Irreducible representations in group theory

Question

I am currently working through a chapter of my quantum mechanics textbook about how group theory relates to quantum mechanics (I considered asking this question on the physics exchange, but figured you all would know more). In the discussion, they prove the following theorem: if the representation of a group $G$ of order $h$ is a direct sum of irreducible representations $i$ with size $l_i$, then $$\sum_il_i^2=h.$$When I tried to test this theorem on some common groups, though, it seems to fall apart. Specifically, consider the group of symmetry transformations $C_{3v}$, the symmetry transformations of an equilateral triangle (three reflections, two rotations, and the identity). If we select the basis functions $f_1=x$, $f_2=y$ and $f_3=z$, the text showed that the representation of this group is the direct sum of a two-dimensional irreducible representation relating $f_1$ and $f_2$ and a one-dimensional irreducible representation for $f_3$ (this is pretty obvious, since transformations in the $xy$-plane don't change $z$). However, it seems like the application of the formula mentioned above should yield that $1^2+2^2=h$ ($1$ and $2$ being the dimensions of the irreducible representations). But $1^2+2^2$ is $5$, whereas $h$ is $6$ in this case, so the theorem seems to fail. What am I doing wrong here?

Answer 1

I think you are getting confused about what the theorem is saying. It says that the sum of the squares of all the irreducible representations equals the order of the group.

You took a single representation of $C_{3v}$ and decomposed it into 2 irreducible representations of dimensions 1 and 2. However, these are not all the irreducible representations of $C_{3v}$! Besides the two you have listed, there is one more irreducible representation of $C_{3v}$. It is the one-dimensional representation which sends rotations to 1 and reflections to -1 (called the alternating representation). So in total we have $1^2 + 1^2 + 2^2 = 6$ as expected.