dan Barbasch Semisimple Lie Algebras Math 649, 2013
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
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