Math 222 notes for Feb. 17
Alison Miller
Last time, Arul showed
Theorem 0.1 (Lie’s Second Theorem). Let G and H be Lie groups with Lie algebras g and h
respectively. Suppose that G is simply connected. Then any Lie algebra homomorphism f : g → h
is equal to Lie φ for a unique Lie group homomorphism φ : Lie G → LieH.
One more result on the relationship between Lie groups and Lie algebras, before we
move on to the next topic.
If g is a Lie algebra over R, does g have to be the Lie algebra of a Lie group? Clearly
if this is the case, g must be finite-dimensional; and in fact, this is sufficient:
Theorem 0.2 (Lie’s Third Theorem). Let g be a finite-dimensional Lie algebra. Then there
∼ g.
exists a Lie group G with Lie G =
Remark. There’s also a Lie’s First Theorem – but nobody talks about it anymore because
it uses obsolete terminology.
The proof of this involves the following, rather hard, theorem, which we won’t prove
(the proof requires a lot of structure theory of Lie algebras, more than we will do in this
class).
Theorem 0.3 (Ado). Let g be a finite-dimensional Lie algebra. Then g is isomorphic to a subalgebra of gln (R) for some n.
Proof of Lie’s Third Theorem. By Ado’s theorem, there exists a homomorphism g ,→ gln (R).
Hence g is the Lie algebra of some G ⊂ gln (R).
1
Representations of Lie groups and Lie algebras
Recall the definition we gave on Day 1:
Definition. A (real or complex) representation ρ of a Lie group G is a homomorphism
ρ : G → GL(V) where V is a (real or complex) vector space.. If V is finite-dimensional (as
it will usually be in this class), we also require that ρ is a homomorphism of Lie groups,
that, is, ρ is smooth.
1
If ρ : G → GL(V) is a representation, it induces a homomorphism of Lie algebras
Lie ρ : Lie G → gl(V).
∼ End(V) is the space of linear
Here gl(V) = Lie(GL(V)). As a vector space, gl(V) =
maps V → V, with Lie bracket [A, B] = A ◦ B − B ◦ A. If we pick a basis for V, we
∼ Mn×n (R) (if V is a vector space over R) or gl (C) =
∼
can identify gl(V) with gln (R) =
n
Mn×n (C) if V is a vector space over C.
This motivates the definition
Definition. A (real or complex) representation µ of a Lie algebra g is a Lie group homomorphism µ : g → gl(V) where V is a (real or complex) vector space.
Note that if G is a simply connected Lie group, then Lie’s second theorem tells us
that representations ρ : G → GL(V) are in bijective correspondence with representations
µ : g → gl(V), by the map ρ 7→ Lie ρ. Hence classifiying the representations of a simply
connected Lie group G is equivalent to classifying representations of the lie algebra
Lie G.
Furthermore, even if G is not simply connected, a representation ρ : G → GL(V) is
still determined by the representation Lie ρ : g → gl(V); this is a result we proved last
Wednesday. (I didn’t actually say the above in class, but it’s worth pointing out.)
But when G is not simply connected, it’s not necessarily the case than any Lie algebra
representation µ : g → gl(V) is equal to Lie ρ for some such representation ρ of G.
However, we can determine whether this is the case as follows: let G̃ be the simply
connected universal cover of G, and let K be the kernel of the covering homomorphism
π : G̃ → G. Then Lie G̃ = g, and G̃ is simply connected, so there must exist ρ̃ : G̃ → GL(V)
such that Lie ρ̃ = µ. If ker ρ̃ contains K, then ρ̃ induces a map ρ : G̃ → G such that the
diagram
G̃
ρ̃
π
G
ρ
GL(V).
commutes. Conversely, if such a ρ existed with Lie ρ = µ, we would necessarily have
˜
ρ ◦ π = ρ̃, and so K ⊂ ker rho.
Hence we have here a criterion for telling which representations of g come from
representations of G.
Example. Let’s classify the 1-dimensional complex representations of G = U(1) = {z ∈
C | |z| = 1} = {e2πiθ | t ∈ R/Z}.
First of all g = Lie(G) is a 1-dimensional Lie algebra; it is spanned by any nonzero X ∈
∂
∈ T1 (G). Since g is 1-dimensional,
Lie(G). We’ll take X to correspond to the element ∂θ
the Lie bracket must be always 0.
Now a 1-dimensional representation of g is a Lie algebra homomorphism µ : g →
∼ C with the trivial Lie bracket. Such a homomorphism is determined
gl1 (C); here gl1 (C) =
2
by µ(X) ∈ gl1 (C); write µ(X) = y ∈ C, so µ(tX) = ty. This clearly defines a Lie algebra
homomorphism (as all Lie brackets are 0).
Now here, the simply connected cover of G is G̃ = R, with covering map π : G̃ → G
given by π(θ) = e2πiθ .
We now find the 1-dimensional representations ρ : G̃ → GL1 (C). We know that for
˜
any µ : g → gl1 (C), there is a unique ρ̃ : G̃ → GL1 (C) with Lie ρ̃ = µ. We can find rho
using the commutative diagram
ρ̃
R
GL1 (C)
expGL
expR
1 (C)
µ
g
gl1 (C).
We start with an arbitrary element of g, which we can write as tX, and chase around
the diagram
t
expR
ρ̃
ety
expGL
1 (C)
µ
tX
ty.
Since the diagram commutes, this tells us that ρ̃ is defined by by ρ̃(t) = ety for all t ∈ R.
This gives us all the representations of G̃.
Now the representation ρ̃ of G̃ induces a representation ρ of G if and only if ker ρ
contains K = {θ | e2πiθ = 1} = Z. That is, y must satisfy ety = 1 for all t ∈ Z, so y = 2πik
for some k ∈ Z.
Since the covering map G̃ → G is given by θ 7→ e2πiθ , the representation ρ of G must
satisfy ρ(e2πiθ ) = e2πikθ for all θ ∈ R. That is, ρ : U(1) → GL1 (C) is given by ρ(z) = zk .
Hence we’ve shown all representations ρ : U(1) → GL1 (C) are of the form ρ(z) = zk .
2
The Adjoint Representation
Every Lie group G automatically comes with a natural representation, the adjoint representation.
Definition. Let G be a Lie group with g = Lie(G). The adjoint representation of G, Ad :
G → GL(Lie(G)), is defined as follows: for g ∈ G, Ad g = Lie(Inn(g)) : g → g where
Inn(g) : G → G is defined by Inn(g)(h) = ghg−1 .
This also gives us a representation of the Lie algebra g.
Definition. The adjoint representation ad : g → gl(g) is given by ad = Lie Ad.
3
Proposition 2.1. For X, Y ∈ g, ad(X)(Y) = [X, Y].
Proof. Because ad = Lie Ad, we have ad(X) =
d
dt
Ad(exp(tX))|t=0 , and so
d
(Ad(exp(tX))(Y)) |t=0 .
(1)
dt
To evaluate this, we now need to evaluate Ad(exp(tX))(Y) for any g ∈ G. Now for
any g ∈ G, s 7→ (g exp(sY)g−1 ) is an integral curve for (Ad g)Y. Hence
ad(X)(Y) =
(Ad gY)1 =
d
(g exp(sY)g−1 )|s=0 = (Ad gY).
ds
(2)
Now, setting g = exp(tX) in (2) and plugging into (1) we obtain
((Ad s)Y)t=1 =
d d
exp(tX) exp(sY) exp(−tX)
dt ds
(3)
Now
exp(tX) exp(sY) exp(−tX) = (exp(tX) exp(sY) exp(−tX) exp(−sY))(exp(sY))
= exp(1 + st[X, Y] + . . . ) exp(1 + sY + . . . )
= exp(1 + sY + st[X, Y] + . . . )
where . . . denotes terms of degree 3 or higher.
∼ T1 G → T1 g =
∼ g is the identity.
The result then follows, since d(exp) : g =
4
(4)