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I am unable to understand why the gauge field must be periodic along the compactified direction in finite temperature field theory. I get that whatever quantity that is measurable and gauge invariant along that direction must be periodic. But that means that the gauge field must be periodic only upto a gauge transformation and not strictly periodic. This is actually the condition used by 't Hooft when formulating gauge theory on $T^4$, where all the four directions are compactified. So, why is the condition different when one direction is compactified?

In terms of equations, I'm saying that the boundary condition must be $$A_{\mu}(\vec{x},\tau + \beta)=\omega A_{\mu}(\vec{x},\tau)\omega^{-1}+\omega\partial_{\mu}\omega^{-1}$$ instead of $$A_{\mu}(\vec{x},\tau +\beta)=A_{\mu}(\vec{x},\tau)$$ where $\omega$ belongs to the gauge group under consideration, say $SU(2)$. It would be great if someone could help me clear up this confusion.

For reference, I am studying from the paper by Harrington and Shepard called "Periodic Euclidean solution and finite temperature Yang-Mills gas". They also have another paper called "Euclidean solutions and finite temperature gauge theory" that talks about this aspect more.

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't Hooft, in "A property of electric and magnetic flux in non-Abelian gauge theories'', explains how to compute the free energy of a configuration with $(n,m)$ units of electric/magnetic flux by considering twisted boundary conditions.

In thermal field theory, we wish to compute the partition function $$ Z = \mathrm{Tr}[\exp(-\beta H)] $$ in the zero flux sector (in particular, with Gauss law imposed on the states in the trace). As explained, for example, in the classic paper by Gross, Pisarski, and Yaffe: "QCD and instantons at finite temperature", this corresponds to strictly periodic gauge fields.

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  • $\begingroup$ Would it be accurate to say that the periodicity is what implements the trace? $\endgroup$ Apr 10, 2020 at 16:20
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    $\begingroup$ @ChiralAnomaly Sort of (that's waht I say if I teach a course), but it's a little more subtle. To compute Z we need to fix the gauge and construct H. Say we pick temporal gauge, so the "coordinates" are $\vec{A}$. Then indeed $\vec{A}$ should be periodic, but what about $A_0$? The answer is that $A_0$ appears as the lagrange multiplier for implementing Gauss law, and to see that it has to be periodic we really have to write this down carefully. $\endgroup$
    – Thomas
    Apr 10, 2020 at 17:21

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