I‘m just learning about representation theory so I‘m probably missing something trivial, but I‘d still be very happy if someone could help.
In class we looked at the following corollary:
Every finite group has a finite amount of equivalence classes of irreducible representations. If $\rho_1, \dots,\rho_k$ is a list of non equivalent irreducible representations then: $\sum_{i=1}^k \dim(\rho_i)=|G|$ Edit: this should be: $\sum_{i=1}^k \dim(\rho_i)^2=|G|$, where $\dim(\rho_i)$ is the dimension of the vector space on which $\rho(g)$ acts.
I tried to find all the irreducible representations of $\mathbb{Z}/4 \mathbb{Z} = \{[0],[1],[2],[3]\}$
I could think of:
$\rho_0: \mathbb{Z}/4 \mathbb{Z} \rightarrow GL(\mathbb{C}), \rho_0(g)=1$ with $\dim(\rho_0)=1$
$\rho_1: \mathbb{Z}/4 \mathbb{Z} \rightarrow GL(\mathbb{C}), \rho_1(g)=e^{i\frac{2 \pi g}{4}}$ with $\dim(\rho_1)=1$
$\rho_2: \mathbb{Z}/4 \mathbb{Z} \rightarrow GL(\mathbb{R}^2), \rho_2(g)= \begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}^g$ with $\dim(\rho_2)=2$
But the squared sum of the dimensions of these representations is already larger than 4, so what am I missing?
Best Answer
If $F$ is a field of characteristic $0$, the group algebra $FG$ of a finite group $G$ is semi-simple. This means it is a direct sum
$$FG=\bigoplus_i \mathrm{End}_{D_i}(V_i)$$ in which $i$ runs over an index set for the irreducible $FG$-modules $V_i$, and $D_i$ is the endomorphism ring of $V_i$ as an $FG$ module (thus each $D_i$ is a division algebra containing $F$). In case $F=\mathbf{C}$ is the field of complex numbers, the only division algebra is $\mathbf{C}$ itself, and it follows that the dimension of $FG$ is the sum of the squares of the dimensions of the irreducible $FG$-modules, as you write.
However, this is not true in general, as your example of $G=\mathbf{Z}/4 \mathbf{Z}$ with $F=\mathbf{R}$ shows. There are $3$ irreducible $\mathbf{R} G$-modules: two one-dimensional ones ("split" over $\mathbf{R}$) and one two-dimensional irreducible representation, on which the generator of $G$ acts by rotations through an angle of $\pi/2$ and which splits upon extending scalars to $\mathbf{C}$ into a direct sum of two conjugate irreducible representations. The sum of the squares of their dimensions is then
$$1+1+4>4.$$
Incidentally, the equality fails in a different way and for a different reason if the characteristic of $F$ divides the order of the group. In that case, the group algebra is not semi-simple, and there are fewer irreducible representations of smaller dimensions, and so over an algebraically closed field of order dividing the order of the group, the sum of the squares of the dimensions of the irreducible representations will be smaller than the order of the group.