III Symmetries, Fields and Particles (Full)

Part III — Symmetries, Fields and Particles

Based on lectures by N. Dorey

Notes taken by Dexter Chua

Michaelmas 2016

These notes are not endorsed by the lecturers, and I have modified them (often

significantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

This course introduces the theory of Lie groups and Lie algebras and their applications

to high energy physics. The course begins with a brief overview of the role of symmetry

in physics. After reviewing basic notions of group theory we define a Lie group as a

manifold with a compatible group structure. We give the abstract definition of a Lie

algebra and show that every Lie group has an associated Lie algebra corresponding to

the tangent space at the identity element. Examples arising from groups of orthogonal

and unitary matrices are discussed. The case of

(2), the group of rotations in three

dimensions is studied in detail. We then study the representations of Lie groups and

Lie algebras. We discuss reducibility and classify the finite dimensional, irreducible

representations of

(2) and introduce the tensor product of representations. The

next part of the course develops the theory of complex simple Lie algebras. We

define the Killing form on a Lie algebra. We introduce the Cartan-Weyl basis and

discuss the properties of roots and weights of a Lie algebra. We cover the Cartan

classification of simple Lie algebras in detail. We describe the finite dimensional,

irreducible representations of simple Lie algebras, illustrating the general theory for the

Lie algebra of

(3). The last part of the course discusses some physical applications.

After a general discussion of symmetry in quantum mechanical systems, we review the

approximate

(3) global symmetry of the strong interactions and its consequences

for the observed spectrum of hadrons. We introduce gauge symmetry and construct a

gauge-invariant Lagrangian for Yang-Mills theory coupled to matter. The course ends

with a brief introduction to the Standard Model of particle physics.

Pre-requisites

Basic finite group theory, including subgroups and orbits. Special relativity and

quantum theory, including orbital angular momentum theory and Pauli spin matrices.

Basic ideas about manifolds, including coordinates, dimension, tangent spaces.

Contents

1 Introduction

2 Lie groups

2.1 Definitions

2.2 Matrix Lie groups

2.3 Properties of Lie groups

3 Lie algebras

3.1 Lie algebras

3.2 Differentiation

3.3 Lie algebras from Lie groups

3.4 The exponential map

4 Representations of Lie algebras

4.1 Representations of Lie groups and algebras

4.2 Complexification and correspondence of representations

4.3 Representations of su(2)

4.4 New representations from old

4.5 Decomposition of tensor product of su(2) representations

5 Cartan classification

5.1 The Killing form

5.2 The Cartan basis

5.3 Things are real

5.4 A real subalgebra

5.5 Simple roots

5.6 The classification

5.7 Reconstruction

6 Representation of Lie algebras

6.1 Weights

6.2 Root and weight lattices

6.3 Classification of representations

6.4 Decomposition of tensor products

7 Gauge theories

7.1 Electromagnetism and U(1) gauge symmetry

7.2 General case

8 Lie groups in nature

8.1 Spacetime symmetry

8.2 Possible extensions

8.3 Internal symmetries and the eightfold way

1 Introduction

In this course, we are, unsurprisingly, going to talk about symmetries. Unlike

what the course name suggests, there will be relatively little discussion of fields

or particles.

So what is a symmetry? There are many possible definitions, but we are

going to pick one that is relevant to physics.

Definition

(Symmetry)

A symmetry of a physical system is a transformation

of the dynamical variables which leaves the physical laws invariant.

Symmetries are very important. As Noether’s theorem tells us, every sym-

metry gives rise to a conserved current. But there is more to symmetry than

this. It seems that the whole physical universe is governed by symmetries. We

believe that the forces in the universe are given by gauge fields, and a gauge

field is uniquely specified by the gauge symmetry group it works with.

In general, the collection of all symmetries will form a group.

Definition

(Group)

A group is a set

of elements with a multiplication rule,

obeying the axioms

(i) For all g

, g

∈ G, we have g

∈ G. (closure)

(ii)

There is a (necessarily unique) element

e ∈ G

such that for all

g ∈ G

, we

have eg = ge = g. (identity)

(iii)

For every

g ∈ G

, there exists some (necessarily unique)

−1

∈ G

such that

−1

= g

−1

g = e. (inverse)

(iv) For every g

, g

∈ G, we have g

) = (g

. (associativity)

Physically, these mean

(i) The composition of two symmetries is also a symmetry.

(ii) “Doing nothing” is a symmetry.

(iii) A symmetry can be “undone”.

(iv) Composing functions is always associative.

Note that the set of elements G may be finite or infinite.

Definition

(Commutative/abelian group)

A group is abelian or commutative

if g

= g

for all g

, g

∈ G. A group is non-abelian if it is not abelian.

In this course, we are going to focus on smooth symmetries. These are

symmetries that “vary smoothly” with some parameters. One familiar example

is rotation, which can be described by the rotation axis and the angle of rotation.

These are the symmetries that can go into Noether’s theorem. These smooth

symmetries form a special kind of groups known as Lie groups.

One nice thing about these smooth symmetries is that they can be studied by

looking at the “infinitesimal” symmetries. They form a vector space, known as

a Lie algebra. This is a much simpler mathematical structure, and by reducing

the study of a Lie group to its Lie algebra, we make our lives much easier. In

particular, one thing we can do is to classify all (simple) Lie algebras. It turns

out this isn’t too hard, as the notion of (simple) Lie algebra is very restrictive.

After understanding Lie algebras very well, we will move on to study gauge

theories. These are theories obtained when we require that our theories obey

some sort of symmetry condition, and then magically all the interactions come

up automatically.

2 Lie groups

2.1 Definitions

A Lie group is a group and a manifold, where the group operations define smooth

maps. We already know what a group is, so let’s go into a bit more details about

what a manifold is.

The canonical example of a manifold one might want to keep in mind is the

sphere

. Having lived on Earth, we all know that near any point on Earth,

looks like

. But we also know that

is not

. We cannot specify a point

on the surface of the Earth uniquely by two numbers. Longitude/latitude might

be a good start, but adding 2

to the longitude would give you the same point,

and things also break down at the poles.

Of course, this will not stop us from producing a map of Cambridge. One

can reasonably come up with consistent coordinates just for Cambridge itself.

So what is true about

is that near each point, we can come up with some

coordinate system, but we cannot do so for the whole space itself.

Definition

(Smooth map)

We say a map

→ R

is smooth if all partial

derivatives of all orders exist.

Definition

(Manifold)

A manifold (of dimension

) is a set

together with

the following data:

(i) A collection U

of subsets of M whose union is M;

(ii)

A collection of bijections

→ V

, where

is an open subset of

These are known as charts.

The charts have to satisfy the following compatibility condition:

–

For all

α, β

, we have

(

∩U

) is open in

, and the transition function

◦ ϕ

−1

: ϕ

∩ U

) → ϕ

∩ U

)

is smooth.

We write (M, ϕ

) for the manifold if we need to specify the charts explicitly.

−1

Definition

(Smooth map)

Let

be a manifold. Then a map

M → R

is smooth if it is smooth in each coordinate chart. Explicitly, for each chart

, ϕ

), the composition f ◦ ϕ

−1

: ϕ

) → R is smooth (in the usual sense)

ϕ(p)

f ◦ ϕ

−1

More generally, if

M, N

are manifolds, we say

M → N

is smooth if for

any chart (

U, ϕ

) of

and any chart (

V, ξ

) of

, the composition

ξ ◦f ◦ ϕ

−1

ϕ(U) → ξ(V ) is smooth.

ξ ◦f ◦ ϕ

−1

Finally, we note that if

M, N

are manifolds of dimensions

and

respectively,

then

M × N

is a manifold of dimension

, where charts (

U, ϕ

U →

)

(

V, ξ

V → R

) of

and

respectively give us a chart

ϕ × ξ

U × V →

× R

= R

m+n

of M × N .

All those definitions were made so that we can define what a Lie group is:

Definition

(Lie group)

A Lie group is a group

whose underlying set is given

a manifold structure, and such that the multiplication map

G ×G → G

and

inverse map

G → G

are smooth maps. We sometimes write

(

) for the

underlying manifold of G.

Example. The unit 2-sphere

= {(x, y, z) ∈ R

: x

+ y

+ z

= 1}

is a manifold. Indeed, we can construct a coordinate patch near

= (0

1).

Near this point, we have

1 − x

− y

. This works, since near the north pole,

the

-coordinate is always positive. In this case,

and

are good coordinates

near the north pole.

However, it is a fact that the 2-sphere S

has no Lie group structure.

In general, most of our manifolds will be given by subsets of Euclidean space

specified by certain equations, but note that not all subsets given by equations

are manifolds! It is possible that they have some singularities.

Definition

(Dimension of Lie group)

The dimension of a Lie group

is the

dimension of the underlying manifold.

Definition

(Subgroup)

A subgroup

is a subset of

that is also a group

under the same operations. We write H ≤ G if H is a subgroup of G.

The really interesting thing is when the subgroup is also a manifold!

Definition

(Lie subgroup)

A subgroup is a Lie subgroup if it is also a manifold

(under the induced smooth structure).

We now look at some examples.

Example.

Let

= (

+) be the

-dimensional Euclidean space with addition

as the group operation. The inverse of a vector

−x

, and the identity is

This is obviously locally homeomorphic to

, since it is

, and addition and

negation are obviously smooth.

This is a rather boring example, since

is a rather trivial manifold, and

the operation is commutative. The largest source of examples we have will be

matrix Lie groups.

2.2 Matrix Lie groups

We write

Mat

(

) for the set of

n×n

matrices with entries in a field

(usually

). Matrix multiplication is certainly associative, and has an identity, namely

the identity I. However, it doesn’t always have inverses — not all matrices are

invertible! So this is not a group (instead, we call it a monoid). Thus, we are

led to consider the general linear group:

Definition (General linear group). The general linear group is

GL(n, F) = {M ∈ Mat

(F) : det M 6= 0}.

This is closed under multiplication since the determinant is multiplicative,

and matrices with non-zero determinant are invertible.

Definition (Special linear group). The special linear group is

SL(n, F) = {M ∈ Mat

(F) : det M = 1} ≤ GL(n, F).

While these are obviously groups, less obviously, these are in fact Lie groups!

In the remainder of the section, we are going to casually claim that all our favorite

matrix groups are Lie groups. Proving these take some work and machinery,

and we will not bother ourselves with that too much. However, we can do it

explicitly for certain special cases:

Example. Explicitly, we can write

SL(2, R) =



a b

c d



: a, b, c, d ∈ R, ad − bc = 1



The identity is the matrix with a = d = 1 and b = c = 0. For a 6= 0, we have

d =

1 + bc

This gives us a coordinate patch for all points where

a 6

= 0 in terms of

b, c, a

which, in particular, contains the identity

. By considering the case where

b 6

= 0, we obtain a separate coordinate chart, and these together cover all of

SL(2, R), as a matrix in SL(2, R) cannot have a = b = 0.

Thus, we see that SL(2, R) has dimension 3.

In general, by a similar counting argument, we have

dim(SL(n, R)) = n

− 1 dim(SL(n, C)) = 2n

− 2

dim(GL(n, R)) = n

, dim(GL(n, C)) = 2n

Most of the time, we are interested in matrix Lie groups, which will be subgroups

of GL(n, R).

Subgroups of GL(n, R)

Lemma. The general linear group:

GL(n, R) = {M ∈ Mat

(R) : det M 6= 0}

and orthogonal group:

O(n) = {M ∈ GL(n, R) : M

M = I}

are Lie groups.

Note that we write O(

) instead of O(

n, R

) since orthogonal matrices make

sense only when talking about real matrices.

The orthogonal matrices are those that preserve the lengths of vectors. Indeed,

for v ∈ R

, we have

|Mv|

= v

Mv = v

v = |v|

We notice something interesting. If M ∈ O(n), we have

1 = det(I) = det(M

M) = det(M)

det

(

) =

1. Now

det

is a continuous function, and it is easy to see that

det

takes both

1. So O(

) has (at least) two connected components. Only one of

these pieces contains the identity, namely the piece

det M

= 1. We might expect

this to be a group on its own right, and indeed it is, because

det

is multiplicative.

Lemma. The special orthogonal group SO(n):

SO(n) = {M ∈ O(n) : det M = 1}

is a Lie group.

Given a frame

, ··· , v

}

(i.e. an ordered basis), any orthogonal

matrix

M ∈

) acts on it to give another frame

∈ R

7→ v

7→ M v

∈ R

Definition

(Volume element)

Given a frame

, ··· , v

}

, the volume

element is

Ω = ε

...i

···v

By direct computation, we see that an orthogonal matrix preserves the sign

of the volume element iff its determinant is +1, i.e. M ∈ SO(n).

We now want to find a more explicit description of these Lie groups, at

least in low dimensions. Often, it is helpful to classify these matrices by their

eigenvalues:

Definition

(Eigenvalue)

A complex number

is an eigenvalue of

M ∈ M

(

)

if there is some (possibly complex) vector v

6= 0 such that

= λv

Theorem.

Let

be a real matrix. Then

is an eigenvalue iff

∗

is an eigenvalue.

Moreover, if M is orthogonal, then |λ|

= 1.

Proof.

Suppose

λv

. Then applying the complex conjugate gives

∗

= λ

∗

Now suppose

is orthogonal. Then

λv

for some non-zero

. We

take the norm to obtain

|Mv

|λ||v

. Using the fact that

|Mv

, we

have |λ| = 1. So done.

Example.

Let

M ∈ SO

(2). Since

det M

= 1, the eigenvalues must be of the

form λ = e

iθ

, e

−iθ

. In this case, we have

M = M (θ) =



cos θ −sin θ

sin θ cos θ



where θ is the rotation angle in S

. Here we have

M(θ

)M(θ

) = M(θ

)M(θ

) = M(θ

+ θ

So we have M(SO(2)) = S

Example.

Consider

(3). Suppose

M ∈ SO

(3). Since

det M

= +1, and

the eigenvalues have to come in complex conjugate pairs, we know one of them

must be 1. Then the other two must be of the form e

iθ

, e

−iθ

, where θ ∈ S

We pick a normalized eigenvector

for

= 1. Then

, and

n · n

= 1.

This is known as the axis of rotation. Similarly,

is the angle of rotation. We

write M (n, θ) for this matrix, and it turns out this is

M(n, θ)

= cos θδ

+ (1 − cos θ)n

− sin θε

ijk

Note that this does not uniquely specify a matrix. We have

M(n, 2π − θ) = M(−n, θ).

Thus, to uniquely specify a matrix, we need to restrict the range of

to 0

≤ θ ≤ π

with the further identification that

(n, π) ∼ (−n, π).

Also note that M(n, 0) = I for any n.

Given such a matrix, we can produce a vector

θn

. Then

lies in the

region

= {w ∈ R

: kwk ≤ π} ⊆ R

This has a boundary

∂B

= {w ∈ R

: kwk = π}

∼

Now we identify antipodal points on

∂B

. Then each vector in the resulting

space corresponds to exactly one element of SO(3).

Subgroups of GL(n, C)

We can similarly consider subgroups of GL(n, C). Common examples include

Definition (Unitary group). The unitary group is defined by

U(n) = {U ∈ GL(n, C) : U

†

U = I}.

These are important in physics, because unitary matrices are exactly those

that preserve the norms of vectors, namely kvk = kUvk for all v.

Again, if

†

= 1, then

|det

(

)

= 1. So

det U

iδ

for some

δ ∈ R

Unlike the real case, the determinant can now take a continuous range of values,

and this no longer disconnects the group. In fact, U(n) is indeed connected.

Definition (Special unitary group). The special unitary group is defined by

SU(n) = {U ∈ U(n) : det U = 1}.

It is an easy exercise to show that

dim[U(n)] = 2n

− n

= n

For

(

), the determinant condition imposes an additional single constraint,

so we have

dim[SU(n)] = n

− 1.

Example. Consider the group G = U(1). This is given by

U(1) = {z ∈ C : |z| = 1}.

Therefore we have

M[U(1)] = S

However, we also know another Lie group with underlying manifold

, namely

SO(2). So are they “the same”?

2.3 Properties of Lie groups

The first thing we want to consider is when two Lie groups are “the same”. We

take the obvious definition of isomorphism.

Definition

(Homomorphism of Lie groups)

Let

G, H

be Lie groups. A map

J : G → H is a homomorphism if it is smooth and for all g

, g

∈ G, we have

J(g

) = J(g

)J(g

(the second condition says it is a homomorphism of groups)

Definition

(Isomorphic Lie groups)

An isomorphism of Lie groups is a bijective

homomorphism whose inverse is also a homomorphism. Two Lie groups are

isomorphic if there is an isomorphism between them.

Example. We define the map J : U(1) → SO(2) by

J(e

iθ

) 7→



cos θ −sin θ

sin θ cos θ



∈ SO(2).

This is easily seen to be a homomorphism, and we can construct an inverse

similarly.

Exercise. Show that M(SU(2))

∼

We now look at some words that describe manifolds. Usually, manifolds that

satisfy these properties are considered nice.

The first notion is the idea of compactness. The actual definition is a bit

weird and takes time to get used to, but there is an equivalent characterization

if the manifold is a subset of R

Definition

(Compact)

A manifold (or topological space)

is compact if every

open cover of X has a finite subcover.

If the manifold is a subspace of some

, then it is compact iff it is closed

and bounded.

Example.

The sphere

is compact, but the hyperboloid given by

−y

−z

1 (as a subset of R

) is not.

Example.

The orthogonal groups are compact. Recall that the definition of

an orthogonal matrix requires

. Since this is given by a polynomial

equation in the entries, it is closed. It is also bounded since each entry of

has

magnitude at most 1.

Similarly, the special orthogonal groups are compact.

Example.

Sometimes we want to study more exciting spaces such as Minkowski

spaces. Let

, and consider the matrices that preserve the metric on

of signature (p, q), namely

O(p, q) = {M ∈ GL(n, R) : M

ηM = η},

where

η =



0 −I



For

p, q

both non-zero, this group is non-compact. For example, if we take

SO(1, 1), then the matrices are all of the form

M =



cosh θ sinh θ

sinh θ cosh θ



where θ ∈ R. So this space is homeomorphic to R, which is not compact.

Another common example of a non-compact group is the Lorentz group

O(3, 1).

Another important property is simply-connectedness.

Definition

(Simply connected)

A manifold

is simply connected if it is

connected (there is a path between any two points), and every loop

→ M

can be contracted to a point. Equivalently, any two paths between any two

points can be continuously deformed into each other.

Example.

The circle

is not simply connected, as the loop that goes around

the circle once cannot be continuously deformed to the loop that does nothing

(this is non-trivial to prove).

Example.

The 2-sphere

is simply connected, but the torus is not.

(3) is

also not simply connected. We can define the map by

l(θ) =

(

θn θ ∈ [0, π)

−(2π − θ)n θ ∈ [π, 2π)

This is indeed a valid path because we identify antipodal points.

The failure of simply-connectedness is measured by the fundamental group.

Definition

(Fundamental group/First homotopy group)

Let

be a manifold,

and

∈ M

be a preferred point. We define

(

) to be the equivalence classes

of loops starting and ending at

, where two loops are considered equivalent if

they can be continuously deformed into each other.

This has a group structure, with the identity given by the “loop” that stays

at x

all the time, and composition given by doing one after the other.

Example. π

(

) =

{

}

and

(

) =

Z ×Z

. We also have

(

(3)) =

We will not prove these results.

3 Lie algebras

It turns out that in general, the study of a Lie group can be greatly simplified by

studying its associated Lie algebra, which can be thought of as an infinitesimal

neighbourhood of the identity in the Lie group.

To get to that stage, we need to develop some theory of Lie algebra, and also

of differentiation.

3.1 Lie algebras

We begin with a rather formal and weird definition of a Lie algebra.

Definition

(Lie algebra)

A Lie algebra

is a vector space (over

) with

a bracket

[ ·, ·] : g × g → g

satisfying

(i) [X, Y ] = −[Y, X] for all X, Y ∈ g (antisymmetry)

(ii)

[

αX

βY, Z

] =

[

X, Z

] +

[

Y, Z

] for all

X, Y, Z ∈ g

and

α, β ∈ F

((bi)linearity)

(iii)

[

Y, Z

]] + [

[

Z, X

]] + [

[

X, Y

]] = 0 for all

X, Y, Z ∈ g

.(Jacobi identity)

Note that linearity in the second argument follows from linearity in the first

argument and antisymmetry.

Some (annoying) pure mathematicians will complain that we should state

anti-symmetry as [

X, X

] = 0 instead, which is a stronger condition if we are

working over a field of characteristic 2, but I do not care about such fields.

There isn’t much one can say to motivate the Jacobi identity. It is a property

that our naturally-occurring Lie algebras have, and turns out to be useful when

we want to prove things about Lie algebras.

Example.

Suppose we have a vector space

with an associative product (e.g.

a space of matrices with matrix multiplication). We can then turn

into a Lie

algebra by defining

[X, Y ] = XY − Y X.

We can then prove the axioms by writing out the expressions.

Definition

(Dimension of Lie algebra)

The dimension of a Lie algebra is the

dimension of the underlying vector space.

Given a finite-dimensional Lie algebra, we can pick a basis B for g.

B = {T

: a = 1, ··· , dim g}.

Then any X ∈ g can be written as

X = X

a=1

where X

∈ F and n = dim g.

By linearity, the bracket of elements X, Y ∈ g can be computed via

[X, Y ] = X

, T

In other words, the whole structure of the Lie algebra can be given by the bracket

of basis vectors. We know that [

, T

] is again an element of

. So we can write

, T

] = f

where f

∈ F are the structure constants.

Definition

(Structure constants)

Given a Lie algebra

with a basis

}

the structure constants are f

given by

, T

] = f

By the antisymmetry of the bracket, we know

Proposition.

= −f

By writing out the Jacobi identity, we obtain

Proposition.

+ f

= 0.

As before, we would like to know when two Lie algebras are the same.

Definition

(Homomorphism of Lie algebras)

A homomorphism of Lie algebras

g, h is a linear map f : g → h such that

[f(X), f(Y )] = f([X, Y ]).

Definition

(Isomorphism of Lie algebras)

An isomorphism of Lie algebras is a

homomorphism with an inverse that is also a homomorphism. Two Lie algebras

are isomorphic if there is an isomorphism between them.

Similar to how we can have a subgroup, we can also have a subalgebra

Definition

(Subalgebra)

A subalgebra of a Lie algebra

is a vector subspace

that is also a Lie algebra under the bracket.

Recall that in group theory, we have a stronger notion of a normal subgroup,

which are subgroups invariant under conjugation. There is an analogous notion

for subalgebras.

Definition

(Ideal)

An ideal of a Lie algebra

is a subalgebra

such that

[X, Y ] ∈ h for all X ∈ g and Y ∈ h.

Example. Every Lie algebra g has two trivial ideals h = {0} and h = g.

Definition (Derived algebra). The derived algebra of a Lie algebra g is

i = [g, g] = span

{[X, Y ] : X, Y ∈ g},

where F = R or C depending on the underlying field.

It is clear that this is an ideal. Note that this may or may not be trivial.

Definition (Center of Lie algebra). The center of a Lie algebra g is given by

ξ(g) = {X ∈ g : [X, Y ] = 0 for all Y ∈ g}.

This is an ideal, by the Jacobi identity.

Definition

(Abelian Lie algebra)

A Lie algebra

is abelian if [

X, Y

] = 0 for

all X, Y ∈ g. Equivalently, if ξ(g) = g.

Definition

(Simple Lie algebra)

A simple Lie algebra is a Lie algebra

that is

non-abelian and possesses no non-trivial ideals.

is simple, then since the center is always an ideal, and it is not

since

is not abelian, we must have

(

) =

{

}

. On the other hand, the derived algebra

is also an ideal, and is non-zero since it is not abelian. So we must have

(

) =

We will later see that these are the Lie algebras on which we can define a non-

degenerate invariant inner product. In fact, there is a more general class, known

as the semi-simple Lie algebras, that are exactly those for which non-degenerate

invariant inner products can exist.

These are important in physics because, as we will later see, to define the

Lagrangian of a gauge theory, we need to have a non-degenerate invariant inner

product on the Lie algebra. In other words, we need a (semi-)simple Lie algebra.

3.2 Differentiation

We are eventually going to get a Lie algebra from a Lie group. This is obtained

by looking at the tangent vectors at the identity. When we have homomorphisms

G → H

of Lie groups, they are in particular smooth, and taking the derivative

will give us a map from tangent vectors in

to tangent vectors in

, which

in turn restricts to a map of their Lie algebras. So we need to understand how

differentiation works.

Before that, we need to understand how tangent vectors work. This is

completely general and can be done for manifolds which are not necessarily Lie

groups. Let

be a smooth manifold of dimension

and

p ∈ M

a point. We

want to formulate a notion of a “tangent vector” at the point

. We know how

we can do this if the space is

— a tangent vector is just any vector in

By definition of a manifold, near a point p, the manifold looks just like R

. So

we can just pretend it is R

, and use tangent vectors in R

However, this definition of a tangent vector requires us to pick a particular

coordinate chart. It would be nice to have a more “intrinsic” notion of vectors.

Recall that in

, if we have a function

→ R

and a tangent vector

, then we can ask for the directional derivative of

along

. We have a

correspondence

v ←→

∂

∂v

This directional derivative takes in a function and returns its derivative at a

point, and is sort-of an “intrinsic” notion. Thus, instead of talking about

, we

will talk about the associated directional derivative

∂

∂v

It turns out the characterizing property of this directional derivative is the

product rule:

∂

∂v

(fg) = f (p)

∂

∂v

g + g(p)

∂

∂v

So a “directional derivative” is a linear map from the space of smooth functions

M → R to R that satisfies the Leibnitz rule.

Definition

(Tangent vector)

Let

be a manifold and write

∞

(

) for the

vector space of smooth functions on

. For

p ∈ M

, a tangent vector is a linear

map v : C

∞

(M) → R such that for any f, g ∈ C

∞

(M), we have

v(fg) = f(p)v(g) + v(f)g(p).

It is clear that this forms a vector space, and we write

for the vector space

of tangent vectors at p.

Now of course one would be worried that this definition is too inclusive, in

that we might have included things that are not genuinely directional derivatives.

Fortunately, this is not the case, as the following proposition tells us.

In the case where

is a submanifold of

, we can identify the tangent

space with an actual linear subspace of

. This is easily visualized when

is a

surface in

, where the tangent vectors consists of the vectors in

“parallel to”

the surface at the point, and in general, a “direction” in

is also a “direction”

, and tangent vectors of

can be easily identified with elements of

the usual way.

This will be useful when we study matrix Lie groups, because this means the

tangent space will consist of matrices again.

Proposition.

Let

be a manifold with local coordinates

}

i=1,··· ,D

for some

region U ⊆ M containing p. Then T

M has basis



∂

∂x



j=1,··· ,D

In particular, dim T

M = dim M .

This result on the dimension is extremely useful. Usually, we can manage to

find a bunch of things that we know lie in the tangent space, and to show that

we have found all of them, we simply count the dimensions.

One way we can obtain tangent vectors is by differentiating a curve.

Definition

(Smooth curve)

A smooth curve is a smooth map

R → M

More generally, a curve is a C

function R → M.

Since we only want the first derivative, being C

is good enough.

There are two ways we can try to define the derivative of the curve at time

= 0

∈ R

. Using the definition of a tangent vector, to specify

˙γ

(0) is to tell how

we can differentiate a function

M → R

(0) in the direction of

˙γ

(0).

This is easy. We define

˙γ(0)(f) =

f(γ(t)) ∈ R.

If this seems too abstract, we can also do it in local coordinates.

We introduce some coordinates

}

near

p ∈ M

. We then refer to

coordinates (at least near p), by

γ : t ∈ R 7→ {x

(t) ∈ R : i = 1, ··· , D}.

By the smoothness condition, we know

(

) is differentiable, with

(0) = 0.

Then the tangent vector of the curve γ at p is

= ˙x

(0)

∂

∂x

∈ T

(M), ˙x

(t) =

It follows from the chain rule that this exactly the same thing as what we

described before.

More generally, we can define the derivative of a map between manifolds.

Definition

(Derivative)

Let

M → N

be a map between manifolds. The

derivative of f at p ∈ M is the linear map

: T

M → T

f(p)

given by the formula

(Df

)(v)(g) = v(g ◦ f )

for v ∈ T

M and g ∈ C

∞

(N).

This will be useful later when we want to get a map of Lie algebras from a

map of Lie groups.

3.3 Lie algebras from Lie groups

We now try to get a Lie algebra from a Lie group G, by considering T

(G).

Theorem.

The tangent space of a Lie group

at the identity naturally admits

a Lie bracket

[ ·, ·] : T

G × T

G → T

such that

L(G) = (T

G, [ ·, ·])

is a Lie algebra.

Definition

(Lie algebra of a Lie group)

Let

be a Lie group. The Lie algebra

, written

(

) or

, is the tangent space

under the natural Lie bracket.

The general convention is that if the name of a Lie group is in upper case

letters, then the corresponding Lie algebra is the same name with lower case

letters in fraktur font. For example, the Lie group of SO(n) is so(n).

Proof.

We will only prove it for the case of a matrix Lie group

G ⊆ Mat

(

Then

can be naturally identified as a subspace of

Mat

(

). There is then

an obvious candidate for the Lie bracket — the actual commutator:

[X, Y ] = XY − Y X.

The basic axioms of a Lie algebra can be easily (but painfully) checked.

However, we are not yet done. We have to check that if we take the bracket

of two elements in

(

), then it still stays within

(

). This will be done by

producing a curve in G whose derivative at 0 is the commutator [X, Y ].

In general, let

be a smooth curve in

with

(0) =

. Then we can Taylor

expand

γ(t) = I + ˙γ(0)t + ¨γ(0)t

+ O(t

Now given

X, Y ∈ T

, we take curves

, γ

such that

˙γ

(0) =

and

˙γ

(0) =

Consider the curve given by

γ(t) = γ

−1

(t)γ

−1

(t)γ

(t) ∈ G.

We can Taylor expand this to find that

γ(t) = I + [X, Y ]t

+ O(t

This isn’t too helpful, as [

X, Y

] is not the coefficient of

. We now do the slightly

dodgy step, where we consider the curve

˜γ(t) = γ(

√

t) = I + [X, Y ]t + O(t

3/2

Now this is only defined for

t ≥

0, but it is good enough, and we see that its

derivative at

= 0 is [

X, Y

]. So the commutator is in

(

). So we know that

L(G) is a Lie algebra.

Example.

Let

(

n, F

), where

. Then

(

n, F

)) =

(

n, F

) =

Mat

(

) because we know it must be an

n × n

-dimensional sub-

space of Mat

(F).

More generally, for a vector space

, say of dimension

, we can consider

the group of invertible linear maps

V → V

, written

(

). By picking a basis

, we can construct an isomorphism

(

)

∼

(

n, F

), and this gives us a

smooth structure on

(

) (this does not depend on the basis chosen). Then

the Lie algebra gl(V ) is the collection of all linear maps V → V .

Example. If G = SO(2), then the curves are of the form

g(t) = M(θ(t)) =



cos θ(t) −sin θ(t)

sin θ(t) cos θ(t)



∈ SO(2).

So we have

˙g(0) =



0 −1

1 0



θ(0).

Since the Lie algebra has dimension 1, these are all the matrices in the Lie

algebra. So the Lie algebra is given by

so(2) =



0 −c

c 0



, c ∈ R



Example.

More generally, suppose

(

), and we have a path

(

)

∈

SO(n).

By definition, we have

(t)R(t) = I.

Differentiating gives

(t)R(t) + R

(t)

R(t) = 0.

for all t ∈ R. Evaluating at t = 0, and noting that R(0) = I, we have

+ X = 0,

where

(0) is a tangent vector. There are no further constraints from

demanding that

det R

= +1, since this is obeyed anyway for any matrix in O(

)

near I.

By dimension counting, we know the antisymmetric matrices are exactly the

matrices in L(O(n)) or L(SO(n)). So we have

o(n) = so(n) = {X ∈ Mat

(R) : X

= −X}.

Example.

Consider

(

). Suppose we have a path

(

)

∈ SU

(

), with

U(0) = I. Then we have

†

(t)U(t) = I.

Then again by differentiation, we obtain

†

+ Z = 0,

where Z =

U(0) ∈ su(n). So we must have

su(n) ⊂ {Z ∈ Mat

†

= −Z}.

What does the condition

det U

(

) = 1 give us? We can do a Taylor expansion by

det U(t) = 1 + tr Z · t + O(t

So requiring that det U(t) = 1 gives the condition

tr Z = 0.

By dimension counting, we know traceless anti-Hermitian matrices are all the

elements in the Lie algebra. So we have

su(n) = {Z ∈ Mat

(C), Z

†

= −Z, tr Z = 0}.

Example.

We look at

(2) in detail. We know that

(2) is the 2

2 traceless

anti-Hermitian matrices.

These are given by multiples of the Pauli matrices

, for

= 1

3, satisfying

= δ

I + iε

ijk

They can be written explicitly as



0 1

1 0



, σ



0 −i

i 0



, σ



1 0

0 −1



One can check manually that the generators for the Lie algebra are given by

= −

iσ

Indeed, each

is in

(2), and they are independent. Since we know

dim su

(2) =

3, these must generate everything.

We have

, T

] = −

[σ

, σ

] = −

iε

abc

= f

where the structure constants are

= ε

abc

Example.

Take

(3). Then

(3) is the space of 3

3 real anti-symmetric

matrices, which one can manually check are generated by





0 0 0

0 0 −1

0 1 0









0 0 1

0 0 0

−1 0 0









0 −1 0

1 0 0

0 0 0





We then have

(

)

= −ε

abc

Then the structure constants are

[

] = f

where

= ε

abc

Note that the structure constants are the same! Since the structure constants

completely determine the brackets of the Lie algebra, if the structure constants

are the same, then the Lie algebras are isomorphic. Of course, the structure

constants depend on which basis we choose. So the real statement is that if

there are some bases in which the structure constants are equal, then the Lie

algebras are isomorphic.

So we get that

(3)

∼

(2), but

(3) is not isomorphic to

(2). Indeed,

the underlying manifold is

(2) is the 3-sphere, but the underlying manifold of

(3) has a fancy construction. They are not even topologically homeomorphic,

since SU(2) is simply connected, but SO(3) is not. More precisely, we have

(SO(3)) = Z/2Z

(SU(2)) = {0}.

So we see that we don’t have a perfect correspondence between Lie algebras and

Lie groups. However, usually, two Lie groups with the same Lie algebra have

some covering relation. For example, in this case we have

SO(3) =

SU(2)

Z/2Z

where

Z/2Z = {I, −I} ⊆ SU(2)

is the center of SU(2).

We can explicitly construct this bijection as follows. We define the map

d : SU(2) → SO(3) by

d(A)

tr(σ

Aσ

†

) ∈ SO(3).

This is globally a 2-to-1 map. It is easy to see that

d(A) = d(−A),

and conversely if

(

) =

(

), then

−B

. By the first isomorphism theorem,

this gives an isomorphism

SO(3) =

SU(2)

Z/2Z

where Z/2Z = {I, −I} is the center of SU(2).

Geometrically, we know

(

(2))

∼

. Then the manifold of

(3) is

obtained by identifying antipodal points of the manifold.

3.4 The exponential map

So far, we have been talking about vectors in the tangent space of the identity

. It turns out that the group structure means this tells us about the tangent

space of all points. To see this, we make the following definition:

Definition

(Left and right translation)

For each

h ∈ G

, we define the left and

right translation maps

: G → G

g 7→ hg,

: G → G

g 7→ gh.

These maps are bijections, and in fact diffeomorphisms (i.e. smooth maps with

smooth inverses), because they have smooth inverses

−1

and

−1

respectively.

In general, there is no reason to prefer left translation over right, or vice

versa. By convention, we will mostly talk about left translation, but the results

work equally well for right translation.

Then since

(

) =

, the derivative D

gives us a linear isomorphism

between

and

, with inverse given by D

−1

. Then in particular, if we

are given a single tangent vector

X ∈ L

(

), we obtain a tangent vector at all

points in G, i.e. a vector field.

Definition (Vector field). A vector field V of G specifies a tangent vector

V (g) ∈ T

at each point

g ∈ G

. Suppose we can pick coordinates

}

on some subset of

and write

v(g) = v

(g)

∂

∂x

∈ T

The vector field is smooth if

(

)

∈ R

are all differentiable for any coordinate

chart.

As promised, given any

X ∈ T

, we can define a vector field by using

∗

:= D

to move this to all places in the world. More precisely, we can define

V (g) = L

∗

(X).

This has the interesting property that if

is non-zero, then

is non-zero

everywhere, because L

∗

is a linear isomorphism. So we found that

Proposition.

Let

be a Lie group of dimension

0. Then

has a nowhere-

vanishing vector field.

This might seem like a useless thing to know, but it tells us that certain

manifolds cannot be made into Lie groups.

Theorem

(Poincare-Hopf theorem)

Let

be a compact manifold. If

has

non-zero Euler characteristic, then any vector field on M has a zero.

The Poincare-Hopf theorem actually tells us how we can count the actual

number of zeroes, but we will not go into that. We will neither prove this, nor

use it for anything useful. But in particular, it has the following immediate

corollary:

Theorem

(Hairy ball theorem)

Any smooth vector field on

has a zero.

More generally, any smooth vector field on S

has a zero.

Thus, it follows that

can never be a Lie group. In fact, the full statement

of the Poincare-Hopf theorem implies that if we have a compact Lie group of

dimension 2, then it must be the torus! (we have long classified the list of all

possible compact 2-dimensional manifolds, and we can check that only the torus

works)

That was all just for our own amusement. We go back to serious work.

What does this left-translation map look like when we have a matrix Lie group

G ⊆ Mat

(

)? For all

h ∈ G

and

X ∈ L

(

), we can represent

as a matrix in

Mat

(F). We then have a concrete representation of the left translation by

∗

X = hX ∈ T

Indeed,

acts on

by left multiplication, which is a linear map if we view

a subset of the vector space

Mat

(

). Then we just note that the derivative of

any linear map is “itself”, and the result follows.

Now we may ask ourselves the question — given a tangent vector

X ∈ T

can we find a path that “always” points in the direction

? More concretely,

we want to find a path γ : R → T

G such that

˙γ(t) = L

∗

γ(t)

Here we are using

γ(t)

to identify

γ(t)

with

(

). In the case of a

matrix Lie group, this just says that

˙γ(t) = γ(t)X.

We also specify the boundary condition γ(0) = e.

Now this is just an ODE, and by general theory of ODE’s, we know a solution

always exists and is unique. Even better, in the case of a matrix Lie group, there

is a concrete construction of this curve.

Definition

(Exponential)

Let

M ∈ Mat

(

) be a matrix. The exponential is

defined by

exp(M) =

∞

`=0

!

∈ Mat

(F).

The convergence properties of these series are very good, just like our usual

exponential.

Theorem.

For any matrix Lie group

, the map

exp

restricts to a map

(

)

→

Proof.

We will not prove this, but on the first example sheet, we will prove this

manually for G = SU(n).

We now let

g(t) = exp(tX).

We claim that this is the curve we were looking for.

To check that it satisfies the desired properties, we simply have to compute

g(0) = exp(0) = I,

and also

dg(t)

∞

`=1

( − 1)!

`−1

= exp(tX)X = g(t)X.

So we are done.

We now consider the set

= {exp(tX) : t ∈ R}.

This is an abelian Lie subgroup of G with multiplication given by

exp(tX) exp(sX) = exp((t + s)X)

by the usual proof. These are known as one-parameter subgroups.

Unfortunately, it is not true in general that

exp(X) exp(Y ) = exp(X + Y ),

since the usual proof assumes that

and

commute. Instead, what we’ve got

is the Baker–Campbell–Hausdorff formula.

Theorem (Baker–Campbell–Hausdorff formula). We have

exp(X) exp(Y ) = exp



X + Y +

[X, Y ] +

([X, [X, Y ]] − [Y, [X, Y ]]) + ···



It is possible to find the general formula for all the terms, but it is messy.

We will not prove this.

By the inverse function theorem, we know the map

exp

is locally bijective. So

we know

(

) completely determines

in some neighbourhood of

. However,

exp

is not globally bijective. Indeed, we already know that the Lie algebra

doesn’t completely determine the Lie group, as

(3) and

(2) have the same

Lie algebra but are different Lie groups.

In general,

exp

can fail to be bijective in two ways. If

is not connected,

then

exp

cannot be surjective, since by continuity, the image of

exp

must be

connected.

Example.

Consider the groups O(

) and

(

). Then the Lie algebra of O(

)

o(n) = {X ∈ Mat

(R) : X + X

= 0}.

So if X ∈ o(n), then tr X = 0. Then we have

det(exp(X)) = exp(tr X) = exp(0) = +1.

So any matrix in the image of

exp

has determinant +1, and hence can only lie

inside SO(n). It turns out that the image of exp is indeed SO(n).

More generally, we have

Proposition.

Let

be a Lie group, and

be its Lie algebra. Then the image

of g under exp is the connected component of e.

On the other hand,

exp

can also fail to be injective. This happens when

has a U(1) subgroup.

Example. Let G = U(1). Then

u(1) = {ix, x ∈ R}

We then have

exp(ix) = e

This is certainly not injective. In particular, we have

exp(ix) = exp(i(x + 2π))

for any x.

4 Representations of Lie algebras

4.1 Representations of Lie groups and algebras

So far, we have just talked about Lie groups and Lie algebras abstractly. But we

know these groups don’t just sit there doing nothing. They act on things. For

example, the group

(

n, R

) acts on the vector space

in the obvious way. In

general, the action of a group on a vector space is known as a representation.

Definition

(Representation of group)

Let

be a group and

be a (finite-

dimensional) vector space over a field

. A representation of

is given

by specifying invertible linear maps

(

) :

V → V

(i.e.

(

)

∈ GL

(

)) for each

g ∈ G such that

D(gh) = D(g)D(h)

for all

g, h ∈ G

. In the case where

is a Lie group and

, we require

that the map D : G → GL(V ) is smooth.

The space

is known as the representation space, and we often write the

representation as the pair (V, D).

Here if we pick a basis

, ··· , e

}

for

, then we can identify

(

) with

(

n, F

), and this obtains a canonical smooth structure when

. This

smooth structure does not depend on the basis chosen.

In general, the map D need not be injective or surjective.

Proposition.

Let

be a representation of a group

. Then

(

) =

and

D(g

−1

) = D(g)

−1

Proof. We have

D(e) = D(ee) = D(e)D(e).

Since D(e) is invertible, multiplying by the inverse gives

D(e) = I.

Similarly, we have

D(g)D(g

−1

) = D(gg

−1

) = D(e) = I.

So it follows that D(g)

−1

= D(g

−1

Now why do we care about representations? Mathematically, we can learn

a lot about a group in terms of its possible representations. However, from a

physical point of view, knowing about representations of a group is also very

important. When studying field theory, our fields take value in a fixed vector

space

, and when we change coordinates, our field will “transform accordingly”

according to some rules. For example, we say scalar fields “don’t transform”,

but, say, the electromagnetic field tensor transforms “as a 2-tensor”.

We can describe the spacetime symmetries by a group

, so that specifying

a “change of coordinates” is equivalent to giving an element of

. For example,

in special relativity, changing coordinates corresponds to giving an element of

the Lorentz group O(3, 1).

Now if we want to say how our objects in

transform when we change

coordinates, this is exactly the same as specifying a representation of

So understanding what representations are available lets us know what kinds of

fields we can have.

The problem, however, is that representations of Lie groups are very hard.

Lie groups are very big geometric structures with a lot of internal complexity.

So instead, we might try to find representations of their Lie algebras instead.

Definition

(Representation of Lie algebra)

Let

be a Lie algebra. A represen-

tation ρ of g on a vector space V is a collection of linear maps

ρ(X) ∈ gl(V ),

for each

X ∈ g

, i.e.

(

) :

V → V

is a linear map, not necessarily invertible.

These are required to satisfy the conditions

[ρ(X

), ρ(X

)] = ρ([X

, X

])

and

ρ(αX

+ βX

) = αρ(X

) + βρ(X

The vector space

is known as the representation space. Similarly, we often

write the representation as (V, ρ).

Note that it is possible to talk about a complex representation of a real Lie

algebra, because any complex Lie algebra (namely

(

) for a complex vector

space

) can be thought of as a real Lie algebra by “forgetting” that we can

multiply by complex numbers, and indeed this is often what we care about.

Definition

(Dimension of representation)

The dimension of a representation

is the dimension of the representation space.

We will later see that a representation of a Lie group gives rise to a represen-

tation of the Lie algebra. The representation is not too hard to obtain — if we

have a representation

G → GL

(

) of the Lie group, taking the derivative of

this map gives us the confusingly-denoted D

G → T

(

)), which is

a map

g → gl

(

). To check that this is indeed a representation, we will have

to see that it respects the Lie bracket. We will do this later when we study the

relation between representations of Lie groups and Lie algebras.

Before we do that, we look at some important examples of representations of

Lie algebras.

Definition

(Trivial representation)

Let

be a Lie algebra of dimension

The trivial representation is the representation

g → F

given by

(

) = 0

for all X ∈ g. This has dimension 1.

Definition

(Fundamental representation)

Let

(

) for

G ⊆ Mat

(

). The

fundamental representation is given by d

: g → Mat

(F) given by

(X) = X

This has dim(d

) = n.

Definition

(Adjoint representation)

All Lie algebras come with an adjoint

representation

Adj

of dimension

dim

(

) =

. This is given by mapping

X ∈ g

to the linear map

: g → g

Y 7→ [X, Y ]

By linearity of the bracket, this is indeed a linear map g → gl(g).

There is a better way of thinking about this. Suppose our Lie algebra

comes from a Lie group

. Writing

Aut

(

) for all the isomorphisms

G → G

, we

know there is a homomorphism

Φ : G → Aut(G)

g 7→ Φ

given by conjugation:

(x) = gxg

−1

Now by taking the derivative, we can turn each Φ

into a linear isomorphism

g → g, i.e. an element of GL(g). So we found ourselves a homomorphism

Ad : G → GL(g),

which is a representation of the Lie group

! It is an exercise to show that

the corresponding representation of the Lie algebra

is indeed the adjoint

representation.

Thus, if we view conjugation as a natural action of a group on itself, then

the adjoint representation is the natural representation of g over itself.

Proposition. The adjoint representation is a representation.

Proof.

Since the bracket is linear in both components, we know the adjoint

representation is a linear map g → gl(g). It remains to show that

[ad

, ad

] = ad

[X,Y ]

But the Jacobi identity says

[ad

, ad

](Z) = [X, [Y, Z]] − [Y, [X, Z]] = [[X, Y ], Z] = ad

[X,Y ]

(Z).

We will eventually want to find all representations of a Lie algebra. To do

so, we need the notion of when two representations are “the same”.

Again, we start with the definition of a homomorphism.

Definition

(Homomorphism of representations)

Let (

, ρ

)

(

, ρ

) be rep-

resentations of

. A homomorphism

: (

, ρ

)

→

(

, ρ

) is a linear map

f : V

→ V

such that for all X ∈ g, we have

f(ρ

(X)(v)) = ρ

(X)(f(v))

for all v ∈ V

. Alternatively, we can write this as

f ◦ ρ

= ρ

◦ f.

In other words, the following diagram commutes for all X ∈ g:

(X) ρ

(X)

Then we can define

Definition

(Isomorphism of representations)

Two

-vector spaces

, V

are

isomorphic if there is an invertible homomorphism f : V

→ V

In particular, isomorphic representations have the same dimension.

If we pick a basis for

and

, and write the matrices for the representations

(

) and

(

), then they are isomorphic if there exists a non-singular

matrix S such that

(X) = SR

(X)S

−1

for all X ∈ g.

We are going to look at special representations that are “indecomposable”.

Definition

(Invariant subspace)

Let

be a representation of a Lie algebra

with representation space

. An invariant subspace is a subspace

U ⊆ V

such

that

ρ(X)u ∈ U

for all X ∈ g and u ∈ U.

The trivial subspaces are U = {0} and V .

Definition

(Irreducible representation)

An irreducible representation is a rep-

resentation with no non-trivial invariant subspaces. They are referred to as

irreps.

4.2 Complexification and correspondence of representa-

tions

So far, we have two things — Lie algebras and Lie groups. Ultimately, the thing

we are interested in is the Lie group, but we hope to simplify the study of a

Lie group by looking at the Lie algebra instead. So we want to understand

how representations of Lie groups correspond to the representations of their Lie

algebras.

If we have a representation

G → GL

(

) of a Lie group

, then taking

the derivative at the identity gives us a linear map

G → T

(

), i.e. a

map

g → gl

(

). To show this is a representation, we need to show that it

preserves the Lie bracket.

Lemma.

Given a representation

G → GL

(

), the induced representation

ρ : g → gl(V ) is a Lie algebra representation.

Proof.

We will again only prove this in the case of a matrix Lie group, so that

we can use the construction we had for the Lie bracket.

We have to check that the bracket is preserved. We take curves

, γ

R → G

passing through I at 0 such that ˙γ

(0) = X

for i = 1, 2. We write

γ(t) = γ

−1

(t)γ

−1

(t)γ

(t) ∈ G.

We can again Taylor expand this to obtain

γ(t) = I + t

, X

] + O(t

Essentially by the definition of the derivative, applying D to this gives

D(γ(t)) = I + t

ρ([X

, X

]) + O(t

On the other hand, we can apply D to (∗) before Taylor expanding. We get

D(γ) = D(γ

−1

)D(γ

−2

)D(γ

So as before, since

D(γ

) = I + tρ(X

) + O(t

it follows that

D(γ)(t) = I + t

[ρ(X

), ρ(X

)] + O(t

So we must have

ρ([X

, X

]) = [ρ(X

), ρ(X

)].

How about the other way round? We know that if

g → gl

(

) is induced

by D : G → GL(V ), then have

D(exp(X)) = I + tρ(X) + O(t

while we also have

exp(ρ(X)) = I + rρ(X) + O(t

So we might expect that we indeed have

D(exp(X)) = exp(ρ(X))

for all X ∈ g.

So we can try to use this formula to construct a representation of

. Given

an element

g ∈ G

, we try to write it as

exp

(

) for some

X ∈ g

. We then

define

D(g) = exp(ρ(X)).

For this to be well-defined, we need two things to happen:

(i) Every element in G can be written as exp(X) for some X

(ii) The value of exp(ρ(X)) does not depend on which X we choose.

We first show that if this is well-defined, then it indeed gives us a representation

of G.

To see this, we use the Baker-Campbell-Hausdorff formula to say

D(exp(X) exp(Y )) = exp(ρ(log(exp(X) exp(Y ))))

= exp



log



exp



X + Y +

[X, Y ] + ···



= exp



X + Y +

[X, Y ] + ···



= exp



ρ(X) + ρ(Y ) +

[ρ(X), ρ(Y )] + ···



= exp(ρ(X)) exp(ρ(Y ))

= D(exp(X))D(exp(Y )),

where

log

is the inverse to

exp

. By well-defined-ness, it doesn’t matter which

log we pick. Here we need to use the fact that

preserves the Lie bracket, and

all terms in the Baker-Campbell-Hausdorff formula are made up of Lie brackets.

So when are the two conditions satisfied? For the first condition, we know

that

is connected, and the continuous image of a connected space is connected.

So a necessary condition is that

must be a connected Lie group. This rules out

groups such as O(

). It turns out this is also sufficient. The second condition is

harder, and we will take note of the following result without proof:

Theorem.

Let

be a simply connected Lie group with Lie algebra

, and let

g → gl

(

) be a representation of

. Then there is a unique representation

D : G → GL(V ) of G that induces ρ.

So if we only care about simply connected Lie groups, then studying the

representations of its Lie algebra is exactly the same as studying the representa-

tions of the group itself. But even if we care about other groups, we know that

all representations of the group give representations of the algebra. So to find a

representation of the group, we can look at all representations of the algebra,

and see which lift to a representation of the group.

It turns out Lie algebras aren’t simple enough. Real numbers are terrible,

and complex numbers are nice. So what we want to do is to look at the

complexification of the Lie algebra, and study the complex representations of

the complexified Lie algebra.

We will start with the definition of a complexification of a real vector space.

We will provide three different definitions of the definition, from concrete to

abstract, to suit different people’s tastes.

Definition

(Complexification I)

Let

be a real vector space. We pick a basis

}

. We define the complexification of

, written

as the complex

linear span of {T

}, i.e.

: λ

∈ C

There is a canonical inclusion

V → V

given by sending

for

∈ R.

This is a rather concrete definition, but the pure mathematicians will not

be happy with such a definition. We can try another definition that does not

involve picking a basis.

Definition

(Complexification II)

Let

be a real vector space. The complex-

ification of

has underlying vector space

V ⊕ V

. Then the action of a

complex number λ = a + bi on (u

, u

) is given by

λ(u

, u

) = (au

− bu

, au

+ bu

This gives

the structure of a complex vector space. We have an inclusion

V → V

by inclusion into the first factor.

Finally, we have a definition that uses some notions from commutative algebra,

which you may know about if you are taking the (non-existent) Commutative

Algebra course. Otherwise, do not bother reading.

Definition

(Complexification III)

Let

be a real vector space. The com-

plexification of

is the tensor product

V ⊗

, where

is viewed as an

(R, C)-bimodule.

To define the Lie algebra structure on the complexification, we simply declare

that

[X + iY, X

+ iY

] = [X, X

] + i([X, Y

] + [Y, X

]) − [Y, Y

]

for X, Y ∈ V ⊆ V

Whichever definition we decide to choose, we have now a definition, and we

want to look at the representations of the complexification.

Theorem.

Let

be a real Lie algebra. Then the complex representations of

are exactly the (complex) representations of g

Explicitly, if

g → gl

(

) is a complex representation, then we can extend

it to g

by declaring

ρ(X + iY ) = ρ(X) + iρ(Y ).

Conversely, if

→ gl

(

), restricting it to

g ⊆ g

gives a representation of

Proof. Just stare at it and see that the formula works.

So if we only care about complex representations, which is the case most of

the time, we can study the representations of the complexification instead. This

is much easier. In fact, in the next chapter, we are going to classify all simple

complex Lie algebras and their representations.

Before we end, we note the following definition:

Definition

(Real form)

Let

be a complex Lie algebra. A real form of

is a

real Lie algebra h such that h

= g.

Note that a complex Lie algebra can have multiple non-isomorphic real forms

in general.

4.3 Representations of su(2)

We are now going to study the complex representations of

(2), which are

equivalently the representations of the complexification

(2). In this section,

we will write

(2) when we actually mean

(2) for brevity. There are a

number of reasons why we care about this.

(i)

We’ve already done this before, in the guise of “spin”/“angular momentum”

in quantum mechanics.

(ii)

The representations are pretty easy to classify, and form a motivation for

our later general classification of representations of complex simple Lie

algebras.

(iii)

Our later work on the study of simple Lie algebras will be based on our

knowledge of representations of su(2).

Recall that the uncomplexified su(2) has a basis

su(2) = span



= −

iσ

: a = 1, 2, 3



where

are the Pauli matrices. Since the Pauli matrices are independent over

C, they give us a basis of the complexification of su(2):

(2) = span

{σ

: a = 1, 2, 3}.

However, it is more convenient to use the following basis:

H = σ



1 0

0 −1



, E

(σ

± iσ

) =



0 1

0 0





0 0

1 0



We then have

[H, E

] = ±2E

, [E

, E

−

] = H.

We can rewrite the first relation as saying

) = ±2E

Together with the trivial result that

(

) = 0, we know

has eigenvalues

±2 and 0, with eigenvectors E

, H. These are known as roots of su(2).

Now let’s look at irreducible representations of

(2). Suppose (

V, ρ

) is a

representation. By linear algebra, we know ρ(H) has an eigenvector v

, say

ρ(H)v

= λv

The eigenvalues of

(

) are known as the weights of the representation

. The

operators E

are known as step operators. We have

ρ(H)ρ(E

= (ρ(E

)ρ(H) + [ρ(H), ρ(E

)])v

= (ρ(E

)ρ(H) + ρ([H, E

])v

= (λ ± 2)ρ(E

(

)

are also eigenvectors of

(

) with eigenvalues

λ ±

2, provided

(

)

are non-zero. This constrains what can happen in a finite-dimensional

representation. By finite-dimensionality, we cannot have infinitely many eigen-

values. So at some point, we have to stop. In other words, a finite-dimensional

representation must have a highest weight Λ ∈ C, with

ρ(H)v

= Λv

, ρ(E

= 0.

Now if we start off with

we can keep applying

−

to get the lower weights.

Indeed, we get

Λ−2n

= (ρ(E

−

))

for each

. Again, this sequence must terminate somewhere, as we only have

finitely many eigenvectors. We might think that the irrep would consist of the

basis vectors

, v

Λ−2

, v

Λ−4

, ··· , v

Λ−2n

However, we need to make sure we don’t create something new when we act on

these by

(

) again. We would certainly get that

(

)

Λ−2n

is an eigenvector

of eigenvalue Λ

−

+ 2, but we might get some other eigenvector that is not

Λ−2n+2

. So we have to check.

We can compute

ρ(E

Λ−2n

= ρ(E

)ρ(E

−

Λ−2n+2

= (ρ(E

−

)ρ(E

) + [ρ(E

), ρ(E

−

)])v

Λ−2n+2

= ρ(E

−

)ρ(E

Λ−2n+2

+ (Λ − 2n + 2)v

Λ−2n+2

using the fact that [

, E

−

] =

. We now get a recursive relation. We can

analyze this by looking at small cases. When n = 1, the first term is

ρ(E

−

)ρ(E

Λ−2n+2

= ρ(E

−

)ρ(E

= 0,

by definition of v

. So we have

ρ(E

Λ−2

= Λv

When n = 2, we have

ρ(E

Λ−4

= ρ(E

−

)ρ(E

Λ−2

+ (Λ − 2)v

Λ−2

= (2Λ − 2)v

Λ−2

In general, ρ(E

Λ−2n

is some multiple of v

Λ−2n+2

, say

ρ(E

Λ−2n

= r

Λ−2n+2

Plugging this into our equation gives

= r

n−1

+ Λ − 2n + 2,

with the boundary condition

(

)

= 0, i.e.

= Λ. If we solve this recurrence

relation, we determine an explicit relation for r

given by

= (Λ + 1 − n)n.

Now returning to the problem that the sequence must terminate, we figure that

we also need to have a lowest weight of the form Λ

−

. Hence we must have

some non-zero vector v

Λ−2n

6= 0 such that

ρ(E

−

Λ−2n

= 0.

For this to be true, we need

N+1

= 0 for some non-negative integer

. So we

have

(Λ + 1 − (N + 1))(N + 1) = 0.

This is equivalent to saying

(Λ − N)(N + 1) = 0.

Since

+ 1 is a non-negative integer, we must have Λ

− N

= 0, i.e. Λ =

. So

in fact the highest weight is an integer!

In summary, we’ve got

Proposition.

The finite-dimensional irreducible representations of

(2) are

labelled by Λ ∈ Z

≥0

, which we call ρ

, with weights given by

{−Λ, −Λ + 2, ··· , Λ − 2, Λ}.

The weights are all non-degenerate, i.e. each only has one eigenvector. We have

dim(ρ

) = Λ + 1.

Proof.

We’ve done most of the work. Given any irrep, we can pick any eigenvector

(

) and keep applying

to get a highest weight vector

, then the above

computations show that

, v

Λ−2

, ··· , v

−Λ

}

is a subspace of the irrep closed under the action of

. By irreducibility, this

must be the whole of the representation space.

The representation

is the trivial representation;

is the fundamental one,

and ρ

is the adjoint representation.

Now what do these tell us about representations of the Lie group

(2)? We

know

(2) is simply connected, so by our discussion previously, we know the

complex representations of

(2) are exactly the representations of

(2). This

is not too exciting.

We know there are other Lie groups with Lie algebra

(2), namely

(3).

Now that

(3) is not simply connected, when does a representation of

(2)

give us a representation of SO(3)? Recall that we had

SO(3)

∼

SU(2)

Z/2Z

So an element in

(3) is a pair

{A, −A}

of elements in

(2). Given a

representation

(2), we obtain a corresponding representation

(2). Now we get a representation of

(3) if and only if

respects the

identification between A and −A. In particular, we need

(−I) = D

(I). (∗)

On the other hand, if this were true, then multiplying both sides by

(

) we

get

(−A) = D

(A).

So (

∗

) is a necessary and sufficient condition for

to descend to a representation

of SO(3).

We know that

−I = exp(iπH).

So we have

(−I) = exp(iπρ

(H)).

We know that ρ

has eigenvalues Λ ∈ S

. So D

(−I) has eigenvalues

exp(iπλ) = (−1)

= (−1)

since

and Λ differ by an even integer. So we know

(

−I

) =

(

) if and

only if Λ is even. In other words, we get a representation of

(3) iff Λ is even.

In physics, we have already seen this as integer and half-integer spin. We

have integer spin exactly when we have a representation of

(3). The fact that

half-integer spin particles exist means that “spin” is really about

(2), rather

than SO(3).

4.4 New representations from old

We are now going to look at different ways of obtaining new representations

from old ones. We start with a rather boring one.

Definition

(Conjugate representation)

Let

be a representation of a real Lie

algebra g on C

. We define the conjugate representation by

¯ρ(X) = ρ(X)

∗

for all X ∈ g.

Note that this need not be an actual new representation.

To obtain more interesting new representations, we recall the following

definitions from linear algebra:

Definition

(Direct sum)

Let

V, W

be vector spaces. The direct sum

V ⊕W

given by

V ⊕W = {v ⊕ w : v ∈ V, w ∈ W }

with operations defined by

⊕ w

) + (v

⊕ w

) = (v

+ v

) ⊕ (w

+ w

)

λ(v ⊕ w) = (λv) ⊕ (λw).

We often suggestively write v ⊕ w as v + w. This has dimension

dim(V ⊕W ) = dim V + dim W.

Definition

(Sum representation)

Suppose

and

are representations of

with representation spaces

and

of dimensions

and

. Then

⊕ V

a representation space with representation ρ

⊕ ρ

given by

(ρ

⊕ ρ

)(X) · (v

⊕ v

) = (ρ

(X)(v

)) ⊕ (ρ

(X)(v

)).

In coordinates, if

is the matrix for

, then the matrix of

⊕ ρ

is given by

⊕ R

)(X) =



(X) 0

0 R

(X)



The dimension of this representation is d

+ d

Of course, sum representations are in general not irreducible! However, they

are still rather easy to understand, since they decompose into smaller, nicer bits.

In contrast, tensor products do not behave as nicely.

Definition

(Tensor product)

Let

V, W

be vector spaces. The tensor product

V ⊗W

is spanned by elements

v ⊗w

, where

v ∈ V

and

w ∈ W

, where we identify

v ⊗ (λ

+ λ

) = λ

(v ⊗ w

) + λ

(v ⊗ w

)

(λ

+ λ

) ⊗ w = λ

⊗ w) + λ

⊗ w)

This has dimension

dim(V ⊗W ) = (dim V )(dim W ).

More explicitly, if

, ··· , e

is a basis of

and

, ··· , f

is a basis for

then {e

⊗ f

: 1 ≤ i ≤ n, 1 ≤ j ≤ m} is a basis for V ⊗ W .

Given any two maps

V → V

and

W → W

, we define

F ⊗ G

V ⊗W → V

⊗ W

(F ⊗ G)(v ⊗ w) = (F (v)) ⊗ (G(w)),

and then extending linearly.

The operation of tensor products should be familiar in quantum mechanics,

where we combine two state spaces to get a third. From a mathematical point

of view, the tensor product is characterized by the fact that a bilinear map

V ×W → U is the same as a linear map V ⊗ W → U .

Definition

(Tensor product representation)

Let

be a Lie algebra, and

, ρ

be representations of

with representation spaces

, V

. We define the tensor

product representation ρ

⊗ ρ

with representation space V

⊗ V

given by

(ρ

⊗ ρ

)(X) = ρ

(X) ⊗ I

+ I

⊗ ρ

(X) : V

⊗ V

→ V

⊗ V

where I

and I

are the identity maps on V

and V

Note that the tensor product representation is not given by the “obvious”

formula ρ

(X) ⊗ ρ

(X). This does not work.

Now suppose (

ρ, V

) is a reducible representation. Then it has a non-trivial

invariant subspace

U ⊆ V

. Now we pick a basis of

and extend it to

. Then

ρ sends any member of U to itself, so the matrix representation of ρ looks like

R(X) =



A(X) B(X)

0 C(X)



However, there is no a priori reason why

(

) should vanish as well. When

this happens, we say the representation is completely reducible.

Definition

(Completely reducible representation)

If (

ρ, V

) is a representation

such that there is a basis of V in which ρ looks like

ρ(X) =







(X) 0 ··· 0

0 ρ

(X) ··· 0

0 0 ··· ρ

(X)







then it is called completely reducible. In this case, we have

ρ = ρ

⊕ ρ

⊕ ··· ⊕ ρ

Here is an important fact:

Theorem.

for

= 1

, ··· , m

are finite-dimensional irreps of a simple Lie

algebra

, then

⊗ ··· ⊗ ρ

is completely reducible to irreps, i.e. we can find

˜ρ

, ··· , ˜ρ

such that

⊗ ··· ⊗ ρ

= ˜ρ

⊕ ˜ρ

⊕ ··· ⊕ ˜ρ

We will not prove this.

4.5 Decomposition of tensor product of su

(2)

representa-

tions

We now try to find an explicit description of the decomposition of tensor products

of irreps of su(2).

We let

and

be irreps of

(2), where Λ

∈ N

. We call the represen-

tation spaces V

and V

We can form the tensor product ρ

⊗ ρ

with representation space

⊗ V

= span

{v ⊗ v

: v ∈ V

, v

∈ V

By definition, for X ∈ su(2), the representation is given by

(ρ

⊗ ρ

)(X)(v ⊗ v

) = (ρ

(X)v) ⊗ v

+ v ⊗ (ρ

(X)v

This gives us a completely reducible representation of su(2) of dimension

dim(ρ

⊗ ρ

) = (Λ + 1)(Λ

+ 1).

We can then write

⊗ ρ

∈Z,Λ

≥0

Λ,Λ

where

Λ,Λ

are some non-negative integers we want to find out. These coefficients

are usually known as Littlewood-Richardson coefficients in general.

Recall that V

has a basis {v

}, where

λ ∈ S

= {−Λ, Λ − 2, ··· , +Λ}.

Similarly, V

has a basis {v

Then we know that the tensor product space has basis

B = {v

⊗ v

: λ ∈ S

, λ

∈ S

We now see what H does to our basis vectors. We have

(ρ

⊗ ρ

)(H)(v

⊗ v

) = (ρ

(H)v

) ⊗ v

+ v

⊗ (ρ

(H)v

)

= (λ + λ

)(v

⊗ v

We thus see that the weights of the tensor product are just the sum of the

weights of the individual components. In other words, we have

Λ,Λ

= {λ + λ

: λ ∈ S

, λ

∈ S

}

Note that here we count the weights with multiplicity, so that each weight can

appear multiple times.

We see that the highest weight is just the sum of the largest weights of the

irreps, and this appears with multiplicity 1. Thus we know

Λ+Λ

Λ,Λ

= 1,

i.e. we have one copy of

Λ+Λ

in the decomposition of the tensor product. We

write

⊗ ρ

= ρ

Λ+Λ

⊕ ˜ρ

Λ,Λ

where ˜ρ

Λ,Λ

has weight set

Λ,Λ

satisfying

Λ,Λ

= S

Λ+Λ

∪

Λ,Λ

We now notice that there is only one Λ + Λ

−

2 term in

Λ,Λ

. So there must be

a copy of ρ

Λ+Λ

−2

as well. We keep on going.

Example. Take Λ = Λ

= 1. Then we have

= {−1, +1}.

So we have

1,1

= {−2, 0, 0, 2}.

We see that the highest weight is 2, and this corresponds to a factor of

. In

doing so, we write

1,1

= {−2, 0, +2} ∪ {0} = S

∪ S

So we have

⊗ ρ

= ρ

⊕ ρ

From the above, one can see (after some thought) that in general, we have

Proposition.

⊗ ρ

= ρ

|N−M|

⊕ ρ

|N−M|+2

⊕ ··· ⊕ ρ

N+M

5 Cartan classification

We now move on to the grand scheme of classifying all complex simple Lie

algebras. The starting point of everything is that we define a natural inner

product on our Lie algebra

. We will find a subalgebra

that plays

the role of the

we had when we studied

(2). The remainder of

will be

controlled by things known as roots, which live in

∗

. We will see that the Killing

form induces a inner product on

∗

, which allows us to think of these roots as

“geometric” objects that live in

. We can then find some strong conditions

that restrict what these roots and their inner products can be, and it turns out

these completely characterize our possible Lie algebras.

5.1 The Killing form

The first thing we want to figure out is an “invariant” inner product on our Lie

algebra

. We will do so by writing down a formula, and then checking that

for a (semi-)simple Lie algebra, it is non-degenerate. Having a non-degenerate

inner product will be very useful. Amongst many things, it provides us with a

bijection between a vector space and its dual.

We recall the following definitions:

Definition

(Inner product)

Given a vector space

over

, an inner product

is a symmetric bilinear map i : V ×V → F.

Definition

(Non-degenerate inner product)

An inner product

is said to be

non-degenerate if for all v ∈ V non-zero, there is some w ∈ V such that

i(v, w) 6= 0.

The question we would like to ask is if there is a “natural” inner product on

g. We try the following:

Definition

(Killing form)

The Killing form of a Lie algebra

is the inner

product κ : g × g → F given by

κ(X, Y ) = tr(ad

◦ ad

where

is the usual trace of a linear map. Since

is linear, this is bilinear in

both arguments, and the cyclicity of the trace tells us this is symmetric.

We can try to write this more explicitly. The map

◦ad

g → g

is given

Z 7→ [X, [Y, Z]].

We pick a basis {T

}

a=1,...,D

for g. We write

X = X

, Y = Y

, Z = Z

We again let f

be the structure constants satisfying

, T

] = f

We then have

[X, [Y, Z]] = X

, [T

, T

]]

= X

= M(X, Y )

where

M(X, Y )

= X

So the trace of this thing is

κ(X, Y ) = tr(M (X, Y )) = κ

, κ

= f

Why is this a natural thing to consider? This is natural because it obeys an

invariance condition:

Definition

(Invariant inner product)

An inner product

on a Lie algebra

invariant if for any X, Y, Z ∈ g, we have

κ([Z, X], Y ) + κ(X, [Z, Y ]) = 0.

Equivalently, we have

κ(ad

X, Y ) + κ(X, ad

Y ) = 0.

What does this condition actually mean? If one were to arbitrarily write

down a definition of invariance, they might try

κ(ad

X, Y ) = κ(X, ad

Y )

instead. However, this is not the right thing to ask for.

Usually, we think of elements of the Lie algebra as some sort of “infinitesimal

transformation”, and as we have previously discussed, the adjoint representa-

tion is how an element

Z ∈ g

naturally acts on

. So under an infinitesimal

transformation, the elements X, Y ∈ g transform as

X 7→ X + ad

Y 7→ Y + ad

What is the effect on the Killing form? It transforms infinitesimally as

κ(X, Y ) 7→ κ(X + ad

X, Y + ad

Y ) ≈ κ(X, Y ) + κ(ad

X, Y ) + κ(X, ad

Y ),

where we dropped the “higher order terms” because we want to think of the

terms as being infinitesimally small (justifying this properly will require going

to the more global picture involving an actual Lie group, which we shall not go

into. This is, after all, just a motivation). So invariance of the Killing form says

it doesn’t transform under this action.

So we now check that the Killing form does satisfy the invariance condition.

Proposition. The Killing form is invariant.

Proof. We have

κ([Z, X], Y ) = tr(ad

[Z,X]

◦ ad

)

= tr([ad

, ad

] ◦ ad

)

= tr(ad

◦ ad

− ad

◦ ad

)

= tr(ad

◦ ad

) − tr(ad

◦ ad

)

Similarly, we have

κ(X, [Z, Y ]) = tr(ad

◦ ad

) − tr(ad

◦ ad

Adding them together, we obtain

κ([Z, X], Y ) + κ(X, [Z, Y ]) = tr(ad

◦ ad

) − tr(ad

◦ ad

By the cyclicity of tr, this vanishes.

The next problem is to figure out when the Killing form is degenerate. This

is related to the notion of simplicity of Lie algebras.

Definition

(Semi-simple Lie algebra)

A Lie algebra is semi-simple if it has no

abelian non-trivial ideals.

This is weaker than the notion of simplicity — simplicity requires that there

are no non-trivial ideals at all!

In fact, it is true that

being semi-simple is equivalent to

being the direct

sum of simple Lie algebras. This is on the second example sheet.

Theorem

(Cartan)

The Killing form of a Lie algebra

is non-degenerate iff

is semi-simple.

Proof.

We are only going to prove one direction — if

is non-degenerate, then

g is semi-simple.

Suppose we had an abelian ideal

a ⊆ g

. We want to show that

(

A, X

) = 0

for all

A ∈ a

and

X ∈ g

. Indeed, we pick a basis of

, and extend it to a basis of

. Then since [

X, A

]

∈ a

for all

X ∈ g

and

A ∈ a

, we know the matrix of

must look like



∗ ∗

0 ∗



Also, if

A ∈ a

, then since

is an abelian ideal,

kills everything in

, and

(X) ∈ a for all X ∈ g. So the matrix must look something like



0 ∗

0 0



So we know

◦ ad



0 ∗

0 0



and the trace vanishes. So

(

A, X

) = 0 for all

X ∈ g

and

A ∈ a

. So

= 0. So

is trivial.

Now if

is non-degenerate, then

is “invertible”. So we can find a

such that

= δ

We can then use this to raise and lower indices.

Note that this definition so far does not care if this is a real or complex

Lie algebra. From linear algebra, we know that any symmetric matrix can be

diagonalized. If we are working over

, then the diagonal entries can be whatever

we like if we choose the right basis. However, if we are working over

, then

the number of positive (or negative) diagonal entries are always fixed, while the

magnitudes can be whatever we like, by Sylvester’s law of inertia.

Thus in the case of a real Lie algebra, it is interesting to ask when the matrix

always has the same sign. It turns out it is the case with always negative sign is

the interesting case.

Definition

(Real Lie algebra of compact type)

We say a real Lie algebra is of

compact type if there is a basis such that

= −κδ

for some κ ∈ R

By general linear algebra, we can always pick a basis so that κ = 1.

The reason why it is called “compact” is because these naturally arise when

we study compact Lie groups.

We will note the following fact without proof:

Theorem.

Every complex semi-simple Lie algebra (of finite dimension) has a

real form of compact type.

We will not use it for any mathematical results, but it will be a helpful thing

to note when we develop gauge theories later on.

5.2 The Cartan basis

From now on, we restrict to the study of finite-dimensional simple complex Lie

algebras. Every time we write the symbol

or say “Lie algebra”, we mean a

finite-dimensional simple complex Lie algebra.

Recall that we have already met such a Lie algebra

(2) = span

{H, E

, E

−

}

with the brackets

[H, E

] = ±2E

, [E

, E

−

] = H.

These are known as the Cartan basis for

(2). We will try to mimic this

construction in an arbitrary Lie algebra.

Recall that when we studied

(2), we used the fact that

is a diagonal

matrix, and

acted as step operators. However, when we study Lie algebras

in general, we want to think of them abstractly, rather than as matrices, so it

doesn’t make sense to ask if an element is diagonal.

So to develop the corresponding notions, we look at the

map associated to

them instead. Recall that the adjoint map of

is also diagonal, with eigenvectors

given by

) = ±2E

(H) = 0.

This is the structure we are trying to generalize.

Definition

(

-diagonalizable)

Let

be a Lie algebra. We say that an element

X ∈ g is ad-diagonalizable if the associated map

: g → g

is diagonalizable.

Example. In su

(2), we know H is ad-diagonalizable, but E

is not.

Now we might be tempted to just look at all

-diagonalizable elements.

However, this doesn’t work. In the case of

(2), each of

, σ

diagonalizable, but we only want to pick one of them as our

. Instead, what

we want is the following:

Definition

(Cartan subalgebra)

A Cartan subalgebra

is a maximal

abelian subalgebra containing only ad-diagonalizable elements.

A Cartan subalgebra always exists, since the dimension of

is finite, and the

trivial subalgebra

{

} ⊆ g

is certainly abelian and contains only

-diagonalizable

elements. However, as we have seen, this is not necessarily unique. Fortunately,

we will later see that in fact all possible Cartan subalgebras have the same

dimension, and the dimension of h is called the rank of g.

From now on, we will just assume that we have fixed one such Cartan

subalgebra.

It turns out that Cartan subalgebras satisfy a stronger property.

Proposition.

Let

be a Cartan subalgebra of

, and let

X ∈ g

. If [

X, H

] = 0

for all H ∈ h, then X ∈ h.

Note that this does not follow immediately from

being maximal, because

maximality only says that ad-diagonalizable elements satisfy that property.

Proof. Omitted.

Example. In the case of su

(2), one possible Cartan subalgebra is

h = span

{H}.

However, recall our basis is given by

H = σ

(σ

± iσ

where the

are the Pauli matrices. Then, by symmetry, we know that

−

gives an equally good Cartan subalgebra, and so does

. So we have

many choices, but they all have the same dimension.

Now we know that everything in

commutes with each other, i.e. for any

H, H

∈ h, we have

[H, H

] = 0.

Since ad is a Lie algebra representation, it follows that

◦ ad

− ad

◦ ad

= 0.

In other words, all these

maps commute. By assumption, we know each

is diagonalizable. So we know they are in fact simultaneously diagonalizable. So

is spanned by simultaneous eigenvectors of the

. Can we find a basis of

eigenvectors?

We know that everything in

is a zero-eigenvector of

for all

H ∈ h

since for H, H

∈ h, we have

) = [H, H

] = 0.

We can arbitrarily pick a basis

: i = 1, ··· , r},

where

is the rank of

. Moreover, by maximality, there are no other eigenvectors

that are killed by all H ∈ h.

We are now going to label the remaining eigenvectors by their eigenvalue.

Given any eigenvector E ∈ g and H ∈ h, we have

(E) = [H, E] = α(H)E

for some constant

(

) depending on

(and

). We call

h → C

the root of

E. We will use the following fact without proof:

Fact.

The non-zero simultaneous eigenvectors of

are non-degenerate, i.e. there

is a unique (up to scaling) eigenvector for each root.

Thus, we can refer to this eigenvector unambiguously by

, where

designates the root.

What are these roots

? It is certainly a function

h → C

, but it is actually a

linear map! Indeed, we have

α(H + H

)E = [H + H

, E]

= [H, E] + [H

, E]

= α(H)E + α(H

= (α(H) + α(H

))E,

by linearity of the bracket.

We write Φ for the collection of all roots. So we can write the remaining

basis eigenvectors as

: α ∈ Φ}.

Example. In the case of su(2), the roots are ±2, and the eigenvectors are E

We can now define a Cartan-Weyl basis for g given by

B = {H

: i = 1, ··· , r} ∪ {E

: α ∈ Φ}.

Recall that we have a Killing form

κ(X, Y ) =

tr(ad

◦ ad

where

X, Y ∈ g

. Here we put in a normalization factor

for convenience later

on. Since g is simple, it is in particular semi-simple. So κ is non-degenerate.

We are going to evaluate κ in the Cartan-Weyl basis.

Lemma. Let H ∈ h and α ∈ Φ. Then

κ(H, E

) = 0.

Proof. Let H

∈ h. Then

α(H

)κ(H, E

) = κ(H, α(H

)

= κ(H, [H

, E

])

= −κ([H

, H], E

)

= −κ(0, E

)

= 0

But since α 6= 0, we know that there is some H

such that α(H

) 6= 0.

Lemma. For any roots α, β ∈ Φ with α + β 6= 0, we have

κ(E

, E

) = 0.

Proof. Again let H ∈ h. Then we have

(α(H) + β(H))κ(E

, E

) = κ([H, E

], E

) + κ(E

, [H, E

]),

= 0

where the final line comes from the invariance of the Killing form. Since

does not vanish by assumption, we must have κ(E

, E

) = 0.

Lemma. If H ∈ h, then there is some H

∈ h such that κ(H, H

) 6= 0.

Proof.

Given an

, since

is non-degenerate, there is some

X ∈ g

such that

(

H, X

)

= 0. Write

, where

∈ h

and

is in the span of the

0 6= κ(H, X) = κ(H, H

) + κ(H, E) = κ(H, H

What does this tell us?

started life as a non-degenerate inner product

. But now we know that

is a non-degenerate inner product on

. By

non-degeneracy, we can invert it within h.

In coordinates, we can find some κ

such that

κ(e

, e

) = κ

for any

. The fact that the inner product is non-degenerate means that we

can invert the matrix κ, and find some (κ

−1

)

such that

(κ

−1

)

= δ

Since

−1

is non-degenerate, this gives a non-degenerate inner product on

∗

In particular, this gives us an inner product between the roots! So given two

roots α, β ∈ Φ ⊆ h

∗

, we write the inner product as

(α, β) = (κ

−1

)

where

(

). We will later show that the inner products of roots will

always be real, and hence we can talk about the “geometry” of roots.

We note the following final result:

Lemma. Let α ∈ Φ. Then −α ∈ Φ. Moreover,

κ(E

, E

−α

) 6= 0

This holds for stupid reasons.

Proof. We know that

κ(E

, E

) = κ(E

, H

) = 0

for all

β 6

−α

and all

. But

is non-degenerate, and

, H

}

span

. So

there must be some E

−α

in the basis set, and

κ(E

, E

−α

) 6= 0.

So far, we know that

, H

] = 0

, E

] = α

for all α ∈ Φ and i, j = 1, ··· , r. Now it remains to evaluate [E

, E

Recall that in the case of su

(2), we had

, E

−

] = H.

What can we get here? For any H ∈ h and α, β ∈ Φ, we have

[H, [E

, E

]] = −[E

, [E

, H]] − [E

, [H, E

]]

= (α(H) + β(H))[E

, E

Now if α + β 6= 0, then either [E

, E

] = 0, or α + β ∈ Φ and

, E

] = N

α,β

α+β

for some N

α,β

What if

= 0? We claim that this time, [

, E

−α

]

∈ h

. Indeed, for any

H ∈ h, we have

[H, [E

, E

−α

]] = [[H, E

], E

−α

] + [[E

−α

, H], E

]

= α(H)[E

, E

−α

] + α(H)[E

−α

, E

]

= 0.

Since

was arbitrary, by the (strong) maximality property of

, we know that

, E

−α

] ∈ h.

Now we can compute

κ([E

, E

−α

], H) = κ(E

, [E

−α

, H])

= α(H)κ(E

, E

−α

We can view this as an equation for [

, E

−α

]. Now since we know that

[

, E

−α

]

∈ h

, and the Killing form is non-degenerate when restricted to

, we

know [E

, E

−α

] is uniquely determined by this relation.

We define the normalized

, E

−α

]

κ(E

, E

−α

)

Then our equation tells us

κ(H

, H) = α(H).

Writing H

and H in components:

= e

, H = e

the equation reads

= α

Since the e

are arbitrary, we know

= (κ

−1

)

So we know

= (κ

−1

)

We now have a complete set of relations:

Theorem.

, H

] = 0

, E

] = α

, E

] =











α,β

α+β

α + β ∈ Φ

κ(E

, E

α + β = 0

0 otherwise

Now we notice that there are special elements

in the Cartan subalgebra

associated to the roots. We can compute the brackets as

, E

] = (κ

−1

)

, E

]

= (κ

−1

)

= (α, β)E

where we used the inner product on the dual space

∗

induced by the Killing

form κ.

Note that so far, we have picked

and

arbitrarily. Any scalar multiple

of them would have worked as well for what we did above. However, it is often

convenient to pick a normalization such that the numbers we get turn out to

look nice. It turns out that the following normalization is useful:

(α, α)κ(E

, E

−α

)

(α, α)

Here it is important that (

α, α

)

= 0, but we will only prove it next chapter where

the proof fits in more naturally.

Under this normalization, we have

, h

] = 0

, e

] =

2(α, β)

(α, α)

, e

] =











αβ

α+β

α + β ∈ Φ

α + β = 0

0 otherwise

Note that the number of roots is

d −r

, where

is the dimension of

and

is its

rank, and this is typically greater than

. So in general, there are too many of

them to be a basis, even if they spanned

(we don’t know that yet). However,

we are still allowed to talk about them, and the above relations are still true.

It’s just that they might not specify everything about the Lie algebra.

5.3 Things are real

We are now going to prove that many things are real. When we do so, we can

then restrict to a real form of h, and then we can talk about their geometry.

The first result we want to prove in this section is the following:

Theorem.

(α, β) ∈ R

for all α, β ∈ Φ.

To prove this, we note that any Lie algebra contains many copies of

(2).

α ∈

Φ, then

−α ∈

Φ. For each pair

±α ∈

Φ, we can consider the subalgebra

with basis {h

, e

−α

}. Then we have

, e

±α

] = ±2e

±α

, e

−α

] = h

These are exactly the relations for the Lie algebra of

(2) (or rather,

(2),

but we will not write the

every time). So this gives a subalgebra isomorphic

to su(2). We call this su(2)

It turns out we also get a lot of representations of su(2)

Definition

(String)

For

α, β ∈

Φ, we define the

-string passing through

α,β

= {β + ρα ∈ Φ : ρ ∈ Z}.

We will consider the case where

is not proportional to

. The remaining

case, where we may wlog take

= 0, is left as an exercise in the example sheet.

We then have a corresponding vector subspace

α,β

= span

β+ρα

: β + ρα ∈ S

α,β

Consider the action of su(2)

on V

α,β

. We have

, e

β+ρα

] =

2(α, β + ρα)

(α, α)

β+ρα



2(α, β)

(α, α)

+ 2ρ



β+ρα

. (∗)

We also have

±α

, e

β+ρα

] ∝

(

β+(ρ±1)α

β + (ρ ± 1)α ∈ Φ

0 otherwise

∈ V

α,β

is invariant under the action of

(2)

. So

α,β

is the representation

space for some representation of su(2)

Moreover, (∗) tells us the weight set of this representation. We have

S =



2(α, β)

(α, α)

+ 2ρ : β + ρα ∈ Φ : ρ ∈ Z



This is not to be confused with the string S

α,β

itself!

Since the Lie algebra itself is finite-dimensional, this representation must

be finite-dimensional. We also see from the formula that the weights are non-

degenerate and are spaced by 2. So it must be an irreducible representation of

su(2). Then if its highest weight is Λ ∈ Z, we have

S = {−Λ, −Λ + 2, ··· , +Λ}.

What does this tell us? We know that the possible values of

are bounded

above and below, so we can write

α,β

= {β + ρα ∈ Φ : n

−

≤ ρ ≤ n

for some n

∈ Z. In particular, we know that

Proposition. For any α, β ∈ Φ, we have

2(α, β)

(α, α)

∈ Z.

Proof. For ρ = n

, we have

2(α, β)

(α, α)

+ 2n

−

= −Λ and

2(α, β)

(α, α)

+ 2n

= Λ.

Adding the two equations yields

2(α, β)

(α, α)

= −(n

+ n

−

) ∈ Z.

We are next going to prove that the inner products themselves are in fact

real. This will follow from the following lemma:

Lemma. We have

(α, β) =

δ∈Φ

(α, δ)(δ, β),

where N is the normalization factor appearing in the Killing form

κ(X, Y ) =

tr(ad

◦ ad

Proof. We pick the Cartan-Weyl basis, with

, E

] = δ

for all i = 1, ··· , r and δ ∈ Φ. Then the inner product is defined by

= κ(H

, H

) =

tr[ad

◦ ad

But we know that these matrices

are diagonal in the Cartan-Weyl basis,

and the non-zero diagonal entries are exactly the δ

. So we can write this as

δ∈Φ

Now recall that our inner product was defined by

(α, β) = α

(κ

−1

)

= κ

where we define

= (κ

−1

)

Putting in our explicit formula for the κ

, this is

(α, β) =

δ∈Φ

(α, δ)(δ, β).

Corollary.

(α, β) ∈ R

for all α, β ∈ Φ.

Proof. We write

α,β

2(α, β)

(α, α)

∈ Z.

Then the previous formula tells us

(β, β)

α,β

δ∈Φ

α,δ

β,δ

We know that

α,β

are all integers, and in particular real. So (

β, β

) must be

real as well. So it follows that (α, β) is real since R

α,β

is an integer.

5.4 A real subalgebra

Let’s review what we know so far. The roots

α ∈

Φ are elements of

∗

. In

general, the roots will not be linearly independent elements of

∗

, because there

are too many of them. However, it is true that the roots span h

∗

Proposition. The roots Φ span h

∗

. In particular, we know

|Φ| ≥ dim h

∗

Proof.

Suppose the roots do not span

∗

. Then the space spanned by the roots

would have a non-trivial orthogonal complement. So we can find

λ ∈ h

∗

such

that (λ, α) = 0 for all α ∈ Φ. We now define

= λ

∈ h.

Then as usual we have

, H] = 0 for all H ∈ h.

Also, we know

, E

] = (λ, α)E

= 0.

for all roots

α ∈

Φ by assumption. So

commutes with everything in the Lie

algebra. This would make

a non-trivial ideal, which is a contradiction since

g is simple.

Since the

α ∈

Φ span, we can find a basis

{α

(i)

∈

Φ :

= 1

, ··· , r}

of roots.

Again, this choice is arbitrary. We now define a real vector subspace

∗

⊆ h

∗

= span

{α

(i)

: i = 1, ··· , r}.

One might be worried that this would depend on our choice of the basis

(i)

which was arbitrary. However, it does not, since any choice will give us the same

space:

Proposition. h

∗

contains all roots.

So h

∗

is alternatively the real span of all roots.

Proof.

We know that

∗

is spanned by the

(i)

as a complex vector space. So

given an β ∈ h

∗

, we can find some β

∈ C such that

β =

i=1

(i)

Taking the inner product with α

(j)

, we know

(β, α

(j)

) =

i=1

(α

(i)

, α

(j)

We now use the fact that the inner products are all real! So

is the solution to

a set of real linear equations, and the equations are non-degenerate since the

(i)

form a basis and the Killing form is non-degenerate. So

must be real. So

β ∈ h

∗

Now the inner product of any two elements of

∗

is real, since an element in

∗

is a real linear combination of the

(i)

, and the inner product of the

(i)

always real.

Proposition.

The Killing form induces a positive-definite inner product on

∗

Proof.

It remains to show that (

λ, λ

)

≥

0 for all

, with equality iff

= 0. We

can write

(λ, λ) =

δ∈Φ

(λ, δ)

≥ 0.

If this vanishes, then (

λ, δ

) = 0 for all

δ ∈

Φ. But the roots span, so this implies

that λ kills everything, and is thus 0 by non-degeneracy.

Now this

∗

is just like any other real inner product space we know and love

from IA Vectors and Matrices, and we can talk about the lengths and angles

between vectors.

Definition (Norm of root). Let α ∈ Φ be a root. Then its length is

|α| =

(α, α) > 0.

Then for any α, β, there is some “angle” ϕ ∈ [0, π] between them given by

(α, β) = |α||β|cos ϕ.

Now recall that we had the quantization result

2(α, β)

(α, α)

∈ Z.

Then in terms of the lengths, we have

2|β|

|α|

cos ϕ ∈ Z.

Since the quantization rule is not symmetric in

α, β

, we obtain a second quanti-

zation constraint

2|α|

|β|

cos ϕ ∈ Z.

Since the product of two integers is an integer, we know that

4 cos

ϕ ∈ Z.

cos ϕ = ±

√

, n = {0, 1, 2, 3, 4}.

So we have some boring solutions

ϕ = 0,

, π,

and non-boring ones

ϕ =

2π

3π

5π

These are the only possibilities!

5.5 Simple roots

Now we have a lot of roots that span

∗

, but they obviously do not form a basis.

For example, whenever

α ∈ h

∗

, then so is

−α

. So it might be helpful to get rid

of half of these things. The divide is again rather arbitrary.

To do so, we pick a hyperplane in

∗

∼

, i.e. a subspace of dimension

r −

Since there are finitely many roots, we can pick it so that it doesn’t contain any

root. Then the plane divides

∗

into 2 sides. We then pick a distinguished side,

and say that α is “positive” if it lies in that side, and negative otherwise.

Then if

is positive,

−α

is negative. More interestingly, if

α, β

are positive

roots, then

is also positive. Similarly, if they are both negative, then the

sum is also negative.

It turns out restricting to positive roots is not enough. However, the following

trick does the job:

Definition

(Simple root)

A simple root is a positive root that cannot be written

as a sum of two positive roots. We write Φ

for the set of simple roots.

We can immediately deduce a few properties about these simple roots.

Proposition.

Any positive root can be written as a linear combination of simple

roots with positive integer coefficients. So every root can be written as a linear

combination of simple roots.

Proof.

Given any positive root, if it cannot be decomposed into a positive sum

of other roots, then it is simple. Otherwise, do so, and further decompose the

constituents. This will have to stop because there are only finitely many roots,

and then you are done.

Corollary. The simple roots span h

∗

To show that they are independent, we need to do a bit of work that actually

involves Lie algebra theory.

Proposition. If α, β ∈ Φ are simple, then α − β is not a root.

Proof.

Suppose

α − β

were a root. By swapping

and

if necessary, we may

wlog assume that α − β is a positive root. Then

α = β + (α − β)

is a sum of two positive roots, which is a contradiction.

Proposition. If α, β ∈ Φ

, then the α-string through β, namely

α,β

= {β + nα ∈ Φ},

has length



αβ

= 1 −

2(α, β)

(α, α)

∈ N.

Proof. Recall that there exists n

such that

α,β

= {β + nα : n

−

≤ n ≤ n

We have shown before that

+ n

−

= −

2(α, β)

(α, α)

∈ Z.

In the case where

α, β

are simple roots, we know that

β − α

is not a root. So

−

≥ 0. But we know β is a root. So we know that n

−

= 0 and hence

= −

2(α, β)

(α, α)

∈ N.

So there are

+ 1 = 1 −

2(α, β)

(α, α)

things in the string.

From this formula, we learn that

Corollary. For any distinct simple roots α, β, we have

(α, β) ≤ 0.

We are now in a position to check that we do get a basis this way.

Proposition. Simple roots are linearly independent.

Proof.

Suppose we have a non-trivial linear combination

of the simple roots.

We write

λ = λ

− λ

−

i∈I

(i)

−

j∈I

−

(j)

where

, b

≥

0 and

, I

−

are disjoint. If

−

is empty, then the sum is a positive

root, and in particular is non-zero. Similarly, if

is empty, then the sum is

negative. So it suffices to focus on the case c

, b

> 0.

Then we have

(λ, λ) = (λ

, λ

) + (λ

−

, λ

−

) − 2(λ

, λ

−

)

> −2(λ

, λ

−

)

= −2

i∈I

j∈I

−

(α

(i)

, α

(j)

)

≥ 0,

since (

(i)

, α

(j)

)

≤

0 for all simple roots

(i)

, α

(j)

. So in particular

is non-

zero.

Corollary. There are exactly r = rank g simple roots roots, i.e.

|Φ

| = r.

5.6 The classification

We now have a (not so) canonical choice of basis of h

∗

B = {α ∈ Φ

} = {α

(i)

: i = 1, ··· , r}.

We want to re-express the Lie algebra in terms of this basis.

Definition

(Cartan matrix)

The Cartan matrix

is defined as the

r × r

matrix

2(α

(i)

, α

(j)

)

(α

(j)

, α

(j)

)

Note that this is not a symmetric matrix. We immediately know that

∈ Z

For each root α

(i)

, we have an su(2) subalgebra given by

= h

(i)

, e

= e

±α

(i)

We then have

, e

] = ±2e

, [e

, e

−

] = h

Looking at everything in our Lie algebra, we have the relations

, h

] = 0

, e

] = ±A

, e

−

] = δ

with no summation implied in the second row. Everything here we’ve seen,

except for the last expression [

, e

−

]. We are claiming that if

i 6

, then the

bracket in fact vanishes! This just follows from the fact that if

(i)

and

(j)

are

simple, then α

(i)

− α

(j)

is not a root.

Note that this does not a priori tell us everything about the Lie algebra. We

have only accounted for step generators of the form

, and also we did not

mention how, say, [e

, e

] behaves.

We know

, e

] = ad

) ∝ e

(i)

+α

(j)

(i)

(j)

∈

Φ. We know that if

(i)

(j)

∈

Φ, then it belongs to a string.

In general, we have

) ∝ e

(j)

+nα

(i)

if α

(j)

+ nα

(i)

∈ Φ, and we know how long this string is. So we have

(ad

)

1−A

= 0.

This is the Serre relation. It turns out this is all we need to completely charac-

terize the Lie algebra.

Theorem

(Cartan)

Any finite-dimensional, simple, complex Lie algebra is

uniquely determined by its Cartan matrix.

We will not prove this.

To achieve the Cartan classification, we need to classify all Cartan matrices,

and then reconstruct g form the Cartan matrix.

We first do the first part. Recall that

2(α

(i)

, α

(j)

)

(α

(j)

, α

(j)

)

∈ Z.

We first read off properties of the Cartan matrix.

Proposition. We always have A

= 2 for i = 1, ··· , r.

This is not a symmetric matrix in general, but we do have the following:

Proposition. A

= 0 if and only if A

= 0.

We also have

Proposition. A

∈ Z

≤0

for i 6= j.

We now get to a less trivial fact:

Proposition. We have det A > 0.

Proof. Recall that we defined our inner product as

(α, β) = α

−1

β,

and we know this is positive definite. So we know

det κ

−1

0. We now write

the Cartan matrix as

A = κ

−1

where we have

(α

(j)

, α

(j)

)

Then we have

det D =

(α

(j)

, α

(j)

)

> 0.

So it follows that

det A = det κ

−1

det D > 0.

It is an exercise to check that if the Cartan matrix is reducible, i.e. it looks

A =



(1)

0 A

(2)



then in fact the Lie algebra is not simple.

So we finally have

Proposition. The Cartan matrix A is irreducible.

So in total, we have the following five properties:

Proposition.

(i) A

= 2 for all i.

(ii) A

= 0 if and only if A

= 0.

(iii) A

∈ Z

≤0

for i 6= j.

(iv) det A > 0.

(v) A is irreducible.

These are hugely restrictive.

Example.

For a rank 1 matrix, there is only one entry, and it is on the diagonal.

So we must have

A =





This corresponds to the simple Lie algebra su(2).

Example.

For a rank 2 matrix, we know the diagonals must be 2, so we must

have something of the form

A =



2 m

 2



We know that the off-diagonals are not all zero, but if one is non-zero, then the

other also is. So they must be both non-zero, and we have

m <

4. We thus

have

(m, ) = (−1, −1), (−1, −2), (−1, −3).

Note that we do not write out the cases where we swap the two entries, because

we will get the same Lie algebra but with a different ordering of the basis.

Now we see that we have a redundancy in the description of the Cartan

matrix given by permutation. There is a neat solution to this problem by drawing

Dynkin diagrams.

Definition

(Dynkin diagram)

Given a Cartan matrix, we draw a diagram as

follows:

(i) For each simple root α

(i)

∈ Φ

, we draw a node

(ii) We join the nodes corresponding to α

(i)

, α

(j)

with A

many lines.

(iii)

If the roots have different lengths, we draw an arrow from the longer root

to the shorter root. This happens when A

6= A

Note that we wouldn’t have to draw too many lines. We have

2|α

(i)

|α

(j)

cos ϕ

, A

2|α

(j)

|α

(i)

cos ϕ

where ϕ

is the angle between them. So we have

cos

But we know cos

∈ [0, 1]. So we must have

∈ {0, 1, 2, 3}.

So we have to draw at most 3 lines, which isn’t that bad. Moreover, we have the

following information:

Proposition. A simple Lie algebra has roots of at most 2 distinct lengths.

Proof. See example sheet.

It is an exercise to see that all the information about

can be found from

the Dynkin diagram.

We can now revisit the case of a rank 2 simple Lie algebra.

Example. The Cartan matrices of rank 2 are given by



2 −1

−1 2





2 −2

−1 2





2 −3

−1 2



These correspond to the Dynkin diagrams

The conditions on the matrices

now translate to conditions on the Dynkin

diagrams which we will not write out, and it turns out we can classify all the

allowed Dynkin diagrams as follows:

Theorem

(Cartan classification)

The possible Dynkin diagrams include the

following infinite families (where n is the number of vertices):

And there are also five exceptional cases:

Example.

The infinite families

, B

, C

, D

correspond to well-known com-

plex Lie groups by

Family Lie Group

(SU(n + 1))

(SO(2n + 1))

(Sp(2n))

(SO(2n))

where the Sp(2n) are the symplectic matrices.

Note that there is some repetition in our list. For example, we have

and

. Also,

does not give a simple Lie algebra, since

it is disconnected. We have D

∼

⊕ A

. So we have

(SO(4))

∼

(SU(2)) ⊕ L

(SU(2)).

Finally, we also have D

= A

, and this reflects an isomorphism

(SU(4))

∼

(SO(6)).

Hence, a list without repetitions is given by

for n ≥ 1,

for n ≥ 2,

for n ≥ 3,

for n ≥ 4.

This classification is very important in modern theoretical physics, since in

many theories, we need to pick a Lie group as, say, our gauge group. So knowing

what Lie groups are around lets us know what theories we can have.

5.7 Reconstruction

Now given the Cartan matrix, we want to reconstruct the Lie algebra itself.

Recall that we had a Cartan-Weyl basis

, E

: i = 1, ··· , r : α ∈ Φ}.

The first thing to do in the reconstruction is to figure out what the set of roots

is.

By definition, the Cartan matrix determines simple roots

(i)

for

= 1

, ··· , r

We can read off the inner products from the Cartan matrix as

2(α

(i)

, α

(j)

)

(α

(j)

, α

(j)

)

2|α

(i)

|α

(j)

cos ϕ

This allows us to find the simple roots.

How about the other roots? We can find them by considering the root strings,

as we know that the length of the α

(i)

-string through α

(j)

is given by



= 1 − A

∈ N.

So we can work out the length of the root string from each simple root. By

Cartan’s theorem, this gives us all roots.

Instead of going through a formal and general reconstruction, we do an

example.

Example. Consider g = A

. We have a Dynkin diagram

So we see that the Cartan matrix is

A =



2 −1

−1 2



So we know that g has two simple roots α, β ∈ Φ, and we know

= A

2(α, β)

(α, α)

2(β, α)

(β, β)

= −1.

So we find that (α, α) = (β, β), and ϕ

αβ

2π

. So we can display the roots as

2π

Since α, β ∈ Φ

, we know ±(α − β) 6∈ Φ, and also we have



α,β

= 

β,α

= 1 −

2(α, β)

(α, α)

= 2.

So we have roots

nα, α

˜nβ

for

n, ˜n ∈ {

}

. So in fact our list of roots

contains

α, β, α + β ∈ Φ.

These are the positive ones. We know each of these roots has a negative

counterpart. So we have more roots

−α, −β, −α − β ∈ Φ.

Then by our theorem, we know these are all the roots. We can find the length

of α + β by

(α + β, α + β) = (α, α) + (β, β) + 2(α, β)

= (α, α) + (β, β) − (β, β)

= (α, α).

So in fact all roots have the same length. So if we draw all our roots, we get

α + β

−α

−β−α − β

So the Cartan-Weyl basis consists of

= {H

, H

, E

±α

, E

±β

, E

±(α+β)

This is the complete Cartan-Weyl basis of

. So this has dimension 2 + 6 = 8.

If we want to find out the brackets of these things, we look at the ones we

have previously written down, and use the Jacobi identity many times.

6 Representation of Lie algebras

We now try to understand the possible representations of Lie algebras. It turns

out they again are pretty restricted.

6.1 Weights

Let

be a representation of

. Then it is completely determined by the

images

7→ ρ(H

) ∈ gl(V )

7→ ρ(E

) ∈ gl(V ).

Again, we know that

[ρ(H

), ρ(H

)] = ρ([H

, H

]) = 0.

So all

(

) commute, and by linear algebra, we know they have a common

eigenvector v

. Again, for each H ∈ h, we know

ρ(H)v

= λ(H)v

for some λ(H), and λ lives in h

∗

Definition

(Weight of representation)

Let

g → gl

(

) be a representation

. Then if

∈ V

is an eigenvector of

(

) for all

H ∈ h

, we say

λ ∈ h

∗

is a

weight of ρ, where

ρ(H)v

= λ(H)v

for all H ∈ h.

The weight set S

of ρ is the set of all weights.

For a weight

, we write

for the subspace that consists of vectors

such

that

ρ(H)v = λ(H)v.

Note that the weights can have multiplicity, i.e.

need not be 1-dimensional.

We write

= dim V

≥ 1

for the multiplicity of the weight.

Example. By definition, we have

, E

] = α

So we have

= α

In terms of the adjoint representation, this says

adj

= α

So the roots α are the weights of the adjoint representation.

Recall that for

(2), we know that the weights are always integers. This

will be the case in general as well.

Given any representation

, we know it has at least one weight

with an

eigenvector

v ∈ V

. We’ll see what happens when we apply the step operators

ρ(E

) to it, for α ∈ Φ. We have

ρ(H

)ρ(E

)v = ρ(E

)ρ(H

)v + [ρ(H

), ρ(E

)]v.

We know that

[ρ(H

), ρ(E

)] = ρ([H

, E

]) = α

ρ(E

So we know that if v ∈ V

, then

ρ(H

)ρ(E

)v = (λ

+ α

)ρ(E

)v.

So the weight of the vector has been shifted by

. Thus, for all vectors

v ∈ V

we find that

ρ(E

)v ∈ V

λ+α

However, we do not know a priori if

λ+α

is a thing at all. If

λ+α

{

}

, i.e.

λ + α is not a weight, then we know that ρ(E

)v = 0.

So the Cartan elements

preserve the weights, and the step operators

increment the weights by α.

Consider the action of our favorite

(2)

subalgebra on the representation

space. In other words, we consider the action of

{ρ

(

)

, ρ

(

)

, ρ

(

−α

)

}

Then

becomes the representation space for some representation

for

(2)

Now we can use what we know about the representations of

(2) to get

something interesting about V . Recall that we defined

(α, α)

(κ

−1

)

So for any v ∈ V

, after some lines of algebra, we find that

ρ(h

)(v) =

2(α, λ)

(α, α)

So we know that we must have

2(α, λ)

(α, α)

∈ Z

for all

λ ∈ S

and

α ∈

Φ. So in particular, we know that (

(i)

, λ

)

∈ R

for all

simple roots

(i)

. In particular, it follows that

λ ∈ h

∗

(if we write

as a sum

of the

(i)

, then the coefficients are the unique solution to a real system of

equations, hence are real).

Again, from these calculations, we see that

λ∈S

is a subrepresentation

, so by irreducibility, everything is a linear combination of these simultaneous

eigenvectors, i.e. we must have

V =

λ∈S

In particular, this means all ρ(H) are diagonalizable for H ∈ h.

6.2 Root and weight lattices

Recall that the simple roots are a basis for the whole root space. In particular,

any positive root can be written as a positive sum of simple roots. So for any

β ∈ Φ, we can write

β =

i=1

(i)

where β

∈ Z for i = 1, ··· , r. Hence all roots lie in a root lattice:

Definition

(Root lattice)

Let

be a Lie algebra with simple roots

(i)

. Then

the root lattice is defined as

L[g] = span

{α

(i)

: i = 1, ··· , r}.

Note that here we are taking the integer span, not the real span.

What we want to do is to define an analogous weight lattice, where all possible

weights lie.

We start by defining some funny normalization of the roots:

Definition (Simple co-root). The simple co-roots are

ˇα

(i)

2α

(i)

(α

(i)

, α

(i)

)

Similarly, we can define the co-root lattice:

Definition

(Co-root lattice)

Let

be a Lie algebra with simple roots

(i)

Then the co-root lattice is defined as

L[g] = span

{ˇα

(i)

: i = 1, ··· , r}.

Definition

(Weight lattice)

The weight lattice

[

] is the dual to the co-root

lattice:

[g] =

∗

[g] = {λ ∈ h

∗

: (λ, µ) ∈ Z for all µ ∈

L[g]}.

So we know λ ∈ L

[g] iff for all α

(i)

, we have

(λ, ˇα

(i)

) =

2(α

(i)

, λ)

(α

(i)

, α

(i)

)

∈ Z.

So by what we have previously computed, we know that all weights

λ ∈ S

are

in L

[g].

Given the basis

B = {ˇα

(i)

: i = 1, ··· , r}

for

L[g], we define the dual basis

∗

= {ω

(i)

: i = 1, ··· , r}

by demanding

(ˇα

(i)

, ω

(j)

) =

2(α

(i)

, ω

(j)

)

(α

(i)

, α

(i)

)

= δ

These

(i)

are known as the fundamental weights. As the simple roots span

∗

we can write

(i)

j=1

(j)

for some B

∈ R. Plugging this into the definition, we find

k=1

2(α

(i)

, α

(k)

)

(α

(i)

, α

(i)

)

= δ

So we know

k=1

= δ

i.e. B is the inverse of the Cartan matrix A. Thus we can write

(i)

j=1

(j)

Example. Consider g = A

. Then we have

A =



2 −1

−1 2



Then we have

α = α

(1)

= 2ω

(1)

− ω

(2)

β = α

(2)

= −ω

(1)

+ 2ω

(2)

So we have

(1)

(2α + β)

(2)

(α + 2β).

We can then draw the weight lattice:

(2)

(1)

Note that the roots are also in the weight lattice, because they are weights of

the adjoint representation.

Given any weight λ ∈ S

⊆ L

[g], we can write

λ =

i=1

(i)

for some λ

∈ Z. These {λ

} are called Dynkin labels of the weight λ.

6.3 Classification of representations

Suppose we have a representation

. We can start with any eigenvector

v ∈ V

. We then keep applying things of the form

(

), where

α ∈

is a

positive weight. Doing so will bring us further away from the hyperplane dividing

the positive and negative weights, and since there are only finitely many weights,

we must eventually stop. This is known as a highest weight.

Definition

(Highest weight)

A highest weight of a representation

is a weight

Λ ∈ S

whose associated eigenvector v

is such that

ρ(E

= 0

for all α ∈ Φ

The Dynkin labels (or indices) of a representation are the Dynkin labels of

its highest weight.

Now if v ∈ V

, then we know

ρ(E

)v ∈ V

λ+α

α ∈ S

(and vanishes otherwise). So acting with the step operators

translates us in the weight lattice by the vector corresponding to the root.

In the case of

(2), we found out an explicit formula for when this will stop.

In this general case, we have the following result which we will not prove:

Theorem. For any finite-dimensional representation of g, if

λ =

i=1

(i)

∈ S

then we know

λ − m

(i)

∈ S

for all m

(i)

∈ Z and 0 ≤ m

(i)

≤ λ

If we know further that

is irreducible, then we can in fact obtain all weights

by starting at the highest weight and applying this procedure.

Moreover, for any

Λ =

(i)

∈ L

[g],

this is the highest weight of some irreducible representation if and only if Λ

≥

for all i.

This gives us a very concrete way of finding, at least the weights, of an

irreducible representation. In general, though, we don’t immediately know the

multiplicities of the weights.

Definition

(Dominant integral weight)

A dominant integral weight is a weight

Λ =

(i)

∈ L

[g],

such that Λ

≥ 0 for all i.

Example.

Take the example of

. It is a fact that the fundamental

representation f has Dynkin labels (1, 0). In other words,

Λ = ω

(1)

To get the remaining weights, we subtract roots:

(i) Λ = ω

(1)

∈ S

. So we can subtract by α

(1)

exactly once.

(ii) So

λ = ω

(1)

− (2ω

(1)

− ω

(2)

)

= −ω

(1)

+ ω

(2)

∈ S

This has Dynkin labels (

−

1). So we can do one further subtraction to

get a new weight

λ − α

(2)

= −ω

(1)

+ ω

(2)

− (2ω

(2)

− ω

(1)

)

= −ω

(2)

In the weight diagram, these are given by

(2)

(1)

(1, 0)

(−1, 1)

(0, −1)

In general, for a dominant integral weight of the form

Λ = Λ

(1)

+ Λ

(2)

∈ L

]

for Λ

, Λ

∈ Z

≥0

. We then get an irrep ρ

(Λ

,Λ

)

of A

One can show, via characters, that the dimension of this representation is

given by

dim ρ

(Λ

,Λ

)

(Λ

+ 1)(Λ

+ Λ

+ 2).

This is symmetric in Λ

and Λ

. So if Λ

= Λ

, then we get a pair of distinct

representations of the same dimension. It is a fact that

λ ∈ S

(Λ

,Λ

)

⇔ −λ ∈ S

(Λ

,Λ

)

and we have

(Λ

,Λ

)

= ¯ρ

(Λ

,Λ

)

On the other hand, if Λ

= Λ

, then this representation is self-conjugate.

The first few representations of A

are given by

Highest weight Dimension Name

(0,0)

1 Trivial

(1,0)

3 Fundamental

(0,1)

3 Anti-fundamental

(2,0)

(0,2)

(1,1)

8 Adjoint

We can figure out the weights of the adjoint representation with Λ = (1

∈ S

Since both labels are positive, we can subtract both α

(1)

and α

(2)

to get

Λ − α

(1)

= (−1, 2)

Λ − α

(2)

= (2, −1)

(2)

(1)

(1, 1)(−1, 2)

(2, −1)

Now we have a +2 in the Dynkin labels. So we can subtract twice. So we have

the following weights:

Λ − α

(1)

− α

(2)

= (0, 0)

Λ − α

(1)

− 2α

(2)

= (1, −2)

Λ − 2α

(1)

− α

(2)

= (−2, 1)

(2)

(1)

(1, 1)(−1, 2)

(2, −1)

(0, 0)

(1, −2)

(−2, 1)

Finally, we can subtract α

(1)

from the second or α

(2)

from the third to get

Λ − 2α

(1)

− 2α

(2)

= (−1, −1) ∈ S

(2)

(1)

(1, 1)(−1, 2)

(2, −1)

(0, 0)

(−2, 1)

(−1, −1)

Now notice that we only have seven weights, not eight. This is because our

algorithm did not tell us the multiplicity of the weights. However, in this

particular case, we know, because this is the adjoint representation, and the

Cartan generators give us two things of weight 0. So we can plot all weights

with multiplicity as

It is a general fact that the weight lattice traces out a polygon like this, and

the vertices at the perimeter are always non-degenerate. So in this case, we

could also have figured out the multiplicty by using this general fact.

6.4 Decomposition of tensor products

The last thing we need to know how to do is how we can decompose tensor

products of irreps. This is not hard. It is just what we did for

(2) but in a

more general context.

We let

and

be irreps of

with corresponding representation spaces

and V

. Then we can write

λ∈S

, V

∈S

If v

∈ V

and v

∈ V

, then for H ∈ h, we have

(ρ

⊗ ρ

)(H)(v

⊗ v

) = ρ

(H)v

⊗ v

+ v

⊗ ρ

(H)v

= λ(H)v

⊗ v

+ λ

(H)v

⊗ v

= (λ + λ

)(H)v

⊗ v

Since things of the form v

⊗ v

span V

⊗ V

, we know

Λ⊗Λ

= {λ + λ

: λ ∈ S

, λ

∈ S

}

with multiplicities. To figure out the decomposition, we find out what the highest

weight of the tensor product is, and then subtract off the factors corresponding

to the highest weight, and then repeat.

Example.

Consider

, and the fundamental representation

(1,0)

. We

consider

(1,0)

⊗ ρ

(1,0)

Recall that the weight set is given by

(1,0)

= {ω

(1)

, −ω

(1)

+ ω

(2)

, −ω

(2)

We can then draw the collection of all weights in ρ

(1,0)

⊗ ρ

(1,0)

as follows:

So we see that we have a highest weight of (2

0). We then use the algorithm to

compute all the weights of such an irrep, remove it from the weight lattice, and

be left with

This is just the anti-fundamental representation. So we have

3 ⊗ 3 = 6 ⊕

7 Gauge theories

We are now going to do something rather amazing. We are going to show that

Maxwell’s equation is “forced” upon us by requiring that our (charged) fields

possess something known as a U(1) gauge symmetry. We are going to write

down essentially the only possible theory under the assumption of this gauge

symmetry, and unwrapping it gives us Maxwell’s equation.

When we do it for Maxwell’s equation, it is not entirely clear that what we

wrote down was “the only possible thing”, but it will be when we try to do it

for general gauge theories.

Before we begin, it is important to note that everything we do here has a very

nice geometric interpretation in terms of connections on principal

-bundles,

and there are very good geometric reasons to pick the definitions (and names) we

use here. Unfortunately, we do not have the time to introduce all the necessary

geometry in order to do so.

7.1 Electromagnetism and U(1) gauge symmetry

In electromagnetism, we had two fields

and

. There are four Maxwell’s

equations governing how they behave. Two of them specify the evolution of the

field and how they interact with matter, while the other two just tells us we can

write the field in terms of a scalar and a vector potential Φ

, A

. Explicitly, they

are related by

E = −∇Φ +

∂A

∂t

B = ∇ × A.

The choice of potentials is not unique. We know that

and

are invariant

under the transformations

Φ 7→ Φ +

∂α

∂t

A 7→ A + ∇α

for any gauge function α = α(x, t).

Now we believe in relativity, so we should write things as 4-vectors. It so

happens that there are four degrees of freedom in the potentials, so we can

produce a 4-vector





We can then succinctly write the gauge transformations as

7→ a

+ ∂

α.

We can recover our fields via something known as the electromagnetic field tensor,

given by

µν

= ∂

− ∂

It is now easy to see that

is invariant under gauge transformations. We can

expand the definitions out to find that we actually have

µν







0 E

−E

0 −B

−E

0 −B

−E

−B







So we do recover the electric and magnetic fields this way.

The free field theory of electromagnetism then has a Lagrangian of

= −

µν

− B

For reasons we will see later on, it is convenient to rescale the a

= −ia

∈ iR = L(U(1))

and write

µν

= −if

µν

= ∂

− ∂

Now the magic happens when we want to couple this to matter.

Suppose we have a complex scalar field φ : R

3,1

→ C with Lagrangian

= ∂

∗

∂

φ − W (φ

∗

φ).

The interesting thing is that this has a global U(1) symmetry by

φ 7→ gφ

∗

7→ g

−1

∗

for g = e

iδ

∈ U(1), i.e. the Lagrangian is invariant under this action.

In general, it is more convenient to talk about infinitesimal transformations

Consider an element

g = exp(X) ≈ 1 + X,

where we think of X as “small”. In our case, we have X ∈ u(1)

∼

iR, and

φ 7→ φ + δ

φ,

∗

7→ φ

∗

+ δ

∗

where

φ = Xφ

∗

= −Xφ

∗

Now this is a global symmetry, i.e. this is a symmetry if we do the same

transformation at all points in the space. Since we have

= 0,

we get a conserved charge. What if we want a local symmetry? We want to have

a different transformation at every point in space, i.e. we now have a function

g : R

3,1

→ U(1),

and we consider the transformations given by

φ(x) 7→ g(x)φ(x)

∗

(x) 7→ g

−1

(x)φ

∗

(x).

This is in general no longer a symmetry. Under an infinitesimal variation

X : R

3,1

→ u(1), we have

φ = Xφ.

So the derivative transforms as

(∂

φ) = ∂

(δ

φ) = (∂

X)φ + X∂

φ.

This is bad. What we really want is for this to transform like φ, so that

∂

φ 7→ g(x)∂

φ.

Then the term ∂

∗

∂

φ will be preserved.

It turns out the solution is to couple

with

. Recall that both of these

things had gauge transformations. We now demand that under any gauge

transformation, both should transform the same way. So from now on, a gauge

transformation X : R

3,1

→ u(1) transforms both φ and A

φ 7→ φ + Xφ

7→ A

− ∂

However, this does not fix the problem, since everything we know so far that

involves the potential

is invariant under gauge transformation. We now do

the funny thing. We introduce something known as the covariant derivative:

= ∂

+ A

Similar to the case of general relativity (if you are doing that course), the

covariant derivative is the “right” notion of derivative we should use whenever

we want to differentiate fields that are coupled with

. We shall now check that

this derivative transforms in the same way as φ. Indeed, we have

φ) = δ

(∂

φ + A

φ)

= ∂

(δ

φ) + A

φ − ∂

Xφ

= X∂

φ + XA

= XD

φ.

This implies that the kinetic term

φ)

∗

is gauge invariant. So we can put together a gauge-invariant Lagrangian

L = −

µν

+ (D

φ)

∗

φ) − W (φ

∗

φ).

So what the electromagnetic potential

gives us is a covariant derivative D

which then allows us to “gauge” these complex fields to give them a larger

symmetry group.

7.2 General case

We now want to do what we had above for a general Lie group

. In particular,

this can be non-abelian, and thus the Lie algebra has non-trivial bracket. It

turns out in the general case, it is appropriate to insert some brackets into what

we have done so far, but otherwise most things follow straight through.

In the case of electromagnetism, we had a Lie group U(1), and it acted on

in the obvious way. In general, we start with a Lie group

, and we have

to pick a representation

on a vector space

. Since we want to have a

real Lagrangian, we will assume our vector space

also comes with an inner

product.

Given such a representation, we consider fields that take values in

, i.e.

functions

3,1

→ V

. We can, as before, try to write down a (scalar)

Lagrangian

= (∂

φ, ∂

φ) − W ((φ, φ)).

Writing out the summation explicitly, we have

µ=0

(∂

φ, ∂

φ) − W ((φ, φ)),

i.e. the sum is performed outside the inner product.

This is not always going to be invariant under an arbitrary action of the

group

, because the inner product is not necessarily preserved. Fortunately, by

definition of being unitary, what we require is that our representation is unitary,

i.e. each D(g) is unitary for all g ∈ G.

It is a theorem that any representation of a compact Lie group is equivalent

to one that is unitary, so we are not losing too much by assuming this (it is also

that a non-compact Lie group cannot have a non-trivial unitary representation).

Again, it will be convenient to think about this in terms of infinitesimal

transformations. Near the identity, we can write

g = exp(X)

for some X ∈ g. Then we can write

D(g) = exp(ρ(X)) ∈ gl(n, C).

We write

g → gl

(

n, C

) for the associated representation of the Lie algebra

. If

is unitary, then

will be anti-Hermitian. Then the infinitesimal transformation

is given by

φ 7→ φ + δ

φ = φ + ρ(X)φ.

In general, an infinitesimal Gauge transformation is then given by specifying a

function

X : R

3,1

→ g.

Our transformation is now

φ = ρ(X(x))φ.

Just as in the abelian case, we know our Lagrangian is no longer gauge invariant

in general.

We now try to copy our previous fix. Again, we suppose our universe comes

with a gauge field

: R

3,1

→ L(G).

Again, we can define a covariant derivative

φ = ∂

φ + ρ(A

)φ.

In the case of a non-abelian gauge symmetry, our gauge field transformed under

X as

= −∂

This expression still makes sense, but it turns out this isn’t what we want, if

we try to compute

). Since the gauge field is non-abelian, there is one

extra thing we can include, namely [

X, A

]. It turns out this is the right thing

we need. We claim that the right definition is

= −∂

X + [X, A

To see this, we show that

Proposition. We have

φ) = ρ(X)D

φ.

The proof involves writing out the terms and see that it works.

Proof. We have

φ) = δ

(∂

φ + ρ(A

)φ)

= ∂

(δ

φ) + ρ(A

)δ

φ + ρ(δ

)φ

= ∂

(ρ(X)φ) + ρ(A

)ρ(X)φ − ρ(∂

X)φ + ρ([X, A

])φ

= ρ(∂

X)φ + ρ(X)∂

φ + ρ(X)ρ(A

)φ

+ [ρ(A

), ρ(X)]φ − ρ(∂

X)φ + ρ([X, A

])φ

= ρ(X)(∂

φ + ρ(A

)φ)

= ρ(X)D

φ,

as required.

Thus, we know that (D

φ,

) is gauge invariant. We can then write down

a gauge invariant “matter” part of the action

= (D

φ, D

φ) − W [(φ, φ)].

Finally, we need to produce a gauge invariant kinetic term for

3,1

→ L

(

For the case of an abelian gauge theory, we had

µν

= ∂

− ∂

∈ L(G).

In this case, there is one extra term we can add, namely [

, A

]

∈ g

, and it

turns out we need it for things to work out. Our field strength tensor is

µν

= ∂

− ∂

+ [A

, A

How does the field strength tensor transform? In the abelian case, we had that

µν

was invariant. Let’s see if that is the case here. It turns out this time the

field strength tensor transforms as the adjoint representation.

Lemma. We have

µν

) = [X, F

µν

] ∈ L(G).

Proof. We have

µν

) = ∂

(δ

) − ∂

(δ

) + [δ

, A

] + [A

, δ

]

= ∂

∂

X + ∂

([X, A

]) − ∂

∂

X − ∂

([X, A

]) − [∂

X, A

]

− [A

, ∂

X] + [[X, A

], A

] + [A

, [X, A

]]

= [X, ∂

] − [X, ∂

] + ([X, [A

, A

]]

= [X, F

µν

where we used the Jacobi identity in the last part.

So to construct a scalar quantity out of

µν

, we need an inner product

invariant under an adjoint representation. We know one of these — the Killing

form! So we can just pick

κ(F

µν

, F

µν

This is known as the Yang-Mills Lagrangian. Note that for each fixed

µ, ν

, we

have that F

µν

∈ g. So we should read this as

µ,ν

κ(F

µν

, F

µν

Putting all these together, we get ourselves an invariant Lagrangian of the system

with a G gauge symmetry.

So the final Lagrangian looks like

L =

µ,ν

κ(F

µν

, F

µν

) + (D

φ, D

φ) + W ((φ, φ)).

Now if we are further told that we actually have a simple complex Lie algebra,

then it is a fact that we can find a real form of compact type. So in particular

we can find a basis B = {T

, a = 1, ··· , dim g} with

= κ(T

, T

) = −κδ

In this basis, we have

−κ

a=1

µν,a

and this does look like a copy of d many electromagnetic fields.

So to construct (a sensible) gauge theory, we need to get a (semi-)simple

Lie algebra, and then find some representations of

. These are things we have

already studied well before!

According to our current understanding, the gauge group of the universe is

G = U(1) × SU(2) × SU(3),

and, for example, the Higgs boson has representation

φ = (+1, ρ

, ρ

8 Lie groups in nature

We are now going to look around and see how the things we have done exhibit

themselves in nature. This chapter, by design, is very vague, and one shall not

expect to learn anything concrete from here.

8.1 Spacetime symmetry

In special relativity, we have a metric

= −dt

+ dx

+ dy

+ dz

The group of (orientation-preserving and) metric-preserving symmetries gives

us the Lorentz group

1). However, in certain cases, we can get something

more (or perhaps less) interesting. Sometimes it makes sense to substitute

so that

= dτ

+ dx

+ dy

+ dz

This technique is known as Wick rotation. If we put it this way, we now have a

symmetry group of SO(4) instead.

It happens that when we complexify this, this doesn’t really matter. The

resulting Lie algebra is

(SO(3, 1)) = L

(SO(4)) = D

Its Dynkin diagram is

This is not simple, and we have

= A

⊕ A

In other words, we have

so(4) = su(2) ⊕ su(2).

At the Lie group level,

(4) does not decompose as

(2)

× SU

(2). Instead,

SU(2) × SU(2) is a double cover of SO(4), and we have

SO(4) =

SU(2) × SU(2)

Now in general, our fields in physics transform when we change coordinates.

There is the boring case of a scalar, which never transforms, and the less boring

case of a vector, which transforms like a vector. More excitingly, we have objects

known as spinors. The weird thing is that spinors do have an

1) action,

but this does not lift to an action of

1). Instead, we need to do something

funny with double covers, which we shall not go into.

In general, spinors decompose into “left-handed” and “right-handed” com-

ponents, known as Weyl fermions, and these correspond to two different rep-

resentations of

. We can summarize the things we have in the following

table:

Field so(3, 1) A

⊕ A

Scalar 1 (ρ

, ρ

)

Dirac Fermion LH ⊕ RH (ρ

, ρ

) ⊕ (ρ

, ρ

)

Vector 4 (ρ

, ρ

)

However, the Lorentz group

1) is not all the symmetries we can do to

Minkowski spacetime. These are just those symmetries that fix the origin. If we

add in the translations, then we obtain what is known as the Poincar´e group.

The Poincar´e algebra is not simple, as the translations form a non-trivial ideal.

8.2 Possible extensions

Now we might wonder how we can expand the symmetry group of our theory

to get something richer. In very special cases, we will get what is known as

conformal symmetry, but in general, we don’t. To expand the symmetry further,

we must consider supersymmetry.

Conformal field theory

If all our fields are massless, then our theory gains conformal invariance. Before

we look into conformal invariance, we first have a look at a weaker notion of

scale invariance.

A massless free field φ : R

3,1

→ V has a Lagrangian that looks like

L = −

(∂

φ, ∂

φ).

This is invariant under scaling, namely the following simultaneous transformations

parameterized by λ:

x 7→ x

= λ

−1

φ(x) 7→ λ

∆

φ(x

where ∆ = dim V is the dimension of the field.

Often in field theory, whenever we have scale invariance, we also get conformal

invariance. In general, a conformal transformation is one that preserves all angles.

So this, for example, includes some mixture and scaling, and is more general

than Lorentz invariance and scale invariance.

In the case of 4 (spacetime) dimensions, it turns out the conformal group

2), and has dimension 15. In general, members of this group can be

written as







0 D

−D 0

+ K

µν







Here the

are the special conformal generators, and the

are translations.

We will not go deep into this.

Conformal symmetry gives rise to conformal field theory. Theoretically, this

is very important, since they provide “end points” of renormalization group flow.

This will be studied more in depth in the Advanced Quantum Field Theory

course.

Supersymmetry

Can we add even more symmetries? It turns out there is a no-go theorem that

says we can’t.

Theorem

(Coleman-Mondula)

In an interactive quantum field theory (sat-

isfying a few sensible conditions), the largest possible symmetry group is the

Poincar´e group times some internal symmetry that commutes with the Poincar´e

group.

But this is not the end of the world. The theorem assumes that we are working

with traditional Lie groups and Lie algebras. The idea of supersymmetry is to

prefix everything we talk about with “super-”, so that this theorem no longer

applies.

To being with, supersymmetry replaces Lie algebras with graded Lie algebras,

or Lie superalgebras. This is a graded vector space, which we can write as

g = g

⊕ g

where the elements in

as said to have grade 0, and elements in

have grade

1. We write

|X|

= 0

1 for

X ∈ g

, g

respectively. These will correspond to

bosonic and fermionic operators respectively.

Now the Lie bracket respects a graded anti-commutation relation

[X, Y ] = −(−1)

|X||Y |

[Y, X].

So for Bosonic generators, they anti-commute as usual, but with we have

two fermionic things, they commute. We can similarly formulate a super-

Jacobi identity. These can be used to develop field theories that have these

“supersymmetries”.

Unfortunately, there is not much experimental evidence for supersymmetry,

and unlike string theory, we are already reaching the scales we expect to see

supersymmetry in the LHC. . .

8.3 Internal symmetries and the eightfold way

Finally, we get to the notion of internal symmetries. Recall that for a complex

scalar field, our field was invariant under a global U(1) action given by phase

change. More generally, fields can carry some representation of a Lie group

In general, if this representation has dimension greater than 1, this means that

we have multiple different “particles” which are related under this symmetry

transformation. Then by symmetry, all these particles will have the same mass,

and we get a degeneracy in the mass spectrum.

When we have such degeneracies, we would want to distinguish the different

particles of the same mass. These can be done by looking at the weights of the

representation. It turns out the different weights correspond to the different

quantum numbers of the particles.

Often, we do not have an exact internal symmetry. Instead, we have an

approximate symmetry. This is the case when the dominant terms in the

Lagrangian are invariant under the symmetry, while some of the lesser terms are

not. In particular, the different particles usually have different (but very similar)

masses.

These internal symmetries are famously present in the study of hadrons, i.e.

things made out of quarks. The basic examples we all know are the nucleons,

namely protons and neutrons. We can list them as

Charge (Q) Mass (M)

p +1 938 MeV

n 0 940 MeV

Note that they have very similar masses. Later, we found, amongst many other

things, the pions:

Charge (Q) Mass (M)

+1 139 MeV

0 135 MeV

−

-1 139 MeV

Again, these have very similar masses. We might expect that there is some

approximate internal symmetry going on. This would imply that there is some

conserved quantity corresponding to the weights of the representation. Indeed,

we later found one, known as isospin:

Charge (Q) Isospin (J) Mass (M)

p +1 +

938 MeV

n 0 −

940 MeV

+1 +1 139 MeV

0 0 135 MeV

−

-1 -1 139 MeV

Isospin comes from an approximate

(2)

symmetry, with a generator given by

H = 2J.

The nucleons then have the fundamental ρ

representation, and the pions have

the ρ

representation.

Eventually, we got smarter, and discovered an extra conserved quantum

number known as hypercharge. We can plot out the values of the isospin and

the hypercharge for our pions and some other particles we discovered, and we

found a pattern:

−

At firsts, physicists were confused by the appearance of this pattern, and tried

very hard to figure out generalizations. Of course, now that we know about Lie

algebras, we know this is the weight diagram of a representation of su(3).

However, the word

(3) hasn’t resolved all the mystery. In reality, we only

observed representations of dimension 1

8 and 10. We did not see anything

else. So physicists hypothesized that there are substructures known as quarks.

Each quark (

) carry a

representation (flavour), and antiquarks (

¯q

) carry a

representation.

The mesons correspond to

q¯q

particles, and the representation decompose as

3 ⊗

3 = 1 ⊕ 8.

Bosons correspond to qqq particles, and these decompose as

3 ⊗ 3 ⊗ 3 = 1 ⊕ 8 ⊕ 8 ⊕ 10.

Of course, we now have to explain why only

q¯q

and

qqq

appear in nature,

and why we don’t see quarks appearing isolated in nature. To do so, we have

to go very deep down in QCD. This theory says that quarks have a

(3)

gauge symmetry (which is a different

(3)) and the quark again carries the

fundamental representation

. More details can be found in the Lent Standard

Model course.