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  The group generated by CPT and the full Lorentz group

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We know the Lorentz group is O(3,1) in 4 dimensional spacetime.

We know that there are 4 disconnected components in the Lorentz group O(3,1), and https://math.stackexchange.com/questions/2204349/difference-between-the-lorentz-group-and-the-restricted-lorentz-group

$$\pi_0(\mathrm{O}(1,3))  \cong \mathbb{Z}_2\times\mathbb{Z}_2.$$

My question is that in QFT we have a discrete charge conjugation symmetry C, parity P and time reversal T.

We know the P and T flips the 4 disconnected components of O(3,1) into each other.

But the charge conjugation C seems not to be in the Lorentz group. Is the group generated by CPT and the Lorentz group O(3,1) the group $$\mathrm{O}(1,3)) \times \mathbb{Z}_2 ?$$

where the new $\mathbb{Z}_2$ is a charge conjugation C symmetry?

How do we understand the total group of CPT and the Lorentz group O(3,1)?

asked Sep 23 in Theoretical Physics by annieheart (15 points) [ revision history ]
edited Sep 25 by Dilaton

If P and T operations "act" on space-time variables $\vec{x}$ and $t$, the charge conjugation operation C acts on solutions, i.e., on more complicated combinations of $\vec{x}$ and $t$. So do the Lorentz transformations, by the way.

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