Table Of ContentRepresentation Theory
Spring 2013, taught by Joe Harris.
Contents
1 Basic Definitions 2
2 Examples of Lie Groups 2
3 Isogenies of Lie Groups 3
4 Maps Between Lie Groups 4
5 Lie Algebras 6
6 The Exponential Map 7
7 Classes of Lie Algebras 9
8 Low-Dimensional Simple Lie Algebras 13
8.1 sl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2
8.2 Plethysm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
8.3 sl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3
9 Simple Lie Algebras 20
10 Analysis of sl 23
n
11 Geometric Plethysm 25
12 The Symplectic Groups Sp 27
2n
13 Orthogonal Lie Algebras 30
13.1 Clifford Algebras and Spin Representations . . . . . . . . . . . . . . . . . . . . . . . 34
14 Classification Theorem 35
15 Construction of g and G 40
2 2
1
16 Complex Lie Groups Associated to Simple Lie Algebras 41
16.1 Forms of sl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
n+1
16.2 Forms of sp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
2n
16.3 Forms of so . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
m
17 Representation Rings and Characters 43
18 The Weyl Character Formula 46
19 Real Forms 48
1 Basic Definitions
A C∞ manifold is a set X with the structure of a topological space and an atlas: an open cover
{U } and homeomorphisms ϕ : U −→∼ ∆ ∼= Rn such that ϕ ◦ϕ−1 is C∞ where defined.
α α α α β
We can talk about C∞ functions on X and C∞ maps X → Y via working with atlases.
Recall that a group is a set X with the structure of e ∈ X, m : X ×X → X, and i : X → X
satisfying appropriate axioms.
A Lie group is a C∞ manifold G with a group structure such that m and i are C∞. We say that
a map G → H is a Lie group homomorphism if it is a group homomorphism which is also a C∞
map.
A Lie subgroup H of G is an H ⊆ G which is a subgroup and a closed submanifold.
As a nonexample, consider a line through the origin in R2 having an irrational slope. The map
R2 → R2/Z2 takes this line to a dense subset of the torus, called an immersed submanifold of the
torus.
A representation of G on a vector space V is a map G → GL(V), the set of invertible linear
maps on V.
2 Examples of Lie Groups
• GL (R) ⊆ Rn2 is a Lie group because of the explicit descriptions of matrix multiplication and
n
inversion. This makes GL(V) a Lie group for any finite dimensional vector space V.
• SL (R) is a submanifold of GL (R) because det : M (R) → R) has nonvanishing differential.
n n n
(It’senoughtocheckattheidentity.)WecanalsolookatSL(V),thesetofϕpreservingavol-
ume form. That is, for some nonzero α ∈ ΛnV∗, we have α(ϕ(v ),...,ϕ(v )) = α(v ,...,v ).
1 n 1 n
• A flag V in V is a nested sequence of subspaces
0 (cid:40) V (cid:40) ··· (cid:40) V (cid:40) V (2.1)
1 k
2
Define B(V) to be the set of automorphisms of V preserving V; that is, ϕ such that ϕ(V ) ⊆ V
i i
for every i (in this case, ϕ(V ) necessarily equals V .)
i i
A full flag is a V where V has dimension i. In an appropriate basis for V, B(V) consists of
i
upper triangular matrices. More generally, B(V) consists of block upper triangular matrices
in an appropriate basis.
• Let N(V) be the subgroup of B(V) consisting of those ϕ such that (ϕ−I)(V ) ⊆ V . That
i i−1
is, ϕ is the identity on successive quotients.
• Let Q be a symmetric bilinear form on V. Q : V × V → R is an element of Sym2V∗.
Assume that Q is nondegenerate. Define O(V,Q) to be the set of ϕ preserving Q; that is,
Q(ϕ(v),ϕ(w)) = Q(v,w). In matrix form, express Q as a symmetric n × n matrix M, so
Q(v,w) = vTMw. Then
O(V,Q) = {A ∈ GL (R) : ATMA = M}. (2.2)
n
If M = I, then O(V,Q) is called O(n). More generally, given Q, we can find a basis such that
M is diagonal with diagonal entries 1 and −1. We say that M has signature (k,n−k), and
O(V,Q) is called O .
k,n−k
SO(V,Q) is the further subgroup of determinant-1 transformations.
• Take Q to be a skew-symmetric nondegenerate bilinear form, so Q ∈ Λ2V∗. Q can be specified
by a skew-symmetric M, and under an appropriate basis, we can write
(cid:18) (cid:19)
O I
M = (2.3)
−I 0
The group of matrices preserving Q is called Sp .
2n
• If V is 7-dimensional and α is a general skew-symmetric trilinear form (α ∈ Λ3V∗), then the
group of automorphisms of V preserving α is nontrivial and called G .
2
3 Isogenies of Lie Groups
Proposition 3.1. If G is a connected Lie group and U is an open subset containing e, then U
generates G.
Proof. ByreplacingU byU∩U−1,wecanwithoutlossofgeneralityassumeU = U−1.LetH bethe
subgroup generated by U. Suppose H (cid:54)= G. Then there exists g ∈ ∂H = H\H. Then gU ∩H (cid:54)= ∅;
choose h in the intersection. If h = gu for u ∈ U ⊆ H, then g = hu−1 ∈ H, a contradiction.
Let H be a Lie group and Γ ⊆ Z(H) discrete. Set G = H/Γ.
Claim. G has the unique structure of a Lie group such that the quotient map H → G is a map of
Lie groups. (In fact, the map is a covering map.)
On the other hand, now start with G a Lie group and let H be a connected topological space
such that H → G is a covering map. Clearly H inherits the structure of a manifold. Choose an
element e(cid:48) ∈ H mapping to e. Then:
3
Theorem 3.2. There exists a unique group structure on H such that H is a Lie group and H → G
is a map of Lie groups.
Proof. First consider the case where H is the universal cover of G. Consider the group law on G as
a map G×G → G. We can uniquely lift
H ×H mH H
(3.1)
G×G mG G
We get a unique map H ×H → H mapping (e(cid:48),e(cid:48)) to e(cid:48). Inversion is similar.
Toseethatm isassociative,considera(bc)((ab)c)−1.ThisisacontinuousmapH×H×H → H
H
lying in the fiber of e. So this map is a constant, identically e(cid:48).
Proposition 3.3. If G is a connected Lie group and Γ ⊆ G is discrete and normal, then Γ ⊆ Z(G).
Proof. For h ∈ Γ, the map G → Γ by g (cid:55)→ ghg−1 is continuous, so must be constant. The value at
e is h, so ghg−1 = g for every g ∈ G.
We say that connected groups G,H are isogenous if there exists a Lie group map G (cid:16) H with
discrete kernel (that is, a covering space map). The isogeny classes are the equivalence classes of
the equivalence relation generated by isogenies. In each isogeny class, there exists a unique initial
member,whichwe’llcallH,thesimplyconnectedform:theuniversalcoveringspaceofanymember.
If Z(H) is discrete, then there also exists a final object H/Z(H), called the adjoint form.
Our basic approach is to describe all representations of the simply connected form, and then say
which ones descend to a given member of the isogeny class.
4 Maps Between Lie Groups
Here is a problem: given G,H, describe all Lie group maps G → H. (The case H = GL(V)
corresponds to representation theory.) We’ll assume that G is simply connected. In fact, ρ : G → H
is determined not just by its values on any open neighborhood of e, but even dρ : T G → T H. In
e e e
other words we have an inclusion
Hom (G,H) (cid:44)→ Hom(T G,T H). (4.1)
Lie e e
We would like to know which linear maps T G → T H actually arise as differentials of Lie group
e e
maps.
If ρ is a Lie group map, then ρ(gh) = ρ(g)ρ(h). In other words, the following diagram commutes
for every g:
4
ρ
G H
mg mρ(g) (4.2)
ρ
G H
where m is multiplication by g. To focus our attention on G,H near the identity, consider
g
instead ψ which is conjugation by g. We also have the commutative diagram
g
ρ
G H
(4.3)
ψg ψρ(g)
ρ
G H
But now ψ (e) = e so we can consider (dψ ) : T G → T G. A get a map G → GL(T G) by
g g e e e e
g (cid:55)→ (dψ ) , called the adjoint representation Ad, a distinguished representation of G. Taking the
g e
differential of Ad, we get a map ad : T G → End(T G). By taking transpose, we can also view ad
e e
as a map T G×T G → T G by (X,Y) (cid:55)→ ad(X)(Y) = [X,Y].
e e e
The point is that if ρ is any Lie group map, then (dρ) respects the binary operation ad on T G.
e e
That is,
dρ×dρ
T G×T G T H ×T H
e e e e
(4.4)
[ , ] [ , ]
dρ
T G T H
e e
In fact, if G is simply connected, then any linear map ϕ : T G → T H is the differential of a Lie
e e
group map ρ : G → H if and only if [ϕ(X),ϕ(Y)] = ϕ([X,Y]).
Fact. We can describe the map [ , ] explicitly.
Start with G = GL(V). Then T G = End(V). Observe that ψ : G → G by h (cid:55)→ ghg−1
e G
extends naturally to End(V), and this extension is the differential of ψ , since ψ is linear. We have
g g
dψ (X) = gXg−1.
g
To describe ad, we start with a given X ∈ End(V) and choose an arc γ : (−(cid:15),(cid:15)) → G with
γ(0) = e and γ(cid:48)(0) = X. Then
[X,Y] = ad(X)(Y) (4.5)
d (cid:12)
= (cid:12) γ Yγ−1 (4.6)
dt(cid:12)t=0 t t
= γ(cid:48)Yγ−1+γ Y(γ−1)(cid:48) (4.7)
0 0 0 0
= γ(cid:48)Yγ−1−γ Y(γ−1γ(cid:48)γ−1) (4.8)
0 0 0 0 0 0
= XY −YX. (4.9)
This formula also applies to any subgroup of GL(V). This has two consequences:
5
• [X,Y] = −[Y,X]; that is, the bracket is skew-symmetric.
• The Jacobi identity holds: for X,Y,Z ∈ T G,
e
[[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y] = 0. (4.10)
5 Lie Algebras
We define a Lie algebra g to be a vector space with a skew-symmetric bilinear map g × g → g
satisfying the Jacobi identity.
So one of our above claims is that for G simply connected and H any Lie group, if g = T G and
e
h = T H, then
e
Hom (G,H) = Hom (g,h). (5.1)
Liegrp Liealg
We will split this assertion into the following two theorems:
Theorem 5.1. If G is connected, any Lie group map ρ : G → H is determined by its differential
dρ : g → h.
Theorem 5.2. If G is simply connected, a linear map ϕ : g → h is the differential of a Lie group
map ρ : G → H if and only if ϕ is a Lie algebra map.
Whycouldn’twejustdefine[X,Y] = XY −YX?TheproblemisthatXY isnotalwaysdefined:
given G (cid:44)→ GL(V) and X,Y ∈ T G viewed as endomorphisms of V, XY depends on the embedding
e
of G into GL(V), and is not canonical. In fact, it may not even lie in the image of T G.
e
We may sometimes write End(V) as gl(V) if we are interested in the Lie algebra structure.
ALiealgebramorphismfromgtohisalinearmapϕ : g → hsuchthatϕ([X,Y]) = [ϕ(X),ϕ(Y)].
A representation of g on V is a Lie algebra morphism g → gl(V). This can be viewed as an action
of g on V such that for every X,Y ∈ g, X(Y(v))−Y(X(v)) = [X,Y](v).
Observe thatgiven agroupGactingon vectorspacesV andW,weget anactionofGonV ⊗W
by g(v ⊗w) = g(v)⊗g(w). In other words, given ρ : G → GL(V) and ρ : G → GL(W), we
V W
obtain a map ρ = ρ ⊗ρ .
V⊗W V W
How do the corresponding Lie algebra maps relate to this? Suppose ϕ : g → gl(V) and ϕ :
V W
g → gl(W)aregiven.Letγ beanarcinGwithγ = eandγ(cid:48) = X.Wehaveγ (v⊗w) = γ (v)⊗γ (w).
0 0 t t t
Differentiating at t = 0,
X(v⊗w) = X(v)⊗w+v⊗X(w). (5.2)
In general, given two Lie algebra representations, the tensor product representation will be
defined by the above.
Here are some other examples:
• If G acts on V, then G correspondingly acts on Sym2V characterized by g(v2) = g(v)2. The
corresponding representation on g is given by X(v2) = 2vX(v).
6
• Given ϕ : V → W, recall that the transpose map ϕT : W∗ → V∗ is the map described by
composition: [ϕT(λ)](v) = λ(ϕ(v)).
Now if G acts on V, then G acts on V∗ by ρ (g) = ρ (g−1)T. The corresponding represen-
V∗ V
tation of g is given by ϕ (X) = −ϕ (X)T.
V∗ V
Remark. • The definition of a Lie algebra doesn’t require g to be finite dimensional, but we’ll
only deal with finite dimensional ones.
• The definition also doesn’t specify the base field. We’ll assume that the base field is either R
or C, and will use R by default.
• In fact, every Lie algebra can be embedded in gl (R) for some n. This implies that every Lie
n
algebra arises from a Lie group.
(cid:94)
• But not every Lie group is a subgroup of GL (R). An example is the universal cover SL (R)
n 2
of SL (R). Another example is given by
2
1 a b (cid:44) 1 0 b
0 1 c 0 1 0 : b ∈ Z . (5.3)
0 0 1 0 0 1
6 The Exponential Map
Suppose G is any Lie group, and X ∈ g = T G. We can define a vector field V = V on G by
e X
V(g) = (dm ) (X). (We may write α instead of dα for a map α on manifolds. If the domain is a
g e ∗
subset of R, then α(cid:48) may be used.)
Now recall from differential geometry the following
Theorem 6.1. If M is any manifold, p ∈ M, and V is a vector field on M, then there exists a
unique germ of an arc γ : (−(cid:15),(cid:15)) → M such that γ(0) = p and γ(cid:48)(t) = v(γ(t)).
Applythistheoremtotheabovesituation.GivenX ∈ g = T GandV thecorrespondingvector
e X
field, let γ : (−(cid:15),(cid:15)) → G be the corresponding integral curve with γ(0) = e.
Claim. γ is a group homomorphism where defined. That is, γ(s+t) = γ(s)γ(t).
Proof. Fix s, and set α(t) = γ(s+t) and β(t) = γ(s)γ(t). Then α(cid:48)(t) = V(γ(s+t)) while β(cid:48)(t) =
(dm ) V(γ(t)). Since V is a left invariant vector field, α(t) and β(t) are both integral curves
γ(s) γ(t)
for V starting at γ(s), so they must be the same arc.
Now using the claim, we can extend γ to R.
In summary, for each X ∈ g, we get a Lie group morphism ϕ : R → G such that ϕ(cid:48) (0) = X.
X X
These properties uniquely determine ϕ . Such a map is called a 1-parameter subgroup.
X
By uniqueness, we get the following naturality: if ρ : G → H is any Lie group map and X ∈ g,
then
ρ◦ϕ = ϕ . (6.1)
X dρ(X)
7
Remark. • ϕ does not need to be injective. However, if X (cid:54)= 0, the kernel will be a nontrivial
X
closed subgroup of R, therefore either {0} or rZ for some r > 0. In these cases, the image of
ϕ will be isomorphic to R or S1.
• If ϕ is injective, the image may be a closed subgroup or simply an immersed subgroup.
X
• We have ϕ (t) = ϕ (λt).
λX X
We now define the exponential map exp : g → G by exp(X) = ϕ (1). So exp restricted to each
X
line through 0 ∈ g is a 1-parameter subgroup.
Fact. The map exp is the unique C∞ map g → G such that the differential at 0 is the identity and
the restriction to lines is a 1-parameter subgroup.
By this characterization of exp, given a Lie group map ρ : G → H, we have
ρ
G H
exp exp (6.2)
dρ
g h
We’ll now show that exp is C∞. Explicitly, if G = GL(V) and g = gl(V), then
X2 X3
exp(X) = 1+X + + +··· (6.3)
2 6
This is also true for any subgroup of GL(V), and therefore any Lie group whose Lie algebra
embeds into gl(V) for some V.
On a small neighborhood of e ∈ G, we can define an inverse to exp, called log. In the case of
GL(V),
(g−I)2 (g−I)3
log(g) = (g−I)− + −··· (6.4)
2 3
Given X,Y ∈ g sufficiently close to 0, we will define X(cid:63)Y = log(exp(X)exp(Y)). We then have
(cid:18)(cid:18) X2 (cid:19)(cid:18) Y2 (cid:19)(cid:19)
X (cid:63)Y = log 1+X + +··· 1+Y + +··· (6.5)
2 2
(cid:18) (cid:18)X2 Y2(cid:19) (cid:19)
= log 1+(X +Y)+ +XY + +··· (6.6)
2 2
(cid:18) (cid:18)X2 Y2(cid:19) (cid:19) 1 (cid:18) (cid:18)X2 Y2(cid:19)(cid:19)
= X +Y + +XY + +··· − X +Y + +XY + +··· (6.7)
2 2 2 2 2
(cid:18)X2 Y2 (X +Y)2(cid:19)
= X +Y + +XY + − +··· (6.8)
2 2 2
[X,Y]
= X +Y + +··· (6.9)
2
The final formula of Campbell-Hausdorff-Dynkin-Baker-etc. is
8
[X,Y] 1
X (cid:63)Y = X +Y + + ([X,[X,Y]]−[Y,[X,Y]])+··· (6.10)
2 12
The important fact is that this expression involves only addition and Lie brackets.
Corollary 6.2. Given h a vector subspace of g, the subgroup H of G generated by exp(h) is an
immersed subgroup with tangent space T H = h if and only if h is a Lie subalgebra of G.
e
We get a bijection between connected immersed subgroups of G and Lie subalgebras of g.
We conclude by proving Theorems 5.1 and 5.2.
Theorem5.1followsfromthenaturalityofexp:asexpislocallyadiffeomorphism,dρdetermines
ρ on a neighborhood of the identity in G.
Proof of Theorem 5.2. Given ϕ : g → h, let j ⊆ g × h be the graph. This is a Lie subalgebra if
and only if ϕ is a Lie algebra map. In this case, there exists J ⊆ G×H an immersed subgroup.
Composing with the projection G × H → G gives a covering map J → G. G simply connected
implies J = G, so J is the graph of a Lie group map G → H.
7 Classes of Lie Algebras
Assume G is connected, and let g be its Lie algebra. g is said to be abelian if [X,Y] = 0 for every
X,Y ∈ g. More generally, for any g, we define the center
Z(g) = {X ∈ g : [X,Y] = 0∀Y ∈ g}. (7.1)
Connected subgroups H ⊆ G correspond to subalgebras h of g. Now suppose H is normal.
Then gHg−1 = H for every G ∈ G. Choose γ : I → G with γ(0) = e and γ(cid:48)(0) = X ∈ g. Then
γ Hγ−1 = H for every t, so ad(X) = Ad(γ ) takes h to h.
t t t
We say that h is an ideal if [g,h] ⊆ h. Then H is normal if and only if h is an ideal.
Remark. If h ⊆ g is a subalgebra, then [ , ] descends to g/h (meaning there exists a Lie algebra
structure on the vector space g/h such that g → g/h is a Lie algebra map) if and only if h is an
ideal.
We say that g is simple if g has no proper nontrivial ideals.
Given a Lie algebra g, we define two sequences of ideals in g:
1. The lower central series: start with D g = g, and inductively define D g = [g,D g]. (In
0 i i−1
particular, D g = [g,g]. We get a sequence of subalgebras
1
D g ⊇ D g ⊇ D g ⊇ ··· (7.2)
0 1 2
2. The derived series: start with D0g, and this time Dig = [Di−1g,Di−1g]. The Jacobi relation
can be used to show that the Dig are in fact ideals.
9
We say that g is nilpotent if D g = 0 for some k, and that g is solvable if Dkg = 0 for some k.
k
We say that g is semisimple if g has no nonzero solvable ideals. (We’ll see that this is equivalent
to being a direct sum of simple algebras.)
AkeyexampleofanilpotentLiealgebraistheLiealgebranofstrictlyuppertriangularmatrices
in gl . This is the Lie algebra of the subgroup of unipotent upper triangular matrices. We will prove
n
that any nilpotent Lie algebra can be realized as a subalgebra of n (for some n).
Another example of a solvable Lie algebra is b, the space of all upper triangular matrices in gl .
n
In fact, every solvable Lie algebra is a subalgebra of b (for some n).
Observe that g is solvable if and only if there exists a sequence of subalgebras
0 ⊆ g ⊆ g ⊆ ··· ⊆ g ⊆ g (7.3)
1 2 k
such that g is an ideal in g and g /g is abelian. As a consequence, if h ⊆ g is an ideal,
i i+1 i+1 i
then g is solvable if and only if h and g/h are solvable.
Remark. Ifa,b ⊆ garesolvableideals,thena+bisalsosolvable.Thisisbecause(a+b)/b = a/(a∩b),
which is solvable.
Corollary 7.1. There exists a maximal solvable ideal in g. This is called the radical of g and denoted
Rad(g).
We now have an exact sequence
0 → Rad(g) → g → g/Rad(g) → 0 (7.4)
with g/Rad(g) semisimple. We will denote g/Rad(g) by g .
ss
Theorem 7.2 (Engel). If ϕ : g → gl(V) is a representation of g such that ∀X ∈ g, ϕ(X) is
nilpotent, then there exists 0 (cid:54)= v ∈ V such that X(v) = 0 for every X ∈ g.
Observe that in this case, there exists a basis v ,...,v of V such that ϕ(g) ⊆ n.
1 n
Remark. If X ∈ gl(V) is nilpotent, then there is a sequence of spaces
V ⊇ V ⊇ V ⊇ ··· ⊇ V ⊇ 0 (7.5)
1 2 k
suchthatX mapsV intoV .Thenad(X)viewedasanendomorphismofgl(V)isalsonilpotent.
i i+1
Proof of Engel’s theorem. Induct on dimg.
Lemma 7.3. g contains an ideal h of codimension 1.
Proof. Lethbeanymaximalpropersubalgebraofg.Weclaimthathisanidealandthatcodim(h ⊆
g) = 1.
Look at the adjoint action ad(h) ⊆ gl(g). Since ad(h) carries h into itself, ad(h) acts on g/h.
Every element acts nilpotently. For if Xk = 0 on V, then ad(X)2k−1Y = 0 for every Y ∈ g.
By the inductive hypothesis, there exists Y ∈ g/h nonzero such that [h,Y] ⊆ h. By maximality,
we must have h+Y = g, so h has codimension 1.
10
