NOTES FOR LIE GROUPS & REPRESENTATIONS

INSTRUCTOR: ANDREI OKOUNKOV

HENRY LIU & MICHAEL ZHAO

Abstract. These are originally live-texed notes for the Fall 2016 offering of MATH GR6343
Lie Groups & Representations, by Henry. They also include additions from the Fall 2018
version of the course, live-texed by Michael.

1. Definition and Examples 1

2. Differentiable structure of Lie groups 3
3. Lie group actions 5
4. Proper actions 7
5. Some Lie group properties 11
6. Symplectic matrices 12
7. Fundamental groups of Lie groups 13
8. From Lie groups to Lie algebras 13
9. The Lie functor 15
10. Lie algebra to Lie group 16
11. Exponential map 16
12. Digression: classical mechanics 19
13. Universal enveloping algebra (lectured by Henry Liu!) 21
14. Poisson algebras and Poisson manifolds 25
15. Baker–Campbell–Hausdorff formula 26
16. Peter–Weyl theorem 29
17. Compact operators 31
18. Complexifications 32
19. Symmetric spaces 33
20. Solvable and nilpotent Lie algebras 35
21. Parabolic and Borel subgroups 38
22. Maximal tori 41
23. More Borel subgroups 43
24. Levi–Malcev decomposition 44
25. Roots and weights 46
26. Root systems 48

1. Definition and Examples

Definition 1.1. A Lie group over a field k (generally R or C) is a group G that is also a
differentiable manifold over k such that the multiplication map G × G → G is differentiable.
Date: October 30, 2018.
1
Remark 1.2. We will see later that x 7→ x−1 on a Lie group G is also differentiable.
Remark 1.3. There are complex Lie groups and real Lie groups. Every complex Lie group
is a real Lie group, since being a complex manifold is stricter than being a real manifold.
Remark 1.4. Besides genus 1, genus g surfaces are not real Lie groups. Looking at left (or
right) translation, we get a map
G × g → Tg G.
This gives a trivialization of the tangent bundle, as we have a map G × g → T G. But the
higher genus surfaces don’t have trivial tangent bundle. If we regard Σ as embedded in the
zero section of T Σ, then the self-intersection is [Σ]·[Σ] = 2−2g, which in general is non-zero.
Example 1.5. Some examples of Lie groups:
(1) k n as a vector space with additive group structure;
(2) T := {z ∈ C∗ : |z| = 1};
(3) k ∗ , the multiplicative group of the field k;
(4) GL(V ), the group of matrices with non-zero determinant;
(5) any finite group, or countable group with discrete topology;
(6) SLn (k), the group of matrices with det = 1;
(7) GL+
n (k), the group of matrices with det > 0;

(8) On (k), the group of matrices with AAT = AT A;

(9) SOn (k) := On (k) ∩ SLn (k);
( )
0 I
(10) Spn (k) := {S : S ΩS = Ω} where Ω :=
T
;
−I 0
(11) Un (k), the group of matrices with U U ∗ = U ∗ U .
Note that Un (k) is not a complex Lie group, since its defining equation contains complex
conjugation, which is not holomorphic.
Definition 1.6. A subgroup of a Lie group is a Lie subgroup if it is a submanifold.
Example 1.7. Consider the torus T2 := S 1 × S 1 = R2 /Z2 , and pick a line R in R2 of
irrational slope. Clearly R is a Lie group and is a subgroup of T2 , but it is definitely not a
Lie subgroup. What went wrong: R needs to be a submanifold, not just a manifold in its own
right.
Example 1.8. Examples of Lie subgroups:
(1) any discrete subgroup is a Lie subgroup;
(2) diagonal matrices in GL(V );
We have to be careful about which field Lie subgroups are taken over. For example, GL(Cn )
is both a complex and real Lie group, but U (n) ⊂ GL(Cn ) is only a real Lie subgroup (since
it is not a complex Lie group).
2
Proposition 1.9. Let G1 , G2 be Lie groups over k. Then G1 × G2 is also a Lie group over
k with the standard structure of a product of groups and a product of manifolds.
Definition 1.10. A group homomorphism m : G1 → G2 of Lie groups is a Lie group
homomorphism if it is differentiable.
Example 1.11. Some examples of Lie group homomorphisms:
(1) the identity map id, or more generally embeddings of Lie subgroups;
(2) any linear map;
(3) the determinant map det;
(4) the conjugation map a(g) : x 7→ gxg −1 ;
(5) the exponential map R → S 1 given by x 7→ eix .
Note that the map which is multiplication by a fixed group element g is not a Lie group
homomorphism, since it is not a group homomorphism.
Definition 1.12. A Lie group homomorphism from p : G → GL(V ) is a linear represen-
tation of G.
Example 1.13. Some examples of linear representations:
exp
(1) R −−→ S 1 ,→ GL(R2 ) given by rotations;
(2) given R, S linear representations of G, we can construct R ⊕ S, R ⊗ S, etc.
Remark 1.14. A representation of a Lie group is its action on a vector space, but we want
to talk about actions in general.

2. Differentiable structure of Lie groups

We establish some notation: let µ : G × G → G be the multiplication map, and la : G → G

(resp. rb : G → G) be left-multiplication by a (resp. right multiplication by b).
Lemma 2.1. The maps la and rb are diffeomorphisms, and therefore have full rank.

Proof. We assumed the multiplication µ is differentiable, so in particular la = µ(a, −) and

rb = µ(−, b) are differentiable. Their inverses la−1 and rb−1 are also differentiable. □

Proposition 2.2. For Xa ∈ Ta G and Yb ∈ Tb G, the differential of multiplication is

dµ|(a,b) (Xa , Yb ) = drb |a (Xa ) + dla |b (Yb ).

Proof. Pick X, Y ∈ Te G such that (dla )|0 (X) = Xa and (dlb )|0 (Y ) = Yb . Then γ : (−1, 1) →
G × G given by (a exp(tX), b exp(tY )) is a curve passing through ((a, b), (Xa , Yb )). We
3
compute
d
dµ|(a,b) (Xa , Yb ) = µ(a exp(tX), b exp(tY ))
dt
( t=0
)
d d
= µ(s, b exp(0Y )) a exp(tX)
ds s=a dt t=0
( )
d d
+ µ(a exp(0X), s) b exp(tY )
ds s=b dt t=0
= (drb )|a (dla )|0 X + (dla )|b (dlb )|0 Y
= (drb )|a Xa + (dla )|b Yb . □
Remark 2.3. To compute even further we would actually need explicit data for the group
multiplication written in terms of la and rb .
Corollary 2.4. Multiplication corresponds to addition in the Lie algebra.

Proof. At (a, b) = (e, e), we have dµ|(e,e) (X, Y ) = X + Y . A different way to interpret this
is that given two paths γ1 (t) and γ2 (t) through 0, we have γ1′ (0) + γ2′ (0) = (γ1 γ2 )′ (0). □
Proposition 2.5. Given that multiplication µ : G × G → G is differentiable, so is the
inversion i : G → G.

Proof. We first show that µ−1 (e) = {(g, g −1 ) : g ∈ G} is an embedded submanifold of G × G.

This comes from an application of the implicit function theorem: the differential
dµ|(a,b) (Xa , Yb ) = (drb )|a (Xa ) + (dla )|b (Yb )
is clearly surjective at every (a, b) ∈ G×G (where Xa ∈ Ta G and Yb ∈ Tb G and la , rb : G → G
are left and right multiplication by a and b respectively), hence in particular it is surjective
for every (g, g −1 ) ∈ µ−1 (e).
The projections π1 , π2 : G × G → G and the embedding ι : µ−1 (e) → G × G are differentiable.
Then π1 ◦ι : µ−1 (e) → G given by (g, g −1 ) 7→ g is differentiable, and in fact a homeomorphism
since every group element has an inverse. By the inverse function theorem, it is a local
diffeomorphism, but bijective local diffeomorphisms are global diffeomorphisms. Hence
(π1 ◦ι)−1
G −−−−−→ µ−1 (e) − g 7→ g −1
ι π
→ G × G −→
2
G,
is differentiable because it is a composition of differentiable maps. □

The following lemma, in conjunction with the constant rank theorem, will be very important
in proving that certain subgroups are Lie groups.
Lemma 2.6. Let f : G → H be a morphism of Lie groups. Then f has constant rank across
G.

Proof. Let g ∈ G be an arbitrary element. Since f is a group homomorphism,

(f ◦ lg )(g0 ) = f (gg0 ) = f (g)f (g0 ) = (lf (g) ◦ f )(g0 ).
4
Taking differentials at e, the identity, dfg ◦ (dlg )e = (dlf (g) )f (e) ◦ dfe . But by lemma 2.1, both
(dlg )e and (dlf (g) )f (e) are full rank, hence dfg and dfe have the same rank. □
Theorem 2.7 (Constant rank theorem). A differentiable map f : M → N of constant rank
k is linearizable in a neighborhood of any point on M .
Corollary 2.8. Let f : M → N be a differentiable map from M an m-fold to N an n-fold.
If the rank of f is constant on M , then:
(1) if f is an immersion (i.e. df is injective, or rankf = m) then locally (in a chart) f
is of the form
f (x1 , . . . , xm ) = (x1 , . . . , xm , 0, . . . , 0);
(2) if f is a submersion (i.e. df is surjective, or rankf = n) then locally (in a chart) f
is of the form
f (x1 , . . . , xn , xn+1 , . . . , xm ) = (x1 , . . . , xn ).
Corollary 2.9. If q ∈ N is a regular value, i.e. f has full rank on every point in f −1 (q),
then f −1 (q) is a regular submanifold.

Proof. On f −1 (q), we know rankf is constant. Hence given q = (x1 , . . . , xn ) on N , we can

write f in the form
f (x1 , . . . , xn , xn+1 , . . . , xm ) = q.
Hence f −1 (q) is given by varying the coordinates xn+1 , . . . , xm . □

If a complex Lie group is compact, then it is abelian and Cn /Λ. The argument goes that
if we have G → GLn (C) is trivial, since there are no global holomorphic functions on a
compact complex manifold. In particular, the adjoint action gives a map G → GL(T1 G).
But if a group has a fixed point, its action is equivalent to an action on the tangent space of a
neighborhood. But it’s trivial on the tangent space at the identity, so it has a neighborhood
on which it is trivial.

3. Lie group actions

Let G be a Lie group (or algebraic group) and let X be a manifold in the same category.
Definition 3.1. A Lie group action of G on X is a differentiable group action G×X → X
given by (g, x) 7→ g · x. Here group action means it satisfies
e · x = x, g1 · (g2 · x) = (g1 g2 ) · x.
Remark 3.2. Note that this may not be a Lie group homomorphism, since for an arbitrary
differentiable manifold X we cannot say anything about whether Diff(X) is a Lie group.
Example 3.3. A linear representation is an action on a vector space by linear operators,
i.e. G → GL(V ). For any group G, we have a few canonical actions:
(1) the left (resp. right) regular action where X = G, and G × G → G is just the
multiplication (g1 , g2 ) 7→ g1 g2 (resp. (g1 , g2 ) 7→ g2 g1−1 );
(2) the adjoint action Ad : G × G → G given by (g, h) 7→ ghg −1 .
5
A homomorphism φ : G → H induces an action of G on H by (g, h) 7→ φ(g)h.

Some typical questions include

(1) Understanding the orbit, e.g. Jordan normal form of matrices is a representative of
the action of G = GLn on g = Mn via adjoint action.
(2) Understanding the space of orbits X/G. For example, the Grassmanian Gr(k, n) is
the quotient of k × n matrices of maximal rank by GLn (k).
Definition 3.4. For x ∈ X, the set Gx ⊂ X is the orbit. The set of orbits is the quotient
X/G. The stabilizer Gx is the set of elements g ∈ G fixing x.
Proposition 3.5. Let G act on X with x ∈ X. Then:
(1) Gx is a Lie subgroup in G;
(2) there is some open set U containing the identity e ∈ G such that U ·x is a submanifold.
In this setting, dim U · x + dim Gx = dim G.

Proof. Define αx : G → X by g 7→ g · x. It has constant rank. Hence Gx = αx−1 (x) is a

regular submanifold by the constant rank theorem, and is also clearly a subgroup.
Similarly, by the constant rank theorem, for each g ∈ G there is some neighborhood U ∋ g
such that its image αx (U ) is a submanifold in X. For g = e, we get that U ·x is a submanifold.
To see that dim U · x + dim Gx = dim G, note that rank-nullity holds for the differential dαx
at x. □
Remark 3.6. Some general questions we can ask about actions:
(1) what are the orbits of the action?
(2) what does the set of orbits X/G look like?
Lemma 3.7. A Lie subgroup H ⊂ G is closed.

Proof. Suppose H ⊂ G is a Lie subgroup. Then its closure H is a subgroup of G. In

particular, H is H-invariant. By definition, H is a submanifold of G. Hence H is open in H.
−1
Right-multiplication is continuous
⊔ so Hx = rx−1 (H) is open in H̄ too. But H̄ is the disjoint
union of cosets, i.e. H̄ \ H = x̸=e Hx is open, i.e. H is also closed in H̄. Since H̄ is the
closure, H = H̄ by definition. □
Remark 3.8. Note that naturally X/G is a topological space. The natural (set-theoretic)
map X → X/G induces a topology on X/G via the quotient topology; however, this topology
is usually non-Hausdorff.
Example 3.9. Here’s an example of a non-Hausdorff topology on the quotient. Let X = R2
and let G = (R, +). There are two possible actions, and the first is non-Hausdorff:
( t ) ( )
e 0 1 t
, .
0 e−t 0 1
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The orbits of the first action look like hyperbolas, along with the four pieces of axes and the
origin. The axes are not separable from the origin. In the second one, the remaining vertical
line in the quotient is not separable from x-axis.
Here’s another example. Take the defining representation of GL1 (R). Take the closure of the
orbit of 1. This is the entire space.
Another example is to embed R → R2 as a line with irrational slope and then quotient by Z2 .

This example shows that the closure of an orbit can have higher dimension. But this doesn’t
happen for algebraic maps (contains a dense open subset of its closure?).
Definition 3.10. A function X/G → R is regular if its lift to X → R is a morphism in
the category of X.
Example 3.11. Let X = C and G = {±1} acting via multiplication. Then a function on
X/G is a function f such that f (z) = f (−z). In other words it is a function g(z 2 ). Hence
z 2 : X/G → C = X is an isomorphism because the sets of regular functions on X/G and X
are the same.
Example 3.12. Let X = C2 and G = {±1} acting via multiplication (x, y) 7→ ±(x, y).
Regular functions here are even functions in (x, y). Any such function factors through
x2 , xy, y 2 , i.e. there is a map from X/G to a cone. Here the image is a cone because there
is the non-trivial relation (x2 )(y 2 ) = (xy)2 . (This is actually a diffeomorphism, not just a
homeomorphism.)
Remark 3.13. Really, X/G is a topological space equipped with a sheaf of functions. The
question is under what conditions is it a nicely behaved space.
Example 3.14. Consider the map
( ita )
e 0
R ∋ t 7→ ∈ U (1)2 ⊂ GL(2).
0 eitb
If a/b ∈ Q, then the image of this map is closed. However if a/b ∈
/ Q, then the image is
dense.

4. Proper actions

Definition 4.1. An action is proper if the following map is proper (as a map of topological
spaces, i.e. the preimage of compact sets is compact):
A : G × X → X × X, (g, x) 7→ (gx, x)

In particular, A−1 ((x, x)) = Stab(x) × {x} must be compact, and many of things we’ve seen
fail this.
Example 4.2. A few examples of proper actions:
(1) the left regular action gives (g1 , g2 ) 7→ (g1 g2 , g2 ), which is an isomorphism, so clearly
it is proper;
7
 (2) if H ⊂ G be a Lie subgroup, the restriction to H of any proper action of G is still
proper;
(3) any action of a compact group is proper.
The “irrational flow” of R on T2 given in 1.7 is not a proper action of R on T2 .
Lemma 4.3. Fix x ∈ X. The evaluation map αx : G → X given by g 7→ gx is proper, and
therefore also closed.

Proof. Let K ⊂ X be a compact set. Then A−1 ({x} × K) = B × {x} for some B. But
B × {x} is compact since A is proper, so B = αx−1 (K) is also compact. Recalling that
proper maps between locally compact Hausdorff spaces (every manifold is locally Rn , which
is locally compact by Heine–Borel) are closed, αx is also closed. □
Proposition 4.4. For a proper action, the stabilizer Gx is compact for all x. Hence the
adjoint action is never proper unless G is compact.

Proof. The evaluation map αx : G → X is proper, so αx−1 ({x}) = Gx is compact. For the
adjoint action (g, h) 7→ (h, ghg −1 ), note that Ge = G must therefore be compact. □
Proposition 4.5. Orbits of a proper action are closed embedded submanifolds, not just
immersed submanifolds.
Remark 4.6. This prevents pathologies like the “irrational flow” of R on T2 .

Proof. Fix x ∈ X. It is clear that Gx is closed since the evaluation map αx : G → X given
by g 7→ gx is closed (by lemma 4.3), so αx (G) = Gx is closed.
Consider the differential dαg : Tg G → Tgx X. The important thing to note is that the rank
of dα is constant, since the differential looks the same everywhere.
To show Gx is an embedded submanifold, it suffices to show it locally. Take a compact ball
B around x. Let A : G × X → X × X denote the map (g, x) 7→ (gx, x). Since the action is
proper, A is proper, i.e. A−1 ((B, x)) = {g ∈ G : gx ∈ B} is compact.
We use compactness to get finiteness restrictions. By the constant rank theorem applied to
the constant rank map g 7→ gx, for each g ∈ G there is an open neighborhood U such that
U x is an embedded submanifold of X. By compactness, A−1 ((B, x)) has a finite cover by
such open sets U , i.e. B ∩ Gx is a finite union of embedded submanifolds. We can shrink B
until B ∩ Gx is contained within just one embedded submanifold. Hence Gx is an embedded
submanifold. □
Proposition 4.7. For a proper action G on X, the quotient X/G is Hausdorff.
Remark 4.8. Suppose R ⊂ X × X is an equivalence relation. The general fact is that X/R
is Hausdorff if and only if R is closed.

Proof. Using the remark, for us, the equivalence relation is precisely the map G×X → X ×X
given by (g, x) 7→ (gx, x). The image of this map is closed because G acts properly on X,
and so we are done. □
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Proposition 4.9. Assume the action of G on X is proper and free, i.e. Gx = {1} for every
x ∈ X. Then X/G is a smooth manifold. (Even more strongly, it is a Hausdorff ringed
space.)

Proof. Pick a point x̄ ∈ X/G, which corresponds to an orbit G · x. The orbit is a smooth
manifold. Let a : G × X → X be the group action, so that da : g ⊕ Tx X → Tx X is just
addition of vectors (ξ, v) 7→ (ξ + v). Pick a small transverse slice S so that we have a map
G × S → X. The claim is that S can be chosen small enough such that this map is an
isomorphism with a neighborhood of the orbit Gx .
(1) Locally near x this map is a diffeomorphism by the inverse function theorem.
(2) It is a local diffeomorphism everywhere since G moves the diffeomorphism around in
the orbits..
(3) Hence we must show G × S → X is bijective with its image (because local diffeo-
morphisms may not be bijective, e.g. covering maps). So suppose ∩ g1 s1 = g2 s2 , i.e.
gs1 = s2 . Choose a sequence S̄1 ⊃ S̄2 ⊃ · · · compact, such that Si = {x}. There
exists a neighborhood U ∋ e ∈ G such that for any g ∈ U , if gS ∩ S ̸= ∅, then
g = e (by looking the differential of such a map would be given by addition by 0,
i.e. g = e). Now look at Gn := {g ∈ G∩ \ U : g S̄n ∩ S̄n ̸= ∅}. This is compact by
properness and G1 ⊃ G2 ⊃ · · · , so that n Gn ̸= ∅, i.e. there is some element g in
the intersection such that g · x = x.
Hence for every S open in the quotient X/G, we have found a neighborhood of orbits. For
every such neighborhood, we have a notion of regular functions: smooth functions which are
G-invariant. This gives S a smooth structure. □
Remark 4.10. In particular, G/H is a manifold for any Lie subgroup H.
Remark 4.11. What if the action is proper but not free? Then there is a point x ∈ X with
non-trivial stabilizer Gx ̸= {1}. The orbit is still a smooth manifold, but now Gx = G/Gx .
Now we can choose the slice S to be Gx -invariant: find a Gx -invariant Riemannian metric
(see below) and then take S to be geodesics through (Tx Gx)⊥ , i.e. S ∼
= (Tx Gx)⊥ .
Proposition 4.12. Every compact Lie group G has a G-invariant finite-measure regular
measure dg.
Remark 4.13. Note that the tangent bundle of any Lie group is trivial, since given a basis
at Te G we can move it around via dLg where Lg is left multiplication by g.

Proof idea. Since T G is trivial, G is orientable, and the left-invariant differential forms cor-
respond to the tangent space Te G. Hence there exists a unique left-invariant top form;
explicitly, it is given by ∧i (gi−1 dgi ). (For manifolds this is a lot easier, because measures are
represented by differential forms, and the Lebesgue measure is the only translation-invariant
measure on Rn .) □
Remark 4.14. Left and right Haar measures both exist, and for compact Lie groups they
coincide. Right translations act on the space of left-invariant Haar measures (which is R+ ),
so for the left and right Haar measures to coincide, we require G has no homomorphism to
9
the positive reals R+ . Sufficient conditions include when G is compact, or simple, or has no
1-dimensional representations at all.
Corollary 4.15. Let ∥·∥0 be an arbitrary Riemannian metric. We can construct an invariant
metric from it using ∫
∥v∥2 := ∥gv∥20 dg.
Gx

Proposition 4.16. Let G compact act on V an affine space, and suppose it preserves a
convex set S in V . Then there exists a vector v ∈ S fixed by G.
∫
Proof. Pick an arbitrary vector v0 ∈ S, and set v := G
µ(dg) g ·v0 . (View v as the barycenter
of the orbit Gv0 .) □
Proposition 4.17. Let G compact act on X a manifold. Then X has a G-invariant Rie-
mannian metric.
Remark 4.18. This is a generalization of the previous proposition.
Theorem 4.19. Let Gx be the stabilizer of a point x ∈ X a manifold. Let S be a Gx -invariant
slice, isomorphic to (Tx Gx)⊥ as a Gx -manifold. Then
GS ∼= G ×G S := (G × S)/Gx
x

as G-manifolds, i.e. manifolds with an action of G. (Here A ×H B := (A × B)/H, where

h(a, b) 7→ (ah−1 , hb) is the standard fiber product.)

Proof. (Did we do this in class?) □

Corollary 4.20. X/G ∼
= S/Gx near Gx.
Corollary 4.21. X has a G-invariant Riemannian metric because G × S has a G × Gx -
invariant metric.

Any finite-dimensional representation

⊕ of a compact group is semi-simple, i.e. if we have a
representation W , then W = i Wi where each Wi is simple. (This comes from how there
is always a quadratic form that is G-invariant; given W ′ ⊂ W , we can always decompose
W = W ′ ⊕ (W ′ )⊥ .)
( )
2 1 a
Example 4.22. Let R act on R by . It has two sub-representations that are trivial,
0 1
but it is not the direct sum of two trivial representations.
Example 4.23 (Grassmannian). Let G = GL(n, R) and H of upper triangular matrices with
the first block being k × k. Then G/H = Gr(n, k). Note that a matrix preserves the span of
the first k basis vectors if and only if it is of the form given by H. Hence G acts on Gr(n, k)
with H stabilizing span(e1 , . . . , ek ).
Alternatively, Gr(n, C) = U (n)/(U (k) × U (n − k)), because U (n) acts transitively on orthog-
onal bases for k-dimensional subspaces, and if an element fixes a k-dimensional subspace it
also fixes the (n − k)-dimensional complement. This decomposition shows that Gr(n, k) is
compact.
10
A chart near L ∈ Gr(n, k) is formed by linear maps( L ) → V /L; the graph of a map is a
subspace. The Grassnammian Gr(n, k) is covered(by) k charts of the form “n × k matrices
n

with prescribed minor being non-zero” (there are nk such minors). This is a generalization
of what we do for projective space, where k = 1 and we have just an n-tuple of numbers.
Hence Gr(n, k) = Mn,k /GL(k) as well, where Mn,k is the set of all n × k matrices.
Remark 4.24. These ways of expressing Gr(n, k) hold over every field (except for U (n)/(U (k)×
U (n − k))). The question we should ask ourselves in general is if G is a linear (i.e. closed
subspace of GL(n)) algebraic group and H ⊂ G is a subgroup, we want to make G/H an
algebraic variety.
The way to do this for Grassmannians is to use the Plücker embedding: if we have L ⊂ V
where dim L = k and dim V = n, then
Λk L ⊂ Λk V
( )
where Λk L is a line and Λk V has a basis of nk elements. The coordinates of L we now
define to be the coordinates of the line Λk L inside Λk V , i.e. precisely the values of the minors
in the n × k matrix representing L in Gr(n, k) = Mn,k /GL(k). To recover the line L, let α
represent Λk L, and take the kernel
V → Λk+1 V, v 7→ v ∧ α.
The kernel is precisely L because e1 ∧ β = 0 iff β = e1 ∧ β ′ .

Overall, we had the following theorems.

Theorem 4.25. For a proper action there exists U a neighborhood of the orbit of x such
that U ∼
= G ×Gx Nx / ∼ where Nx is the normal to Tx of the orbit, i.e. Nx = Tx X/Tx Gx , and
where (gh, v) ∼ (g, hv) for h ∈ Gx .
Corollary 4.26. On X there is a G-invariant Riemannian metric.

Something about partition of unity.

Corollary 4.27. Stabilizers of nearby points are conjugate to a subgroup in Gx .

There’s an action of SL2 on cubic polynomials in x1 and x2 , by acting on the (x1 , x2 )-plane.

5. Some Lie group properties

Let G be a real Lie groups. Then the set of left-invariant measures on G is acted on by
G by right translation. By uniqueness up to a positive scalar, this is a bijection with the
positive real numbers, which is acted upon by the determinant of the adjoint action of G on
g. But this action is trivial if G is abelian, if G is compact, if G has no characters, e.g. if it
is simple, if G is a complexification of compact Lie group.
GLn (R) SLn (R) On (R) SOn (R) Un SUn Sp2n (R)
dim n2
n −1
2 n(n−1)
2
n(n−1)
2
n n − 1 n(2n + 1)
2 2

π0 Z2 1 Z2 1 1 1 1
π1 Z2 Z2 Z2 Z2 Z 1 Z.
11
We used the following facts (some of which are explained in the following subsections) in
populating the table.
(1) There is a surjective continuous map det : GL(n, R) → R× , but R× is not connected.
Hence GL(n, R) and even O(n, R) is not connected. Given M ∈ GL+ (n, R), construct
a path from M to I as follows: given a basis v1 , . . . , vn , Gram–Schmidt provides an
orthogonal basis
⟨v2 , w1 ⟩ ∑ ⟨vn , wi ⟩
w1 = v1 , w2 = v2 − t w1 , . . . , wn = vn − t wn
⟨w1 , w1 ⟩ i<n
⟨wi , wi ⟩
where we added the parameter t to obtain a homotopy to O(n, R); then use the
homotopy (cos θ)e1 +(sin θ)w to move basis vectors to the standard basis while staying
in O(n, R). For the other groups, a similar argument works, except there is no
obstruction arising from positive/negative determinant.
(2) U (n) = O(2n) ∩ Sp(2n, R) (complex vs real picture). This is useful because Sp(2n, R)
retracts onto U (n): given A ∈ Sp(2n, R), there is a polar decomposition A = SU
where S := (AT A)1/2 is symmetric and symplectic, and U is unitary, so by a preceding
lemma, A(t) = S t U is the homotopy.
(3) Using the long exact sequence of homotopy coming from the fibration SU(n − 1) →
SU(n) → S 2n−1 , we get
π1 (SU(n)) = π1 (SU(n − 1)) = · · · = π1 (SU(2)) = π1 (S 3 ) = 0.
Similarly, SO(n − 1) → SO(n) → S n−1 shows πi (SO(n)) = π1 (SO(3)) = Z/2Z.
Everything else retracts onto SO and SU.

6. Symplectic matrices
( )
T 0 I
Definition 6.1. A matrix M is symplectic if M JM = J, where J = . The
−I 0
collection of 2n × 2n symplectic matrices is denoted Sp(n, k) (over a field k).
Definition 6.2. The Pfaffian of a skew-symmetric matrix ω is given by taking the associated
2-form ω = aij ei ∧ ej , then computing 1/n!ω n = Pf(ω)e1 ∧ · · · ∧ e2n .
Lemma 6.3. Pf 2 (A) = det(A) for any skew-symmetric matrix A.
Lemma 6.4. Symplectic matrices have determinant 1.

Proof. Use the Pfaffian argument: Pf(Ω) = Pf(M T ΩM ) = det(M ) Pf(Ω), and since Pf(Ω) ̸=
0, we have det(M ) = 1. □
Proposition 6.5. Let S ∈ Sp(2n, R) be positive definite. Then it can be diagonalized using
a unitary change of basis, i.e. there exists U ∈ U (2n, R) such that S = U T DU where D is
diagonal.
Remark 6.6.
( Here )
U (2n, R) is the image of U (n) inside M (2n, R), under the identification
A B
A + iB 7→ . In particular, if U ∈ U (2n, R), we have U T U = I.
−B A
12
Corollary 6.7. If M is a symmetric symplectic matrix, then M α ∈ Sp(2n, R) for α > 0.

Proof. Diagonalize M = U T DU and note that M α = U T Dα U , which is still in Sp(2n, R).

We require symmetric so that taking the α power makes sense (i.e. diagonalizing and taking
each eigenvalue to the α power). □

7. Fundamental groups of Lie groups

Proposition 7.1. Let π : G̃ → G be the universal cover of the Lie group G. Let ẽ ∈ π −1 (e).
Then there exists a unique multiplicative structure on G̃ (with ẽ the identity), that makes π
a homomorphism of Lie groups.

Proof. Consider the commutative diagram

G̃ × G̃ −−−→ G̃
 
 
y py

µ
G × G −−−
G
→ G.
Let α : G̃ × G̃ → G be the diagonal map. Then im (α∗ ) lies in p∗ (π∗ (G̃)), so we have a unique
lift of α to µ̃. Associativity follows from uniqueness. Facts:
(1) the kernel of p is discrete and normal;
(2) a discrete normal subgroup of a path connected Lie group is central. □
Corollary 7.2. π1 (G) is abelian.

Proof. (I zoned out. Help?) □

Remark 7.3. It turns out that for Lie groups, π2 (G) = 0 and π3 (G) is torsion-free.

8. From Lie groups to Lie algebras

Recall that we have a smooth transitive action of G on itself via Lg (h) := gh.
Definition 8.1. A vector field X on G is left invariant if (Lg )∗ X = X, i.e. (dLg )h (Xh ) =
Xgh .

For a left invariant vector field, because the action of G is transitive, the vector field is fully
determined by Xe , its value at the identity.
Proposition 8.2. For X and Y vector fields on a smooth manifold M , the commutator
[X, Y ]f = X(Y f ) − Y (Xf ) is a vector field on M .
Proposition 8.3. If M = G is a Lie group, and X, Y are left-invariant, then so is [X, Y ].
Proposition 8.4. If F : G → H and X is a left invariant vector field on G, then there is a
unique left invariant vector field on H such that
dFg (Xg ) = YF (g) , ∀g ∈ G.
13
Definition 8.5. The Lie algebra g of a Lie group G is the set of left-invariant vector fields
with the bracket [·, ·]. A representation of a Lie algebra g is a Lie algebra homomorphism
g → gl(V ) for some vector space V .
Proposition 8.6. Given a Lie group representation ρ : G → GL(V ), the differential dρ : g →
gl(V ) is a Lie algebra representation.
Example 8.7. Let φg (h) = ghg −1 . Then φg (e) = e, so we can differentiate at e to get
dφg : g → g given by X 7→ gXg −1 called Ad : G → GL(g). Differentiating once more we get
ad : g → gl(g).
Example 8.8. Consider det : GLn (R) → R× . We find that de (det)(X) = tr(X).
Example 8.9. The tensor product of two representations of a Lie group G is g · (v ⊗ w) =
(g · v) ⊗ (g · w). Differentiating,
(d/dt)(g(t)v ⊗ g(t)w)|t=0 = Xv ⊗ w + v ⊗ Xw,
giving the tensor product of two Lie algebra representations.
Theorem 8.10 (Existence). Let G, H be Lie groups with G simply connected. Then for any
Lie algebra homomorphism φ : g → h, there exists a map f : G → H such that df = φ.

Proof sketch. Take a path g(t) in G from e to g and define a path ξ(t) in Te (G) by g ′ (t) =
dLg(t) ξ(t). Consider a solution h(t) in H of the differential equation
h′ (t) = dLh(t) φ(ξ(t))h(t).
Define f (g) := h(1). We need to check this is well-defined.
Suppose g0 , g1 are two paths in G with gi (0) = e and gi (1) = g. Since G is simply con-
nected, these paths are homotopic; call the square given by the homotopy g. Define maps
A, B : [0, 1] × [0, 1] → g by taking A(t, s0 ) to be the velocity path for g(t, s0 ), and B(t0 , s) to
be the velocity path for g(t0 , s), i.e.
∂g(t, s)/∂t = A(t, s)g(t, s), ∂g(t, s)/∂s = B(t, s)g(t, s).
Hence (∂B/∂t − ∂A/∂s)g = ABg − BAg = [A, B]g. Define a map h : [0, 1] × [0, 1] → H to
be a solution
∂h(t, s)/∂t = φ(A(t, s))h(t, s).
If we can show that h(1, s) does not depend on s, we are done. Look at the equation
∂h/∂s = B̃(t, s)h(t, s), ∂ B̃/∂t = ∂(φ(A))/∂s = [φ(A), B̃].
This differential equation in t is satisfied by φ(B) and B̃(0, s) = 0. By uniqueness of solutions,
B̃(1, s) = φ(B(1, s)) = 0, i.e. h(1, s) is independent of s. □

Theorem 8.11 (Uniqueness). If G is a connected Lie group, then any map f : G → H is

determined by its differential df : g → h.

Proof. (I zoned out. Help?) □

14
 9. The Lie functor

There is a functor from the category of (real or complex) connected 1-connected Lie groups
to the category of Lie algebras (over real or complex), given by
f df
G 7→ g := Te G, G1 −
→ G2 7→ g1 −
→ g2 .
For every given df , there is a unique f determined by solving the relevant differential equa-
tion. The hard part is, given g, find a Lie group G whose Lie algebra is g.
The amazing thing is that this is very nearly an equivalence.
For any G, there is an exact sequence
γ7→γ(1)=g
1 → H → Ĝ −−−−−−→ G → 1
where Ĝ is the universal cover, and H is a normal discrete subgroup (isomorphic to π1 (G),
f fˆ
which is abelian). Any map of Lie groups G1 −
→ G2 induces a map Ĝ1 −
→ Ĝ2 which preserves
the kernels of Ĝ1 → G1 and Ĝ2 → G2 .
If H = Gx for a G-action on X, then the Lie algebra of H is ker(g → Tx X) where this map
is the differential of g 7→ gx.
Definition 9.1. A Poisson algebra is a commutative algebra and a Lie algebra, but with
bracket {·, ·}, satisfying the Leibniz rule
{a, bc} = {a, b}c + {a, c}b.
In other words, a 7→ {a, ·} is a map A → Der(A, {·, ·}). (This is the Hamiltonian vector
flow.) Analogously, ad : ξ 7→ [ξ, ·] is also a map g 7→ Der(g, [·, ·]).
Remark 9.2. If one has a family of associative products ∗ℏ such that
(a ∗ℏ b)|ℏ=0 = ab,
then define
a ∗ℏ b − b ∗ℏ a
{a, b} = lim .
ℏ→0 ℏ
Since the numerator is the commutator, it satisfies the Jacobi identity, and therefore so
does {a, b}. Hence we should view Poisson algebras as first-order approximations to non-
commutative algebras, at the point where they are commutative.
Example 9.3. Take R2n with coordinates p1 , . . . , pn , q1 , . . . , qn . We make it a Poisson algebra
by declaring {pi , qj } = δij . What non-commutative algebra is this the first-order approxima-
tion of? Take Pi = ℏ∂qi , which satisfies [Pi , qj ] = ℏδij . In fact, Sp(2n) has a very concrete
description: it consists of polynomials in pi , qj of degree 2, under the Poisson bracket {·, ·}.

Recall that π1 (SO(n)) = Z/2 for n ≥ 3, and Z for n = 2. Hence we can construct the
universal cover of SO(n) as follows. Take a quadratic form Q on a vector space V , and
define the Clifford algebra by v · v = Q(v).
Example 9.4. If we take V = R and Q(x) = −x2 , then the Clifford algebra is C. If instead
we take Q(x) = x2 , we get R ⊕ R.
15
Example 9.5. Take ei ej + ej ei = δij , and note that [e1 e2 , ej ] is linear in e and preserves
e2j = Q(ej ). Hence the dimension of the Clifford algebra Cl associated to this quadratic
form Q is 2dim V . The space of quadratic vectors in Cl is the Lie algebra of SO(n). The
corresponding Lie group, called the Spin group Spin(Q), is the set of invertible elements
x ∈ Cl that preserve V under v 7→ xvx−1 . Clearly this map is in SO(V, Q) since it preserves
the quadratic form Q, and is a two-fold cover with kernel ±1.

10. Lie algebra to Lie group

How do we get from the Lie algebra to the Lie group? Let g be a Lie algebra. Step 1 is to
apply Ado’s theorem.
Theorem 10.1 (Ado). Any finite-dimensional Lie algebra has a faithful linear representation
g → gl(V ).

ad
Proof sketch. One representation we have is g −→ gl(g). The kernel is given by the center,
so we must deal with it. We have a faithful representation of g/Z(g), so by inducting on the
dimension of the center, we can move this faithful representation up to g. □

Then look for G ⊂ GL(V ) (which need not be a Lie subgroup). The Lie algebra g sits in the
tangent space Te GL(V ). Using the local triviality of the tangent bundle T GL(V ), we can
make the foliation by G in GL(V ) have tangent space (dlh )g at the point h of GL(V ). These
tangent spaces form an involutive distribution, and are therefore integrable by Frobenius.
Theorem 10.2 (Frobenius). A field of k-planes is integrable if and only if the subspace of
vector fields tangent to any field of k-planes is a Lie algebra.

Proof. Choose a local frame ∂xi for the distribution and check that the commutator of two
basis vectors is zero. So we can change coordinates such that ∂ei is the local frame. □

Hence we can lift the Lie algebra g to a manifold G by integrating the distribution. That G
is a subgroup follows from exponentiating the addition map on tangent vectors.
Example 10.3. We can apply this machinery to find all connected commutative Lie groups
G, i.e. the commutator is 0. Hence the Lie group G must have universal cover Rn , with
kernel a discrete subgroup Zk . It follows that G = Rn−k × (S 1 )k .
(We can actually use this to prove the fundamental theorem of algebra: if [F : C] > 1, then
F × = R2d \ {0} ∼
= S 2d−1 × R, which is not commutative by the above result.)

11. Exponential map

There is a Lie algebra map from R (as a Lie algebra) to any other Lie algebra. Hence we have
a Lie algebra map R ∋ 1 → ξ ∈ g that can be integrated to give a map exp : (R, +) → G,
which satisfies the differential equation ∂t etξ = ξetξ . In particular if g ⊂ gl(V ), then exp is
exactly the matrix exponential.
Proposition 11.1. ea eb ̸= ea+b unless [a, b] = 0.
16
Proof. If [a, b] = 0, then there is a Lie algebra homomorphism R2 ∋ 1 7→ (a, b) ∈ g, which
lifts to a Lie group homomorphism (R2 , +) → G. That this is a homomorphism gives
ea eb = ea+b . □
Proposition 11.2. The exponential map a 7→ ea is a diffeomorphism near e because d expe =
id.
Proposition 11.3 (Trotter product formula). ea+b = limn→∞ (ea/n eb/n )n .

Proof. Without loss of generality, we can arbitrarily scale a + b. So suppose a is very small,
where ea = 1 + a + O(a2 ). Then we are done. □
Remark 11.4. There is a formula due to Baker–Campbell–Hausdorff of ln(ea eb ) in terms a
convergent series involving only commutators. Then in a chart near the identity, multipli-
cation is analytic in that chart. Hence a Lie group is actually a real analytic manifold.

What is the differential of the exponential map in general? This tells us when exp fails to
be a diffeomorphism.
Theorem 11.5. d exp(ξ)e−ξ = F (adξ )dξ where
∑ xk
F (x) = (ex − 1)/x = .
k≥0
(k + 1)!

Proof. Assume gl ⊂ GL(n). Then

∑
exp(x) = 1 + x + x2 /2 + · · · = xn /n!
n≥0

is the usual power series. When we differentiate, we must be careful because x is not
necessarily commutative:
∑ xa dx xb
d(ex ) = .
a≥0, b≥0
(a + b + 1)!
Trick: write this series as a product, by noting that
∑ ∫ 1
xa dx xb
= esx dx e(1−s)x ds (1)
a≥0, b≥0
(a + b + 1)! 0

by observing that
∫ 1
a!b!
sa (1 − s)b ds = .
0 (a + b + 1)!
To extract an exp(x), we commute the ds term past the dx term (by conjugating the dx by
e−sx :)
∫ 1 (∫ 1 )
sx (1−s)x s adx
e dx e ds = ds e (dx) ex .
0 0
∫1
Hence F (x) = 0
sx
e dx, which is indeed the expression we want. □
17
Remark 11.6. Equation (1) is a very general formula. Let X be a manifold and v(x, t)
be a time-dependent vector field on X. Let G(t0 , t1 ) : X → X be the flow from time t = t0
to t = t1 . If we vary the field, i.e. v 7→ v + δv, what will happen to the flow? We don’t
know anything about G, but we can take the interval [t0 , t1 ] and partition it into [ti/n , t(i+1)/n ],
which gives a product
G(t0 , t1 ) = · · · G(t1/n , t2/n )G(t0 , t1/n ).
Taking the variation with respect to v, of course we get a sum:
∑
n
δv G(t0 , t1 ) = G(t(n−1)/n , t1 ) · · · δv G(ti/n , t(i+1)/n ) · · · G(t0 , t1/n ).
i=1

But what is the flow G(t, t + ϵ) for a very short time? Well, it is just G(t, t + ϵ) = 1 +
ϵv(x, t) + O(ϵ2 ). Hence if n is large,
dG(ti/n , t(i+1)/n ) = dv(x, t)|ti/n − ti+1 /n|.
Then for n → ∞, we get a sum corresponding to the Riemann integral
∫ t1
δv G(t0 , t1 ) = G(t, t1 ) dv G(t0 , t) dt.
t0

Corollary 11.7. exp is a local isomorphism if 2πik for k ̸= 0 is not an eigenvalue of the
adjoint.

Proof. exp is not a local isomorphism if the differential kills something, which happens if 0
is an eigenvalue of F (ad ξ), i.e. 2πik is an eigenvalue of ad ξ. □
Example 11.8. If ad(ξ) is nilpotent for every ξ, then exp is a covering. For example,
take the Lie group consisting of upper triangular matrices. (Such Lie algebras are called
nilpotent.)
Theorem 11.9 (Cartan). A closed subgroup H ⊂ G of a Lie group G is a Lie subgroup,
and the Lie algebra h of H is
h = {ξ ∈ g : etξ ∈ H ∀t}.

Proof. Define h this way; we will show it is the Lie algebra.

(1) It is a linear subspace: ea+b = limn→∞ (ea/n eb/n )n , and the right hand side lies in H
for all n, so the limit lies in H because H is closed.
(2) It is a Lie subalgebra (i.e. closed under bracket) because Ad(etξ ) = t ad(ξ) + O(t2 )
preserves H.
Write g = h ⊕ p where p is the complementary linear subspace. Since exp is a local isomor-
phism, G = eh ep locally (where eh and ep are submanifolds and we are taking their pointwise
product).
Claim: H = eh locally. Suppose not. Then no matter how small we make our neighborhood,
there exists pn ∈ p such that pn → 0 and epn ∈ H. (If these points are not on p, of course
we can “project” them onto p by multiplying by elements of H.) But this is impossible,
18
since then we can find a convergent subsequence among pn /∥pn ∥ (where we literally take any
norm), which we suppose converges to ξ ∈ p. Then
etξ = lim et(pn /∥pn ∥) = lim epn [t/∥pn ∥]+pn {t/∥pn ∥} ∈ H
n→∞ n→∞

since epn [t/∥pn ∥] ∈ H but pn {t/∥pn ∥} → 0. (Here [x] denotes integral part and {x} fractional
part.) □
Example 11.10. We have the formula
( a ) ( )
e c a c eaa−b
−e b
log =
0 eb 0 b
so there is a singularity when a = b + 2πik for k ̸= 0. In other words, when there is a zero
in exp, there is a singularity in log.
Proposition 11.11. Let G be a compact Lie group, so that G has a Haar measure. Then
the geodesics in this metric are getξ , i.e. et Ad(g)ξ g. More generally, for any Lie group G,
( ) ( ) ( )
left-invariant ∼ right-invariant ∼ metrics
= = .
metrics on G metrics on G on g

Right translations act on left-invariant metrics via the Ad action on g. If G is compact, then
this action preserves some metric on g (because the set of metrics is convex).

12. Digression: classical mechanics

Example 12.1. Left-invariant metrics on SO(3) generalize Euler’s equations for rigid bodies.
The configuration space of a rigid body in R3 is R3 × SO(3) (for center of mass and rotation).
We can always work in a coordinate system where the center of mass is at rest, so only SO(3)
remains. Given a rotation g(t), we can view ġ as ġ = gξ for some angular velocity vector
ξ ∈ g, i.e. “in the body.” Alternatively, we can find a vector ω such that ωg = ġ, where ω
is some angular velocity in the space. Here the kinetic energy is the metric on g, i.e. some
bilinear form on ξ, satisfying
1 1 1
∥ġ∥2 = ∥g −1 ġ∥2 = ∥ξ∥2 .
2 2 2
The
∫ motion of the rigid body will be a geodesic under this metric. The Lagrangian
∫ here is
dt ∥ġ∥ /2. Note however that this is not the length of the geodesic, which is dt ∥ġ∥/2. It
2

is better to integrate ∥ġ∥2 even though length is reparametrization invariant.

∫
Remark 12.2. More generally, Lagrangians are written dt L(x(t), ẋ(t), t), and physical
paths x(t) are extremals of this functional. To find extremals, we vary x 7→ x + δx, to get
∫ ∫
dt (∂x Lδx + ∂ẋ Lδ ẋ) = dt (∂x L − ∂t ∂ẋ L) δx.

Since x is an extremal, this variation must vanish, i.e. ∂t ∂ẋ L = ∂x L, the Euler–Lagrange
equation. The description of classical mechanics in this manner allows us to easily work
in moving coordinate systems.
19
 i
∫
There is a quantum mechanical version of this where we look at e ℏ L dt
. Then as ℏ → 0 we
get the classical action.
Definition 12.3. We can rewrite the Lagrangian as a function H(p, x) where p is now a
cotangent vector by E + U where E is the kinetic energy and U is the potential energy, or,
H(p, x) = max(⟨p, ẋ⟩ − L(x, ẋ, t)).
ẋ
This is a special case of the general procedure to make a function on the tangent space a
function on the cotangent space, i.e. the Legendre transform.
The maximum is achieved when p = ∂ẋ L. The equations
q̇ = ∂p H, ṗ = −∂q H
where q := x are Hamilton’s equations. This says there is a Poisson algebra structure
{pi , qj } = δij on the space of functions, so that ∂t f (p, q) = {H, f }. (Note: ∂t H = {H, H} =
0, so energy is conserved.) Derivation of Hamilton’s equations (noting that δ q̇ = 0 because
we are at an extremal for q̇):
dH = d max(⟨p, q̇⟩ − L(q, q̇, t))
q̇
= q̇δp − ∂q Lδq − ∂t Lδt
= q̇δp − ṗδq − ∂t Lδt.
Hence we are done.
Remark 12.4. Note that all of this discussion comes from looking at the tautological 1-form
θ on the tangent space of a manifold. The Jacobi identity for the Poisson bracket that the
two-form dθ gives to functions on T ∗ M is the statement that d2 θ = 0.
Remark 12.5 (Remark on Legendre Transform). Legendre transform is connected to two
different areas of mathematics. One is planar duality. Consider P2 with coordinates [x0 : x1 :
x2 ]. Then the set of hyperplanes are the global sections of OP2 (1). These form a projective
space, so we have a dual projective space, and hyperplanes in one correspond to points in the
other.
Now if we take a convex set, the hyperplane tangent to it at a point will give something that
is the boundary of a convex set in the dual projective space. In particular, given F : A1 → A1 ,
we can look at the convex hull of its graph in A2 . Then for fixed p, which we think of as the
slope, looking at the maximum of ⟨p, x⟩ − F (x) over x corresponds to finding the maximal
intercept on the p-axis.
The Legendre transform for y ∈ T in general is
F (y) 7→ max(⟨p, y⟩ − F (y)),
y∈T
∗
where p ∈ T , some dual space, on which we have a pairing between T and T ∗. An example
of this is the Fourier transform
∫
i
F (y) 7→ e ℏ (⟨p,y⟩−F (y)) dy

when we take ℏ → 0. In the limit, the asymptotics of this are determined by the maximum
value of the exponent.
20
 13. Universal enveloping algebra (lectured by Henry Liu!)

Associated to a Lie algebra g we will define an associative algebra U g such that the category
of finite-dimensional representations of g is equivalent to the category of finite-dimensional
representations of U g. Our goal is to find a basis for this algebra U g. First we recall some
constructions in linear algebra. Let k be a commutative field; we consider Lie algebras over
this field for this section.
Definition 13.1.⊕ For k any field and V a vector space over k, we can define the tensor
algebra T ∗ V := m T m V where T m (V ) := V ⊗m . We can also define it using a universal
property: it is the algebra with a map V → T ∗ V such that any other map V → A factors
through T ∗ V . It comes equipped with the inclusion i : V → T V .
Definition 13.2. From the tensor algebra, we get the symmetric algebra S ∗ (V ) =
T ∗ (V )/I, where I is the ideal generated by all elements of the form x ⊗ y − y ⊗ x for
any x, y ∈ V .
One can also describe it by the universal property that any linear map to a commutative
algebra factors uniquely through SV , which comes from the universal property of quotients.
It comes equipped with a map i : V → SV . From the explicit construction, as we only
quotient by degree 2 elements, we see that this is still injective.
If V has a basis x1 , . . . , xn , then S ∗ V ∼
= k[x1 , . . . , xn ]. In particular, the quotient map
∗ ∗
σ : T (V ) → S (V ) is injective on T V = k and T 1 V = V , since the generators of the ideal
0

I are degree 2. By the universal property of the tensor algebra, S i (V ) = σ(T i V ).

Definition 13.3. The universal enveloping algebra U g of a Lie algebra g is a pair
(i, U g) where U g is an associative algebra with unit, and i : g → U g satisfying the following
universal property:
for any associative algebra A with unit, any algebra homomorphism ϕ : g → A
with ϕ(x)ϕ(y) − ϕ(y)ϕ(x) = ϕ([x, y]) factors through i : g → U g.
As usual, with any definition via universal properties, U g must be unique up to unique
isomorphism. Its explicit construction, to show existence, is to take U g := T ∗ (g)/J where J
is the ideal generated by x ⊗ y − y ⊗ x − [x, y] for all x, y ∈ g. Let π : T ∗ (g) → U g be the
quotient map.
Example 13.4. There is a preferred basis of sl2 consisting of raising and lower operators
and the commutator of them. Explicitly, they are the matrices
( ) ( ) ( )
0 1 0 0 1 0
e= ,f = ,h = .
0 0 1 0 0 −1
We can compute the commutators as [h, e] = 2e, [h, f ] = −2f, [e, f ] = h. Then U (sl2 ) is the
free algebra generated by the symbols e, f, h modulo the relations he − eh = 2e, hf − f h =
−2f, ef − f e = h. However, we don’t know yet that i : g → U g is injective. At least, we
know the scalars are in U g.
Remark 13.5. Note that elements in the ideal J are not homogeneous: x ⊗ y and y ⊗ x
have degree 2, but [x, y] has degree 1. So it is not obvious that π|g is injective, which was the
case for the symmetric algebra. (Actually, it turns out π|g is injective, which we will prove
21
later.) However it is clear that π|k is injective. In particular, at least U g contains scalars
and is non-empty.
Definition 13.6. There is a filtration on the tensor algebra, given by Tm := T 0 ⊕ T 1 ⊕ · · · ⊕
T m (where the T i (V ) are the graded components). We get an induced filtration Un := π(Tn )
on the universal enveloping algebra.
Definition 13.7. Whenever⊕we have a filtration, we can consider the associated graded
m
algebra Gr := Gr(U g) := m≥0 Gr where Grm := Um /Um−1 . Clearly it has an algebra
structure, because there is an induced multiplication
Grm × Grn = Um /Um−1 × Un /Un−1 → Um+n /Um+n−1 = Grm+n .
So Gr is a graded associative algebra with unit 1. We have a surjective map T m → Um →
Gm = Um /Um−1 for each graded component, so we get a surjective map ϕ : T ∗ (g) → Gr.
Remark 13.8. The algebra structure on the associated graded algebra may be different from
the original algebra.
Lemma 13.9. ϕ is an algebra homomorphism, and ϕ(I) = 0 where I is generated by
x ⊗ y − y ⊗ x for x, y ∈ g.

Proof. That ϕ is an algebra homomorphism is easy, because it is induced by an algebra

homomorphism. It suffices to check ϕ(I) = 0. But π(x ⊗ y − y ⊗ x) = π([x, y]) by the
construction of the universal enveloping algebra. Then because ϕ arises from π : T ∗ (g) →
U (g),
ϕ(x ⊗ y − y ⊗ x) ∈ U1 /U1 = 0. □
Theorem 13.10 (Poincaré–Birkhoff–Witt (PBW)). Since I ⊂ ker(ϕ), we have an induced
map Sg = T ∗ g/I → Gr(U g). This is an isomorphism of associative algebras, i.e. Gr(U g) is
just a polynomial algebra on the Lie algebra
Corollary 13.11. Let W be a subspace of T m g, and suppose the map T m → S m g is an
isomorphism on W . Then π(W ) is a complement to Um−1 in Um .

Proof. Consider the map from the graded piece:

π
Tm −
→ Um → Grm = Um /Um−1 .
∼
=
We have a different map T m → S m g −
→ Grm (where the isomorphism is by PBW) which
makes a commutative diagram. Since W ⊂ T m is sent isomorphically to S m g, we know
W ∼
= Grm = Um /Um−1 . Hence in Um , we see Grm is a complement to Um−1 . □
Corollary 13.12. The map i : g → U g is injective.

Proof. This is trivial: take S 1 g = g, and PBW says it maps isomorphically to Gr1 = U1 /U0 .
□
Corollary 13.13. Let (x1 , x2 , . . .) be a basis for the Lie algebra g. Then the elements
xi(1) · · · xi(m) := π(xi(1) ⊗ · · · ⊗ xi(m) ) m ∈ Z≥0 , i(1) ≤ i(2) ≤ · · · ≤ i(m)
form a basis for U g, along with 1, as a vector space.
22
Proof. Recall that U g has a filtration U0 ⊂ U1 ⊂ · · · . So if we can give a basis for every
Um /Um−1 , we can put them together to get a basis of the whole space U g. Let W be the
subspace of T m spanned by elements of the form xi(1) ⊗ · · · ⊗ xi(m) . It satisfies the conditions
of an earlier corollary, i.e. it is mapped isomorphically into S m . By that corollary, the images
of these elements form a basis for the complement of Um−1 . Putting these elements together,
we get a basis for all of U g. □
Corollary 13.14. Let h ⊂ g be a Lie subalgebra. Extend a basis (h1 , h2 , . . .) of h to an
ordered basis (h1 , h2 , . . . , x1 , x2 , . . .) of g. Then the map U h → U g is injective and U g is a
free U h-module with basis {xi(1) · · · xi(m) } ∪ {1}.

Proof of PBW. We are in the situation of the following diagram, where J = ⟨x ⊗ y − y ⊗

x − [x, y] | x, y ∈ g⟩ and I = ⟨x ⊗ y − y ⊗ x | x, y ∈ g⟩.

Sm = T m /I
z φ

π
Tm Um = Tm /J Grm = Um /Um−1
We already know φ is surjective, so it suffices to prove injectivity. In other words, we must
show that if t ∈ Tm g such that π(t) ∈ Um−1 , then t ∈ I.
(Setup) Fix a basis {xλ }λ∈Ω of g. Write S ∗ g = k[zλ ] for λ ∈ Ω. For each sequence Σ =
(λ1 , . . . , λn ) of indices with λ1 ≤ · · · ≤ λn , let
zΣ := zλ1 · · · zλn ∈ S m g
xΣ := xλ1 ⊗ · · · ⊗ xλm ∈ T m g.
More generally, for x ∈ T g, we have an induced element in Sg called z(x). Write λ ≤ Σ to
mean λ ≤ µ for every µ ∈ Σ.
Assume there exists a Lie algebra representation ρ : g → End(Sg) satisfying:
(1) ρ(xλ )zσ = zλ zσ if λ ≤ Σ;
(2) ρ(xλ )zΣ ≡ zλ zΣ mod Sm if |Σ| = m;
(3) if we extend ρ to ρ : T ∗ g → End(S ∗ g), then ker ρ ⊃ J.
What this means is that ρ(xi ⊗ xj )zJ = ρ(xj ⊗ xi )zJ + ρ([xi , xj ])zJ .
We show the following result: if t ∈ Tm ∩ J, written t = tm + tm−1 + · · · where ti ∈ T i g are
the homogeneous components, then tm ∈ I. The representation ρ : g → End(S ∗ g) extends
to a representation ρ : T ∗ g → End(S ∗ g), so ρ(t) = 0 for t ∈ Tm ∩ J. Then using property
2 above, the highest degree component of ρ(t)1 = z(sm ) (mod Sm−1 ) is determined by tm ,
and is actually 0. Hence tm ∈ I.
Now we proceed with the proof of PBW. Let t ∈ T m g and π(t) ∈ Um−1 . We want to show
t ∈ I. If π(t) ∈ Um−1 = π(Tm−1 ), we know π(t) = π(t′ ) for t′ ∈ Tm−1 . Hence π(t − t′ ) = 0,
and we are in the situation of the preceding result: t − t′ ∈ Tm ∩ J, so we know the highest
degree part of t − t′ , i.e. t itself, lies in I. Hence t ∈ I.
23
Finally, we need to construct the representation ρ : g → End(S ∗ g). Equivalently, for every
m, we need a map fm : g ⊗ S m → Sm+1 g satisfying the three properties we want:
(1) fm (xλ ⊗ zΣ ) = zλ zΣ if λ ≤ Σ and zΣ ∈ S m ;
(2) fm (xλ ⊗ zΣ ) − zλ zΣ ∈ S k for k ≤ m and zΣ ∈ S k ;
(3) fm (xλ ⊗ fm (xµ ⊗ zτ )) = fm (xµ ⊗ fm (xλ ⊗ zτ )) + fm ([xλ , xµ ] ⊗ zτ ).
Just do it. We construct
fm (xλ ⊗ zi(1) ⊗ · · · ⊗ zi(m) ) = zλ ⊗ zi(1) ⊗ · · · , λ ≤ i(1).
If i(1) < λ, then we can swap two terms using the third property:
fm (xλ ⊗ zi(1) ⊗ · · · ⊗ zi(m) ) = fm (xi(1) ⊗ zλ ⊗ zi(1) ⊗ · · · ) + fm ([xλ, xi(1) ] ⊗ zi(2) ⊗ · · · )
which is well-defined because [xλ , xi(1) ] lies in g and the remainder lies in S m−1 .
So we could use induction: if we defined fm−1 , we have defined fm . Formally, induct on
m. For m = 0, let f0 (xi ⊗ 1) = zi for all i. Now we use the commutator relation to push
computations with fm onto fm−1 . Explicitly we have fm (xλ ⊗zΣ ) = zλ zΣ if λ ≤ Σ. Otherwise
if Σ = (µ, τ ) for µ < λ, and τ a multi-index, then
fm (xλ ⊗ zΣ ) = fm (xλ ⊗ fm−1 (xµ ⊗ zτ ))
Since µ < λ, we know by the third property that this is equal to
fm (xµ ⊗ fm (xλ ⊗ zΣ )) + fm−1 ([xλ , xµ ] ⊗ zτ ).
The hard part is to compute
fm (xλ ⊗ zτ ) = fm−1 (xλ ⊗ zτ ) ≡ zλ zτ mod Sm−1 .
Hence now everything is well-defined, because we’ve pushed everything into lower degree,
and verifies property (2) and propery (1) by construction.
By construction, we automatically satisfy property (1) and (2). Look at property (3),
which we abbreviate as
xi xj zJ = xj xi zJ + [xi , xj ]zJ .
By construction, we satisfy this when j < i and j ≤ J (because we took j = i1 ). When
j > i and i ≤ J we also satisfy this by rearranging terms and using [xi , xj ] = −[xj , xi ].
Clearly j = i is also fine. So we only need to check it when neither i ≤ J nor j ≤ J are
true. Write J = (k, L), so that we have k ≤ L and k < i and k < j. We have
xi xj zJ = xi xj xk zL = xi (xk xj zL + [xj , xk ]zL )
= xk xi xj zL + [xi , xk ](xj zL ) + [xj , xk ](xi zL ) + [xi , [xj , xk ]]zL
where the last equality used the induction hypothesis that (3) holds for fm−1 . Note that
this equality goes through even if we swap i and j. Interchanging them and subtracting,
xi xj zJ − xj xi zJ = (xk xi xj zL − xk xj xi zL ) + ([xi , [xj , xk ]]zL − [xj , [xi , xk ]]zL )
= xk [xi , xj ]zL + ([xi , [xj , xk ]]zL + [xj , [xk , xi ]]zL )
= [xi , xj ](xk zL ) + ([xk , [xi , xj ]] + [xi , [xj , xk ]] + [xj , [xk , xi ]])zL = [xi , xj ]xJ .
24
Finally, the check that this construction satisfies the third property is a computation using
the Jacobi identity for the bracket (which we haven’t used yet). We enforced property (3)
only in a special case. It says that
xi xj zJ = xj xi zJ + [xi , xj ]zJ .
The special case was when j < i and j ≤ J. The case where j > i and j ≤ J is checked by
symmetry in i and j, and the fact that [xi , xj ] = −[xj , xi ]. There’s also nothing interesting
in the case j = i and j ≤ J. Because i, j are interchangeable, really the only case to check
is neither i ≤ J nor j ≤ J. The way to check this is to use the Jacobi identity, by extracting
a third index out of J. □

14. Poisson algebras and Poisson manifolds

A Poisson algebra A has two products: one as a commutative, associative algebra, and
another as a Lie algebra. These products are compatible by the Leibniz rule
{f, g1 g2 } = {f, g1 }g2 + {f, g2 }g1 ,
i.e. the bracket {f, −} is a derivation for the commutative associative algebra. Recall that
{−, −} arises as the commutative limit of non-commutative algebras ∗ℏ :
f ∗ℏ g − g ∗ℏ f
{f, g} = lim .
ℏ→0 ℏ
This limit is called the classical limit. The process in reverse is called quantization and
is much more difficult.
Any commutative associative algebra can be thought of as a collection of functions on some-
thing. For example, if the ring of functions on a manifold has the structure of a Poisson
algebra, we call it a Poisson manifold.
Example 14.1. Let X = T ∗ M . Then functions on X consist of pullbacks of functions on
M , and also vector fields on M . We also have the algebra of differential operators of M whose
lower-order bits are these two types of objects, where if coordinates on M are (q1 , . . . , qn ),
then there is the commutation relation [∂qi , qi ] = δij . If we denote pi := ℏ∂qi (by rescaling by
ℏ along fibers), then [pi , qj ] = ℏδij . The corresponding Poisson bracket is {pi , qj } = δij .
Remark 14.2. Consider the maximal ideal mx = {f : f (x) = 0} in the algebra of functions
on X. Then {c, −} = 0 where c is a constant, but we also have
{m2x , −}|x = 0,
since {f, −}|x is determined by the class of f − f (x) in mx /m2x , which is the cotangent space.
Hence the Poisson bracket goes from differentials to functions, and therefore is a tensor.
Example 14.3. A Lie algebra g is not a Poisson manifold, but its dual g∗ is. Functions on g∗
include constants k, and linear functions g, and so on: k ⊕ g ⊕ S 2 g ⊕ · · · , denoted S • g. What
is the non-commutative algebra whose limit is this? It is the universal enveloping algebra
U gℏ , with a parameter ℏ: in the universal enveloping algebra U g, we had ξη − ηξ = [ξ, η],
but for U gℏ we define ξη − ηξ = ℏ[ξ, η], with ℏ of degree 1.
25
Example 14.4. The intersection of the previous two examples is called the Heisenberg
Lie algebra, where [pi , qj ] = eδij , where e is a central element. (We can always mod out by
central elements.)

Fix H a function on X, called the Hamiltonian. Then Hamilton’s equation says

d
f = {H, f }.
dt
As discussed, {H, −} is a derivation of a commutative product, i.e. a vector field on X, which
specifies dynamics. (Not every dynamical system is Hamiltonian though.) For example, the
geodesic flow we discussed earlier on is an example of Hamiltonian dynamics, with X = T ∗ M
and H(p, q) = (1/2)∥p∥2 . (Of course, this corresponds to the Lagrangian formulation
∫
1 t1
L(q, q̇, t) dt, L(q, q̇, t) := ∥q̇(t)∥2 ,
2 t0
since H(p, q) = maxq̇ (⟨p, q⟩ − L(q, q̇)).) The Legendre transform is the classical limit of the
Fourier transform.
Lemma 14.5. The following are equivalent:
(1) {H, G} = 0 for some function G;
(2) G is preserved by the flow of H;
(3) H is preserved by the flow of G.

If H = (1/2)∥ξ∥2 , then we get geodesics in a left-invariant metrics. Then H is preserved

by left translations by G, but there is dim G worth of flows. We call preserved quantities
integrals, so there are dim G many integrals. For a rigid body, we write the phase space
T ∗ SO(3) as either g × G (with coordinates (ω, g)) or G × g (with coordinates (g, ξ)), and it
turns out these integrals are precisely the angular momentum ω.
So we understand ω, and we want to look at the time-evolution of ξ. By general principles,
d 1
ξ = {∥ξ∥2 , ξ}.
dt 2
We know the Poisson bracket {ξ1 , ξ2 } = −[ξ1 , ξ2 ] (the minus sign is because the ξ are left
invariant). Hence we re-interpret {∥ξ∥2 , ξ} as a bracket on T ∗ G as {ξ, ∥ξ∥2 } a bracket on g∗ :
d 1 1
ξ = {∥ξ∥2 , ξ} = {ξ, ∥xi∥2 }.
dt 2 2
Because the metric is both left and right invariant, ξ is Ad-invariant, fixed by the action of
G, i.e. {η, ∥ξ∥2 } = 0 for every η ∈ g. Hence ξ is a constant.

15. Baker–Campbell–Hausdorff formula

In a neighborhood of the identity, exp : g → G is a diffeomorphism. What does multiplication

look like in this chart? In other words, what is log(eX eY )? We know the first-order terms
are X + Y .
26
Warmup: start with a matrix Lie group, where eX = 1 + X + X 2 /2 + · · · and log(1 + X) =
X − X 2 /2 + · · · . Then
log(eX eY ) = log(1 + X + Y + X 2 /2 + XY + Y 2 /2 + · · · )
= X + Y + (X 2 /2 + XY + Y 2 /2 − (X + Y )2 /2) + · · · = X + Y + [X, Y ]/2 + · · · .

Let g be the free Lie algebra generated by variables X and Y . Then it is graded by the number
of generators: g = g0 ⊕ g1 ⊕ · · · , where for example g3 contains [x, [x, y]] and [y, [x, y]]. What
is the dimension of gn ? The∑ universal enveloping algebra U g is a free associative algebra and
is also graded. If we take d≥0 td dim(U g)d to be the generating function of the dimensions,
it is equal to (1 − 2t)−1 . From this we can compute the dimensions of the grading on g.
Consider exp : g → U cg (completion with respect to the grading) given by X 7→ ∑ n
n≥0 X /n!.
This is an isomorphism between series 0 + · · · in U cg and series 1 + · · · in U
cg. (Sidenote:
completion means we take a series to converge if the degree of its terms goes to infinity.)
Then we will show log(eX eY ) lies in b g, i.e. that all the terms in the resulting series involve
only (nested) commutators.
Suppose G is finite. Then it has a group algebra
⊕
A := CG ∼ = End(V ).
irreps V

The map from G to CG does not remember the group, e.g. think when G is abelian. How can
we reconstruct the group from the group algebra? Well, there is a (coassociative) diagonal
map
∆
G− → G × G, g 7→ (g, g)
which is a group homomorphism. By linearity, this extends to an algebra homomorphism
∆ ∑
A− → A → A. This map remembers the multiplication on irreps V1 ⊕ V2 = mi12 Vi . Hence
the group is the set of solutions in A to ∆(x) = x ⊕ x, which is a non-linear equation.
(Elements x satisfying this equation are called group-like.)
Definition 15.1. Such an algebra A with a coassociative comultiplication is called a bialge-
bra. A bialgebra is a Hopf algebra if in addition it has an anti-automorphism S : A → A
called the antipode. In our case, we take S(g) := g −1 .

Let G be a Lie group. Then take A = CG, i.e. finite linear combinations, which can
be viewed as measures with finite support (where multiplication is precisely convolution).
Define a map
∆ : U g → U g ⊗ U g, X 7→ X ⊗ 1 + 1 ⊗ X.
This is the differential of ∆ : G → G ⊗ G. We can sanity-check:
[∆(X), ∆(Y )] = [X ⊗ 1 + 1 ⊗ X, Y ⊗ 1 + 1 ⊗ Y ] = [X, Y ] ⊗ 1 + 1 ⊗ [X, Y ] = ∆([X, Y ]).
Hence we have a Hopf algebra structure on U g.
Proposition 15.2. If k is a field of characteristic 0, then the set of primitive elements
{solutions to ∆y = y ⊗ 1 + 1 ⊗ y} ⊂ U g
is equal to g.
27
Remark 15.3. This is no longer true in characteristic p, since
∆(X p ) = ∆(X)p = (X ⊗ 1 + 1 ⊗ X)p = X p ⊗ 1 + 1 ⊗ X p
shows that X p is also primitive.

Proof. Filter U g by degree (as in PBW). Denote the associated graded by Gr U g, which is just
Sg, the symmetric algebra. View Sg as the polynomial algebra on g∗ . If y is primitive, then
the top degree term of y is primitive for Sg. But comultiplication on Sg is just ∆ : C[g∗ ] →
C[g∗ × g∗ ] = C[g∗ ] ⊗ C[g∗ ]. In other words,
y(λ + µ) = y(λ) + y(µ), λ, µ ∈ g∗ .
Hence the top degree term of y is additive, and therefore linear. So y itself is linear, and
therefore y ∈ g. (This is where we need characteristic 0: in characteristic p, it is not true
that if a polynomial is additive, it is linear.) □
Lemma 15.4. An element X ∈ g is primitive if and only if eX := 1 + Y is group-like. In
other words, ∆X = X ⊗ 1 + 1 ⊗ X if and only if ∆eX = eX ⊗ eX .

Proof. This is a statement about a 1-dimensional Lie algebra g generated by X. Then U g

really just is polynomials on g∗ , and ea+b = ea eb . □
Theorem 15.5. log(eX eY ) ∈ g.

Proof. If we have a Lie algebra g freely generated by X, Y , then X and Y are primitive. By
the lemma, eX and eY are group-like. Then their product eX eY is group-like, since
∆(g1 g2 ) = ∆(g1 )∆(g2 ) = (g1 ⊗ g1 )(g2 ⊗ g2 ) = (g1 g2 ) ⊗ (g1 g2 ).
But then log(eX eY ) is primitive, by the lemma. □

So how do we actually write log(eX eY ) as a sum of (nested) commutators? Consider the

map Φ : U g → ĝ which takes a monomial in U g and replaces the (free) multiplication with
the Lie bracket, e.g.
xyx3 7→ [[[[x, y], x], x], x].
Another example: [x, y] ∈ g2 goes to [x, y] − [y, x] = 2[x, y] ∈ ĝ.
Lemma 15.6. An element A ∈ gk ⊂ g ⊂ U g satisfies Φ(A) = kA. In particular, A can be
written in terms of (nested) commutators.

Hence, using this lemma, we can convert the expression in U g for log(eX eY ) into a sum
of (nested) commutators, sometimes called the Baker–Campbell–Hausdorff series in
Dynkin form. This series has a radius of convergence 1.
Corollary 15.7. Lie groups are actually real analytic.
Example 15.8. Some examples of compact Lie groups: S 1 = R/Z, SU(n), U (n), O(n, R).
Some examples of non-compact Lie groups: GL(n, R), SL(n, R), O(n, C).

If G is a compact Lie group, then it has the following nice properties.

28
 (1) G has a left and right invariant measure µHaar , which is finite. (This comes from the
fact that any homomorphism G → (R>0 , ∗) is trivial.)
(2) (Averaging) Using this measure, we can take a vector to another vector fixed by the
action of the group G: ∫
v 7→ g · v µ(dg);
G

(3) (Complete reducibility) Any complex finite-dimensional ⊕representation V of G has

a positive definite Hermitian metric, and therefore V = Vi where the Vi are irre-
ducible.
(4) G has a left and right invariant Riemannian metric, which induces a positive-definite
bilinear form (·, ·) on g which is invariant, i.e. (Ad(g)ξ, Ad(g)η) = (ξ, η). This can
be differentiated to give ([ξ, γ], η) = (ξ, [γ, η]). Equivalently, ad(γ) is skew-symmetric.
Proposition 15.9. If g has a positive-definite invariant metric, then the universal cover Ĝ
of its Lie group is Rn times some compact Lie group.

Proof.
⊕First, apply complete reducibility to the adjoint representation of G on g, to get
g = i gi where the gi are simple. A simple Lie algebra can either be R or a simple non-
abelian Lie algebra. So it suffices to show that if g is simple non-abelian with positive-definite
invariant metric (·, ·), then Ĝ is compact.
Given ξ ∈ g, the exponential etξ is a geodesic. Claim: there is some constant c such that
it fails to be a minimal geodesic for ∥tξ∥ > c. We know ad(tξ) is skew-symmetric, so its
eigenvalues are purely imaginary. By rescaling ξ, which gives us the constant c, we can make
sure its eigenvalues are not a subset of (−2πi, 2πi). (Not all its eigenvalues can be zero,
otherwise it commutes with everything.) Hence the volume of Ĝ is bounded. □

16. Peter–Weyl theorem

We now look at a generalization of Fourier’s theorem, which says that there is an isometry
⊕
d
L2 (R/Z, dx) ∼= Ce2πikx .
k
⊕
(Here c means to take the direct sum of the subspaces first, and then to take the com-
pletion.) From the perspective of Lie theory, the summands Ce2πikx are 1 × 1 irreducible
representations of G.
Definition 16.1. If V is a representation of G, then there is a function
ϕℓ,v (g) := ℓ(g · v), v ∈ V, ℓ ∈ V ∗
called a matrix element. (We will prove soon that matrix elements are orthogonal.)
Theorem 16.2 (Peter–Weyl). If V ranges over all irreducible complex representations of
G, then
⊕
d
L2 (G, µHaar ) = (V ∗ ⊗ V, (A, B) := (tr A∗ B)/ dim)
V
where V ∗ ⊗ V are the matrix elements.
29
Remark 16.3. There is an action of G × G on L2 (G, µHaar ) by left and right translation:
(Lg f )h := f (g −1 h), (Rg f )h = f (hg).
What are the left and right actions of G on matrix elements? Well,
(Lg ϕℓ,v )h = ℓ(g −1 hv) = ϕgℓ,v , (Rg ϕℓ,v )h = ℓ(hgv) = ϕℓ,gv .
Hence the embedding V ∗ ⊗ V → {matrix elements} is (G × G)-equivariant. In fact, matrix
elements of V are precisely functions that transform in a representation V under Rg . The
space V ∗ ⊗ V = End(V ) has a natural Hermitian form (A, B) := tr A∗ B, i.e. the elementary
matrices Eij are orthonormal.
Theorem 16.4. Matrix elements of inequivalent irreducible representations are orthogonal.
Matrix elements ϕij of a representation V are orthogonal and
1
∥ϕij ∥2L2 (G) = .
dim V
∑
Hence ∥ ϕii ∥ = 1.

∫ W be irreducible representations of G, and let A : V → W be any operator.

Proof. Let V,
Then Ā := gAg −1 : V → W commutes with all g ∈ G. Schur’s lemma says that:
(1) if W ̸= V , then Ā = 0;
(2) if W = V , then Ā = λI where λ = tr A/ dim V .
If we choose an invariant Hermitian form for V then g −1 = (ḡ)T (i.e. g ∈ U (V )). Taking
A = Eij , the integral becomes
(∫ )
−1
gEij g dµ(g) = (ϕℓj , ϕki )L2 .
kl
□

Hence we have shown that

⊕
(V ∗ ⊗ V, ∥ · ∥2 / dim V ) → L2 (G, µ)
irreps V

is an injection, and the left hand side is (G × G)-equivariant. The image consists of G-
finite vectors in L2 (G), i.e. vectors that transform in a finite-dimensional representation.
A rephrasing Peter–Weyl is that the image of this map is dense.
Lemma 16.5. Peter–Weyl is equivalent to showing that G has a faithful linear representation.

⊕⊂ GL(W ). Polynomials of GL(W ) are

Proof. If W is a faithful linear representation, then G
⊗n
just matrix elements of W , which decomposes as Vi,n where Vi,n are irreps. But Stone–
Weierstrass says polynomials are dense in continuous functions, and continuous functions
are dense in L2 . □

Hence we have proved Peter–Weyl for all the compact groups we have seen; it is an easy
consequence of Stone–Weierstrass.
30
 17. Compact operators

Let V be a Banach space (though we will work with Hilbert spaces only). Recall that the
unit ball {v : ∥v∥ ≤ 1} is compact if and only if dim V < ∞.
Definition 17.1. An operator A : V → V is compact if it sends bounded sets to pre-compact
sets, i.e. sets whose closures are compact.
∑
Example 17.2. A map A : Cn → Cn is an n × n matrix. We have (Av)i = j aij v j , which
∫
we can write as [Af ](i) = a(i, j)f (j) with the counting measure, on basis vectors {1, . . . , n}.
But we can replace {1, . . . , n} with (X, ν) where ν is a measure. So we consider maps
∫
K := f (x) 7→ K(x, y)f (y).
X

Then K : L (X) → C(X) ⊂ L (X) and takes bounded sets to pre-compact sets; we know pre-
2 2

compact sets (in C(X) with the sup norm) are precisely those whose functions are uniformly
bounded and equi-continuous, so this is not hard to check. For example,
∫ ∫
|Kfn (x1 ) − Kfn (x2 )| ≤ |K(x1 , y) − K(x2 , y)||fn (y)| dy ≤ C |fn (y)|2 dy.

Another proof of the same fact: use that an operator A is compact if and only if it is the limit
of finite rank operators in the operator norm. Such maps are called integral operators and
are a primary example of compact operators.
Remark 17.3. Here is the more general situation. Suppose we have a functor F from
topological spaces to algebras that behaves well with respect to pushforwards and pullbacks.
Then F (X × X) acts on F (X) via
Af := (p1 )∗ (A · p∗2 (f )),
called a Fourier–Mukai kernel.
Theorem 17.4 (Spectral theorem for compact self-adjoint operators). If K = K ∗ is com-
⊕
pact, then V = c i Cvi such that Kvi = λi vi , and limi→∞ |λi | → 0. In general,
∑
K= λi (fi , ·)ei
i

with |λi | → 0, where ∥ei ∥ = ∥fi ∥ = 1.

Example 17.5. Let X = G, and consider the operator K which is the average of left shifts
by g ∈ G: ∫
K := k(g)Lg dg, (Lg f )(h) := f (g −1 h).

Here k is some continuous function on G which we think of as a weight. Explicitly,

∫ ∫
[Kf ](h) = k(g)f (g h) dg = k(hg −1 )f (g) dg.
−1

So if we declare K(h, g) := k(hg −1 ), we have obtained an integral operator. We can make it

self-adjoint by imposing k(g −1 ) = k(g). Hence by the spectral theorem, if λi and vi are the
31
eigenvalues and eigenvectors, respectively, of K, then
⊕
d
L2 (G) = Cvi
i

consists of summands which are clearly finite-dimensional for non-zero eigenvalues. (This
comes from limi→∞ |λi | = ∞, so every non-zero eigenvalue can appear only a finite number
of times.)

This is how we finish off the proof of Peter–Weyl! Note that K commutes with the right-
⊕
action of G. Hence G acts on the right on c i Cvi , and every vector corresponding to λ ̸= 0
is
∫ G-finite. For λ = 0, choose a sequence kn such that kn → δe and kn (g −1 ) = kn (g). Then
kn (hg −1 )f (g) dg → f (h) shows that f is zero.

18. Complexifications
⊕ ⊕
Definition 18.1. The finite part of c V End(V ) is V End(V ). We denote it by L2 (G)fin .

Consider L2 (SU(n)). Its finite part L2 (SU(n))fin is precisely C[SL(n, C)], since the complex-
ification of SU(n) is SL(n, C).
Definition 18.2. Given a 1-connected compact Lie group G with Lie algebra g, its com-
plexification GC is the 1-connected complex Lie group with Lie algebra gC := g ⊗R C.

Hence there is a correspondence between:

(1) finite-dimensional complex representations of G;
(2) finite-dimensional complex representations of g (by Lie’s theorem);
(3) finite-dimensional complex representations of gC ;
(4) finite-dimensional complex representations of GC (by Lie’s theorem again).
Clearly G sits in GC as a totally real submanifold. Matrix elements of GC are complex
analytic, and matrix elements of G are real analytic. The map between the two is by
restriction and by analytic continuation.
While in general L2 (G) is not an algebra (the product of two L2 functions is not necessarily
L2 anymore), matrix elements are analytic and therefore form an algebra:
End(V ) ⊗ End(V ′ ) ⊂ End(V ⊗ V ′ ).
This algebra is finitely generated. (It also clearly has no zero divisors.) So we can make the
analytic variety GC algebraic by producing this finitely generated algebra which separates
points. In other words, GC is automatically a linear algebraic group. Also, because finite-
dimensional complex representations of compact G are semisimple, the same holds for finite-
dimensional complex representations of GC .
Let G be a linear algebraic group, i.e. a closed subgroup of GL(N, k) for k algebraically
closed. It is fairly easy to show that if G is reductive, then the category of representations
32
 ⊕ ∗
of G is semisimple, and also that the analogue k[G] = V V ⊗ V of Peter–Weyl holds.
Reductive Lie groups arise as complexifications of Lie groups.

19. Symmetric spaces

⊕
Let G be a compact Lie group, and H a Lie subgroup. We know L2 (G/H) = irreps V V∗⊗
V H . In general, we can ask: what can we say about V H ?
Definition 19.1. Let X be a compact (for simplicity) Riemannian manifold. We call X
symmetric if for every point x ∈ X, there exists an isometry sx which fixes x and acts by
−1 on Tx X.
Remark 19.2. Since every isometry preserves geodesics, to specify an isometry it suffices
to specify its action on a point and on the tangent bundle.
Example 19.3. The spheres S n are clearly symmetric. We can also mod by {±1} to get
RPn . In fact, any compact Lie group G is symmetric: the isometry around the origin is
g 7→ g −1 .

Suppose any two points on X are connected by a geodesic. Pick two points x, y and let
(x + y)/2 denote the midpoint on the geodesic connecting them. What is τx→y := s(x+y)/2 sx ?
It preserves the geodesic, and on the geodesic it will be a translation by the length from x to y.
It is therefore true that the group of isometries acts transitively. Hence X = Isom(X)/ Stabx .
How do we pick out the stabilizer? Note that Stabx ⊂ Isom(X)sx . By the example below,
we see this may not be an equality.
Example 19.4. Take S n−1 = SO(n)/SO(n−1) with x = e1 . Then sx is diag(1, −1, −1, . . . , −1).
But then
 
 ∗ 0 0 ··· 

 0 

SO(n) = 
sx
0 ∗
 = O(n − 1) ̸= SO(n − 1).


 
 .. 
.

In fact, we see that Stabx ⊃ Isom(X)s0x , the connected component of the identity. In general,
the following proposition is true.
Proposition 19.5. Gs ⊃ Stabx ⊃ (Gs )0 .

Proof. Any isometry that commutes with reflection by sx takes x to a fixed point of sx . □

Let G be a compact Lie group with an automorphism s : G → G of order 2. Then Gs , the

collection of fixed points of s, may not be connected, but we can choose a subgroup H such
that Gs ⊃ H ⊃ (Gs )0 . (Keep in mind the example of the sphere, where Gs = O(n − 1) and
(Gs )0 = SO(n − 1).) Then s descends to X = G/H, and the identity 1 is an isolated fixed
point. So we have shown that symmetric spaces are precisely the quotients of compact Lie
groups G by a subgroup H such that Gs ⊃ H ⊃ (Gs )0 where s2 = 1 is an involution.
33
Example 19.6. If X = G is a compact Lie group, then at least G × G acts transitively via
(g1 , g2 ) · x = g1 xg2−1 . The stabilizer Stab1 of the identity is precisely the diagonal ∆(G). On
G×G, there is an involution that permutes factors. It descends to x 7→ x−1 on X. In this
case, the stabilizer Stab1 is precisely the fixed points (G × G)s .
Example 19.7. The complex Grassmannian Gr(k, n, C) can be written as U (n)/(U (k) ×
U (n−k)). Of course, U (k)×U (n−k) is the matrix commuting with diag(1, 1, . . . , 1, −1, −1, . . . , −1).
It follows that the complex Grassmannian is a symmetric space. In the real case, we can
write Gr(k, n, R) as SO(n)/S(O(k) × O(n − k)). Alternatively, we can also quotient by
SO(k) × SO(n − k) to get the oriented Grassmannian, a double cover of Gr(k, n, R).
∑n
Example 19.8. Equip R2n with a symplectic form ω = i=1 dpi ∧ dqi . A Lagrangian
subspace is an n-dimensional subspace L ⊂ R such that ω|L = 0. It is easy to see that
2n

n is the maximal dimension for which ω|L = 0 can happen, since ω is non-degenerate. The
space of all Lagrangian subspaces is called the Lagrangian Grassmannian L Gr(2n).

This is a homogeneous
√ space, but the way to see this is interesting. Think of R2n ∼ = Cn
pi + −1qi . Then ω is proportional to the imaginary part of the Hermitian form
via zi := ∑
(z, w) := i z̄i wi . By definition, the unitary group U (n) preserves the Hermitian form, and
therefore preserves, separately, its real and imaginary parts. Hence U (n) preserves ω, and
is in fact transitive on L Gr(2n). The stabilizer of a point is O(n), since it is precisely the
stabilizer of Rn ⊂ Cn , i.e. where im z = 0. Note that O(n) = U (n)s where s is complex
conjugation g 7→ ḡ. Alternatively, we can also take U (n)/SO(n) to get the double cover
consisting of oriented Lagrangian subspaces.
Theorem 19.9 (Gelfand lemma). If X = G/H is a symmetric space, then dim V H ∈ {0, 1}
for any irrep V .

⊕ ∗
Proof. We know L2 (H) = W W ⊗ W where the sum is over irreps W . Inside ∫ the sum
is the trivial representation C · 1. Therefore there exists a projector P : f (h) 7→ H f (h) dh
where dh is the normalized Haar measure. This is analogous to the Fourier case:

∑ ∫ 1
f (t) = fˆ(k)e2πikt , fˆ(k) = f (t)e−2πikt dt
k 0

extracts fˆ(k). In our projector, we are just extracting the coefficient associated to the trivial
representation.

Consider L2 (H\G/H), i.e. functions invariant under the H-action on both the left and
the right. This is just P L2 (G)P by the definition of the projector P . Similarly, the same
applies for C(H\G/H), the space of left and right invariant continuous functions on G.
⊕
Hence L2 (H\H/H) = c V (V ∗ )H ⊗ V H since we take invariants on both sides. But each
term is just End(V H ).⊕The statement that dim V H ∈ {0, 1} for every V is equivalent
H
to the statement that V End(V ) is commutative. But this algebra is commutative iff
its completions are commutative, i.e. C(H\G/H) is commutative. So it suffices to prove
C(H\G/H) is commutative.
34
Fact: if an algebra A has an anti-automorphism σ, i.e. a linear map such that σ(ab) =
σ(b)σ(a), such that σ = 1, then A is commutative. This is stupidly obvious but is ap-
parently somewhat deep. Take A = C(H\G/H) = C(H\X). We will define such an
anti-automorphism σ on A by first defining it on G. Define it to be σ : g 7→ s(g −1 ) = s(g)−1
(since s is a group automorphism), so that it is an anti-automorphism of G and therefore of
C(G) and therefore of A. Now we show it is the identity on A. Given g near the identity in
X, we can write it as g = τx→y h. Then
σ(g) = σ(h)σ(τx→y ) = σ(h)τx→y .
Hence σ(g) ∈ HgH, i.e. applying σ does not change the two-sided coset. It follows that σ is
the identity on A = C(H\G/H). □
Remark 19.10. It was important for H to be compact because we needed to integrate over
H, but not so important for G to be compact. Indeed, there are non-compact symmetric
spaces like the Lobachevsky plane.
⊕
Corollary 19.11. L2 (X) = c dim V H =1 V .
Corollary 19.12. G-invariant operators (of any nature) in L2 (X) commute.

Proof. Such operators commute with G and preserve the decomposition of L2 (X), and there-
fore act by scalars in each V . So of course they commute. □

20. Solvable and nilpotent Lie algebras

Let F be any field (of any characteristic, and not necessarily algebraically closed). Through-
out, let L denote the Lie algebra, finite dimensional over the field F .
Definition 20.1. Define the following sequence of ideals:
L(1) := L, L(2) := [L(1) , L(1) ], L(3) := [L(2) , L(2) ], ··· .
We say L is solvable if L(n) = 0 for some n.
Example 20.2. A basic example is the Lie algebra L of upper triangular matrices inside
gl(n, F ). It is easy to check that L is solvable.
Proposition 20.3. (1) If L is solvable, then so are all the subalgebras and homomorphic
images of L.
(2) If I ⊂ L is a solvable ideal such that L/I is solvable, then L is also solvable.
(3) If I, J ⊂ L are solvable ideals, then I + J is also solvable.

Proof. (1) is obvious. (2) follows by noting that L/I is solvable implies (L/I)(n) = 0 for
some n, i.e. L(n) ⊂ I for some n. But I is solvable, so L is therefore also solvable. (3) follows
from the isomorphism (I + J)/J → I/(I ∩ J). Since I is solvable, I/(I ∩ J) is solvable by
(1). But J is also solvable, so by (2), I + J is also solvable. □
Definition 20.4. By (3) in the preceding proposition, there must exist a unique maximal
solvable ideal in L, called the radical rad L of L. We say L is semisimple if rad L = 0.
35
Remark 20.5. For any L, it follows that L/ rad(L) is semisimple.
Definition 20.6. Define another sequence of ideals:
L1 := L, L2 := [L1 , L1 ], L3 := [L1 , L2 ], ··· .
We say L is nilpotent if Ln = 0 for some n.
Example 20.7. The Lie algebra of strictly upper triangular matrices in gl(n, F ) is nilpotent.
Remark 20.8. It is easy to see that L(i) ⊂ Li . Hence nilpotent implies solvable. The
converse is not true.
Proposition 20.9. (1) If L is nilpotent, then so are all the subalgebras and homomorphic
images of L.
(2) If L/Z(L) is nilpotent, so is L.
(3) If L is nilpotent and non-zero, then Z(L) ̸= 0.

Proof. (1) is obvious. (2) comes from (L/Z(L))i = 0 implying Li ⊂ Z(L), so that Li+1 = 0.
(3) comes from 0 = Ln = [L, Ln−1 ] implying 0 ̸= Ln−1 ⊂ Z(L). □
Remark 20.10. Note that L is nilpotent iff for some n, ad x1 ad x2 · · · ad xn (y) = 0 for every
x1 , . . . , xn ∈ L. In particular, (ad x)n = 0. So ad x ∈ gl(L) is a nilpotent matrix.
Theorem 20.11 (Engel). L is nilpotent if and only if all elements of L are ad-nilpotent,
i.e. ad x is a nilpotent matrix for all x ∈ L.
Remark 20.12. Question: given a nilpotent matrix X ∈ gl(V ), is the adjoint ad X also
nilpotent? Yes, because (ad X)Y = XY − Y X is nilpotent. However, the converse is not
true: take X = I, which is not nilpotent, but ad X = 0.
Theorem 20.13. Let L be a subalgebra of gl(V ) (with dim V < ∞). If L consists of nilpotent
endomorphisms and V ̸= 0, then there exists a non-zero vector v ∈ V such that Lv = 0.

Proof. Use induction on the dimension of L. The base cases dim L = 0, 1 are obvious. So
take dim L ≥ 2, and let 0 ̸= K ⊊ L be a subalgebra. By the previous remark, since every
element in K is nilpotent, the adjoint action of K on L is also nilpotent. The adjoint action
of K on L/K (which is well-defined because the action preserves K) is also nilpotent. Hence
there is a homomorphism K → gl(L/K). By the induction hypothesis, there exists a non-
zero element x + K ∈ L/K such that (ad K)(x + K) = 0, i.e. [K, x] ⊂ K with x ∈ / K.
Hence the normalizer NL (K) contains x, and therefore K ⊊ NL (K). So if we take K to be
a maximal proper subalgebra of L, then NL (K) = L because of the maximality of K, and
dim L/K = 1. Write L = K + F · z for some z ∈ L \ K. Define
W = {v ∈ V : K · v = 0},
which is non-zero because x exists. It suffices now to find an element in W annihilated by
z. We have
xzv = [x, z]v + zxv = 0 + zxv
since x ∈ NL (K). Then z commutes with the K action, and therefore we can find v ∈ W
such that zv = 0. □
36
 ad
Proof of Engel’s theorem. Consider the map L −→ gl(L). By hypothesis, the operators ad x
are nilpotent for every x ∈ L. Hence by the preceding theorem, there exists v ∈ L such that
(ad x)v = 0 for all x ∈ L. Engel’s theorem follows by induction on the dimension of L, using
that dim L/Z(L) < dim L and that L/Z(L) nilpotent implies L nilpotent. □
Corollary 20.14. Let L ⊂ gl(V ). If L consists of nilpotent endomorphisms, then there
exists a flag (Vi ) in V such that X · Vi ⊂ Vi−1 for all i and all X ∈ L. In other words, there
exists a basis of V such that all the matrices of L are strictly upper triangular.

Proof. Using the theorem, find v ∈ V such that Lv = 0. Take V1 = F v. Now induct to find
a flag on V /V1 which can be lifted back to V . □

From now on, assume char F = 0, and F = F̄ is algebraically closed. We would like an
analogue of Engel’s theorem for solvable Lie algebras.
Theorem 20.15. If L ⊂ gl(V ) is solvable (with dim V < ∞), then V contains a common
eigenvector for L.

Proof. Again, induct on dim L. We first find a ideal K ⊂ L of codimension 1. Note that
[L, L] ̸= L, and is therefore a proper subalgebra. Let K be the pre-image of a codimension
1 subspace in L/[L, L]. Such a subspace is an ideal because L/[L, L] is abelian. Hence K
is a codimension 1 ideal in L. Now by the induction hypothesis, there exists an eigenvector
v ∈ V for K with associated character λ : K → F (i.e. xv = λ(x)v). Fix such a character λ,
and define
W := {w ∈ V : xw = λ(x)w ∀x ∈ K}.
Since v ∈ W , we know W ̸= 0. Finally, we show L preserves W . Pick x ∈ L, w ∈ W , and
y ∈ K. Then
yxw = [y, x]w + xyw = λ([y, x])w + λ(y)xw
since [y, x] ∈ K (because K is an ideal). So if we can show λ([y, x]) = 0, then xw ∈ W .
Let n be the smallest integer such that w, xw, x2 w, . . . , xn w are linearly dependent. Define
Wi := F w + F xw + · · · + F xi−1 w and W0 := 0, and Wn := Wn+1 := · · · . Check by induction
(using commutators to push terms into Wi ) that for all y ∈ K, we have
yWi ⊂ Wi , yxi w ∼ = λ(y)xi w mod Wi .
Hence trWn y = nλ(y), because the first equation says y is an upper triangular matrix, and
the second equation says the diagonal of y consists of only λ(y). Now we have
nλ([y, x]) = trWn [y, x] = 0
because trWn [y, x] is just the trace of two matrices. Because char F = 0, we can divide by n
to get λ([y, x]) = 0. Hence write L = K + F z, and find an eigenvector in W for z. Then we
are done. □
Corollary 20.16 (Lie). If L ⊂ gl(V ) is solvable (with dim V < ∞), then L stabilizes some
flag (Vi ) in V . In other words, the matrices of L, relative to some basis, are upper triangular.

Proof. Obvious. □
37
Corollary 20.17. If L is solvable, then there exists a chain of ideals of L 0 ⊂ L1 ⊂ · · · ⊂
Ln = L such that dim Li = i.
ad
Proof. Apply the preceding corollary to the adjoint representation L −→ gl(L). □
Corollary 20.18. If L is solvable, then x ∈ [L, L] implies ad x is nilpotent. In particular,
[L, L] is nilpotent.
ad
Proof. Consider the adjoint representation L −→ gl(L). Then ad L consists of upper trian-
gular matrices, and ad[L, L] = [ad L, ad L] consists of strictly upper triangular matrices. By
Engel’s theorem, [L, L] is nilpotent. □
Remark 20.19. Conversely, if [L, L] is nilpotent, then L is solvable. This is because L/[L, L]
is commutative and therefore solvable, and [L, L] is nilpotent and therefore solvable.
Theorem 20.20 (Cartan). Let L ⊂ gl(V ) (with dim V < ∞). If tr xy = 0 for all x ∈ [L, L]
and y ∈ L, then L is solvable.
Lemma 20.21. Let A ⊂ B be two subspaces of gl(V ). Set
M := {x ∈ gl(V ) : [x, B] ⊂ A}.
Suppose x ∈ A satisfies tr xy = 0 for all y ∈ M . Then x is nilpotent.

Proof. This is a statement from Humphrey’s book. We will skip the proof. □

Proof of Cartan’s theorem. We know that L is solvable iff [L, L] is nilpotent. Hence it suffices
to prove [L, L] is nilpotent. By Engel’s theorem, it suffices to show ad[L, L] is nilpotent.
Apply the lemma: let A = [L, L], and B = L, so that M = {x ∈ gl(V ) : [x, L] ⊂ [L, L]}.
In particular, M ⊃ L. For z ∈ M , we have tr([x, y]z) = tr(x[y, z]), but [y, z] ∈ L so by
hypothesis, this trace vanishes. Hence we can apply the lemma, and we are done. □
Corollary 20.22. Let L be a Lie subalgebra such that tr(ad x ad y) = 0 for all x ∈ [L, L]
and y ∈ L. Then L is solvable.

21. Parabolic and Borel subgroups

Definition 21.1. A variety X is complete if for any other variety Y , the projection
pr2
X × Y −−→ Y is a closed morphism.
Proposition 21.2. Let X be complete. Then:
(1) a closed subvariety of X is also complete;
(2) if Y is complete, then so is the product X × Y ;
(3) if ϕ : X → Y is a morphism, then ϕ(X) is closed and complete;
(4) if X is a subvariety of Y , then X is closed;
(5) if X is irreducible, then k[X] = k;
(6) if X is affine, then X is finite;
38
 (7) a projective variety is complete.
Definition 21.3. G is solvable if there exists a series of subgroups {1} = G0 ≤ G1 ≤ · · · ≤
Gn = G such that Gj−1 is normal in Gj an Gj /Gj−1 is abelian. G is nilpotent if there exists
n such that (x1 , (x2 , . . . , (xn , y)) · · · ) = e for all x1 , . . . , xn , y ∈ G, where (x, y) := xyx−1 y −1 .
Definition 21.4. A closed subgroup P is parabolic if G/P is complete.
Example 21.5. Let G = GL(n, k). Take P to be the block-diagonal matrices with a k × k
block and a (n − k) × (n − k) block. Then P is a parabolic subgroup, since G/P is just the
Grassmannian Gr(n, k), which is projective and therefore complete.
Lemma 21.6. If P is parabolic, then G/P is projective.

Proof. We already know G/P is quasi-projective by construction. We also know it is com-

plete. Hence G/P is a closed subset of a projective variety, and therefore projective. □
Lemma 21.7. Let Q ⊂ P ⊂ G be parabolic subgroups of G. Then Q ⊂ G is also parabolic.

Proof. We need to show G/Q is complete, i.e. for any variety Z, the projection G × Z →
G/Q × Z → Z is closed. (Fact: a map X → Y between G-varieties gives an open map
X × Z → Y × Z.) Equivalently, we must show that A ⊂ G × X closed such that (g, x) ∈ A
implies (gQ, x) ∈ A. Consider
α
P × G × X −−−→ G × X
x x
 
 
α−1 A −−−→ A.
Then something happens. (?) □
Lemma 21.8. If P ⊂ G is parabolic, then any Q ⊃ P is parabolic. Also, P is parabolic if
and only if P 0 ⊂ G0 is parabolic (connected components).

Proof. Clearly G/P → G/Q is surjective. But G/P is complete, so the image G/Q is also
complete. The second claim uses the fact that G/G0 is finite, so G0 ⊂ G is automatically
parabolic. This holds for any G, so in particular P 0 ⊂ P is parabolic. If P ⊂ G is parabolic,
P 0 ⊂ G is also parabolic. The map G0 /P 0 ⊂ G/P 0 is closed, so since closed subvarieties
of complete varieties are complete, G0 /P 0 is complete, and therefore P 0 ⊂ G0 is parabolic.
Conversely, if P 0 ⊂ G0 is parabolic, we know G0 ⊂ G is parabolic, so by transitivity, P 0 ⊂ G
is parabolic. But P 0 ⊂ P ⊂ G, so by the first part of the lemma, P ⊂ G is also parabolic. □
Proposition 21.9. A connected group G contains a non-trivial parabolic subgroup if and
only if G is not solvable.

Proof. Fact: if G acts on X, then there exists a closed orbit in X. (If G is a unipotent group,
then every orbit is closed.) Put G ⊂ GL(V ) for dim V sufficiently large. In particular, G
acts on PV . Then there exists a closed orbit Ox , which bijects with G/Gx . Since Ox is
closed, it is projective and therefore complete. Then the stabilizer Gx is parabolic.
39
If Gx = G, then consider the action of G on P(V /kx). By the same argument, we can find
another parabolic subgroup. Hence there are two cases:
(1) there exists a non-trivial parabolic subgroup, i.e. at some point we stop, with Gx ̸= G;
(2) there does not exist a non-trivial parabolic subgroup, i.e. Gx = G at each step,
and therefore G is contained within upper triangular matrices. But upper triangular
matrices are solvable, and subgroups of solvable groups are solvable, so G is solvable.
Conversely, assume G is connected and solvable, and we want to show G has no non-trivial
parabolic subgroup. Assume P ⊂ G is a maximal parabolic subgroup. Consider (G, G),
which is also connected. Define Q = P · (G, G), which is also connected, and contains the
parabolic subgroup P and is therefore parabolic.
(1) If Q = P , then (G, G) ⊂ P (and is a normal subgroup). Then G/P is affine, and
therefore finite. But it is also connected, so P = G.
(2) If Q = G, then G(G/P ) = P (G, G)/P ∼ = (G, G)/((G, G) ∩ P ). But (G, G) ∩ P ⊂
(G, G) is parabolic. By induction on dim G, we can descend to working with (G, G),
and hence P = G.
Hence there is no non-trivial parabolic subgroup P ⊂ G. □
Theorem 21.10 (Borel’s fixed point theorem). Let G be a connected solvable linear algebraic
group. Let X be a complete G-variety. Then there exists a point x ∈ X fixed by G.
Remark 21.11. If G acts on V , then G also acts on PV . If there is a line L ∈ PV fixed by
G, then there is an eigenvector for the group G.
Example 21.12. Note that in characteristic 0, a Lie group G is solvable if and only if
its Lie algebra g is solvable. In characteristic non-zero, the converse is false: g solvable
does not imply G solvable. For example, the Lie algebra sl(2, F ) is solvable over a field of
characteristic 2, because it has the standard basis {e, f, h} satisfying [h, e] = 2e, [h, f ] = 2f ,
and [e, f ] = h, which is nilpotent. They both act on P(F 2 ), but sl(2, F ) does not have a fixed
point.

Proof of Borel’s fixed point theorem. Since G acts on X, there exists a closed orbit Ox ∼
=
G/Gx . We assumed G is complete, so Ox is also complete. Hence Gx is a parabolic subgroup.
But G is connected and solvable, so by the proposition either Gx = G or Gx = {e}. Hence
either x is the desired fixed point, or we get a contradiction. □
Definition 21.13. A Borel subgroup of G is a closed connected solvable subgroup of G
which is maximal among all subgroups with these properties.
Example 21.14. Take GL(n). Then the subgroup of all upper triangular matrices is a Borel
subgroup.
Theorem 21.15. (1) P ⊂ G if parabolic if and only if P contains a Borel subgroup.
(2) Any Borel subgroups are parabolic.
(3) Any two Borel subgroups are conjugate.
40
Proof. (1) Assume P is parabolic. Take any Borel subgroup B. Then B acts on G/P by left
multiplication, so by Borel’s fixed point theorem, there exists gP ∈ G/P fixed by B. Then
g −1 Bg ∈ P is a Borel subgroup, by definition. Conversely, assume G is not solvable. Then
there exists a parabolic subgroup P ⊂ G. Then pick a Borel set B ⊂ P (by the forward
direction). By induction on dim G, we get B is parabolic in P . Since P is parabolic in G, it
follows that B is parabolic in G.
(2) Easy, using the forward direction of (1).
(3) Apply Borel’s fixed point theorem. □
Theorem 21.16 (Lie–Kolchin). Let G be a closed connected and solvable subgroup of GLn .
Then there exists some x ∈ GLn such that xGx−1 is a subset of the upper triangular matrices.

22. Maximal tori

Theorem 22.1 (Kolchin). Let V be a vector space over F , and let G be any subgroup of
GL(V ) that consists of unipotent elements (i.e. all eigenvalues are 1). Then G has a fixed
point.

Proof. We are solving the linear equation g · v = v, so we can assume F = F̄ . We can also
assume V is irreducible. Finally, we can assume the image of the group algebra F [G] in
End(V ) is all of End(V ), by Burnside. It suffices to show g − 1 = 0 for all g ∈ G. Compute
tr((g − 1)g ′ ) = tr gg ′ − tr g ′ = dim V − dim V = 0.
On the other hand, matrices of the form (g − 1)g ′ span End(V ). Since tr(ab) is non-
degenerate, it follows that g − 1 = 0 for all g ∈ G. □

An important use of fixed point theorems in Lie theory is to show conjugacy of certain kinds
of subgroups.
(1) If G is an arbitrary Lie group, then all maximal compact Lie subgroups are conjugate.
(2) If K is a compact Lie group, then all maximal connected abelian subgroups (maximal
tori) are conjugate.
(3) If G is a connected linear algebraic group over k = k̄, then all connected solvable
groups (i.e. Borel subgroups) are conjugate.
The general argument goes as follows: if H, H ′ ⊂ G are two subgroups of a certain kind,
and we want to prove gH ′ g −1 ⊂ H. The subgroup H is the stabilizer of 1 in G/H. So
gH ′ g −1 ⊂ H iff H ′ fixes a point in G/H, namely g −1 H.
For example, to show (2), we need a torus T ′ ∼= (S 1 )m to have a fixed point on K/T . Clearly
we can write (S 1 )m as the closure of a single orbit, because we can pick an irrational orbit.
So this is really a question about whether an operator g ∈ T ′ acting on K/T has a fixed
point. The Lefschetz fixed point theorem says that for g ∈ Diff(M ) with M a manifold,
∑ ∑M
dim
x
(−1) = (−1)i tr g|H i (M,C) .
x∈M g i=0
41
In particular, if g ∈ Diff(M )0 , then since tr g|H i (M,C) depends only on the isotopy class of g,
it behaves the same as the identity, i.e.

∑ ∑M
dim
x
(−1) = (−1)i dim H i (M, C) = χ(M ).
x∈M g i=0

So if the Euler characteristic χ(M ) is non-zero, then g must have a fixed point.
How do we prove Lefschetz’s fixed point theorem? Consider the diagonal ∆ ⊂ M 2 . If Γ
∑the graphx of G, then it is (1 × G)∆ where G acts on the second coordinate. We have
is
x∈M g (−1) = ∆ ∩ Γ. But the Künneth formula says
∑
[∆] = αi ⊗ αi ∈ H middle (M 2 , C)
i
∑
where {αi } and {αi } are Poincaré duals. So the class [Γ] of the graph is just αi ⊗ g(αi ).
But now after applying the pairing, this sum is just the trace of the matrix corresponding
to g.
So it suffices to show χ(K/T ) is non-zero, since we know it is a compact manifold. For
example, let K = U (n) and T be the diagonal matrices inside. Then M = K/T is the space
of complete flags, since U (n) acts on orthonormal frames up to rescaling. Then M T is just
the coordinate flags, which consists of Sn , the symmetric group, acting on the standard flag.
Hence |M T | = χ(M ) = |Sn | ̸= 0. In general, let N (T ) := {g ∈ K : gT g −1 = T } be the
normalizer. Then W = N (T )/T is called the Weyl group.
Lemma 22.2. T is the connected component in N (T ), so W is actually a discrete group.

Proof. There is a map N (T ) → Aut(T ) given by g 7→ (t 7→ gtg −1 ). But Aut(T ) is a discrete

group, since these automorphisms come from its universal cover, which is a lattice. The
connected component of N (T ) is therefore mapped to the connected component of Aut(T ),
which is just the identity. Hence N (T )0 = C(T )0 . But T is maximal connected abelian, so
C(T )0 = T . □

Theorem 22.3. χ(K/T ) = |W |, which in particular is non-zero.

Proof. Consider M = K/N (T ). Then K/T → M is a covering of degree |W |. Hence it

suffices to prove χ(M ) = 1. We do this by computing the fixed points of T on M , and
then applying the Lefschetz fixed point theorem. But T fixes a point iff gT g −1 = T modulo
N (T ), so there is only one fixed point. To get the index (−1)T of this fixed point, consider
the action of T on T1 M = Lie(K)/ Lie(T ). This is just a torus acting on a vector space, so
each (rotation) action is non-trivial (i.e. all weights are non-zero). Hence we have one fixed
point with index 1, since the index of the origin under rotations is 1. Hence χ(M ) = 1. □

Remark 22.4. We really require characteristic 0 here; it turns out not all maximal tori are
conjugate in SL(n, Qp ) or SL(n, Zp ).
42
 23. More Borel subgroups

Let G be either a complex Lie group or an algebraic group. To use fixed point theory, we
assume k = k̄.
Theorem 23.1 (Borel). All Borel subgroups are conjugate.
Example 23.2. Take G = GL(n). Then every Borel subgroup B is conjugate to the subgroup
of upper diagonal matrices, by Lie’s theorem. Actually, we can deduce Lie’s theorem from the
fixed point theorem: G/B is the space of complete flags 0 ⊂ F1 ⊂ F2 ⊂ · · · ⊂ Fn = Cn . This
space is projective, because it is a closed subspace of the Grassmannian. So every solvable
subgroup will preserve a flag, and therefore is upper triangular in the corresponding basis.

Proof. The idea is to fix one Borel subgroup B0 and show that G/B0 is projective. Then
any other Borel subgroup B will have a fixed point on G/B0 = M , so that gBg −1 ⊂ B0 .
Choose a B0 of maximal dimension, i.e. dim B0 = maxB dim B. Choose an embedding
G ⊂ GL(n) (to be made more precise later). Consider the action of G on Fl(n), the space
of flags. A Borel subgroup B acting on Fl(n) will have some fixed point F0 , where F0 is a
flag. So consider the orbit G · F0 ⊂ Fl(n). It is closed, because it is of minimal dimension:
dim G · F0 = dim G − dim StabG F0 , and StabG F0 is solvable, and we chose B0 maximal.
Hence M = G · F0 is projective, and M B ̸= ∅ for any connected solvable B. So there exists
g such that gBg −1 ⊂ (Stab F0 )0 . But (Stab F0 )0 is solvable and connected and contains
B0 . By maximality of B0 , we have (Stab F0 )0 = B0 . We can actually make Stab F0 = B0
by using Chevalley’s theorem to find an embedding G ⊂ GL(n) and a vector e1 such that
B0 = StabG (Ce1 ). □

Remark 23.3. We say P ⊂ G is parabolic if G/P is projective. These G/P are called
homogeneous projective varieties. Note that G/P is projective iff P contains a Borel
subgroup. It is a fact that there are only finitely many such P in G up to conjugacy.
Proposition 23.4. The connected component of the normalizer N (B) = {g ∈ G : gBg −1 ⊂
B} of a Borel subgroup is equal to B itself.

Proof. We know N (B)0 ⊂ B, because otherwise adding g ∈ N (B)0 \ B into B creates a

bigger connected solvable subgroup. Now we show N (B)/B is trivial. Every Borel subgroup
fits into an exact sequence 1 → U → B → T → 1 where U is unipotent and T is semisimple.
(Think of T as the diagonal and U as the strictly upper triangular entries.) Consider the
action of T on G/N (B), which is the space of all Borel subgroups. Then [B] ∈ G/N (B) is
an isolated fixed point of T . We know T[B] G/N (B) = g/b, where b is the Lie algebra of B,
and 0 is the unique fixed point. Hence the variety G/N (B) is a vector space plus something
(the “boundary”) of codimension one. Then π1 (G/N (B)) = 0. Hence there is a fibration

N (B)/B → G/B → G/N (B)

which is a priori a finite cover, i.e. N (B)/B is finite. But G/N (B) is 1-connected, so G/B
is connected, and therefore N (B)/B is trivial. □
43
Theorem 23.5 (B-B decomposition). Let M be projective and smooth ∪ inside P(V ). Let
T ⊂ GL(V ) be a torus acting on M . Then the fixed point locus M T = i Fi is also smooth,
where the Fi are connected components.
Definition 23.6. In the situation of the theorem, given a generic 1-parameter subgroup
σ : GL(1) → T , define the attracting manifold
Attr(Fi ) := {m ∈ M : lim σ(z)m ∈ Fi }.
z→0
×
A map k → M extends uniquely to P → M since M is projective, and limz→0 σ(z)m is
1

just this additional point.

Example 23.7. Let M = GL(n)/B, and T the diagonal. Then
M T = {g ∈ G : gT g −1 ⊂ B}/B = {g ∈ G : gT g −1 ⊂ T }/T = NG (T )/T = W,
the Weyl group, because the normalizer of T inside B is NB (T ) = T . Take w = 1 ∈ W .
The torus T acts on the equivalence class of diag(t1 , . . . , tn ) by ti /tj for i > j on the (i, j)-th
entry. If we take a 1-parameter subgroup such that ti /tj → 0, the attracting manifold Attr(1)
is precisely the group N− , the lower triangular
⊔ B along with 1’s along the diagonal. (We
know via Gaussian elimination that GL(n) = w N− wB.)
⊔
Theorem 23.8. M = i Attr(Fi ), and Attr(Fi ) → Fi is an affine linear bundle.
Remark 23.9. This gives a decomposition of an algebraic variety into pieces, each of which
is a vector bundle over a simpler algebraic variety. This equality is actually structure-
preserving. For example, the Hodge structure on M is equivalent to the Hodge structures on
the Attr(Fi ), shifted appropriately.
Theorem
∏ 23.10 (Borel). Let G be an algebraic group over k = k̄. We can ask for tori
∼
T = GL(1, k). Then all maximal tori are conjugate.

Proof sketch. Since T is commutative, in particular solvable, and connected, there exists B
such that T ⊂ B. All B are conjugate, so it is enough to show that all T ⊂ B are B-
conjugate. In fact, they are conjugate under U ⊂ B, the unipotent radical, by induction on
dim U . □

24. Levi–Malcev decomposition

Theorem⊕ 24.1 (Levi–Malcev). Any Lie algebra g decomposes as ⊕ a semidirect sum g =

r ⊕ gss i gi where r is solvable, called the radical, and gss := i gi is a sum of simple
non-abelians. (We have [gss , r] ⊂ r.)
Remark 24.2. Solvable Lie algebras have non-trivial moduli, but simple Lie algebras are
rigid, i.e. they have no non-trivial deformations.
Remark 24.3. We will construct r as the maximal solvable ideal in g. We must show it is
uniquely determined. This is because if r1 , r2 ⊂ g are solvable, then r1 + r2 are also solvable.

Proof of Levi–Malcev. The radical r of g fits into a short exact sequence 0 → r → g → gss →
0, where gss is semisimple. It remains to show gss is a sum of simples. This we do using
Cartan’s theorem below. □
44
Definition 24.4. A Lie algebra g is semisimple if its radical is zero.
Definition 24.5. If g is a Lie algebra and ρ : g → gl(V ) is a linear representation, define
(a, b)ρ := tr(ρ(a)ρ(b)).
This is invariant in the sense that
ad g ⊂ so(g, (·, ·)ρ ), i.e.(a, [b, c])ρ = ([a, b], c)ρ .
The Killing form is (·, ·)ad .
Theorem 24.6 (Cartan). g is semisimple iff the Killing form is non-degenerate.
⊕
Corollary 24.7. g is semisimple iff g = gi where gi are simple.

Proof. Let g1 ⊂ g be a simple ideal. Then g⊥ ⊥

1 is also an ideal: if ξ ∈ g1 , then

(g1 , [b, ξ]) = ([g1 , b], ξ) = 0

since [g1 , b] ⊂ g1 . Since g1 is simple, g1 ∩ g⊥
1 is g1 or 0. The former cannot happen because
the Killing form is non-degenerate. □

Proof of Cartan’s theorem. If the Killing form is degenerate, then g⊥ ⊂ g is a non-zero ideal,
on which Killing form is identically zero. In particular, (a, [b, c])ad = 0. Hence by the
following theorem, g⊥ is solvable, so g is not semisimple.
Conversely, suppose the radical r is non-zero. Then by taking enough commutators, we get
an abelian ideal a. For any y ∈ g and any a ∈ a,
(ad(y) ad(a))2 x ⊂ ad(y) ad(a) ad(y)a ⊂ ad(y) ad(a)a = 0.
Hence tr(ad(y) ad(a)) = 0. So a ⊂ g⊥ , and the Killing form is degenerate. □
Theorem 24.8. Let g ⊂ gl(V ) be a Lie subalgebra. Then
tr([a, b]c) = 0 ∈ (∧3 gl(V )∗ )GL(V )
identically iff g is solvable.
Remark 24.9. The space (∧3 gl(V )∗ )GL(V ) is one-dimensional, because given a 3-form on
the tangent space gl(V ) of GL(V ), we can extend it to a left and right invariant element of
Ω3 GL(V ). In particular, it restricts to Ω3 U (V ). It is a general principle that H 3 (GL(V ))
is 1-dimensional, coming from H 3 of its maximal compact U (V ), and is represented by an
invariant form.
Remark 24.10. Given X ∈ gl(V ), take its Jordan decomposition X = Xs + Xn where
Xs is semisimple and Xn is nilpotent such that [Xs , Xn ] = 0. Fact: both Xs and Xn are
polynomials in X. In particular, in a linear representation, a tensor is preserved by X iff it
is preserved by Xs and Xn .
Lemma 24.11. If X ∈ g where g is the Lie algebra of an algebraic group, then Xs , Xn ∈ g.
Definition 24.12. Let galg be the intersection of all Lie algebras of algebraic groups that
contain g. It is the Lie algebra of Ḡ, the Zariski closure of G, which sits in the chain of
inclusions GL(V ) ⊃ G ⊃ G.
Proposition 24.13. [g, g] = [galg , galg ].
45
Proof. Consider {x ∈ gl(V ) : [x, g] ⊂ [g, g]}. It is the Lie algebra of the group {h : hgh−1 ∈
[g, g]}. Hence galg is contained in it, i.e. [g, galg ] ⊂ [g, g]. □
Proposition 24.14. Suppose A ⊂ B ⊂ End(V ), and
g = {x : [x, B] ⊂ A} = Lie{g : gBg −1 ≡ B mod A}
Then for any x ∈ g⊥ , with respect to (x, y) := tr(x, y), we have xs = 0.

Proof. Firstly, xs ∈ g, since g is algebraic. If e1 , . . . , en is an eigenbasis with eigenvalues λi ,

then Eij are eigenvectors of ad(xs ) with eigenvalues λi − λj . If we can find a function f on
the set {λi − λj } such that f (λi − λj ) = µi −∑µj , then the operator ad(y) =⊥f (ad(xs )) where
y = diag(µ1 , . . . , µn ). But y ∈ g and hence µi λi = tr yx = 0 (since x ∈ g ). Consider the
Q-vector space V spanned by λi in C. We must show dimQ V = 0. Suppose not. Then there
µ ∑ ∑
exists a non-zero linear function V − → Q. Now apply µ to i µi λi , to get i µ2i , which is 0
iff every µi = 0. □

Proof of theorem. If g is solvable, it consists of upper triangular matrices, and clearly tr([a, b]c) =
0 when a, b, c are upper triangular. Conversely, consider the short exact sequence 0 →
Z(g) → g → ad g → 0. Then g is solvable iff ad g is solvable. Let
g̃ := {w : [w, g] ⊂ [g, g]}.
If w ∈ g̃, then tr w[y, z] = tr[w, y]z. But [w, y] = [x, y] for some x, y ∈ g, by the definition of
g̃. Hence tr[w, y]z = tr[x, y]z = 0. So [y, z] ∈ (g̃)⊥ , i.e. [y, z]s = 0, and [g, g] is nilpotent. □

25. Roots and weights

Example 25.1. Consider g = sl(n), which has a subalgebra of diagonal matrices

∑
h := {diag(a1 , . . . , an ) : ai = 0}
i

called the Cartan subalgebra. We can ask how g decomposes under ad h. It will decompose
as ⊕
g=h⊕ CEij
i̸=j

where Eij is an eigenvalue of weight αij := ai − aj ∈ h∗ , i.e. [h, Eij ] = αij (h)Eij .
Definition 25.2. The roots of g are the elements α ∈ h∗ which are non-zero weights of
ad h. So the above decomposition can be written as
⊕
g=h⊕ gα
α

where gα is the eigenspace corresponding to α.

⊕
Proposition 25.3. Let V be a representation of g, so that V = α Vα . Then gα Vβ ⊂ Vα+β .

Proof. Let e ∈ gα and v ∈ Vβ . Then compute

hev = [h, e]v + ehv = α(h)ev + β(h)ev. □
46
Corollary 25.4. For every root α, there is also a root −α.

Proof. The proposition shows [gα , gβ ] ⊂ gα+β . Then ad(gα ) ad(gβ )gγ ⊂ gγ+α+β . Since there
are only finitely many roots, ad(gα ) ad(gβ ) is nilpotent unless α = −β. Hence the trace of
this operator is 0, i.e. gα ⊥ gβ with respect to the Killing form unless α = −β. So there is
a non-degenerate pairing between gα and g−α given by the Killing form. □
Remark 25.5. We have a map SL2 → Ad(G) given by
( ) ( )
0 1 0 0
7→ Eij , 7→ Eji .
0 0 1 0
( )
0 1
In SL2 , let the matrix map to sα ∈ Ad(G). This will be a permutation of the roots,
1 0
but at the same time also a linear transformation β 7→ β − ℓα (β)α where ℓα is some linear
function.
Definition 25.6. A root system is a finite collection of non-zero vectors spanning a vector
space such that for every α there exists a linear transformation of the form β 7→ β − ℓα (β)α,
where ℓα (β) ∈ Z, that preserves the root system and sends α to −α, i.e. ℓα (α) = 2.
Remark 25.7. These conditions are stronger than they seem. Since a root system is finite,
the permutation group on the vectors in the root system is finite. In particular, the group W
generated by the linear transformations sα is finite, and therefore compact. So it preserves a
positive definite inner product (·, ·). Under this inner product,
(α, β)
sα (β) = β − 2 α,
(α, α)
a reflection. Such groups generated by reflections can be classified: these are the crystallo-
graphic groups.
Definition 25.8. Let g be a Lie algebra. A subalgebra h ⊂ g is a Cartan subalgebra if h
is nilpotent and the normalizer of h is h itself.
Definition 25.9. Let V be a representation of h, e.g. the adjoint action on g. By Lie’s
theorem, h ∈ h goes to a upper triangular matrix with αi (h) on the diagonal. Call the
αi ∈ (h/[h, h])∗ ⊂ h∗ the weights of V . Write Vα for the generalized eigenspace of a weight
α, i.e.
Vα := {v ∈ V : (h − α(h))i v = 0 for some i}.
Clearly Vα is invariant under h.
Remark
⊕ 25.10. Applying this definition to the adjoint representation, we get g = g0 ⊕
α̸=0 α We will show that g0 = h. The proposition we showed earlier gives gα Vβ ⊂ Vα+β .
g .
Definition 25.11. The rank of g is the minimal number of zero eigenvalues of ad x for
x ∈ g. Equivalently, it is the maximum size of a minor in ad x (over the field of rational
functions in x) that is not identically zero. We say x ∈ g is regular if ad x has this generic
rank.
Remark 25.12. The set of regular elements x ∈ g is a Zariski open set, since it is given by
the condition that at least one of the minors is non-zero.
47
Proposition 25.13. Let x be regular and consider
⊕
g = gx0 ⊕ gxα .
α̸=0

Then dim gx0 = rankg and h := gx0 is a Cartan subalgebra.

Proof. We know [h, h] ⊂ h, from the result that [gα , gβ ] ⊂ gα+β . So we can restrict ad h to
h. Then ad(y)|h is nilpotent for every y ∈ h, because otherwise ad(y) will have fewer zero
eigenvalues than x, since
ad(y) = ad(y)|h ⊕ ad(y)|g/h .
Hence h is nilpotent, by definition. Now suppose some element z is in the normalizer of h,
i.e. [x, z] ∈ h. By the nilpotence of h, we know ad(x)N z = 0 for N ≫ 0. Hence z ∈ h, by
the definition of h. □
⊕
Remark 25.14. Let h be an arbitrary Cartan subalgebra. Then g = h ⊕ α̸=0 gα and define
hreg := {h : α(h) ̸= 0 ∀α}
so that for all x ∈ hreg , gx0 = h.
Proposition 25.15. Let g be a simple Lie algebra.
(1) The Cartan subalgebra h is commutative and consists of ad-semisimple elements.
(2) The Killing form restricted to h is non-degenerate.

Proof. We know h nilpotent implies ([x, y], z) = 0 for any x, y, z ∈ h. Since z is arbitrary,
and the Killing form is non-degenerate, [x, y] = 0 for all x, y ∈ h.
⊕
In the decomposition g = h ⊕ α∈h∗ gα , we know gα ⊥ gβ (with respect to the Killing form)
unless α + β = 0. So gα and g−α are dual, leaving h in the direct sum. Hence the Killing
form is also non-degenerate on h. □

26. Root systems

Definition 26.1. A root system ∆ ⊂ Rn \ {0} is a finite subset of non-zero vectors such
that for any α ∈ ∆, the reflection
(α, β)
rα (β) := β − 2 α
(α, α)
preserves ∆ and ⟨α, β⟩ := 2(α, β)/(α, α) is an integer. We say ∆ is
(1) reducible if ∆ = ∆1 ⊕ ∆2 , and
(2) reduced if 2α ∈
/ ∆ for any α ∈ ∆.
Example 26.2 (Root systems for n = 1). Suppose α ∈ ∆. Then −α ∈ ∆ as well. Take
another vector β ∈ ∆. Then 2(α, β)/(α, α) must be an integer, i.e. 2β/α ∈ Z. So there is
only one reduced root system, called A1 , given by {±α}, and one non-reduced root system
{±α, ±2α}. It turns out Lie algebras always have reduced root systems, so {±α} corresponds
to sl(2).
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Example 26.3 (Root systems for n = 2). Suppose there is a vector β forming an angle θ
with α, and this is the smallest θ formed by any vector with α. Then
(α, β)2
⟨α, β⟩⟨β, α⟩ = 4 = 4 cos2 θ
(α, α)(β, β)
must be an integer. So there are five possibilities.
(1) (θ = π/2) This is exactly A1 ⊕ A1 , and corresponds to the root system D2 .
(2) (θ = π/3) Here ⟨α, β⟩ = ⟨β, α⟩ = 1, so α, β are equal length with angle π/3 between
them. By applying reflections, we get the root system A2 , corresponding to sl(3).
(3) (θ = π/4) Here ⟨α, β⟩⟨β, α⟩ = 2, so there is a choice of factorization.
√
(a) If we pick ⟨α, β⟩ = 1 and ⟨β, α⟩ = 2, then β is 2 longer than α. By applying
reflections, we get the root system B2 , corresponding to so(2n + 1).
(b) Alternatively, if we pick ⟨α, β⟩ = 2, then we get the root system C2 , corresponding
to sp(2n).
(4) (4 cos2 θ = 3) This gives the exceptional root system G2 .

Take e ∈ gα . Via the Killing form, g−α = g∗α . We know [e, f ] ∈ h. To know which element
in h, it is enough to pair it using the Killing form:
α(h)
([e, f ], h) = (e, [f, h]) = (e, α(h)f ) = 2 .
(α, α)
If we identify h ∼ = h∗ via the Killing form, we can think of α as an element in h, so that
([e, f ], h) = 2(α, h)/(α, α).
Definition 26.4. Write hα := 2α/(α, α), also sometimes denoted α∨ .
Proposition 26.5. The elements e, f, hα form a copy of sl(2), and up to scalars, hα is the
same vector regardless of the choice of e and f .

Proof. We just computed [e, f ] = hα , and we know that

(α, α)
[hα , e] = α(hα )e = 2 e = 2e, [hα , f ] = −2f. □
(α, α)
Corollary 26.6. The dimension of gα is 1, and if α ∈ ∆, then nα ∈
/ ∆ for n ̸= ±1.
⊕
Proof. Consider the action of sl(2)α := span{e, f, hα } on Chα ⊕ n∈Z̸=0 gnα . Then e is a
raising operator and f is a lowering operator, i.e. [e, gnα ] ⊂ g(n+1)α , and similarly for f . But
this whole thing is a finite-dimensional sl(2)-module with a 1-dimensional space of weight
0 (with respect to hα ) and with even weights. By the representation theory of sl(2), this
representation is irreducible. But it contains sl(2), and is therefore equal to sl(2). □
⊕
Similarly, take β ∈
/ Zα, and look at n∈Z gβ+nα . Then e raises, f lowers, and hα acts by the
scalar ⟨β, α⟩ on gβ . By the corollary, each gβ+mα has dimension either 0 or 1.
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Corollary 26.7. This representation is irreducible, ⟨β, α⟩ ∈ Z, and for any β ∈ ∆, the
vector rα (β) := β − ⟨β, α⟩α is also in ∆.

Proof. Any finite-dimensional sl2 representation has weight spaces symmetric across the
origin. But each weight space here has dimension either 0 or 1, so this representation cannot
split. Also, ⟨β, α⟩ is the scalar that hα acts by on gβ , and we know for finite-dimensional
representations that this is an integer. Finally, rα (β) is precisely the weight corresponding
to reflecting β across the origin. □

We have shown that the set of weights of ad(h) is a root system. It remains to show that it
is reduced.

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