# A question to gauge fixing in nonabelian gauge theories

In quantum gauge theories it is usual to fix the gauge with the equation $\partial^\mu A_\mu = 0$ where $A_\mu$ is the gauge connection. From this gauge fixing condition the remaining gauge degree of freedom can be determined.

Unfortunaitly, there is appearing a determinant factor (Faddeev-Popov determinant) in the Feynman path integral coming from the change in the Haar measure due to this gauge fixing condition.

It is clear that without gauge fixing one integrates infinitely many (in the Feynman path integral) over the same contribution $e^{iS}$ since $S$ doesn't change under a gauge transformation. Is it possible to insert a gauge fixing factor $\delta(g-g_0)$ where $g$ is the gauge group and $g_0$ is a constant gauge group in the path integral, i.e. is it possible to restrict the gauge group to only a fixed function such that any other possibilities are not respected?

Or more precisely: What gauge fixing condition must be applied on $A_\mu$ to impose that the gauge group is fixed to only a single function over the spacetime?

EDIT (ATTEMPT): I am thinking about the gauge fixing condition $n^\mu A_\mu = 0$ with $n^\mu$ as a 4-vector function; can it satisfy above conditions? May be the gauge transformation given explicitely by $A_\mu \mapsto A_\mu + \Xi_\mu$ then it must also hold $n^\mu \Xi_\mu = 0$ and then if $\Xi^\mu$ is a fixed function, the 4-vector $n^\mu$ is chosen such that above gauge condition is satisfied.

• I'm not sure what your question is. The sentence "impose that the gauge group is fixed to only a single function over the spacetime" does not make any sense to me - the gauge group is a group, it is not a function of anything. – ACuriousMind Aug 27 '15 at 14:05
• More precisely, I asked how the gauge fixing condition must be chosen to arrive at a uniquely determined gauge group. The gauge group can vary in spacetime but is uniquely determined by a gauge fixing relation. – kryomaxim Aug 27 '15 at 14:09
• Your terminology is...non-standard. The gauge group does not vary at all, it is just something like $\mathrm{SU}(n)$. It's uniquely determined. Are you asking what the conditions must fulfill to have a unique gauge field $A$ as their solution? (You can't do that in general, search for "Gribov problem" or "Gribov ambiguity") – ACuriousMind Aug 27 '15 at 14:11
• I am asking what gauge fixing function $g(A_\mu)=0$ I have to use to impose the condition that the gauge group is uniquely determined. In the Feynman path integral I should integrate only over 1 gauge equivalence class in the final gauge-fixed version. – kryomaxim Aug 27 '15 at 14:27

As a last remark, a consistency requirement on the gauge condition typically requires it to be of "derivative type", i.e. it has to fulfill a certain differential equation and will typically contain derivatives of the gauge field. A condition like $n^\mu A_\mu = 0$ will usually not even locally choose a unique point in a gauge orbit, and may not be consistently preserved under time evolution.