6Representation of Lie algebras

III Symmetries, Fields and Particles

6.1 Weights

Let

ρ

be a representation of

g

on

V

. Then it is completely determined by the

images

H

i

7→ ρ(H

i

) ∈ gl(V )

E

α

7→ ρ(E

α

) ∈ gl(V ).

Again, we know that

[ρ(H

i

), ρ(H

j

)] = ρ([H

i

, H

j

]) = 0.

So all

ρ

(

H

i

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

eigenvector v

λ

. Again, for each H ∈ h, we know

ρ(H)v

λ

= λ(H)v

λ

for some λ(H), and λ lives in h

∗

.

Definition

(Weight of representation)

.

Let

ρ

:

g → gl

(

V

) be a representation

of

g

. Then if

v

λ

∈ V

is an eigenvector of

ρ

(

H

) for all

H ∈ h

, we say

λ ∈ h

∗

is a

weight of ρ, where

ρ(H)v

λ

= λ(H)v

λ

for all H ∈ h.

The weight set S

ρ

of ρ is the set of all weights.

For a weight

λ

, we write

V

λ

for the subspace that consists of vectors

v

such

that

ρ(H)v = λ(H)v.

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

V

λ

need not be 1-dimensional.

We write

m

λ

= dim V

λ

≥ 1

for the multiplicity of the weight.

Example. By definition, we have

[H

i

, E

α

] = α

i

E

α

So we have

ad

H

i

E

α

= α

i

E

α

.

In terms of the adjoint representation, this says

ρ

adj

(H

i

)E

α

= α

i

E

α

.

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

Recall that for

su

C

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

will be the case in general as well.

Given any representation

ρ

, we know it has at least one weight

λ

with an

eigenvector

v ∈ V

λ

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

ρ(E

α

) to it, for α ∈ Φ. We have

ρ(H

i

)ρ(E

α

)v = ρ(E

α

)ρ(H

i

)v + [ρ(H

i

), ρ(E

α

)]v.

We know that

[ρ(H

i

), ρ(E

α

)] = ρ([H

i

, E

α

]) = α

i

ρ(E

α

).

So we know that if v ∈ V

λ

, then

ρ(H

i

)ρ(E

α

)v = (λ

i

+ α

i

)ρ(E

α

)v.

So the weight of the vector has been shifted by

α

. Thus, for all vectors

v ∈ V

λ

,

we find that

ρ(E

α

)v ∈ V

λ+α

.

However, we do not know a priori if

V

λ+α

is a thing at all. If

V

λ+α

=

{

0

}

, i.e.

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

α

)v = 0.

So the Cartan elements

H

i

preserve the weights, and the step operators

E

α

increment the weights by α.

Consider the action of our favorite

su

(2)

α

subalgebra on the representation

space. In other words, we consider the action of

{ρ

(

h

α

)

, ρ

(

e

α

)

, ρ

(

e

−α

)

}

on

V

.

Then

V

becomes the representation space for some representation

ρ

α

for

su

(2)

α

.

Now we can use what we know about the representations of

su

(2) to get

something interesting about V . Recall that we defined

h

α

=

2

(α, α)

H

α

=

2

(α, α)

(κ

−1

)

ij

α

j

H

i

.

So for any v ∈ V

λ

, after some lines of algebra, we find that

ρ(h

α

)(v) =

2(α, λ)

(α, α)

v.

So we know that we must have

2(α, λ)

(α, α)

∈ Z

for all

λ ∈ S

ρ

and

α ∈

Φ. So in particular, we know that (

α

(i)

, λ

)

∈ R

for all

simple roots

α

(i)

. In particular, it follows that

λ ∈ h

∗

R

(if we write

λ

as a sum

of the

α

(i)

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

equations, hence are real).

Again, from these calculations, we see that

L

λ∈S

ρ

V

λ

is a subrepresentation

of

V

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

eigenvectors, i.e. we must have

V =

M

λ∈S

ρ

V

λ

.

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