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 re-express 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
+
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
n
e
i
+
(e
j
+
) e
α
(j)
+
(i)
if α
(j)
+
(i)
Φ, and we know how long this string is. So we have
e
i
±
)
1A
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 finite-dimensional, 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 off-diagonals are not all zero, but if one is non-zero, then the
other also is. So they must be both non-zero, 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
=
2|α
(i)
|
|α
(j)
|
cos ϕ
ij
, A
ji
=
2|α
(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 well-known 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.