5Cartan classification
III Symmetries, Fields and Particles
5.6 The classification
We now have a (not so) canonical choice of basis of h
∗
R
B = {α ∈ Φ
S
} = {α
(i)
: i = 1, ··· , r}.
We want to reexpress the Lie algebra in terms of this basis.
Definition
(Cartan matrix)
.
The Cartan matrix
A
ij
is defined as the
r × r
matrix
A
ij
=
2(α
(i)
, α
(j)
)
(α
(j)
, α
(j)
)
.
Note that this is not a symmetric matrix. We immediately know that
A
ij
∈ Z
.
For each root α
(i)
, we have an su(2) subalgebra given by
{h
i
= h
α
(i)
, e
i
±
= e
±α
(i)
}.
We then have
[h
i
, e
i
±
] = ±2e
i
±
, [e
i
+
, e
i
−
] = h
i
.
Looking at everything in our Lie algebra, we have the relations
[h
i
, h
j
] = 0
[h
i
, e
j
±
] = ±A
ji
e
j
±
[e
i
+
, e
j
−
] = δ
ij
h
i
,
with no summation implied in the second row. Everything here we’ve seen,
except for the last expression [
e
i
+
, e
j
−
]. We are claiming that if
i 6
=
j
, then the
bracket in fact vanishes! This just follows from the fact that if
α
(i)
and
α
(j)
are
simple, then α
(i)
− α
(j)
is not a root.
Note that this does not a priori tell us everything about the Lie algebra. We
have only accounted for step generators of the form
e
i
±
, and also we did not
mention how, say, [e
i
+
, e
j
+
] behaves.
We know
[e
i
+
, e
j
+
] = ad
e
i
+
(e
j
+
) ∝ e
α
(i)
+α
(j)
if
α
(i)
+
α
(j)
∈
Φ. We know that if
α
(i)
+
α
(j)
∈
Φ, then it belongs to a string.
In general, we have
ad
n
e
i
+
(e
j
+
) ∝ e
α
(j)
+nα
(i)
if α
(j)
+ nα
(i)
∈ Φ, and we know how long this string is. So we have
(ad
e
i
±
)
1−A
ji
e
j
±
= 0.
This is the Serre relation. It turns out this is all we need to completely charac
terize the Lie algebra.
Theorem
(Cartan)
.
Any finitedimensional, simple, complex Lie algebra is
uniquely determined by its Cartan matrix.
We will not prove this.
To achieve the Cartan classification, we need to classify all Cartan matrices,
and then reconstruct g form the Cartan matrix.
We first do the first part. Recall that
A
ij
=
2(α
(i)
, α
(j)
)
(α
(j)
, α
(j)
)
∈ Z.
We first read off properties of the Cartan matrix.
Proposition. We always have A
ii
= 2 for i = 1, ··· , r.
This is not a symmetric matrix in general, but we do have the following:
Proposition. A
ij
= 0 if and only if A
ji
= 0.
We also have
Proposition. A
ij
∈ Z
≤0
for i 6= j.
We now get to a less trivial fact:
Proposition. We have det A > 0.
Proof. Recall that we defined our inner product as
(α, β) = α
T
κ
−1
β,
and we know this is positive definite. So we know
det κ
−1
>
0. We now write
the Cartan matrix as
A = κ
−1
D,
where we have
D
j
k
=
2
(α
(j)
, α
(j)
)
δ
j
k
.
Then we have
det D =
Y
j
2
(α
(j)
, α
(j)
)
> 0.
So it follows that
det A = det κ
−1
det D > 0.
It is an exercise to check that if the Cartan matrix is reducible, i.e. it looks
like
A =
A
(1)
0
0 A
(2)
,
then in fact the Lie algebra is not simple.
So we finally have
Proposition. The Cartan matrix A is irreducible.
So in total, we have the following five properties:
Proposition.
(i) A
ii
= 2 for all i.
(ii) A
ij
= 0 if and only if A
ji
= 0.
(iii) A
ij
∈ Z
≤0
for i 6= j.
(iv) det A > 0.
(v) A is irreducible.
These are hugely restrictive.
Example.
For a rank 1 matrix, there is only one entry, and it is on the diagonal.
So we must have
A =
2
.
This corresponds to the simple Lie algebra su(2).
Example.
For a rank 2 matrix, we know the diagonals must be 2, so we must
have something of the form
A =
2 m
` 2
.
We know that the offdiagonals are not all zero, but if one is nonzero, then the
other also is. So they must be both nonzero, and we have
m` <
4. We thus
have
(m, `) = (−1, −1), (−1, −2), (−1, −3).
Note that we do not write out the cases where we swap the two entries, because
we will get the same Lie algebra but with a different ordering of the basis.
Now we see that we have a redundancy in the description of the Cartan
matrix given by permutation. There is a neat solution to this problem by drawing
Dynkin diagrams.
Definition
(Dynkin diagram)
.
Given a Cartan matrix, we draw a diagram as
follows:
(i) For each simple root α
(i)
∈ Φ
S
, we draw a node
(ii) We join the nodes corresponding to α
(i)
, α
(j)
with A
ij
A
ji
many lines.
(iii)
If the roots have different lengths, we draw an arrow from the longer root
to the shorter root. This happens when A
ij
6= A
ji
.
Note that we wouldn’t have to draw too many lines. We have
A
ij
=
2α
(i)

α
(j)

cos ϕ
ij
, A
ji
=
2α
(j)

α
(i)

cos ϕ
ij
,
where ϕ
ij
is the angle between them. So we have
cos
2
ϕ
ij
=
1
4
A
ij
A
ji
.
But we know cos
2
ϕ
ij
∈ [0, 1]. So we must have
A
ij
A
ji
∈ {0, 1, 2, 3}.
So we have to draw at most 3 lines, which isn’t that bad. Moreover, we have the
following information:
Proposition. A simple Lie algebra has roots of at most 2 distinct lengths.
Proof. See example sheet.
It is an exercise to see that all the information about
A
ji
can be found from
the Dynkin diagram.
We can now revisit the case of a rank 2 simple Lie algebra.
Example. The Cartan matrices of rank 2 are given by
2 −1
−1 2
,
2 −2
−1 2
,
2 −3
−1 2
These correspond to the Dynkin diagrams
The conditions on the matrices
A
ij
now translate to conditions on the Dynkin
diagrams which we will not write out, and it turns out we can classify all the
allowed Dynkin diagrams as follows:
Theorem
(Cartan classification)
.
The possible Dynkin diagrams include the
following infinite families (where n is the number of vertices):
A
n
:
B
n
:
C
n
:
D
n
:
And there are also five exceptional cases:
E
6
:
E
7
:
E
8
:
F
4
:
G
2
:
Example.
The infinite families
A
n
, B
n
, C
n
, D
n
correspond to wellknown com
plex Lie groups by
Family Lie Group
A
n
L
C
(SU(n + 1))
B
n
L
C
(SO(2n + 1))
C
n
L
C
(Sp(2n))
D
n
L
C
(SO(2n))
where the Sp(2n) are the symplectic matrices.
Note that there is some repetition in our list. For example, we have
A
1
=
B
1
=
C
1
=
D
1
and
B
2
=
C
2
. Also,
D
2
does not give a simple Lie algebra, since
it is disconnected. We have D
2
∼
=
A
1
⊕ A
1
. So we have
L
C
(SO(4))
∼
=
L
C
(SU(2)) ⊕ L
C
(SU(2)).
Finally, we also have D
3
= A
3
, and this reflects an isomorphism
L
C
(SU(4))
∼
=
L
C
(SO(6)).
Hence, a list without repetitions is given by
A
n
for n ≥ 1,
B
n
for n ≥ 2,
C
n
for n ≥ 3,
D
n
for n ≥ 4.
This classification is very important in modern theoretical physics, since in
many theories, we need to pick a Lie group as, say, our gauge group. So knowing
what Lie groups are around lets us know what theories we can have.