5Cartan classification

III Symmetries, Fields and Particles

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
su
(2).
If
α
Φ, then
α
Φ. For each pair
±α
Φ, we can consider the subalgebra
with basis {h
α
, e
α
, e
α
}. Then we have
[h
α
, e
±α
] = ±2e
±α
[e
α
, e
α
] = h
α
.
These are exactly the relations for the Lie algebra of
su
(2) (or rather,
su
C
(2),
but we will not write the
C
every time). So this gives a subalgebra isomorphic
to su(2). We call this su(2)
α
.
It turns out we also get a lot of representations of su(2)
α
.
Definition
(String)
.
For
α, β
Φ, we define the
α
-string passing through
β
to
be
S
α,β
= {β + ρα Φ : ρ Z}.
We will consider the case where
β
is not proportional to
α
. The remaining
case, where we may wlog take
β
= 0, is left as an exercise in the example sheet.
We then have a corresponding vector subspace
V
α,β
= span
C
{e
β+ρα
: β + ρα S
α,β
}.
Consider the action of su(2)
α
on V
α,β
. We have
[h
α
, e
β+ρα
] =
2(α, β + ρα)
(α, α)
e
β+ρα
=
2(α, β)
(α, α)
+ 2ρ
e
β+ρα
. ()
We also have
[e
±α
, e
β+ρα
]
(
e
β+(ρ±1)α
β + (ρ ±1)α Φ
0 otherwise
V
α,β
.
So
V
α,β
is invariant under the action of
su
(2)
α
. So
V
α,β
is the representation
space for some representation of su(2)
α
.
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, ··· , }.
What does this tell us? We know that the possible values of
ρ
are bounded
above and below, so we can write
S
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.
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) =
1
N
X
δΦ
(α, δ)(δ, β),
where N is the normalization factor appearing in the Killing form
κ(X, Y ) =
1
N
X
Y
).
Proof. We pick the Cartan-Weyl basis, with
[H
i
, E
δ
] = δ
i
E
δ
for all i = 1, ··· , r and δ Φ. Then the inner product is defined by
κ
ij
= κ(H
i
, H
j
) =
1
N
H
i
H
j
].
But we know that these matrices
H
i
are diagonal in the Cartan-Weyl basis,
and the non-zero diagonal entries are exactly the δ
i
. So we can write this as
κ
ij
=
1
N
X
δΦ
δ
i
δ
j
.
Now recall that our inner product was defined by
(α, β) = α
i
β
j
(κ
1
)
ij
= κ
ij
α
i
β
j
,
where we define
β
j
= (κ
1
)
jk
β
k
.
Putting in our explicit formula for the κ
ij
, this is
(α, β) =
1
N
X
δΦ
α
i
δ
i
δ
j
β
j
=
1
N
X
δΦ
(α, δ)(δ, β).
Corollary.
(α, β) R
for all α, β Φ.
Proof. We write
R
α,β
=
2(α, β)
(α, α)
Z.
Then the previous formula tells us
2
(β, β)
R
α,β
=
1
N
X
δΦ
R
α,δ
R
β
We know that
R
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.