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Under parity, a four-vector $V^{\mu}=(V^0,\boldsymbol{V})$ transforms as

$$(V^0,\boldsymbol{V})\rightarrow(V^0,-\boldsymbol{V})$$

which makes sense as parity only reverses the spatial components. However, a pseudo (or axial) four-vector transforms as

$$(V^0,\boldsymbol{V})\rightarrow(-V^0,\boldsymbol{V})$$

so the time component is reversed. Why does parity, which I thought only involved space, change something to do with time?

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  • $\begingroup$ It does so for relativistic covariance. Recall you call $\bar \psi \gamma^\mu \psi$ a vector and $\bar \psi \gamma^5\gamma^\mu \psi$ an axial. $\endgroup$
    – Cosmas Zachos
    Dec 26, 2021 at 22:31
  • $\begingroup$ Linked. $\endgroup$
    – Cosmas Zachos
    Dec 26, 2021 at 22:37
  • $\begingroup$ If you wrote precisely an example of the four-axial you have in mind, like the spinor bilinear in my comment above, you’d observe its PT transformation properties, not just P. $\endgroup$
    – Cosmas Zachos
    Dec 26, 2021 at 23:43
  • $\begingroup$ Is it obvious that relativistic covariance makes it transform like this then? For example, why can't there be a $\gamma^6$ such that $\bar \psi \gamma^6\gamma^\mu \psi$ is invariant under parity but is still a four-vector under proper orthochronous transformations? (I know that taking antisymmetric combinations of $\gamma$ matrices exhausts all possibilities but I am looking for a more fundamental argument) $\endgroup$
    – Ghorbal
    Dec 27, 2021 at 0:35
  • $\begingroup$ There must be an argument for the nonexixtence of such a γ, linked to the CPT theorem, but I’m not au courant… I’m just describing our world… $\endgroup$
    – Cosmas Zachos
    Dec 27, 2021 at 2:26

1 Answer 1

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One way of looking at it: a pseudovector in 4 dimensions is really a trivector. The space of pseudovectors is spanned not by $\hat t, \hat x, \hat y, \hat z$ but by $\hat x{\wedge}\hat y{\wedge}\hat z,\;\hat y{\wedge}\hat z{\wedge}\hat t,\;\hat z{\wedge}\hat t{\wedge}\hat x,\;\hat t{\wedge}\hat x{\wedge}\hat y$. Parity takes $\hat x{\wedge}\hat y{\wedge}\hat z$ to $(-\hat x){\wedge}(-\hat y){\wedge}(-\hat z) = -(\hat x{\wedge}\hat y{\wedge}\hat z)$, and $\hat y{\wedge}\hat z{\wedge}\hat t$ to $(-\hat y){\wedge}(-\hat z){\wedge}\hat t = \hat y{\wedge}\hat z{\wedge}\hat t$, and so on.

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  • $\begingroup$ A neat summary! Since $\gamma^\mu \gamma^5\propto \epsilon ^{\mu\nu\kappa\lambda} \gamma_\nu \gamma_\kappa \gamma_\lambda$, it is manifest that the spacelike components are invariant under P, and the timelike one flips... $\endgroup$
    – Cosmas Zachos
    Dec 30, 2021 at 16:02

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