5Cartan classification

III Symmetries, Fields and Particles

5.5 Simple roots

Now we have a lot of roots that span

h

∗

R

, but they obviously do not form a basis.

For example, whenever

α ∈ h

∗

R

, 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

h

∗

R

∼

=

R

r

, i.e. a subspace of dimension

r −

1.

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

root. Then the plane divides

h

∗

R

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

α

is positive,

−α

is negative. More interestingly, if

α, β

are positive

roots, then

α

+

β

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

S

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

∗

R

.

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

α

and

β

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

S

, then the α-string through β, namely

S

α,β

= {β + nα ∈ Φ},

has length

`

αβ

= 1 −

2(α, β)

(α, α)

∈ N.

Proof. Recall that there exists n

±

such that

S

α,β

= {β + nα : n

−

≤ n ≤ n

+

},

We have shown before that

n

+

+ 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

n

−

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

−

= 0 and hence

n

+

= −

2(α, β)

(α, α)

∈ N.

So there are

n

+

+ 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

λ

of the simple roots.

We write

λ = λ

+

− λ

−

=

X

i∈I

+

c

i

α

(i)

−

X

j∈I

−

b

j

α

(j)

,

where

c

i

, b

j

≥

0 and

I

+

, I

−

are disjoint. If

I

−

is empty, then the sum is a positive

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

I

+

is empty, then the sum is

negative. So it suffices to focus on the case c

i

, b

j

> 0.

Then we have

(λ, λ) = (λ

+

, λ

+

) + (λ

−

, λ

−

) − 2(λ

+

, λ

−

)

> −2(λ

+

, λ

−

)

= −2

X

i∈I

+

X

j∈I

−

c

i

b

j

(α

(i)

, α

(j)

)

≥ 0,

since (

α

(i)

, α

(j)

)

≤

0 for all simple roots

α

(i)

, α

(j)

. So in particular

λ

is non-

zero.

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

|Φ

S

| = r.