Introduction to Lie Algebras and Their Representations: Michaelmas 2010

63 Pages·2011·0.517 MB·English

by  Ian Grojnowski, notes taken by Elena Yudovina

INTRODUCTION TO LIE ALGEBRAS AND THEIR
REPRESENTATIONS
PART III
MICHAELMAS 2010
IAN GROJNOWSKI
DPMMS, UNIVERSITY OF CAMBRIDGE
[email protected]
TYPESET BY ELENA YUDOVINA
STATISTICAL LABORATORY, UNIVERSITY OF CAMBRIDGE
[email protected]
1. Introduction, motivation, etc.
Lecture 1 For most of this lecture, we will be motivating why we care about Lie algebras
and why they are defined the way they are. If there are sections of it that don’t make sense,
they can be skipped (although if none of it makes sense, you should be worried).
Connected Lie groups are groups which are also differentiable manifolds. The simple
algebraic groups over algebraically closed fields (which is a nice class of Lie groups) are:
SL(n) = {A ∈ Mat |detA = 1}
n
SO(n) = {A ∈ SL |AAT = I}
n
SP(n) = {A ∈ SL |some symplectic form...}
n
and exactly 5 others (as we will actually see!)
(We are ignoring finite centers, so Spin(n), the double cover of SO(n), is missing.)
Remark. Each of these groups has a maximal compact subgroup. For example, SU(n) is the
maximal compact subgroup of SL(n), while SO (R) is the maximal compact subgroup of
n
SO (C). These are themselves Lie groups, of course. The representations of the maximal
n
compactsubgrouparethesameasthealgebraicrepresentationsofthesimplealgebraicgroup
are the same as the finite-dimensional representations of its (the simple algebraic group’s)
Lie algebra – which are what we study.
Lie groups complicated objects: for example, SU (C) is homotopic to a 3-sphere, and SU
2 n
is homotopic to a twisted product of a 3-sphere, a 5-sphere, a 7-sphere, etc. Thus, studying
Lie groups requires some amount of algebraic topology and a lot of algebraic geometry. We
want to replace these “complicated” disciplines by the “easy” discipline of linear algebra.
Therefore, instead of a Lie group G we will be considering g = T G, the tangent space to
I
G at the identity.
Definition. A linear algebraic group is a subgroup of some GL defined by polynomial
n
equations in the matrix coefficients.
We see that the examples above are linear algebraic groups.
Date: Michaelmas 2010.
1
2 IAN GROJNOWSKI
Remark. It is not altogether obvious that this is an intrinsic characterisation (it would
be nice not to depend on the embedding into GL ). The intrinsic characterisation is as
n
“affine algebraic groups,” that is, groups which are affine algebraic varieties and for which
multiplication and inverse are morphisms of affine algebraic varieties. One direction of this
identification is relatively easy (we just need to check that multiplication and inverse really
are morphisms); in the other direction, we need to show that any affine algebraic group has
a faithful finite-dimensional representation, i.e. embeds in GL(V). This involves looking at
the ring of functions and doing something with it.
We will now explain a tangent space if you haven’t met with it before.
Example 1.1. Let’s pretend we’re physicists, since we’re in their building anyway. Let
G = SL , and let |(cid:15)| (cid:28) 1. Then for
2
(cid:18) (cid:19) (cid:18) (cid:19)
1 a b
g = +(cid:15) +higher-order terms
1 c d
to lie in SL , we must have detg = 1, or
2
(cid:18) (cid:19)
1+(cid:15)a (cid:15)b
1 = det +h.o.t = 1+(cid:15)(a+d)+(cid:15)2(junk)+h.o.t
(cid:15)c 1+(cid:15)d
Therefore, to have g ∈ G we need to have a+d = 0.
To do this formally, we define the dual numbers
E = C[(cid:15)]/(cid:15)2 = {a+b(cid:15)|a,b ∈ C}
For a linear algebraic group G, we consider
G(E) = {A ∈ Mat (E)|A satisfies the polynomial equations defining G ⊂ GL }
n n
For example,
(cid:18) (cid:19)
α β
SL (E) = { |α,β,γ,δ ∈ E,αδ −βγ = 1}
2 γ δ
The natural map E → C, (cid:15) 7→ 0 gives a projection π : G(E) → G. We define the Lie algebra
of G as
g ∼= π−1(I) ∼= {X ∈ Mat (C)|I +(cid:15)X ∈ G(E)}.
n
In particular,
(cid:18) (cid:19)
a b
sl = { ∈ Mat (C)|a+d = 0}.
2 c d 2
Remark. I +X(cid:15) represents an “infinitesimal change at I in the direction X”. Equivalently,
the germ of a curve SpecC[[(cid:15)]] → G.
Exercise 1. (Do this only if you already know what a tangent space and tangent bundle
are.) Show that G(E) = TG is the tangent bundle to G, and g = T G is the tangent space
1
to G at 1.
Example 1.2. Let G = GL = {A ∈ Mat |A−1exists}.
n n
Claim:
G(E) := {A˜ ∈ Mat (E)|A˜−1exists}
n
= {A+B(cid:15)|A,B ∈ Mat C,A−1exists}
n
Indeed, (A+B(cid:15))(A−1 −A−1BA−1(cid:15)) = I.
INTRODUCTION TO LIE ALGEBRAS AND THEIR REPRESENTATIONS PART III MICHAELMAS 20103
Remark. ItisnotobviousthatGL isitselfalinearalgebraicgroup(whatarethepolynomial
n
equations for “determinant is non-zero”?). However, we can think of GL ⊂ Mat as
n n+1
(cid:18) (cid:19)
A
{ |det(A)×λ = 1}.
λ
Example 1.3. G = SL C.
n
Exercise 2. det(I +(cid:15)X) = 1+(cid:15)·trace(X)
As a corollary to the exercise, sl = {X ∈ Mat |trace(X) = 0}.
n n
Example 1.4. G = O C = {AAT = I}. Then
n
g := {X ∈ Mat C|(I +(cid:15)X)(I +(cid:15)X)T = I}
n
= {X ∈ Mat C|I +(cid:15)(X +XT) = I}
n
= {X ∈ Mat C|X +XT = 0}
n
the antisymmetric matrices. Now, if X +XT = 0 then 2×trace(X) = 0, and since we’re
working over C and not in characteristic 2, we conclude trace(X) = 0. Therefore, this is
also the Lie algebra of SO .
n
This is not terribly surprising, because topologically O has two connected components
n
corresponding to determinant +1 and −1 (they are images of each other via a reflection).
Since g = T G, it cannot see the det = −1 component, so this is expected.
1
The above example propmts the question: what exactly is it in the structure of g that we
get from G being a group?
The first thing to note is that g does not get a multiplication. Indeed, (I+A(cid:15))(I+B(cid:15)) =
I +(A+B)(cid:15), which has nothing to do with multiplication.
The bilinear operation that turns out to generalize nicely is the commutator, (P,Q) 7→
PQP−1Q−1. Taken as a map G × G → G that sends (I,I) 7→ I, this should give a map
T G×T G → T G by differentiation.
1 1 1
Remark. Generally, differentiation gives a linear map, whereas what we will get is a bilinear
map. This is because we will in fact differentiate each coordinate: i.e. first differentiate the
map f : G → G,Q 7→ PQP−1Q−1 with respect to Q to get a map f˜ : g → g (with P ∈ G
P P
still) and then differentiate again to get a map g×g → g.
What is this map?
Let P = I +A(cid:15), Q = I +Bδ where (cid:15)2 = δ2 = 0 but (cid:15)δ 6= 0. Then
PQP−1Q−1 := (I +A(cid:15))(I +Bδ)(I −A(cid:15))(I −Bδ)
= I +(AB −BA)(cid:15)δ.
Thus, the binary operation on g should be [A,B] = AB −BA.
Definition. The bracket of A and B is [A,B] = AB −BA.
Exercise 3. Show that (PQP−1Q−1)−1 = QPQ−1P−1 implies [A,B] = −[B,A], i.e. the
bracket is skew-symmetric.
Exercise 4. Show that the associativity of multiplication in G implies the Jacobi identity
0 = [[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]
4 IAN GROJNOWSKI
Remark. Themeaningof“implies”intheaboveexercisesisasfollows: wewanttothinkofthe
bracket as the derivative of the commutator, not as the explicit formula [A,B] = AB −BA
(which makes the skew-symmetry obvious, and the Jacobi identity only slightly less so). For
example, we could have started working in a different category.
We will now define our object of study.
Definition. Let K be a field, charK 6= 2,3. A Lie algebra g is a vector space over K
equipped with a bilinear map (Lie bracket) [,] : g×g → g with the following properties:
(1) [X,Y] = −[Y,X] (skew-symmetry);
(2) [[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y] = 0 (Jacobi identity).
Lecture 2 Examples of the Lie algebra definition from last time:
Example 1.5. (1) gl = Mat with [A,B] = AB −BA.
n n
(2) so = {A+AT = 0}
n
(3) sl = {trace(A) = 0} (note that while trace(AB) 6= 0 for A,B ∈ sl , we do have
n n
trace(AB) = trace(BA), so [A,B] ∈ sl even though AB 6∈ sl )
n n
 
1
 ... 
 
 1 
(4) sp = {A ∈ gl |JATJ−1+A = 0}whereJ isthesymplecticform 
2n 2n  −1 
 
 ... 
−1
∗ ∗ ... ∗
∗ ... ∗
 
(5) b = { ... ... ∈ gln} the upper triangular matrices (the name is b for Borel,
∗
but we don’t need to worry about that yet)
0 ∗ ... ∗
0 ... ∗
 
(6) n = { ... ... ∈ gln} the strictly upper triangular matrices (the name is n
0
for nilpotent, but we don’t need to worry about that yet)
(7) For any vector space V, let [,] : V ×V → V be the zero map. This is the “abelian
Lie algebra”.
Exercise 5. (1) Check directly that gl is a Lie algebra.
n
(2) Check that the other examples above are Lie subalgebras of gl , that is, vector
n
subspaces closed under [,].
(cid:18) (cid:19)
∗ ∗
Example 1.6. is not a Lie subalgebra of gl .
∗ 0 2
Exercise 6. Find the algebraic groups whose Lie algebras are given above.
Exercise 7. Classify all Lie algebras of dimension 3. (You might want to start with di-
mension 2. I’ll do dimension 1 for you on the board: skew symmetry of bracket means that
bracket is zero, so there is only the abelian one.)
INTRODUCTION TO LIE ALGEBRAS AND THEIR REPRESENTATIONS PART III MICHAELMAS 20105
Definition. A representation of a Lie algebra g on a vector space V is a homomorphism of
Lie algebras φ : g → gl . We say that g acts on V.
V
Observe that, above, our Lie algebras were defined with a (faithful) representation in tow.
There is always a representation coming from the Lie algebra acting on itself (it is the
derivative of the group acting on itself by conjugation):
Definition. For x ∈ g define ad(x) : g → g as y 7→ [x,y].
Lemma 1.1. ad : g → Endg is a representation (called the adjoint representation).
Proof. We need to check whether ad([x,y]) = ad(x)ad(y) − ad(y)ad(x), i.e. whether the
equality will hold when we apply it to a third element z. But:
ad([x,y])(z) = [[x,y],z]
and
ad(x)ad(y)(z)−ad(y)ad(x)(z) = [x,[y,z]]−[y,[x,z]] = −[[y,z],x]−[[z,x],y].
These are equal by the Jacobi identity. (cid:3)
Definition. The center of g is {x ∈ g : [x,y] = 0, ∀y ∈ g} = ker(ad : g → Endg).
Thus, if g has no center, then ad is an embedding of g into Endg (and conversely, of
course).
Is it true that every finite-dimensional Lie algebra embeds into some gl , i.e. is lin-
V
ear? (Equivalently, is it true that every finite-dimensional Lie algebra has a faithful finite-
dimensional representation?) We see that if g has no center, then the adjoint representation
(cid:18) (cid:19)
0 ∗
makes this true. On the other hand, if we look inside gl , then n = is abelian, so
2 0 0
maps to zero in gl , despite the fact that we started with n ⊂ gl . That is, we can’t always
n 2
just take ad.
Theorem 1.2 (Ado’s theorem; we will not prove it). Any finite-dimensional Lie algebra
over K is a subalgebra of gl , i.e. admits a faithful finite-dimensional representation.
n
Remark. ThisistheLiealgebraequivalentofthestatementwemadelasttimeaboutalgebraic
groups embedding into GL(V). That theorem was “easy” because there was a natural
representation to look at (functions on the algebraic group). There isn’t such a natural
object for Lie groups, so Ado’s theorem is actually hard.
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
0 1 1 0 0 0
Example 1.7. g = sl with basis e = , h = , f = (these are the
2 0 0 0 −1 1 0
standard names). Then [e,f] = h, [h,e] = 2e, and [h,f] = −2f.
Exercise 8. Check this!
A representation of sl is a triple E,F,H ∈ Mat satisfying the same bracket relations.
2 n
Where might we find such a thing?
In this lecture and the next, we will get them as derivatives of representations of SL . We
2
will then rederive them from just the linear algebra.
Definition. If G is an algebraic group, an algebraic representation of G on V is a homomor-
phism of groups ρ : G → GL(V) defined by polynomial equations in the matrix coefficients.
6 IAN GROJNOWSKI
(This can be done so that it is invariant of the embedding into GL . Think about it.)
n
To get a representation of the Lie algebra out of this, we again use the dual numbers E. If
we use E = K[(cid:15)]/(cid:15)2 instead of K, we will get a homomorphism of groups G(E) → GL (E).
V
Moreover, since ρ(I) = I and it commutes with the projection map, we get
ρ(I +A(cid:15)) = I +(cid:15)×(some function of A) = I +(cid:15)dρ(A)
(this is to be taken as the definition of dρ(A)).
Exercise 9. dρ is the derivative of ρ evaluated at I (i.e., dρ : T G → T GL ).
I I V
Exercise 10. The fact that ρ : G → GL was a group homomorphism means that dρ : g →
V
gl is a Lie algebra homomorphism, i.e. V is a representation of g.
V
Example 1.8. G = SL .
2
Let L(n) be the space of homogeneous polynomials of degree n in two variables x,y, with
basis xn,xn−1y,...,xyn−1,yn (so dimL(n) = n+1). Then GL acts on L(n) by change of
2
(cid:18) (cid:19)
a b
coordinates: for g = and f ∈ L(n) we have
c d
((ρ g)f)(x,y) = f(ax+cy,bx+dy)
n
In particular, ρ is the trivial representation of GL , ρ is the usual 2-dimensional represen-
0 2 1
tation K2, and
 
(cid:18) (cid:19) a2 ab b2
a b
ρ2 c d = 2ac ad+bc 2bd
c2 cd d2
Since GL acts on L(n), we see that SL acts on L(n).
2 2
Remark. The proper way to think of this is as follows. GL acts on P1 and on O(n) on P1,
2
hence on the global sections Γ(O(n),P1) = SnK2. That’s the source of these representations.
We can do this sort of thing for all the other algebraic groups we listed previously, using flag
varieties instead of P1 and possibly higher (co)homologies instead of the global sections (this
is a theorem of Borel, Weil, and Bott). That, however, requires algebraic geometry, and gets
harder to do in infinitely many dimensions.
(cid:18) (cid:19)
0 1
Differentiating the above representations, e.g. for e = , we see
0 0
ρ(I +(cid:15)e)xiyj = xi((cid:15)x+y)j = xiyj +(cid:15)jxi+1yj−1
Therefore, dρ(e)xiyj = jxi+1yj−1,
Exercise 11. (1) In the action of the Lie algebra,
e(xiyj) = jxi+1yj−1
f(xiyj) = ixi−1yj+1
h(xiyj) = (i−j)xiyj.
(2) Check directly that these formulae give a representation of sl .
2
(3) Check that L(2) is the adjoint representation.
(4) Show that the formulae e = x ∂ , f = y ∂ , h = x ∂ − y ∂ give an (infinite-
∂y ∂x ∂x ∂y
dimensional!) representation of sl on k[x,y].
2
INTRODUCTION TO LIE ALGEBRAS AND THEIR REPRESENTATIONS PART III MICHAELMAS 20107
(5) Let char(k) = 0. Show that L(n) is an irreducible representation of sl , hence of
2
SL .
2
Lecture 3 Last time we defined a functor
(algebraic representations of a linear algebraic group G) →
(Lie algebra representations of g = Lie(G))
via ρ 7→ dρ.
This functor is not as nice as you would like to believe, except for the Lie algebras we care
about.
Example 1.9. G = C× =⇒ g = Lie(G) = C with [x,y] = 0.
A representation of g = C on V is a matrix A ∈ End(V) (to ρ : C → End(V) corresponds
A = ρ(1)).
W ⊆ V is a submodule iff AW ⊆ W, and ρ is isomorphic to ρ0 : g → End(V0) iff A and
A0 are conjugate as matrices.
Therefore, representations of g correspond to the Jordan normal form of matrices.
As any linear transformation over C has an eigenvector, there’s always a 1D subrep of V.
Therefore, V is irreducible iff dimV = 1. Also, V is completely decomposable (a direct sum
of irreducible representations) iff A is diagonalizable.
 
0 1
0 1
 
For example, if A =  ... ...  then the associated representation is indecom-
 
 ... 1
0
posable, but not irreducible. Invariant subspaces are he i,he ,e i,...,he ,e ,...,e i, and
1 1 2 1 2 n
none of them has an invariant orthogonal complement.
Now let’s look at the representations of G = C×.
Theirreduciblerepresentationsareρ : G → GL = Aut(C)viaz 7→ (multiplication by) zn
n 1
for n ∈ Z. Every finite-dimensional representation is a direct sum of these.
Remark. Proving these statements about representations of G takes some theory of algebraic
groups. However, you probably know this for S1, which is homotopic to G. Recall that to
get complete reducibility in the finite group setting you took an average over the group; you
can do the same thing here, because S1 is compact and so has finite volume. In fact, for the
theory that we study, our linear algebraic groups have maximal compact subgroups, which
have the same representations, which for this reason are completely reducible. We will not
prove that representations of the linear algebraic groups are the same as the representations
of their maximal compact subgroups.
Observe that the representations of G and of g are not the same. Indeed, ρ 7→ dρ will
send ρ 7→ n ∈ C (prove it!). Therefore, irreducible representations of G embed into ir-
n
reducible representations of g, but the map is not remotely surjective (not to mention the
decomposability issue).
The above example isn’t very surprising, since g is also the Lie algebra Lie(C,+), and we
should not expect representations of g to coincide with those of C×. The surprising result is
8 IAN GROJNOWSKI
Theorem 1.3 (Lie). ρ 7→ dρ is an equivalence of categories RepG → Repg if G is a simply
connected simple algebraic group.
Remark. A “simple algebraic group” is not simple in the usual sense: e.g. SL has a center,
n
which is obviously a normal subgroup. However, if G is simply connected and simple in the
above sense, then the center of G is finite (e.g. Z = µ , the nth roots of unity), and the
SLn n
normal subgroups are subgroups of the center.
Exercise 12. If G is an algebraic group, and Z is a finite central subgroup of G, then
Lie(G/Z) = Lie(G). (Morally, Z identifies points “far away”, and therefore does not affect
the tangent space at I.)
We have now seen that the map (algebraic groups) → (Lie algebras) is not injective (see
above exercise, or – more shockingly – G = C,C×). In fact,
Exercise 13. Let G = C× n C where C× acts on C via t · λ = tnλ (so (t,λ)(t0,λ0) =
n
(tt0,t0nλ + λ0)). Show that G ∼= G iff n = ±m. Show that Lie(G ) ∼= Cx + Cy with
n m n
[x,y] = y independently of n.
The map also isn’t surjective (its image are the “algebraic Lie algebras”). This is easily
seen in characteristic p; for example, sl /center cannot be the image of an algebraic group.
p
In general, algebraic groups have a Jordan decomposition – every element can be written as
(semisimple)×(nilpotent), – and therefore the algebraic Lie algebras should have a Jordan
decomposition as well. In Bourbaki, you can find an example of a 5-dimensional Lie algebra,
for which the semisimple and the nilpotent elements lie only in the ambient gl and not in
V
the algebra as well.
2. Representations of sl
2
From now on, all algebras and representations are over C (an algebraically closed field of
characteristic 0). Periodically we’ll mention which results don’t need this.
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
0 1 1 0 0 0
Recallthebasisofsl wase = , h = , andf = withcommutation
2 0 0 0 −1 1 0
relations [e,f] = h, [h,e] = 2e, [h,f] = −2f.
For the next while, we will be proving the following
Theorem 2.1.
(1) For all n ≥ 0 there exists a unique irreducible representation of g = sl of dimension
2
n+1. (Recall L(n) from the previuos lecture; these are the only ones.)
(2) Every finite-dimensional representation of sl is a direct sum of irreducible represen-
2
tations.
Let V be a representation of sl .
2
Definition. The λ-weight space for V is V = {v ∈ V : hv = λv}, the eigenvectors of h with
λ
eigenvalue λ.
Example 2.1. In L(n), we had L(n) = Cxiyj for λ = i−j.
λ
Let v ∈ V , and consider ev. We have
λ
hev = (he−eh+eh)v = [h,e]v +e(hv) = 2ev +eλv = (λ+2)ev.
That is, if v ∈ V then ev ∈ V and similarly fv ∈ V . (These are clearly iff.)
λ λ+2 λ−2
INTRODUCTION TO LIE ALGEBRAS AND THEIR REPRESENTATIONS PART III MICHAELMAS 20109
We will think of this pictorially as a chain
e
... (cid:10) V (cid:10) V (cid:10) V (cid:10) ...
λ−2 λ λ+2
f
E.g., in L(n) we had xiyj sitting in the place of the V ’s, and the chain went from λ = n to
λ
λ = −n:
V ←→ V ←→ V ←→ ... ←→ V ←→ V ←→ V
n n−2 n−4 −(n−4) −(n−2) −n
hxni (cid:10) hxn−1yi (cid:10) hxn−2y2i (cid:10) ... (cid:10) hx2yn−2i (cid:10) hxyn−1i (cid:10) hyni
(Note that the string for L(n) has n+1 elements.)
Definition. If v ∈ V ∩kere, i.e. hv = λv and ev = 0, then we say that v is a highest-weight
λ
vector of weight λ.
Lemma 2.2. Let V be a representation of sl and v ∈ V a highest-weight vector of weight
2
λ. Then W = hv,fv,f2v,...i is an sl -invariant subspace, i.e. a subrepresentation.
2
Proof. We need to show hW ⊆ W, fW ⊆ W, and eW ⊆ W.
Note that fW ⊆ W by construction.
Further, since v ∈ V , we saw above that fkv ∈ V , and so hW ⊆ W as well.
λ λ−2k
Finally,
ev = 0 ∈ W
efv = ([e,f]+fe)v = hv +f(0) = λv ∈ W
ef2v = ([e,f]+fe)fv = (λ−2)fv +f(λv) = (2λ−2)fv ∈ W
ef3v = ([e,f]+fe)f2v = (λ−4)f2v +f(2λ−2)fv = (3λ−6)f2v ∈ W
and so on. (cid:3)
Exercise 14. efnv = n(λ−n+1)fn−1v for v a highest-weight vector of weight λ.
Lemma 2.3. Let V be a finite-dimensional representation of sl and v ∈ V a highest-weight
2
vector of weight λ. Then λ ∈ Z .
≥0
Remark. Somehow, theintegralityconditionistheLiealgebrarememberingsomethingabout
the topology of the group. (Saying what this “something” is more precisely is difficult.)
Proof. Note that fkv all lie in different eigenspaces of h, so if they are all nonzero then they
are linearly independent. Since dimV < ∞, we conclude that there exists a k such that
fkv 6= 0 and fk+rv = 0 for all r ≥ 1. Then by the above exercise,
0 = efk+1v = (k +1)(λ−k)fkv
from which λ = k, a nonnegative integer. (cid:3)
Proposition 2.4. If V is a finite-dimensional representation of sl , it has a highest-weight
2
vector.
Proof. We’re over C, so we can pick some eigenvector of h. Now apply e to it repeatedly:
v,ev,e2v,... belong to different eigenspaces of h, so if they are nonzero, they are linearly
independent. Therefore, there must be some k such that ekv 6= 0 but ek+rv = 0 for all r ≥ 1.
Then ekv is a highest-weight vector of weight (λ+2k). (cid:3)
10 IAN GROJNOWSKI
Corollary 2.5. If V is an irreducible representation of sl then dimV = n+1, and V has
2
a basis v ,v ,...,v on which sl acts as hv = (n−2i)v ; fv = v (with fv = 0); and
0 1 n 2 i i i i+1 n
ev = i(n − i + 1)v . In particular, there exists a unique (n + 1)-dimensional irreducible
i i−1
representation, which must be isomorphic to L(n).
Exercise 15. Work out how this basis is related to the basis we previously had for L(n).
Lecture 4
We will now show that every finite-dimensional representation of sl C is a direct sum of ir-
2
reducible representations, or(equivalently)the category of finite-dimensionalrepresentations
of sl C is semisimple, or (equivalently) every representation of sl C is completely reducible.
2 2
Morally, we’ve shown that every such representation consists of “strings” descending from
the highest weight, and we’ll now show that these strings don’t interact. We’ll first show
that they don’t interact when they have different highest weights (easier) and then that they
also don’t interact when they have the same highest weight (harder).
We will also show that h acts diagonalisably on every finite-dimensional representation,
while e and f are nilpotent. In fact,
Remark.
(cid:18) (cid:19) (cid:18) (cid:19)
a t
Spanh = { } = Lie{ } ⊂ Lie(SL )
−a t−1 2
(cid:18) (cid:19)
t
where { } is the maximal torus inside SL . Like for C×, the representations here
t−1 2
should correspond to integers. In some hazy way, the fact that we are picking out the
representations of a circle and not any other 1-dimensional Lie algebra is a sign of the Lie
algebra remembering something about the group. (Saying what it is remembering and how
is a hard question.)
Example 2.2. Recall that C[x,y] is a representation of sl via the action by differentail
2
operators, and is a direct sum of the representations L(n).
Exercise 16. Show that the formulae e = x ∂ , f = y ∂ , h = x ∂ −y ∂ give a representation
∂y ∂x ∂x ∂y
of sl on xλyµC[x/y,y/x] for any λ,µ ∈ C, and describe the submodule structure.
2
Definition. Let V be a finite-dimensional representation of sl . Define Ω = ef+fe+1h2 ∈
2 2
End(V) to be the Casimir of sl .
2
Lemma 2.6. Ω is central; that is, eΩ = Ωe, fΩ = Ωf, and hΩ = Ωh.
Exercise 17. Prove it. For example,
1 1 1
eΩ = e(ef +fe+ h2) = e(ef −fe)+2(efe)+ (eh−he)h+ heh
2 2 2
1 1 1
= 2efe+ heh = (fe−ef)e+2efe+ h(he−eh)+ heh = Ωe
2 2 2
Observe that we can write Ω = (ef − fe) + 2fe + 1h2 = (1h2 + h) + 2fe. This will be
2 2
useful later.
Corollary 2.7. If V is an irreducible representation of sl , then Ω acts on V by a scalar.
2
Proof. Schur’s lemma. (cid:3)