# Why do covariantly constant spinors on the circle require periodic boundary conditions?

I am asking in the context of a paper of Witten on instability of Kaluza-Klein spacetimes (https://www.sciencedirect.com/science/article/pii/0550321382900074).

The discussion involves applying Witten's positive energy theorem to the space $$\mathbb{R}^{3,1}\times S^1$$. The proof involves (zero-mass) solutions to the Dirac equation on a spacelike hypersurface $$\Sigma \subset \mathbb{R}^{3,1} \times S^1$$-- e.g. the solutions of $$i\gamma^aD_a \varepsilon = 0$$ on the space $$\mathbb{R}^3 \times S^1$$ (where the sum over $$a$$ runs over directions tangent to $$\mathbb{R}^3 \times S^1$$, but the gamma matrices are defined for the entire $$\mathbb{R}^{3,1}\times S^1$$ spacetime).

Witten claims that such a covariantly constant spinor can only exist if the solution $$\varepsilon$$ has periodic boundary conditions on the $$S^1$$ (as opposed to $$4\pi$$ periodic/$$2\pi$$ antiperiodic boundary conditions like we have on the plane). I am failing to see how this is implied by the Dirac equation above.

The claim is certainly true, and I think can be understood by considering supersymmetry. For supersymmetry to be preserved by a compactification $$M^4 \times X$$, there must exist a covariantly constant spinor on $$X$$. From a spacetime perspective, this ensures that the variation of the fermions vanishes in the background. From a worldsheet perspective, supersymmetry is only preserved in the action if the boundary conditions are chosen to be periodic. Hence if supersymmetry is preserved, then boundary conditions must be periodic and there exists a covariantly constant spinor, as desired. My trouble is that this line of reasoning appeals to a supersymmetric theory, while the claim should be true regardless?

How can I show that solutions of $$i\gamma^aD_a \varepsilon = 0$$ on the space $$\mathbb{R}^3 \times S^1$$ require the choice of periodic boundary conditions on the $$S^1$$?

• Either this is trivial, or I am missing the point of the question. – Buzz Jun 9 '20 at 0:51
• It’s not trivial to me—would you mind explaining what you have in mind as a trivial explanation? – hulsey Jun 9 '20 at 3:37
• I’ve edited to clarify that with spinors, there is always a choice of anti periodic versus periodic boundary conditions on the circle. – hulsey Jun 9 '20 at 4:20

Buzz is right; I've realized that covariantly constant solutions on $$S^1$$ are precisely constants, which of course need periodic boundary conditions.