5Cartan classification
III Symmetries, Fields and Particles
5.4 A real subalgebra
Let’s review what we know so far. The roots
α ∈
Φ are elements of
h
∗
. In
general, the roots will not be linearly independent elements of
h
∗
, 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
h
∗
. 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
H
λ
= λ
i
H
i
∈ h.
Then as usual we have
[H
λ
, H] = 0 for all H ∈ h.
Also, we know
[H
λ
, E
α
] = (λ, α)E
α
= 0.
for all roots
α ∈
Φ by assumption. So
H
λ
commutes with everything in the Lie
algebra. This would make
hH
λ
i
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 :
i
= 1
, ··· , r}
of roots.
Again, this choice is arbitrary. We now define a real vector subspace
h
∗
R
⊆ h
∗
by
h
∗
R
= span
R
{α
(i)
: i = 1, ··· , r}.
One might be worried that this would depend on our choice of the basis
α
(i)
,
which was arbitrary. However, it does not, since any choice will give us the same
space:
Proposition. h
∗
R
contains all roots.
So h
∗
R
is alternatively the real span of all roots.
Proof.
We know that
h
∗
is spanned by the
α
(i)
as a complex vector space. So
given an β ∈ h
∗
, we can find some β
i
∈ C such that
β =
r
X
i=1
β
i
α
(i)
.
Taking the inner product with α
(j)
, we know
(β, α
(j)
) =
r
X
i=1
β
i
(α
(i)
, α
(j)
).
We now use the fact that the inner products are all real! So
β
i
is the solution to
a set of real linear equations, and the equations are non-degenerate since the
α
(i)
form a basis and the Killing form is non-degenerate. So
β
i
must be real. So
β ∈ h
∗
R
.
Now the inner product of any two elements of
h
∗
R
is real, since an element in
h
∗
R
is a real linear combination of the
α
(i)
, and the inner product of the
α
(i)
is
always real.
Proposition.
The Killing form induces a positive-definite inner product on
h
∗
R
.
Proof.
It remains to show that (
λ, λ
)
≥
0 for all
λ
, with equality iff
λ
= 0. We
can write
(λ, λ) =
1
N
X
δ∈Φ
(λ, δ)
2
≥ 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
h
∗
R
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| =
p
(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
2
ϕ ∈ Z.
So
cos ϕ = ±
√
n
2
, n = {0, 1, 2, 3, 4}.
So we have some boring solutions
ϕ = 0,
π
2
, π,
and non-boring ones
ϕ =
π
6
,
π
4
,
π
3
,
2π
3
,
3π
4
,
5π
6
.
These are the only possibilities!