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I can't find any reference which mentions the dependence of the theorem on spacetime dimension, but it would be very interesting to know what if any it has!

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With one minor qualification, the answer is yes: the CPT theorem holds for all spacetime dimensions.

The qualification is that the P in CPT should be interpreted as a reflection of an odd number of spatial dimensions. (The simplest choice is to reflect just one spatial dimension.) If the total number of spatial dimensions is odd, then this is the same as reflecting all of them; but if the total number of spatial dimensions is even, the distinction is important. That's because a reflection of an even number of spatial dimensions can be un-done by ordinary rotations that are continuously connected to the identity. The P in CPT should not be continuously connected to the identity.

To address this, Witten suggests using a new symbol $R$ for a reflection of a single spatial dimension, so the CPT theorem would be called the CRT theorem. Here's an excerpt from page 5 in "Fermion Path Integrals And Topological Phases", http://arxiv.org/abs/1508.04715, where $D$ denotes the number of spacetime dimensions:

...symmetry that in $D = 4$ is usually called CPT (where C is charge conjugation and P is parity or spatial inversion). However, this formulation is not valid for odd $D$ (even spatial dimension $d$), since in that case P is contained in the connected part of the spatial rotation group and should be replaced by an operation that reverses the orientation of space. To use a language that is equivalent to the usual CPT statement for even $D$ but is uniformly valid for all $D$, we will refer instead to CRT symmetry, where R is a reflection of one spatial coordinate. CRT symmetry is always valid in a relativistic theory of any dimension...

Some differences between the effect of CPT in $4k$ and $4k+2$ spacetime dimensions are mentioned in section 4 of Alvarez-Gaumé and Witten (1983), "Gravitational anomalies," Nuclear Physics B 234: 269-330.

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    $\begingroup$ Thank you! This question came out of my curiosity for how a 3d dirac field transforms under parity, so your reference was perfect. $\endgroup$ – LucashWindowWasher Jun 27 at 23:02

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