I'm a mathematician interested in abstract QFT. I'm trying to undersand why, under certain (all?) circumstances, we must have $T^2 = -1$ rather than $T^2 = +1$, where $T$ is the time reversal operator. I understand from the Wikipedia article that requiring that energy stay positive forces $T$ to be represented by an anti-unitary operator. But I don't see how this forces $T^2=-1$. (Or maybe it doesn't force it, it merely allows it?)

Here's another version of my question. There are two distinct double covers of the Lie group $O(n)$ which restrict to the familiar $Spin(n)\to SO(n)$ cover on $SO(n)$; they are called $Pin_+(n)$ and $Pin_-(n)$. If $R\in O(n)$ is a reflection and $\tilde{R}\in Pin_\pm(n)$ covers $R$, then $\tilde{R}^2 = \pm 1$. So saying that $T^2=-1$ means we are in $Pin_-$ rather than $Pin_+$. (I'm assuming Euclidean signature here.) My question (version 2): Under what circumstances are we forced to use $Pin_-$ rather than $Pin_+$ here?

(I posted a similar question on physics.stackexchange.com last week, but there were no replies.)

EDIT: Thanks to the half-integer spin hint in the comments below, I was able to do a more effective web search. If I understand correctly, Kramer's theorem says that for even-dimensional (half integer spin) representations of the Spin group, $T$ must satisfy $T^2=-1$, while for the odd-dimensional representations (integer spin), we have $T^2=1$. I guess at this point it becomes a straightforward question in representation theory: Given an irreducible representation of $Spin(n)$, we can ask whether it is possible to extend it to $Pin_-(n)$ (or $Pin_+(n)$) so that the lifted reflections $\tilde R$ (e.g. $T$) act as an anti-unitary operator.

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