7Gauge theories

III Symmetries, Fields and Particles 7.2 General case
We now want to do what we had above for a general Lie group
G
. In particular,
this can be non-abelian, and thus the Lie algebra has non-trivial bracket. It
turns out in the general case, it is appropriate to insert some brackets into what
we have done so far, but otherwise most things follow straight through.
In the case of electromagnetism, we had a Lie group U(1), and it acted on
C
in the obvious way. In general, we start with a Lie group
G
, and we have
to pick a representation
D
of
G
on a vector space
V
. Since we want to have a
real Lagrangian, we will assume our vector space
V
also comes with an inner
product.
Given such a representation, we consider fields that take values in
V
, i.e.
functions
φ
:
R
3,1
V
. We can, as before, try to write down a (scalar)
Lagrangian
L
φ
= (
µ
φ,
µ
φ) W ((φ, φ)).
Writing out the summation explicitly, we have
L
φ
=
3
X
µ=0
(
µ
φ,
µ
φ) W ((φ, φ)),
i.e. the sum is performed outside the inner product.
This is not always going to be invariant under an arbitrary action of the
group
G
, because the inner product is not necessarily preserved. Fortunately, by
definition of being unitary, what we require is that our representation is unitary,
i.e. each D(g) is unitary for all g G.
It is a theorem that any representation of a compact Lie group is equivalent
to one that is unitary, so we are not losing too much by assuming this (it is also
that a non-compact Lie group cannot have a non-trivial unitary representation).
transformations. Near the identity, we can write
g = exp(X)
for some X g. Then we can write
D(g) = exp(ρ(X)) gl(n, C).
We write
ρ
:
g gl
(
n, C
) for the associated representation of the Lie algebra
g
. If
D
is unitary, then
ρ
will be anti-Hermitian. Then the infinitesimal transformation
is given by
φ 7→ φ + δ
X
φ = φ + ρ(X)φ.
In general, an infinitesimal Gauge transformation is then given by specifying a
function
X : R
3,1
g.
Our transformation is now
δ
X
φ = ρ(X(x))φ.
Just as in the abelian case, we know our Lagrangian is no longer gauge invariant
in general.
We now try to copy our previous fix. Again, we suppose our universe comes
with a gauge field
A
µ
: R
3,1
L(G).
Again, we can define a covariant derivative
D
µ
φ =
µ
φ + ρ(A
µ
)φ.
In the case of a non-abelian gauge symmetry, our gauge field transformed under
X as
δ
X
A
µ
=
µ
X.
This expression still makes sense, but it turns out this isn’t what we want, if
we try to compute
δ
X
(D
µ
φ
). Since the gauge field is non-abelian, there is one
extra thing we can include, namely [
X, A
µ
]. It turns out this is the right thing
we need. We claim that the right definition is
δ
X
A
µ
=
µ
X + [X, A
µ
].
To see this, we show that
Proposition. We have
δ
X
(D
µ
φ) = ρ(X)D
µ
φ.
The proof involves writing out the terms and see that it works.
Proof. We have
δ
X
(D
µ
φ) = δ
X
(
µ
φ + ρ(A
µ
)φ)
=
µ
(δ
X
φ) + ρ(A
µ
)δ
X
φ + ρ(δ
X
A
µ
)φ
=
µ
(ρ(X)φ) + ρ(A
µ
)ρ(X)φ ρ(
µ
X)φ + ρ([X, A
µ
])φ
= ρ(
µ
X)φ + ρ(X)
µ
φ + ρ(X)ρ(A
µ
)φ
+ [ρ(A
µ
), ρ(X)]φ ρ(
µ
X)φ + ρ([X, A
µ
])φ
= ρ(X)(
µ
φ + ρ(A
µ
)φ)
= ρ(X)D
µ
φ,
as required.
Thus, we know that (D
µ
φ,
D
µ
φ
) is gauge invariant. We can then write down
a gauge invariant “matter” part of the action
L
φ
= (D
µ
φ, D
µ
φ) W [(φ, φ)].
Finally, we need to produce a gauge invariant kinetic term for
A
µ
:
R
3,1
L
(
G
).
For the case of an abelian gauge theory, we had
F
µν
=
µ
A
ν
ν
A
µ
L(G).
In this case, there is one extra term we can add, namely [
A
µ
, A
ν
]
g
, and it
turns out we need it for things to work out. Our field strength tensor is
F
µν
=
µ
A
ν
ν
A
µ
+ [A
µ
, A
ν
].
How does the field strength tensor transform? In the abelian case, we had that
F
µν
was invariant. Let’s see if that is the case here. It turns out this time the
field strength tensor transforms as the adjoint representation.
Lemma. We have
δ
X
(F
µν
) = [X, F
µν
] L(G).
Proof. We have
δ
X
(F
µν
) =
µ
(δ
X
A
ν
)
ν
(δ
X
A
µ
) + [δ
X
A
µ
, A
ν
] + [A
µ
, δ
X
A
ν
]
=
µ
ν
X +
µ
([X, A
ν
])
ν
µ
X
ν
([X, A
µ
]) [
µ
X, A
ν
]
[A
µ
,
ν
X] + [[X, A
µ
], A
ν
] + [A
µ
, [X, A
ν
]]
= [X,
µ
A
ν
] [X,
ν
A
µ
] + ([X, [A
µ
, A
n
]]
= [X, F
µν
].
where we used the Jacobi identity in the last part.
So to construct a scalar quantity out of
F
µν
, we need an inner product
invariant under an adjoint representation. We know one of these the Killing
form! So we can just pick
L
A
=
1
g
2
κ(F
µν
, F
µν
).
This is known as the Yang-Mills Lagrangian. Note that for each fixed
µ, ν
, we
have that F
µν
g. So we should read this as
L
A
=
1
g
2
X
µ,ν
κ(F
µν
, F
µν
).
Putting all these together, we get ourselves an invariant Lagrangian of the system
with a G gauge symmetry.
So the final Lagrangian looks like
L =
1
g
2
X
µ,ν
κ(F
µν
, F
µν
) + (D
µ
φ, D
µ
φ) + W ((φ, φ)).
Now if we are further told that we actually have a simple complex Lie algebra,
then it is a fact that we can find a real form of compact type. So in particular
we can find a basis B = {T
a
, a = 1, ··· , dim g} with
κ
ab
= κ(T
a
, T
b
) = κδ
ab
.
In this basis, we have
L
A
=
κ
g
2
d
X
a=1
F
µν,a
F
µν,a
,
and this does look like a copy of d many electromagnetic fields.
So to construct (a sensible) gauge theory, we need to get a (semi-)simple
Lie algebra, and then find some representations of
g
. These are things we have