Sum of squares of degrees of irreducible representations equals order of group

Statement
Suppose $$G$$ is a finite group, $$K$$ is a splitting field for $$G$$, and $$\chi_1,\chi_2, \dots, \chi_r$$ are the characters of the irreducible linear representations (up to equivalence) of $$G$$ over $$K$$. Let $$d_i$$ be the degree of $$\chi_i$$. In other words, $$d_i$$ are the fact about::degrees of irreducible representations of $$G$$. Then:

$$\! d_1^2 + d_2^2 + \dots + d_r^2 = |G|$$

This fact is instrumental in defining the Plancherel measure on the set of irreducible representations of a finite group, which assigns a measure of $$d_i^2/|G|$$ to the $$i^{th}$$ irreducible representation.

Note also that the $$d_i$$s (up to rearrangement) are the same for all splitting fields -- see degrees of irreducible representations are the same for all splitting fields.

Alternative formulations

 * Regular representation over splitting field has multiplicity of each irreducible representation equal to its degree
 * Group ring over splitting field is direct sum of matrix rings for each irreducible representation
 * Peter-Weyl theorem: A generalization to compact groups.

Similar facts

 * Number of irreducible representations equals number of conjugacy classes
 * Character orthogonality theorem
 * Column orthogonality theorem
 * Grand orthogonality theorem

See also the many facts about the degrees of irreducible representations under degrees of irreducible representations.

Facts used

 * 1) uses::Maschke's averaging lemma, which we use to say that every representation is completely reducible.
 * 2) uses::Orthogonal projection formula, which in turn uses uses::character orthogonality theorem. See inner product of functions for the notation.

Proof in characteristic zero
Note: We can in fact use this proof to also show that there are only finitely many equivalence classes of irreducible representations, though the formulation below does not quite do that.

Given: A finite group $$G$$ with irreducible representations having characters $$\chi_1, \chi_2,\dots, \chi_r$$ and degrees $$d_1, d_2, \dots, d_r$$.

To prove: $$d_1^2 + d_2^2 + \dots + d_r^2 = |G|$$

Proof: We let $$\rho$$ be the regular representation of $$G$$, i.e., the permutation representation obtained by using the regular group action. Let $$\alpha$$ be the character of $$\rho$$.

Proof in other characteristics
This follows from the characteristic zero proof, and the fact that degrees of irreducible representations are the same for all splitting fields.