III Symmetries, Fields and Particles

5Cartan classification

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.