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According to https://physics.stackexchange.com/a/488388/42982 we can define $C,P,T$ discrete symmetry transformation in even dimensional spacetime.

How could we write $C,P,T$ symmetry transformation in even dimensional $d$ spacetime on a relativistic Weyl fermion? Say $d=2,4,6,8,10?$ (We consider Weyl instead of Dirac fermion because Weyl fermion is a more basic building block. Two Weyl fermions can give one Dirc fermion.)

p.s. In odd dimensional spacetime, we can define $C$, Reflection, $T$ discrete symmetry transformations. But let us focus on even dimensional spacetime here.

ann marie cœur
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    $P$ can be defined in odd dimensional spacetime as well. It's just that in odd dimensions $P$ is not $\vec{x} \to - \vec{x}$, its $x^1 \to - x^1$ with everything else remaining the same. – Prahar May 05 '21 at 07:09
  • @PraharMitra what property does this definition give you? Why would you say that the object that is defined in this way is $P$? – Prof. Legolasov May 05 '21 at 07:31
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    In odd dimensions $\vec{x} \to - \vec{x}$ is NOT a discrete transformation, it is pure rotation (which is connected to the identity element). To describe $P$ which is a discrete transformation, you need to modify its definition. You can modify it to allow an inversion of odd number of spatial directions. The most obvious choice is simply to invert one spatial direction and I chose $x^1$ for that. Obviously, you are free to choose some other definition if you like as long as it fits the constraints I've described here. – Prahar May 05 '21 at 07:34
  • a question here https://math.stackexchange.com/questions/4247620/pinp-q-and-pinq-p-are-not-isomorphic – Марина Marina S Sep 12 '21 at 13:27

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Time reversal $T$ and parity $P$ are the two generators of the $\mathbb{Z}_2\times \mathbb{Z}_2$ in the short exact sequence $\mathbb{Z}_2\times \mathbb{Z}_2\to\mathrm{O}(p,q) \to \mathrm{SO}^+(p,q)$, i.e. they are precisely the "part" of the full orthogonal group $\mathrm{O}$ that we lose when we pass to the Lie algebra $\mathfrak{so}$/connected component of the identity $\mathrm{SO}^+$ (the "proper orthochronous transformations"). Writing down the discrete $C,P,T$ symmetries for fermions tends to be very annoying because we obtain the Dirac representation via the Clifford algebra $\mathrm{Cl}_\mathbb{C}(p,q)$, but the representation of the Pin groups (covers of the $\mathrm{O}(p,q)$ groups like the spin group for the $\mathrm{SO}(p,q)$) depends on the sign convention because unfortunately $\mathrm{Pin}(p,q)$ is not identical to $\mathrm{Pin}(q,p)$.

The Weyl spinors are projective representations of $\mathrm{SO}^+$, but not of $\mathrm{O}$: The Dirac representation of the Clifford algebra is not reducible to the Weyl representations as a representation of $\mathrm{O}$ - the action of parity on a Dirac term is precisely to "exchange" its Weyl spinor components with each other, see e.g. this answer by Nephente, so the Weyl representations are not invariant subspace with respect to parity.

A way to figure out what the correct representation of time reversal on spinors is is outlined in this answer of mine. Both charge conjugation and parity in arbitrary dimension are discussed at length in "Gamma matrices, Majorana fermions, and discrete symmetries in Minkowski and Euclidean signature" by Mike Stone.

ACuriousMind
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  • Thanks so much +1. "depends on the sign convention because unfortunately Pin(,) is not identical to Pin(,)." => – ann marie cœur May 05 '21 at 16:56
  • "unfortunately Pin(,) is not identical to Pin(,)" => Aren't Pin(,) and Pin(,) isomorphic? – ann marie cœur May 05 '21 at 16:57
  • @anniemarieheart No. They are different central extensions of $\mathrm{O}(p,q)\cong \mathrm{O}(q,p)$ in general – ACuriousMind May 05 '21 at 16:58
  • someone asked a new question here https://math.stackexchange.com/questions/4247620/pinp-q-and-pinq-p-are-not-isomorphic – ann marie cœur Sep 12 '21 at 13:29
  • Hi ACuriousMind♦ and ann, thanks for the interesting question and post --- I suspect that you got the exact sequence incorrect, it should be corrected to $ \mathrm{SO}^+(p,q) \to\mathrm{O}(p,q) \to \mathbb{Z}_2\times \mathbb{Z}_2$, and the second map should be related to the determinant signs of each of two sub-blocks. – wonderich Sep 24 '21 at 03:00
  • @wonderich In general, for a group $G$ and a subgroup $H\subset G$ we have the short exact sequence $H\to G \to G/H$. So there's both a sequence $\mathbb{Z}_2\times\mathbb{Z}_2 \to \mathrm{O}\to\mathrm{SO}^+$ and a sequence $\mathrm{SO}^+\to \mathrm{O}\to\mathbb{Z}_2\times\mathbb{Z}_2$, but the former is what I wanted here since I'm talking about $\mathrm{SO}^+$ as a quotient of $\mathrm{O}$. – ACuriousMind Sep 24 '21 at 07:31
  • I am not 100% sure ℤ2×ℤ2→O→SO+ because I only know SO→O→ ℤ2 as in the figure in https://en.wikipedia.org/wiki/Pin_group#Definite_form. I am not sure that ℤ2 is a normal subgroup of O, since that ℤ2 can be ℤ2-time-reversal that anti-commutes with SO? So O = SO $\rtimes$ ℤ2 is what I 100% certain, not what you wrote. – wonderich Sep 25 '21 at 02:47