Irreducible Representations of Complex Semisimple Lie Algebras

17 Pages·2016·0.355 MB·English

by  Rush Brown

Irreducible Representations
of Complex Semisimple Lie Algebras
Rush Brown
September 8, 2016
Abstract
In this paper we give the background and proof of a useful theorem classifying irreducible
representationsofsemisimplealgebrasbytheirhighestweight,aswellasrestrictingwhatthat
highest weight can be.
Contents
1 Introduction 1
2 Lie Algebras 2
3 Solvable and Semisimple Lie Algebras 3
4 Preservation of the Jordan Form 5
5 The Killing Form 6
6 Cartan Subalgebras 7
7 Representations of sl 10
2
8 Roots and Weights 11
9 Representations of Semisimple Lie Algebras 14
1 Introduction
InthispaperIbuilduptoausefultheoremcharacterizingtheirreduciblerepresentationsofsemisim-
ple Lie algebras, namely that finite-dimensional irreducible representations are defined up to iso-
morphism by their highest weight ω and that ω(H ) is an integer for any root α of R. This is
α
the main theorem Fulton and Harris use for showing that the Weyl construction for sl gives all
n
(finite-dimensional) irreducible representations.
1
In the first half of the paper we’ll see some of the general theory of semisimple Lie algebras—
building up to the existence of Cartan subalgebras—for which we will use a mix of Fulton and
Harris and Serre ([1] and [2]), with minor changes where I thought the proofs needed less or more
clarification (especially in the proof of the existence of Cartan subalgebras). Most of them are,
however, copied nearly verbatim from the source. In the second half of the paper we will describe
the roots of a semisimple Lie algebra with respect to some Cartan subalgebra, the weights of
irreducible representations, and finally prove the promised result.
AlthoughI’vetriedtobefairlythorough,acoupleofthetheoremsusedaretoolongortootangential
to include full proofs. For these I’ve included citations. This paper also leaves out some important
parts of the theory that aren’t strictly necessary for proving the final theorem (such as the Weyl
groups of root systems and their relationship to automorphisms of semisimple Lie algebras fixing a
Cartan subalgebra), so this summary should not be considered comprehensive.
In this paper k = C, although any algebraically closed field of characteristic 0 will work for most
of the general theory.
2 Lie Algebras
Definition 2.1. ALiealgebraisavectorspaceg(inthispaperitwillalwaysbefinite-dimensional)
together with a binary operation [·,·] called a Lie bracket which is bilinear, skew-symmetric, and
satisfies the Jacobi identity: [X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0 for X,Y,Z ∈g.
Definition 2.2. An ideal g(cid:48) of a Lie algebra g is a sub-Lie algebra such that [X,Y]∈g(cid:48) for X ∈g
and Y ∈g(cid:48).
Although the bracket [·,·] may not be associative, the first isomorphism theorem still holds for Lie
algebras.
A Lie algebra is called abelian if [·,·]≡0.
For any finite-dimensional vector space W we can form the Lie algebra gl(W) (also denoted gl ,
n
where n=dimW), which has underlying vector space M(W):=End(W) and Lie bracket defined
[X,Y]=XY −YX for X,Y ∈M(W). The traceless endomorphisms of W, denoted sl(W) or sl ,
n
form a Lie subalgebra of gl(W).
Example 2.3. The Lie algebra sl of traceless 2×2-matrices is spanned by the elements
2
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
0 1 1 0 0 0
X = , H = , and Y =
0 0 0 −1 1 0
which have the relations [X,Y]=H, [H,X]=2X, and [H,Y]=−2Y.
Definition 2.4. A representation of a Lie algebra g is a Lie algebra homomorphism ρ:g→gl(W)
for some finite-dimensional W.
We will sometimes drop the ρ and write X ·w or X(w) instead of ρ(X)w for X ∈ g and w ∈ W.
For any g we can construct a representation, called the adjoint representation, ad : g → gl(g) by
taking X (cid:55)→[X,·]∈End(g).
2
Definition 2.5. For any Lie algebra g, the universal enveloping algebra U(g) of g is defined to be
Tg/(cid:104)v⊗w−w⊗v−[v,w]:v,w ∈g(cid:105), where Tg is the tensor algebra of g.
A representation of g is the same thing as a U(g)-module. Note that a representation of g is not
necessarily the same thing as a representation of the image of the adjoint representation, as the
latter might not be faithful. The next theorem provides a handy basis for U(g):
Theorem 2.6 (Poincar´e-Birkhoff-Witt). Let {X ,...X } be a basis for g and let ι:g(cid:44)→U(g) be
1 k
the linear inclusion of g into U(g). If M ={ι(X1)⊗m1 ⊗···⊗ι(Xk)⊗mk :mi ∈N}, then M ∪{1}
forms a C-basis for U(g).
Proof. See theorem 4.3 of [3].
3 Solvable and Semisimple Lie Algebras
Definitions 3.1. The lower central series {D g} of a Lie algebra g is defined inductively D g =
n 1
[g,g], D g=[g,D g]. A Lie algebra is called nilpotent if its lower central series stabilizes at 0.
n+1 n
Similarly,
Definitions 3.2. The derived series {Dng} is defined D1g = [g,g], Dn+1g = [Dng,Dng]. A Lie
algebra is called solvable if its derived series stabilizes at 0; a Lie algebra is called semisimple if it
contains no nonzero solvable ideals.
Notethatcontainingnononzerosolvableidealsisequivalenttocontainingnononzeroabelianideals:
on the one hand an abelian ideal is solvable, and on the other hand if g(cid:48) ⊂g is a nonzero solvable
ideal, then the last nonzero term of the derived series of g(cid:48) is a nonzero abelian ideal of g (the fact
that this is an ideal of g follows by the Jacobi identity and induction on the length of the derived
series).
Semisimple Lie algebras are of interest for two reasons: firstly, every Lie algebra g can be put in a
short exact sequence
0→Rad(g)→g→g/Rad(g)→0,
where Rad(g) is the sum of all solvable ideals in g (and thus itself a solvable ideal) and g/Rad(g)
is semisimple. Secondly, semisimple Lie algebras contain special subalgebras which make their
representations much easier to understand:
Theorem 3.3. Every semisimple Lie algebra g contains a subalgebra h, called a Cartan subalgebra,
which is abelian, acts diagonally on g, and is maximal among subalgebras with this property.
Cartan subalgebras are useful for studying representations of semisimple Lie algebras because they
are well-behaved, but still large enough in g that ρ| determines ρ:g→gl(V). However, to prove
h
theorem 3.3 we need a couple of fundamental facts.
The following two theorems will be useful mainly because they apply to the image of the adjoint
representation:
Theorem 3.4 (Engel). If g ⊂ gl(V) is a Lie subalgebra such that every X ∈ g is a nilpotent
endomorphism of V, then there is a nonzero v ∈V such that X(v)=0 for all X ∈g.
3
Proof. Wewillprovethisbyinductiononthedimensionofg, showingthatgmustcontainanideal
of codimension one—which by the induction hypothesis annihilates some nonzero subspace W of
V—and then showing that an element spanning the rest of g must also kill some element of W.
First we note that if X ∈ gl(V) is any nilpotent endomorphism of V, then ad(X) is a nilpotent
endomorphism (of gl(V)): if X is nilpotent, then XK = 0 for some K; for any other Y ∈ gl(V),
ad(X)2K(Y)∈gl(V)isahomogeneouspolynomialinX andY ofdegree2K+1inwhicheveryterm
has at most one power of Y, so every term must contain a factor of XK = 0, i.e. 0 = ad(X)2K ∈
End(gl(V)).
Nextweproducethepromisedideal. Leth⊂gbeanymaximalpropersubalgebraofg. Sincehisa
subalgebra, the adjoint action of h on g preserves h and so acts on g/h. On the other hand, by the
last paragraph, the adjoint action of each X ∈h on g is nilpotent, so its action on g/h is nilpotent.
By induction, this means that some nonzero Y ∈ g/h is killed by the adjoint action of h, so there
is some nonzero representative Y ∈g\h such that [h,Y]⊂h. Thus h is an ideal of the subalgebra
g(cid:48) of g generated by Y and h (which by the above is spanned by Y and h). By maximality of h, g(cid:48)
must be all of g, so h is an ideal of g of codimension one.
By induction on the dimension of g, there is a nonzero v ∈V such that X(v)=0 for every X ∈h,
so the subspace W := {w ∈ V : X(w) = 0 for every X ∈ h} is nonzero. To prove the theorem it
remains to show that Y(u) = 0 for some nonzero u ∈ W. This is where we use that h is an ideal:
foranyw ∈W andX ∈h, X(Y(w))=Y(X(w))+[X,Y](w). X and[X,Y]areinh, sobothterms
ontheright-handsidearezero. ThusX(Y(w))=0foreveryX ∈handeveryw ∈W, soY carries
W into itself. Since Y acts nilpotently on all of V, this means there must be some nonzero u∈W
such that Y(u)=0.
Corollary 3.5. If g is a Lie algebra such that for every X ∈g, ad(X) is a nilpotent endomorphism
of g, then g is nilpotent.
Theorem 3.6 (Lie). If g⊂gl(V) is a solvable Lie algebra, then there is a nonzero v ∈V such that
v is an eigenvector of g.
For this theorem we first prove the following lemma:
Lemma 3.7. LethbeanidealofaLiealgebrag, letV bearepresentationofgandsupposeλ∈h∗.
If W ={v ∈V :X(v)=λ(X)v for every X ∈h}, then for every Y ∈g, Y(W)⊂W.
Proof of lemma. Suppose w ∈ W, w (cid:54)= 0 and X ∈ h. Then X(Y(w)) = Y(X(w))+[X,Y](w) =
λ(X)Y(w)+λ([X,Y])w, since [X,Y] ∈ h. Thus to show Y(w) ∈ W, it remains to show that
λ([X,Y])=0 for all X ∈h:
Let U ⊂ V be the span of w,Y(w),Y2(w),.... U is certainly preserved by Y, but for any X ∈ h,
X(Yk(w)) = Y(X(Yk−1(w))) + [X,Y](Yk−1(w)), so by induction on k we see that U is also
preserved by X. In fact this shows that the action of X ∈h on U is upper-triangular with respect
tothebasisw,Y(w),Y2(w),...,withdiagonalentriesλ(X),soTr(X)=λ(X)·dimU. Ontheother
hand, Tr([X,Y])=0 for any endomorphisms X,Y of V. Thus λ([X,Y])=0 for any X ∈h.
Proof of theorem 3.6. As in the proof of Engel’s theorem we want an ideal h ⊂ g of codimension
one. Thistimeit’sabiteasier: sincegissolvable,Dg=[g,g](cid:54)=g. ThecommutatorDgisanideal,
so this means the Lie algebra a = g/Dg is nonzero. Any subspace of a is an ideal, so by the first
4
isomorphismtheoremthepreimageofanycodimensiononesubspaceofawillbeacodimensionone
ideal of g.
By induction on the dimension of g, we may assume there is some nonzero v ∈ V that is an
0
eigenvector for every element of h. For X ∈ h let λ(X) be the eigenvalue of X corresponding to
v and let W = {v ∈ V : X(v) = λ(X)v for every X ∈ h}. Again as in the proof of Engel, let
0
Y ∈g\h. By the lemma, Y(W)⊂W, so Y must have some nonzero eigenvector in W, completing
the proof.
Corollary 3.8. If g ⊂ gl(V) is a solvable Lie algebra, then the elements of g can simultaneously
be put in upper triangular form.
Proof. Induct on the dimension of V.
The next theorem, despite its obvious importance (for example, for motivating the study of irre-
duciblerepresentationsofsemisimpleLiealgebrasinthefirstplace),willonlybeusedonceortwice
in the rest of the paper. Because of this and the length of its proof, this is one of the theorems we
will palm off.
Theorem 3.9 (Weyl). For any semisimple Lie algebra g and any finite-dimensional representation
V, V is completely reducible (i.e. can be written as a direct sum of irreducible subrepresentations).
For something like Weyl’s original proof, using integration to prove the Peter-Weyl theorem for
compact Lie groups and thence the Weyl theorem for semisimple Lie algebras, see [4]. For a more
modern algebraic proof, see section 7.8 of [5].
4 Preservation of the Jordan Form
In this section we show what is probably the most useful fact about linear representations of
semisimple Lie algebras: the various definitions of the “nilpotent” and “semisimple” parts of an
element are equivalent, or rather consistent across all representations
Theorem 4.1. Suppose g is a semisimple Lie subalgebra of gl(V). If X ∈ g, then X ,X ∈ g,
s n
where X is the semisimple part of X and X is the nilpotent part of X as an endomorphism of V.
s n
Proof. [X,g]⊂g,andlikewise[p(X),g]⊂gforanyp∈C[T]. BothX andX canbeexpressedas
s n
polynomials in X, so [X ,g]⊂g and [X ,g]⊂g. Letting n be the subalgebra {A∈gl(V):[A,g]⊂
s n
g}, this means X ,X ∈ n. For each g-subrepresentation W of V, we define another subalgebra
s n
s ={Y ∈gl(V):Y(W)⊂W and Tr(Y| )=0}.
W W
Now we claim that
 
(cid:92)
g(cid:48) :=n∩ sW=g.
W⊂V asubrep
Since g is semisimple, Dg = g. Thus every X ∈ g is traceless, so g ⊂ g(cid:48). Since g ⊂ g(cid:48) ⊂ n, g is
an ideal of g(cid:48). By theorem 3.9, g has a complementary g-representation U in g(cid:48), i.e. g(cid:48) = g⊕U,
so it remains to show that U = 0; for this it suffices to show that for any Y ∈ U, Y| = 0 for
W
5
any irreducible g-subrepresentation W ⊂ V, since any Y ∈ g(cid:48) must preserve each such W and by
theorem 3.9 we know that V is a sum of irreducible representations. Now, [g,g(cid:48)]⊂g, so [g,U]=0.
Thus Y commutes with g, so Y| is a g-intertwining endomorphism of W. By Schur’s lemma for
W
Lie algebras (which is just the same as Schur’s lemma for finite groups), this means Y| =λ·I for
W
some λ∈C. But Y| is traceless, so Y| =0.
W W
For any semisimple g we have the adjoint representation ad:g(cid:44)→gl(g), so by the above lemma for
any X ∈g we have X ,X ∈g, whose adjoint actions on g are the semisimple and nilpotent parts
s n
of ad(X) respectively.
Observation4.2. Forg,g(cid:48) semisimpleandρ:g→g(cid:48) aLiealgebrahomomorphism,ρ(X )=ρ(X)
s s
and ρ(X )=ρ(X) .
n n
Proof. By the lemma, ρ(X) ∈ im(ρ), so it suffices to show that the restriction of ad(ρ(X )) to
s s
im(ρ) is diagonal. Thus it suffices to consider the case im(ρ) = g(cid:48), i.e. g(cid:48) = g/ker(ρ); ker(ρ) is
an ideal of g, so ad(X ) must act (diagonally) on ker(ρ) and thus (diagonally) on g/ker(ρ). An
s
identical argument works for X .
n
Corollary 4.3. Foranyrepresentation ρ:g→gl(V)ofasemisimpleLiealgebragandanyX ∈g,
ρ(X ) is the semisimple part of ρ(X) and ρ(X ) is the nilpotent part of ρ(X).
s n
Proof. Bytheabove,ρ(X )isthesemisimplepartofρ(X)regardedasanelementofthesemisimple
s
Lie algebra g(cid:48) :=ρ(g)⊂gl(V), i.e. ρ(X )| is semisimple. On the other hand, by theorem 4.1, the
s g(cid:48)
semisimple part ρ(X) of ρ(X), as an endomorphism of V, is in g(cid:48), so to show that the semisimple
s
part ρ(X) of ρ(X) is ρ(X ) it suffices to show that the adjoint action of the semisimple part of
s s
ρ(X) on g(cid:48) is diagonal.
If any Y ∈gl(V) is diagonal with respect to some basis B of V then ad(Y) is diagonal with respect
to the standard basis1 of gl(V) induced by B, so ad(Y )=ad(Y) . Thus the adjoint action of the
s s
semisimple part of ρ(X) on g is diagonal. However, by the above, the semisimple part of ρ(X) acts
on g(cid:48), so it must act diagonally on g(cid:48). An identical argument works for ρ(X ).
n
Thus for g semisimple, representations respect the Jordan decomposition of elements of g.
5 The Killing Form
Now we introduce a bilinear form B on Lie algebras, called the Killing form:
Definition 5.1. B(X,Y):=Tr(ad(X)◦ad(Y)) for X,Y ∈g any Lie algebra.
Proposition5.2. ForX,Y,Z ∈gaLiealgebraandB definedasabove,B([X,Y],Z)=B(X,[Y,Z])
Proof. For any endomorphisms of a vector space A,B,C,
Tr((AB−BA)C)=Tr((AB−BA)C)+Tr(B(AC)−(AC)B)=Tr(A(BC−CB)).
1Thematriceswithoneentryof1andtherest0,usuallydenoted{Eij}
6
TheKillingformishandyfordetectingsemisimplicityandforseparatingtherootspacesofsemisim-
ple Lie algebras, which we will encounter later in this section. To show these properties we first
define B on gl(V) by B (X,Y) = Tr(X ◦Y); as usual we do this to prove something about the
V V
adjoint action of a general Lie algebra.
Theorem 5.3 (Cartan’s criterion). If g is a subalgebra of gl(V) and B (X,Y) = 0 for every
V
X,Y ∈g, then g is solvable.
Proof. By Engel’s theorem it suffices to show that every element of Dg is nilpotent. Let X ∈Dg,
take a basis B for V that puts X in Jordan form, and let λ ,...,λ be the diagonal entries of X.
1 r
We need to show that the λ are all zero, for which it suffices to show that λ λ +···+λ λ =0.
i 1 1 r r
Let D =X be the semisimple part of X, which by the above is the diagonal transformation with
s
entries λ ,...,λ and an element of g by theorem 4.1. Let D be the diagonal transformation with
1 r
entries λ ,...,λ . Since B (D,X) = λ λ +···+λ λ , we want to show that B (D,X) = 0.
1 r V 1 1 r r V
Since X is a sum of elements of the form [Y,Z], Y,Z ∈g, B (D,X) is a sum of terms of the form
V
B (D,[Y,Z])= B ([D,Y],Z). Our hypothesis is that B (Y,Z)=0 for Y,Z ∈ g, so it suffices to
V V V
show [D,Y]∈g.
For this it suffices to show that ad(D) is a polynomial in ad(D), since ad(D)kY ∈ g. But with
respect to the standard basis of gl(V) induced by B, ad(D) and ad(D) are diagonal complex
conjugate matrices, and thus polynomials in each other.
Corollary 5.4. For any Lie algebra g and B defined on g as above, B is nondegenerate if and only
if g is semisimple.
Proof. Let s = {X ∈ g : B(X,Y) = 0 for all Y ∈ g}. If X ∈ s and Y ∈ g then for any Z ∈ g,
B([X,Y],Z) = B(X,[Y,Z]) = 0, so s is an ideal. Further, by Cartan’s criterion, the image of the
adjoint action of s, ad(s)⊂gl(g) is solvable, so s is solvable. Now suppose g is semisimple. By the
above, s⊂g is a solvable ideal, so s=0.
Going the other way, suppose s = 0. Any Lie algebra containing a nonzero solvable ideal must
contain a nonzero abelian ideal, namely the last nonzero element of the solvable ideal’s derived
series, so it suffices to show that any abelian ideal a ⊂ g must be zero. If X ∈ a and Y ∈ g,
then ad(X) ◦ ad(Y) maps g into a (since the latter is an ideal) and a to zero, so B(X,Y) =
Tr(ad(X)◦ad(Y))=0, so a⊂s=0.
6 Cartan Subalgebras
Proof of Theorem 3.3. ForX ∈g,letP (T)=det(T−ad(X)),i.e. thecharacteristicpolynomialof
X
i=n
ad(X). Ifn=dimg,wecanwritethecoefficientsofP (T)asfunctionsofX: P (T)= (cid:80) a (X)Ti.
X X i
i=0
If X = X e +···+X e for some basis {e } of g, a (X) is a homogeneous polynomial of degree
1 1 n n i i
n−i in X ,...,X . We call the least index l such that a is not identically zero on g the rank of g.
1 n l
7
Definition 6.1. We call an element H ∈ g regular if a (H) (cid:54)= 0; equivalently, an element H ∈ g
l
is regular if the dimension of the nilspace of ad(H) (i.e. the subspace of g killed by some power of
ad(H)) is minimal among the elements of g.
g must contain regular elements,2 and our eventual goal will be to show that for H ∈ g regular,
g (H) ⊂ g is a Cartan subalgebra. First, however, some observations about how the Lie bracket
0
and the Killing form behave on the Jordan blocks of an element H ∈g:
For any H ∈g we can decompose g according to the Jordan blocks of ad(H):
(cid:77)
g= g (H), (1)
λ
λ
where g (H)={X ∈g:∃k st for k(cid:48) ≥k,(ad(H)−λI)k(cid:48)(X)=0}.
λ
Observation 6.2. For any H ∈g, [g (H),g (H)]⊂g (H)
λ µ λ+µ
Proof. For X ∈g (H) and Y ∈g (H), we claim that
λ µ
k (cid:18) (cid:19)
(cid:88) k
(ad(H)−(λ+µ)I)k([X,Y])= [(ad(H)−λI)j(X),(ad(H)−µI)k−j(Y)]. (2)
j
j=0
All the terms on the right vanish for k sufficiently large, so this implies [X,Y] ∈ g (H). We
λ+µ
prove the equation by induction on k. By the induction hypothesis,
(ad(H)−(λ+µ)I)k([X,Y])
k−1(cid:18) (cid:19)
(cid:88) k−1
=(ad(H)−(λ+µ)I) [(ad(H)−λI)j(X),(ad(H)−µI)k−1−j(Y)],
j
j=0
but by the Jacobi identity,
(ad(H)−(λ+µ)I)[(ad(H)−λI)j(X),(ad(H)−µI)k−1−j(Y)]
=−(λ+µ)[(ad(H)−λI)j(X),(ad(H)−µI)k−1−j(Y)]+
[ad(H)(ad(H)−λI)j(X),(ad(H)−µI)k−1−j(Y)]+
[(ad(H)−λI)j(X),ad(H)(ad(H)−µI)k−1−j(Y)]
=[(ad(H)−λI)j+1(X),(ad(H)−µI)k−1−j(Y)]+
[(ad(H)−λI)j(X),(ad(H)−µI)k−j(Y)],
2Infact,theregularelementsofgformaZariskiopensubset.
8
so the above becomes
k−1(cid:18) (cid:19)
(cid:88) k−1
[(ad(H)−λI)j+1(X),(ad(H)−µI)k−1−j(Y)]+
j
j=0
k−1(cid:18) (cid:19)
(cid:88) k−1
[(ad(H)−λI)j(X),(ad(H)−µI)k−j(Y)]
j
j=0
k (cid:18) (cid:19)
(cid:88) k
= [(ad(H)−λI)j(X),(ad(H)−µI)k−j(Y)].
j
j=0
ThusifY ∈g (H)forλ(cid:54)=0thenbytheaboveobservation,ad(Y)takeseachg (H)(µpossibly0)to
λ µ
adifferentg (H);X ∈g (H)preservestheg (H)(notethatthismeansg (H)isasubalgebraofg),
µ(cid:48) 0 µ 0
soad(X)◦ad(Y)alsotakeseachg (H)toadifferentg (H),soB(X,Y)=Tr(ad(X)◦ad(Y))=0,
µ µ(cid:48)
(cid:76)
i.e. g (H) ⊥ g (H). By corollary 5.4, however, B is nondegenerate, so the restriction of B to
0 λ
λ(cid:54)=0
g (H) must be nondegenerate.
0
Now suppose H is a regular element of g.
Claim 6.3. g (H) is nilpotent.
0
Proof. Bycorollary3.5toEngel’stheoremitsufficestoshowthatforeachY ∈g (H),therestriction
0
of ad(Y) to g (H) is a nilpotent endomorphism. Let ad(cid:48)(Y) = ad(Y)| and let ad(cid:48)(cid:48)(Y) be
0 g0(H)
the induced endomorphism of ad(Y) on the vector space quotient g/g (H). We define subsets
0
U ={Y ∈g (H):ad(Y) is not nilpotent} and V ={Y ∈g (H):ad(cid:48)(cid:48)(Y) is invertible}.
0 0
Suppose that U is nonempty. U is an open subset of g (H) and V is a nonempty open subset
0
(since X ∈ V). Further, V is the complement of an algebraic variety, so it’s dense in g and so U
0
and V intersect, say at some element Y. The multiplicity of 0 as an eigenvalue of ad(Y)—i.e., the
dimensionofg (Y)—isthemultiplicityaof0asaneigenvalueofad(cid:48)(Y)plusthemultiplicitybof0
0
as an eigenvalue of ad(cid:48)(cid:48)(Y). Since Y ∈U, a<dimg (H). On the other hand, since Y ∈V, b=0.
0
Thus dimg (Y)<dimg (H), contradicting the regularity of H. Thus U =∅.
0 0
The claim shows that g (H) is solvable, so by Lie’s theorem there is a basis for g in which ad(X)
0
is upper-triangular for every X ∈ g (H). Thus for X,Y,Z ∈ g (H), ad([X,Y]) is strictly upper-
0 0
triangularandad(Z)isupper-triangularwithrespecttothesamebasis,soTr(ad([X,Y])◦ad(Z))=
B([X,Y],Z) = 0. However, B is nondegenerate on g (H), so it must be that [X,Y] = 0 for any
0
X,Y ∈g (H), i.e. g (H) is abelian.
0 0
Next we show that any element of g (H) is semisimple:
0
Claim 6.4. If n(g (H)) is the “normalizer” of g (H), defined as {X ∈ g : [X,g (H)] ⊂ g (H)},
0 0 0 0
then n(g (H))=g (H).
0 0
9
Proof. LetZ ∈n(g (H)),soinparticular[Z,H]∈g (H),meaningthatad(H)k[Z,H]=0forsome
0 0
k ∈N. Thus ad(H)k+1Z =0, so Z ∈g (H).
0
If c(g (H)) := {X ∈ g : [X,g (H)] = 0} (called the commutator of g (H)) then, since g (H)
0 0 0 0
is abelian, g (H) ⊂ c(g (H)). On the other hand, c(g (H)) ⊂ n(g (H)), which by the above is
0 0 0 0
contained in g (H), so c(g (H)) = g (H). Now, if X,Y ∈ g (H) and X is the nilpotent part of
0 0 0 0 n
X, then [X,Y] = 0 so [X ,Y] = 0. Thus X ∈ c(g (H)) = g (H). With respect to the same
n n 0 0
basis of g (H) as we used to show that g (H) is abelian, ad(X ) is strictly upper-triangular, so
0 0 n
B(X ,Y) = 0 for any Y ∈ g (H). However, the restriction of B to g (H) is nondegenerate, so
n 0 0
X = 0, so X is semisimple. Since the elements of g (H) commute, this shows that the adjoint
n 0
action of g (H) on g is diagonal.
0
Since c(g (H))=g (H), maximality is immediate.
0 0
7 Representations of sl
2
InthissectionwetakeabreakfromtheorytolookatasemisimpleLiealgebra,sl ,anduseaCartan
2
subalgebratocharacterizeallofitsirreduciblerepresentations. Thiswillgiveaconcreteexampleof
how Cartan subalgebras are used to study representations by decomposing them into root spaces,
as well as motivating the definition of roots in section 8. Further, the subalgebras s ∼= sl of
α 2
a general semisimple Lie algebra g (which we will construct in section 8) and particularly their
representations will be useful to the study of the roots and irreducible representations of g.
Recall the basis X,Y,H from example 2.3.
Claim 7.1. sl is semisimple.
2
Proof. First suppose that sl has a one-dimensional ideal, spanned by some element Z = aX +
2
bY +cH. Then[Z,X]=−(bH+2cX)isin thespanof Z, sob=0andlikewise [Z,Y]=aH−2cY
is in the span of Z, so a = 0 (the apparent alternatives [Z,X] = 0 and [Z,Y] = 0 are not in fact
alternatives, as they imply b = 0 and a = 0 anyway). Thus Z = cH, but the span of H is not an
ideal.
Thus the only way sl could fail to be semisimple is if it had a two-dimensional abelian ideal a,
2
spannedbysomenonzeroU,V. SinceU andV commute,theycansimultaneouslybeputinupper-
triangular form with respect to some basis B(cid:48) of C2. Thus, defining X(cid:48),Y(cid:48),H(cid:48) according to this
new basis, a must be the span of X(cid:48) and H(cid:48). However, X(cid:48) and H(cid:48) do not commute, so a cannot be
abelian.
In this case we can see that H spans a Cartan subalgebra h, and its adjoint action on sl has
2
eigenvalues 2,0,−2 on the eigenvectors X,H,Y respectively.
Now suppose V is an irreducible sl -representation. By corollary 4.3, H acts semisimply on V, so
2
we can decompose V = (cid:76)V , where each ω is an element of h∗ such that H(v) = ω (H)·v for
ωi i i
i
v ∈V . We call the ω the weights of V.
ωi i
10