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

This fact is related to: linear representation theory
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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 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.

Related facts

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.

Particular cases

Groups

Group/group type	Order	Degrees of irreducible representations	Squares of degrees of irreducible representations
trivial group	1	1	1
abelian group of order $n$	$n$	1 ($n$ times)	1 ($n$ times)
symmetric group:S3	6	1,1,2	1,1,4
dihedral group:D8	8	1,1,1,1,2	1,1,1,1,4
quaternion group	8	1,1,1,1,2	1,1,1,1,4
dihedral group:D10	10	1,1,2,2	1,1,4,4
alternating group:A4	12	1,1,1,3	1,1,1,9
direct product of S3 and Z2	12	1,1,1,1,2,2	1,1,1,1,4,4
dicyclic group:Dic12	12	1,1,1,1,2,2	1,1,1,1,4,4
symmetric group:S4	24	1,1,2,3,3	1,1,4,9,9
special linear group:SL(2,3)	24	1,1,1,2,2,2,3	1,1,1,4,4,4,9

Families

Family of groups	Order	Degrees of irreducible representations	Family-specific explanation for sum of squares
dihedral group of even degree $n$	$2n$	$1$ ($4$ times), $2$ ($(n-2)/2$ times)	algebraic simplification. See also linear representation theory of dihedral groups
dihedral group of odd degree $n$	$2n$	$1$ ($2$ times), $2$ ($(n-1)/2$ times)	algebraic simplification. See also linear representation theory of dihedral groups
symmetric group of degree $n$	$n!$	For each partition $\lambda$ of $n$, an irreducible representation of degree $d_\lambda$, which is the number of Young tableaux of shape $\lambda$	Robinson-Schensted correspondence. See also linear representation theory of symmetric groups
general affine group of degree one over a finite field of size $q$	$q(q - 1)$	$1$ ($q - 1$ times), $q - 1$ ($1$ time)	See linear representation theory of general affine group of degree one over a finite field
general linear group of degree two over a finite field of size $q$	$q(q+1)(q-1)^2$	1 ($q - 1$ times), $q$ ($q - 1$ times), $q + 1$ ($(q - 1)/(q - 2)/2$ times), $q - 1$ ($q(q - 1)/2$ times)	Algebraic simplification. See linear representation theory of general linear group of degree two over a finite field
projective general linear group of degree two over a finite field of size $q$	$q(q+1)(q-1) = q^3 - q$	Case $q$ odd: 1 (2 times), $q - 1$ ($(q - 1)/2$ times), $q$ (2 times), $q + 1$ ($(q - 3)/2$ times) Case $q$ even: 1 (1 time), $q - 1$ ($q/2$ times), $q$ (1 time), $q + 1$ ($(q - 2)/2$ times)	Algebraic simplification. See linear representation theory of projective general linear group of degree two over a finite field
special linear group of degree two over a finite field of size $q$	$q(q+1)(q-1) = q^3 - q$	Case $q$ odd: 1 (1 time), $(q - 1)/2$ (2 times), $(q + 1)/2$ (2 times), $q - 1$ ($(q - 1)/2$ times), $q$ (1 time), $q + 1$ ($(q - 3)/2$ times) Case $q$ even: 1 (1 time), $q - 1$ ($q/2$ times), $q$ (1 time), $q + 1$ ($(q - 2)/2$ times)	Algebraic simplification. See linear representation theory of special linear group of degree two over a finite field
projective special linear group of degree two over a finite field of size $q$	$q(q+1)(q-1)/2 = (q^3 - q)/2$ if $q$ odd; $q^3 - q$ if $q$ even	Case $q$ odd: 1 (1 time), $(q + 1)/2$ (2 times), $q - 1$ ($(q - 1)/4$ or $(q - 3)/4$ times depending on congruence class of $q$ mod $4$), $q$ (1 time), $q + 1$ ($(q - 5)/4$ or $(q - 3)/4$ times depending on congruence class of $q$ mod $4$) Case $q$ even: 1 (1 time), $q - 1$ ($q/2$ times), $q$ (1 time), $q + 1$ ($(q - 2)/2$ times)	Algebraic simplification. See linear representation theory of projective special linear group of degree two over a finite field

Facts used

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

Proof

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 $\rho$.

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$\alpha$ takes the value $|G|$ at the identity element of $G$, and zero elsewhere.				[SHOW MORE] Since $\rho$ is a permutation represenation, the trace of $\rho(g)$, which is $\alpha(g)$, is the number of fixed points under the permutation corresponding to $\rho(g)$. Since multiplication by a non-identity group element has no fixed points, we get $\alpha(g) = 0$ for all non-identity $g$. For the identity element, we get the identity matrix of degree $|G|$, so that's the trace.
2	The inner product $\langle \alpha, \chi_i \rangle$ equals $d_i$ for all $i$.			Step (1)	[SHOW MORE] In the summation for the inner product, all terms are zero except the term at the identity element, which is $|G|d_i$. Dividing by $|G|$ we get $d_i$.
3	$\alpha$ is the sum $\sum_{i=1}^r d_i\chi_i$	Facts (1),(2)	$\chi_i$ are characters of (all the) irreducible representations.	Step (2)	[SHOW MORE] We know by Fact (1) that $\alpha$ is a nonnegative integer combination of the $\chi_i$s, because $\rho$ is a nonnegative integer combination of the corresponding representations. By Fact (2) and Step (2), the coefficients are precisely $\langle \alpha, \chi_i \rangle = d_i$.
4	The value of $\alpha$ at the identity element is $\sum_{i=1}^r d_i^2$.			Step (3)	[SHOW MORE] At the identity element, $\chi_i$ takes the value $d_i$, so plugging in Step (3) gives this.
5	$\sum_{i=1}^r d_i^2 = |G|$			Steps (1), (4)	[SHOW MORE] By Step (1), $\alpha$ at the identity element of $G$ is $|G|$. By Step (4), it is $\sum_{i=1}^r d_i^2$. Combining, we get the result.

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.