Graduate Is the kinetic mixing gauge-invariant for non-Abelian gauge fields?

[Ramtin123]

Consider two non-Abelian gauge fields $A_\mu^a$ and $A_\mu^{'a}$ belonging to the same symmetry group. An example could be the SM electroweak isospin fields and another exotic SU(2) hidden sector where $a=1, \dots 3$.
Is the kinetic mixing of the following form gauge-invariant?
$$ F_{\mu\nu}^a F^{'a\mu\nu} $$
where $F_{\mu\nu}^a$ and $F_{\mu\nu}^{'a}$ denote the corresponding field strength tensors.
What is the difference between this case and a usual non-Abelian kinetic term $F_{\mu\nu}^a F^{a\mu\nu}$?

 

[malawi_glenn]

You need to have a different "gauge index" on the F' tensor than a. I think... But gauge fields always transform under the fundemantal representation, so I think FF' should be gauge invariant. Here is a paper on it https://arxiv.org/abs/1604.00044

We do have kinetic mixing if we have more than one abelian gauge symmetry.

 

[Ramtin123]

You need to have a different "gauge index" on the F' tensor than a. I think... But gauge fields always transform under the fundemantal representation, so I think FF' should be gauge invariant. Here is a paper on it https://arxiv.org/abs/1604.00044

We do have kinetic mixing if we have more than one abelian gauge symmetry.

That paper discusses kinetic mixing of an Abelian U(1) gauge field with the electroweak isospin fields as shown in equations (1.1) and (1.3). The Abelian field strength tensor $X_{\mu\nu}$ is gauge invariant. This is not true for non-Abelian field strength tensor $F^a_{\mu\nu}$. But the bilinear $F_{\mu\nu}^a F^{a\mu\nu}$ is gauge-invariant.
In this paper Arxiv 2104.01871 [hep-ph] , in the introduction, it is claimed that non-Abelian kinetic mixing is not gauge invariant. But the author does not explain why.

• The gauge indices are contracted $tr (F_{\mu\nu}^a T^a F^{ 'b\mu\nu} T^b) = \frac{1}{2} F_{\mu\nu}^a F^{ 'a\mu\nu}$
• I think you mean "gauge fields always transform under the adjoint representation".

 

[malawi_glenn]

gauge fields always transform under the adjoint representation".

Yes I did.

I did some calculations, the coupling constant $g$ is included in the transformation matrices, $U(x) = \text{exp}(- \text{i} g\theta(x)^a T^a)$.

$\text{Tr} ( F^{\mu \nu} B_ {\mu \nu}) \to \text{Tr} (U F^{\mu \nu} U^\dagger V B_ {\mu \nu} V^\dagger) \neq \text{Tr} ( F^{\mu \nu} B_ {\mu \nu})$ because $U^\dagger V \neq I$ and $U^\dagger V \neq I$ because of different gauge-couplings.