Fundamental and Adjoint Representation of Gauge Groups

[shirosato]

Basic question, but nevertheless.

In a non-Abelian gauge theory, the fermions transform in the fundamental representation, i.e. doublets for SU(2), triplets for SU(3), while the gauge fields transform in the adjoint representation, which can be taken straight from the structure constants of the theory. From my understanding, in the adjoint representation, the group transformations can be represented as d-dimensional matrices, where d is the number of generators of the group, i.e. 3x3 for SU(2) and 8x8 for SU(3). Would the fields then be seen as vectors on which these matrices act? I think I'm severely misunderstanding something.

What is the physical relevance in the way fields transform? Is it manifested in the interactions? I have seen a technicolour scenario where the fermions transform in the adjoint representation and that seriously changes things. Anyway, any help would be appreciated.

 

[tom.stoer]

In the adjont rep. the gauge field strength are defined as

$$F_{\mu\nu}(x) = F_{\mu\nu}^a(x) T^a$$

The transformation of field strength in the adjoint rep. is

$$F_{\mu\nu}(x) \to U^\dagger(x) F_{\mu\nu}(x)U(x) = F^a(x) U^\dagger(x) T^aU(x)$$

I think this makes clear that the transformation is nothing else but a "rotation" of the basis vectors, i.e. the generators. The vector space on which the adoint rep. of the gauge group acts is the algebra spanned by the generators itself; that is somehow the definition of the adjoint rep.; in all other reps. the vector space is something different.

So the vectors are the elements of the Lie algebra (we do not care about the fact that there are space-time vectors as well; this is not relevant here). The generators are the basis vectors spanning the algebra. They allow for the construction of a linear vector space where you can add elements of the Lie algebra like

$$\theta(x) = \theta^a(x) T^a$$

The vector space has an inner product; this is defined as

$$\langle\theta|\eta\rangle = tr\theta\eta = \theta^a \eta^b tr[T^a, T^b] = \theta^a \eta^b \frac{1}{2}\delta_{ab} = \theta^a \eta^a$$

The inner product in the Lie algebra can essentially be calculated component-wise. This is the meaning that the Lie algebra is a vector space and that the fields transform as vectors under the Lie group.

The inner product is invariant under "rotations" as can be seen via the trace identity

$$\langle\theta^\prime|\eta^\prime\rangle = tr U^\dagger\theta U U^\dagger \eta U = tr\theta\eta$$

Therefore the gauge field term in the Lagrangian

$$\frac{1}{2}tr F_{\mu\nu}(x) F^{\mu\nu} = \frac{1}{4}F_{\mu\nu}^a(x) F^{\mu\nu}^a(x)$$

is nothing else but the invariant length of the "vector F".

Strictly speaking the gauge potentials live not in the adjoint rep. as they transform not only via a pure rotation but have the additional derivative term; nevertheless they can be expanded locally just like the field strength. This is the mathematical reason why a quadratic mass term for the gauge fields breaks gauge symmetry: the term

$$\frac{1}{2}tr A_{\mu}(x) A^{\mu}(x)$$

is not the "invariant length" of the "vector A"; the gauge potential is simply not a vector in this vector space.

 

[tom.stoer]

Why am I no longer allowed to edit my posts?

In the formula regarding the inner product the commutator $$[T^a, T^b]$$ is wrong, it must read simply $$T^aT^b$$

 

[arivero]

On the same topic, it is a good oportunity to explain the uses of real, complex and conjugate

 

[shirosato]

Thank you, that was pretty helpful. But what about the fundamental representation? I know its a bit odd to ask, but is there any physical intuition about any of this or some basic insight to how this was all deduced?

 

[tom.stoer]

You should read about the "eightfold way" http://en.wikipedia.org/wiki/Eightfold_Way_(physics)