# Why gauge fields are traceless Hermitian?

So I've had a read of this, and I'm still not convinced as to why gauge fields are traceless and Hermitian. I follow the article fine, it's just the section that says "don't worry about this complicated maths, the point is that the gauge field is in the Lie algebra". Well yes exactly, but what I'm unsure about is why the gauge field is in the Lie algebra.

1. Is it true that the set of $N\times N$ matrices that are Hermitian and traceless is equal to $\mathfrak{su}(N)$, given the matrix commutator as the bracket? The article I linked shows that all elements of $\mathfrak{su}(N)$ are Hermitian and traceless, but not that all Hermitian traceless $N \times N$ matrices are necessarily in $\mathfrak{su}(N)$.

2. The article appears to show that the gauge field is in $\mathfrak{su}(N)$ by it's transformation properties, but I don't really understand. To me, to show that that a matrix was in $\mathfrak{su}(N)$ by its transformation properties would mean showing that the matrix generated an element of $SU(N)$ when exponentiated, but the transformation law written in the article looks to me to be that of a connection (eg. in GR but for a gauge group). Can someone explain this to me?

3. And if the gauge field is in the Lie algebra, this means that it generates a gauge transformation. Is this transformation in some sense special? I envisage a set of transformations indexed by $\mu$ something like $M_\mu = \exp \left(-i g A_\mu\right)$

4. I also see that the field strength tensor is traceless Hermitian. I can see that it's Hermitian, but not that it's traceless. Is $[A, B]$ traceless given that $A$ and $B$ are traceless? Also this also means that the field strength tensor is in the Lie algebra (presumably), so does that also generate a transformation?

1. Every Hermitian traceless matrix $H$ is in $\mathfrak{su}(N)$ since $\mathrm{Tr}(H) = 0$ and so $$\exp(\mathrm{tr}(\mathrm{i}H)) = \det(\exp(\mathrm{i}H)) = 1$$ so $\exp(\mathrm{i}H)$ is unitary with determinant $1$, hence in $\mathrm{SU}(N)$.

2. The gauge field is always in the Lie algebra of the gauge group since it is introduced to cancel terms that are Lie algebra-valued, see my answer here. The gauge transformations $M_\mu$ are not special in any way I know of. Very similar looking to this, and indeed special, are the holonomies $h(\gamma)$ along paths $\gamma$ on spacetime, which are essentially the group elements given by $$h(\gamma) = \exp(\mathrm{i}\int_\gamma A)$$

3. Since $$\mathrm{Tr}([A,B]) = \mathrm{Tr}(AB) - \mathrm{Tr}(BA)$$ and $\mathrm{Tr}(AB) = \mathrm{Tr}(BA)$, the trace of every commutator is zero, so you can explicitly see that $F$ is traceless for $\mathrm{SU}(N)$ theories. More generally, $F$ is $\mathfrak{g}$-valued simply because it is the gauge covariant derivative of the gauge field: $$F = \mathrm{d}A + A\wedge A$$ Both summands are $\mathfrak{g}$-valued $2$-forms, so the sum also is.

4. For the connection (pun intended) to GR, have a look at my answer here describing the Christoffel symbols as components of a gauge field. Gauge fields are connections on principal bundles, see my basic description of the build-up of gauge theory here.

(I apologize for the many self-references, but I didn't feel like typing all that again.)

I think it's easier to see this if you start from the matter representations rather than the vector field side.

For example, imagine that you have some matter field $\psi$ that transforms under some simple Lie group $G$ representation according

$$\psi \to g \psi$$ where $g \in G$.

Now, the derivative term is not invariant if $g=g(x)$ as one has

$$\overline \psi \gamma^\mu \partial_\mu \psi \to \overline \psi \gamma ^\mu g \partial_\mu \psi + \overline \psi \gamma^\mu \overline g (\partial_\mu g )\psi$$

where $\overline g$ is such that $\overline \psi \to \overline \psi \overline g$ and $\overline g g = 1$ (i.e., $\overline g$ is $g^\dagger$ for $SU(n)$ and $g^T$ for $SO(n)$).

In order to make the derivative term you have to include a vector field that would assimilate

$$\overline \psi \gamma^\mu \overline g (\partial_\mu g )\psi$$

such that you can define a covariant derivative that would transform

$$D_\mu \psi \to g D_\mu \psi$$

as the matter field.

This is done by considering a vector field $A_\mu$ such that

$$D_\mu = \partial_\mu + i A_\mu$$

if it transforms as $$A_\mu \to g A_\mu \overline g -i(\partial _\mu g) \overline g$$ then the term $$\overline \psi \gamma^\mu D_\mu \psi$$ is invariant.

Now notice that the new field transform as an element of the Lie algebra as

$$Ad_g X = g X \overline g$$

is the action of the action of the group when acting on the adjoint representation (and the non-homogeneous part is the gauge transformation in the sense it maps equivalent configurations in the field space).

As such the, in my opinion, least confusing way to state this is that the gauge field transforms under the adjoint representation of the group acting on the matter fields. Therefore, if the group is $SU(n)$ then the gauge fields' representation is that of $n\times n$ traceless and Hermitian matrices.

Furthermore, you can easily show that if $A_\mu$ lives in the Lie algebra then the Field strength tensor will do so as well almost immediately as

$$F_{\mu\nu} \propto [D_\mu,D_\nu] \to g [D_\mu,D_\nu]\overline g$$ transforms in the adjoint representation.

In my opinion, the sentence

And if the gauge field is in the Lie algebra, this means that it generates a gauge transformation.

is not correct as the gauge field and the matter fields have to transform jointly in order to keep the theory gauge-invariant.

Another thing you mentioned was the parallelism with GR. The best way to understand this (at least classically) is to construct a principle bundle with some group $G$. In this construction, the field $A_\mu$ can be seen as the connection (geometrical quantity) of the bundle. The gauge transformation is then a map between Fibres, to which the gauge field (the connection) has to transform accordingly. In this view you can also see that the gauge field does not generate gauge transformations, but it transforms under gauge transformations.

DISCLAIMER: I normally mess up with signs and $i$ factors when talking about these things. I didn't use a unique reference and did most of it from the top of my head.