Non-unitary gauge transformation

[DuckAmuck]

TL;DR

What happens for non-unitary gauge transformations when it comes to fermion factors?

You see in the literature that the vector potentials in a gauge covariant derivative transform like:
$$A_\mu \rightarrow T A_\mu T^{-1} + i(\partial_\mu T) T^{-1}$$
Where T is not necessarily unitary. (In the case that it is $T^{-1} = T^\dagger$)
My question is then if T is not unitary, how is $\bar{\psi}$ transforming?
Since
$$\psi \rightarrow T\psi$$
$$\bar{\psi} \rightarrow (T\psi)^\dagger \gamma_0 = \psi^\dagger \gamma_0 \gamma_0 T^\dagger \gamma_0 = \bar{\psi} \gamma_0 T^\dagger \gamma_0$$
This seems to necessitate that $$\gamma_0 T^\dagger \gamma_0 = T^{-1}$$
This can only be the case if $T^\dagger = T^{-1}$, or if $T^\dagger$ is a 4x4 matrix that multiplies with $\gamma_0$ matrices to give the inverse.
Otherwise, in general, you are left with something like:
$$\bar{\psi} (\partial_\mu + i A_\mu) \psi \rightarrow \bar{\psi} \gamma_0 T^\dagger \gamma_0 (\partial_\mu + i T A_\mu T^{-1} - (\partial_\mu T)T^{-1} ) T\psi$$
$$=\bar{\psi} (\gamma_0 T^\dagger \gamma_0 T) (\partial_\mu + i A_\mu) \psi$$
So to be gauge invariant, this object $$\gamma_0 T^\dagger \gamma_0 T = 1$$ but that is not the case in general.
Can this be simplified more, maybe for cases where T is a 2x2 matrix that commutes with $\gamma_0$?

 

[malawi_glenn]

What is $\psi$? A Dirac spinor?

The $\gamma_0$ and $T$ are matrices in different spaces, the $\gamma_0$ acts on spinor index and $T$ on group "space" index. So you do not need to worry about the generator matrix for the group commute with $\gamma_0$ or not.

What group do you have in mind? SO(N)? G_2 ?

 

[DuckAmuck]

What is $\psi$? A Dirac spinor?

The $\gamma_0$ and $T$ are matrices in different spaces, the $\gamma_0$ acts on spinor index and $T$ on group "space" index. So you do not need to worry about the generator matrix for the group commute with $\gamma_0$ or not.

What group do you have in mind? SO(N)? G_2 ?

Yes, of course they are acting on different spaces in most cases. Was trying to keep things very generalized in an attempt to "rescue" invariance, but I think that may be overkill.
And, Psi is indeed a dirac spinor.
My question still remains on what to do about T being non-unitary.
As $A_\mu \rightarrow T A_\mu T^{-1} +i (\partial_\mu T) T^{-1}$
But $\bar{\psi} \rightarrow \bar{\psi} T^\dagger$
So it is not clear why literature asserts that the transformation on A is for non-unitary transformations, because $\bar{\psi}$ tranforms by $T^\dagger$ not $T^{-1}$.
The transformation is something like:
$\bar{\psi} \gamma^\mu (\partial_\mu + iA_\mu)\psi \rightarrow \bar{\psi} \gamma^\mu T^\dagger T(\partial_\mu + iA_\mu)\psi$ which is not invariant if T is not unitary. So what can remedy this if anything?

 

[malawi_glenn]

But you have to choose which representation the dirac field must transform under. And most (compact) Lie Groups have complex representations, but not all of them. This is why $E_8$ is never considered as a model for GUT's because it has no complex representations and you can not couple them to dirac fermions.