In the C3v point group, it lists three sigma Vs together in one column (class), and it doesn't list the cVs separately. Consequently there are three irreducible representations.

In the C2v point group, however, it lists 2 separate cVs in separate columns. Consequently there are four irreducible representations.

My question is, why are the cVs grouped together in C3v, but not grouped together in C2v?


1 Answer 1


The $\sigma_\mathrm v$ mirror planes in the $C_\mathrm{3v}$ point group are themselves related by symmetry: note that they can be interchanged via rotation by 120 degrees about the preexisting $C_3$ axis. To be technically precise, they belong to the same conjugacy class, in the sense that applying $C_3$, then one mirror plane, and then the inverse of $C_3$ is the same as applying a different mirror plane:

$$C_3\sigma_{\mathrm v, i}C_3^{-1} = \sigma_{\mathrm v, j}$$

As a result of this, they have to be treated together and not separately, in the sense that the characters in the character table represent how each irrep transforms under all the three mirror planes collectively. Unfortunately, this is hard to explain properly without delving into the actual matrices. But it should be possible to see that the two mirror planes in $C_\mathrm{2v}$ are not related to each other in the same way.

  • $\begingroup$ @KameronShrum, if you think it addressed your queries sufficiently, can I encourage you to accept the answer (green tick near the voting buttons) -- same goes for your other questions, too. See: chemistry.stackexchange.com/help/someone-answers Of course if you don't think it's enough (like if you want a more rigorous explanation), then feel free to not accept it, by all means. I do not desperately need reputation. :-) $\endgroup$ Commented Apr 30, 2020 at 16:21

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