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Let us think of the Euclidean Dirac operator $iD^k \gamma_k$ on the rectangle $[-1,1]^4$ with the periodic boundary conditions.

The covariant derivative $iD^k$ carries a gauge potential term and we assume that the term is continuous and supported on $[-1,1]^4$.

From this information, can we conclude that this operator on $L^2[-1,1]^4 \otimes \mathbb{C}^4$ is self-adjoint? Also, can we compute its spectrum explicitly?

There seems not much reference on the Dirac operator contained within the periodic box. Could anyone please help me?

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  • $\begingroup$ Does "periodic boundary conditions" mean the domain is $(S^1)^4$? $\endgroup$
    – Ryan Budney
    Commented Apr 16, 2022 at 22:43
  • $\begingroup$ For general gauge potential, you'll have a hard time finding explicit eigenfunctions. $\endgroup$
    – Michael Engelhardt
    Commented Apr 17, 2022 at 3:05
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    $\begingroup$ The statement on $S^1 $ does not generalize to higher dimensions. You can only use a gauge transformation to get rid of one of the components of the gauge potential (modulo a zero mode). $\endgroup$
    – Michael Engelhardt
    Commented Apr 18, 2022 at 13:55
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    $\begingroup$ My comment had nothing to do with the asymptotic freedom of QCD. I merely meant that the spectrum does become $2\pi n$ for large $n$ to leading order, since the gauge potential will only give corrections of order one. $\endgroup$
    – Michael Engelhardt
    Commented Apr 18, 2022 at 18:10
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    $\begingroup$ There is not much to say - say you have found an eigenfunction. If you multiply it by the fixed external gauge potential, the result is bounded, as opposed to the contribution from the derivative part of the Dirac operator, which diverges as $n$. $\endgroup$
    – Michael Engelhardt
    Commented Apr 19, 2022 at 1:04

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