5Cartan classification
III Symmetries, Fields and Particles
5.2 The Cartan basis
From now on, we restrict to the study of finitedimensional simple complex Lie
algebras. Every time we write the symbol
g
or say “Lie algebra”, we mean a
finitedimensional simple complex Lie algebra.
Recall that we have already met such a Lie algebra
su
C
(2) = span
C
{H, E
+
, E
−
}
with the brackets
[H, E
±
] = ±2E
±
, [E
+
, E
−
] = H.
These are known as the Cartan basis for
su
C
(2). We will try to mimic this
construction in an arbitrary Lie algebra.
Recall that when we studied
su
C
(2), we used the fact that
H
is a diagonal
matrix, and
E
±
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
ad
map associated to
them instead. Recall that the adjoint map of
H
is also diagonal, with eigenvectors
given by
ad
H
(E
±
) = ±2E
±
ad
H
(H) = 0.
This is the structure we are trying to generalize.
Definition
(
ad
diagonalizable)
.
Let
g
be a Lie algebra. We say that an element
X ∈ g is addiagonalizable if the associated map
ad
X
: g → g
is diagonalizable.
Example. In su
C
(2), we know H is addiagonalizable, but E
±
is not.
Now we might be tempted to just look at all
ad
diagonalizable elements.
However, this doesn’t work. In the case of
su
(2), each of
σ
1
, σ
2
, σ
3
is
ad

diagonalizable, but we only want to pick one of them as our
H
. Instead, what
we want is the following:
Definition
(Cartan subalgebra)
.
A Cartan subalgebra
h
of
g
is a maximal
abelian subalgebra containing only addiagonalizable elements.
A Cartan subalgebra always exists, since the dimension of
g
is finite, and the
trivial subalgebra
{
0
} ⊆ g
is certainly abelian and contains only
ad
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
h
be a Cartan subalgebra of
g
, and let
X ∈ g
. If [
X, H
] = 0
for all H ∈ h, then X ∈ h.
Note that this does not follow immediately from
h
being maximal, because
maximality only says that addiagonalizable elements satisfy that property.
Proof. Omitted.
Example. In the case of su
C
(2), one possible Cartan subalgebra is
h = span
C
{H}.
However, recall our basis is given by
H = σ
3
E
±
=
1
2
(σ
1
± iσ
2
),
where the
σ
i
are the Pauli matrices. Then, by symmetry, we know that
σ
1
=
E
+
+
E
−
gives an equally good Cartan subalgebra, and so does
σ
2
. So we have
many choices, but they all have the same dimension.
Now we know that everything in
h
commutes with each other, i.e. for any
H, H
0
∈ h, we have
[H, H
0
] = 0.
Since ad is a Lie algebra representation, it follows that
ad
H
◦ ad
H
0
− ad
H
0
◦ ad
H
= 0.
In other words, all these
ad
maps commute. By assumption, we know each
ad
H
is diagonalizable. So we know they are in fact simultaneously diagonalizable. So
g
is spanned by simultaneous eigenvectors of the
ad
H
. Can we find a basis of
eigenvectors?
We know that everything in
h
is a zeroeigenvector of
ad
H
for all
H ∈ h
,
since for H, H
0
∈ h, we have
ad
H
(H
0
) = [H, H
0
] = 0.
We can arbitrarily pick a basis
{H
i
: i = 1, ··· , r},
where
r
is the rank of
h
. 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
ad
H
(E) = [H, E] = α(H)E
for some constant
α
(
H
) depending on
H
(and
E
). We call
α
:
h → C
the root of
E. We will use the following fact without proof:
Fact.
The nonzero simultaneous eigenvectors of
h
are nondegenerate, i.e. there
is a unique (up to scaling) eigenvector for each root.
Thus, we can refer to this eigenvector unambiguously by
E
α
, where
α
designates the root.
What are these roots
α
? It is certainly a function
h → C
, but it is actually a
linear map! Indeed, we have
α(H + H
0
)E = [H + H
0
, E]
= [H, E] + [H
0
, E]
= α(H)E + α(H
0
)E
= (α(H) + α(H
0
))E,
by linearity of the bracket.
We write Φ for the collection of all roots. So we can write the remaining
basis eigenvectors as
{E
α
: α ∈ Φ}.
Example. In the case of su(2), the roots are ±2, and the eigenvectors are E
±
.
We can now define a CartanWeyl basis for g given by
B = {H
i
: i = 1, ··· , r} ∪ {E
α
: α ∈ Φ}.
Recall that we have a Killing form
κ(X, Y ) =
1
N
tr(ad
X
◦ ad
Y
),
where
X, Y ∈ g
. Here we put in a normalization factor
N
for convenience later
on. Since g is simple, it is in particular semisimple. So κ is nondegenerate.
We are going to evaluate κ in the CartanWeyl basis.
Lemma. Let H ∈ h and α ∈ Φ. Then
κ(H, E
α
) = 0.
Proof. Let H
0
∈ h. Then
α(H
0
)κ(H, E
α
) = κ(H, α(H
0
)E
α
)
= κ(H, [H
0
, E
α
])
= −κ([H
0
, H], E
α
)
= −κ(0, E
α
)
= 0
But since α 6= 0, we know that there is some H
0
such that α(H
0
) 6= 0.
Lemma. For any roots α, β ∈ Φ with α + β 6= 0, we have
κ(E
α
, E
β
) = 0.
Proof. Again let H ∈ h. Then we have
(α(H) + β(H))κ(E
α
, E
β
) = κ([H, E
α
], E
β
) + κ(E
α
, [H, E
β
]),
= 0
where the final line comes from the invariance of the Killing form. Since
α
+
β
does not vanish by assumption, we must have κ(E
α
, E
β
) = 0.
Lemma. If H ∈ h, then there is some H
0
∈ h such that κ(H, H
0
) 6= 0.
Proof.
Given an
H
, since
κ
is nondegenerate, there is some
X ∈ g
such that
κ
(
H, X
)
6
= 0. Write
X
=
H
0
+
E
, where
H
0
∈ h
and
E
is in the span of the
E
α
.
0 6= κ(H, X) = κ(H, H
0
) + κ(H, E) = κ(H, H
0
).
What does this tell us?
κ
started life as a nondegenerate inner product
on
g
. But now we know that
κ
is a nondegenerate inner product on
h
. By
nondegeneracy, we can invert it within h.
In coordinates, we can find some κ
ij
such that
κ(e
i
H
i
, e
0
j
H
j
) = κ
ij
e
i
e
0
j
for any
e
i
,
e
0
j
. The fact that the inner product is nondegenerate means that we
can invert the matrix κ, and find some (κ
−1
)
ij
such that
(κ
−1
)
ij
κ
jk
= δ
i
k
Since
κ
−1
is nondegenerate, this gives a nondegenerate inner product on
h
∗
.
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
)
ij
α
i
β
j
,
where
α
i
:=
α
(
H
i
). 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
α
, E
−α
) 6= 0
This holds for stupid reasons.
Proof. We know that
κ(E
α
, E
β
) = κ(E
α
, H
i
) = 0
for all
β 6
=
−α
and all
i
. But
κ
is nondegenerate, and
{E
β
, H
i
}
span
g
. So
there must be some E
−α
in the basis set, and
κ(E
α
, E
−α
) 6= 0.
So far, we know that
[H
i
, H
j
] = 0
[H
i
, E
α
] = α
i
E
α
for all α ∈ Φ and i, j = 1, ··· , r. Now it remains to evaluate [E
α
, E
β
].
Recall that in the case of su
C
(2), we had
[E
+
, E
−
] = H.
What can we get here? For any H ∈ h and α, β ∈ Φ, we have
[H, [E
α
, E
β
]] = −[E
α
, [E
β
, H]] − [E
β
, [H, E
α
]]
= (α(H) + β(H))[E
α
, E
β
].
Now if α + β 6= 0, then either [E
α
, E
β
] = 0, or α + β ∈ Φ and
[E
α
, E
β
] = N
α,β
E
α+β
for some N
α,β
.
What if
α
+
β
= 0? We claim that this time, [
E
α
, E
−α
]
∈ h
. Indeed, for any
H ∈ h, we have
[H, [E
α
, E
−α
]] = [[H, E
α
], E
−α
] + [[E
−α
, H], E
α
]
= α(H)[E
α
, E
−α
] + α(H)[E
−α
, E
α
]
= 0.
Since
H
was arbitrary, by the (strong) maximality property of
h
, we know that
[E
α
, E
−α
] ∈ h.
Now we can compute
κ([E
α
, E
−α
], H) = κ(E
α
, [E
−α
, H])
= α(H)κ(E
α
, E
−α
).
We can view this as an equation for [
E
α
, E
−α
]. Now since we know that
[
E
α
, E
−α
]
∈ h
, and the Killing form is nondegenerate when restricted to
h
, we
know [E
α
, E
−α
] is uniquely determined by this relation.
We define the normalized
H
α
=
[E
α
, E
−α
]
κ(E
α
, E
−α
)
.
Then our equation tells us
κ(H
α
, H) = α(H).
Writing H
α
and H in components:
H
α
= e
α
i
H
i
, H = e
i
H
i
.
the equation reads
κ
ij
e
α
i
e
j
= α
i
e
i
.
Since the e
j
are arbitrary, we know
e
α
i
= (κ
−1
)
ij
α
j
.
So we know
H
α
= (κ
−1
)
ij
α
j
H
i
.
We now have a complete set of relations:
Theorem.
[H
i
, H
j
] = 0
[H
i
, E
α
] = α
i
E
α
[E
α
, E
β
] =
N
α,β
E
α+β
α + β ∈ Φ
κ(E
α
, E
β
)H
α
α + β = 0
0 otherwise
.
Now we notice that there are special elements
H
α
in the Cartan subalgebra
associated to the roots. We can compute the brackets as
[H
α
, E
β
] = (κ
−1
)
ij
α
i
[H
j
, E
β
]
= (κ
−1
)
ij
α
i
β
j
E
β
= (α, β)E
β
,
where we used the inner product on the dual space
h
∗
induced by the Killing
form κ.
Note that so far, we have picked
H
i
and
E
α
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:
e
α
=
s
2
(α, α)κ(E
α
, E
−α
)
E
α
h
α
=
2
(α, α)
H
α
.
Here it is important that (
α, α
)
6
= 0, but we will only prove it next chapter where
the proof fits in more naturally.
Under this normalization, we have
[h
α
, h
β
] = 0
[h
α
, e
β
] =
2(α, β)
(α, α)
e
β
[e
α
, e
β
] =
n
αβ
e
α+β
α + β ∈ Φ
h
α
α + β = 0
0 otherwise
Note that the number of roots is
d −r
, where
d
is the dimension of
g
and
r
is its
rank, and this is typically greater than
r
. So in general, there are too many of
them to be a basis, even if they spanned
h
(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.