5Cartan classification

III Symmetries, Fields and Particles

5.1 The Killing form
The first thing we want to figure out is an “invariant” inner product on our Lie
algebra
g
. We will do so by writing down a formula, and then checking that
for a (semi-)simple Lie algebra, it is non-degenerate. Having a non-degenerate
inner product will be very useful. Amongst many things, it provides us with a
bijection between a vector space and its dual.
We recall the following definitions:
Definition
(Inner product)
.
Given a vector space
V
over
F
, an inner product
is a symmetric bilinear map i : V × V F.
Definition
(Non-degenerate inner product)
.
An inner product
i
is said to be
non-degenerate if for all v V non-zero, there is some w V such that
i(v, w) 6= 0.
The question we would like to ask is if there is a “natural” inner product on
g. We try the following:
Definition
(Killing form)
.
The Killing form of a Lie algebra
g
is the inner
product κ : g × g F given by
X
Y
),
where
tr
is the usual trace of a linear map. Since
is linear, this is bilinear in
both arguments, and the cyclicity of the trace tells us this is symmetric.
We can try to write this more explicitly. The map
X
Y
:
g g
is given
by
Z 7→ [X, [Y, Z]].
We pick a basis {T
a
}
a=1,...,D
for g. We write
X = X
a
T
a
, Y = Y
a
T
a
, Z = Z
a
T
a
.
We again let f
ab
c
be the structure constants satisfying
[T
a
, T
b
] = f
ab
c
T
c
.
We then have
[X, [Y, Z]] = X
a
Y
b
Z
c
[T
a
, [T
b
, T
c
]]
= X
a
Y
b
Z
c
f
e
f
bc
d
T
e
= M(X, Y )
c
e
Z
c
T
e
,
where
M(X, Y )
c
e
= X
a
Y
b
f
e
f
bc
d
.
So the trace of this thing is
κ(X, Y ) = tr(M(X, Y )) = κ
ab
X
a
Y
b
, κ
ab
= f
c
f
bc
d
.
Why is this a natural thing to consider? This is natural because it obeys an
invariance condition:
Definition
(Invariant inner product)
.
An inner product
κ
on a Lie algebra
g
is
invariant if for any X, Y, Z g, we have
κ([Z, X], Y ) + κ(X, [Z, Y ]) = 0.
Equivalently, we have
Z
X, Y ) + κ(X, ad
Z
Y ) = 0.
What does this condition actually mean? If one were to arbitrarily write
down a definition of invariance, they might try
Z
X, Y ) = κ(X, ad
Z
Y )
Usually, we think of elements of the Lie algebra as some sort of “infinitesimal
transformation”, and as we have previously discussed, the adjoint representa-
tion is how an element
Z g
naturally acts on
g
. So under an infinitesimal
transformation, the elements X, Y g transform as
Z
X
Z
Y
What is the effect on the Killing form? It transforms infinitesimally as
κ(X, Y ) 7→ κ(X + ad
Z
Z
Y ) κ(X, Y ) + κ(ad
Z
X, Y ) + κ(X, ad
Z
Y ),
where we dropped the “higher order terms” because we want to think of the
Z
terms as being infinitesimally small (justifying this properly will require going
to the more global picture involving an actual Lie group, which we shall not go
into. This is, after all, just a motivation). So invariance of the Killing form says
it doesn’t transform under this action.
So we now check that the Killing form does satisfy the invariance condition.
Proposition. The Killing form is invariant.
Proof. We have
κ([Z, X], Y ) = tr(ad
[Z,X]
Y
)
Z
X
Y
)
Z
X
Y
X
Z
Y
)
Z
X
Y
X
Z
Y
)
Similarly, we have
κ(X, [Z, Y ]) = tr(ad
X
Z
Y
X
Y
Z
).
κ([Z, X], Y ) + κ(X, [Z, Y ]) = tr(ad
Z
X
Y
X
Y
Z
).
By the cyclicity of tr, this vanishes.
The next problem is to figure out when the Killing form is degenerate. This
is related to the notion of simplicity of Lie algebras.
Definition
(Semi-simple Lie algebra)
.
A Lie algebra is semi-simple if it has no
abelian non-trivial ideals.
This is weaker than the notion of simplicity simplicity requires that there
are no non-trivial ideals at all!
In fact, it is true that
g
being semi-simple is equivalent to
g
being the direct
sum of simple Lie algebras. This is on the second example sheet.
Theorem
(Cartan)
.
The Killing form of a Lie algebra
g
is non-degenerate iff
g
is semi-simple.
Proof.
We are only going to prove one direction if
κ
is non-degenerate, then
g is semi-simple.
Suppose we had an abelian ideal
a g
. We want to show that
κ
(
A, X
) = 0
for all
A a
and
X g
. Indeed, we pick a basis of
a
, and extend it to a basis of
g
. Then since [
X, A
]
a
for all
X g
and
A a
, we know the matrix of
X
must look like
X
=
∗ ∗
0
.
Also, if
A a
, then since
a
is an abelian ideal,
A
kills everything in
a
, and
A
(X) a for all X g. So the matrix must look something like
A
=
0
0 0
.
So we know
A
X
=
0
0 0
,
and the trace vanishes. So
κ
(
A, X
) = 0 for all
X g
and
A a
. So
A
= 0. So
a
is trivial.
Now if
κ
is non-degenerate, then
κ
ab
is “invertible”. So we can find a
κ
ab
such that
κ
ab
κ
bc
= δ
c
a
.
We can then use this to raise and lower indices.
Note that this definition so far does not care if this is a real or complex
Lie algebra. From linear algebra, we know that any symmetric matrix can be
diagonalized. If we are working over
C
, then the diagonal entries can be whatever
we like if we choose the right basis. However, if we are working over
R
, then
the number of positive (or negative) diagonal entries are always fixed, while the
magnitudes can be whatever we like, by Sylvester’s law of inertia.
Thus in the case of a real Lie algebra, it is interesting to ask when the matrix
always has the same sign. It turns out it is the case with always negative sign is
the interesting case.
Definition
(Real Lie algebra of compact type)
.
We say a real Lie algebra is of
compact type if there is a basis such that
κ
ab
= κδ
ab
,
for some κ R
+
.
By general linear algebra, we can always pick a basis so that κ = 1.
The reason why it is called “compact” is because these naturally arise when
we study compact Lie groups.
We will note the following fact without proof:
Theorem.
Every complex semi-simple Lie algebra (of finite dimension) has a
real form of compact type.
We will not use it for any mathematical results, but it will be a helpful thing
to note when we develop gauge theories later on.