Orthogonality of characters

Theorem

Orthogonality of characters

T-124Home MU-402Threads structure

Statement

Let \(G\) be a finite group and let all representations be finite-dimensional over \(\mathbb{C}\). Define the Hermitian pairing on class functions \(\varphi,\psi:G\to\mathbb{C}\) by \[ \langle \varphi,\psi\rangle \;=\; \frac{1}{|G|}\sum_{g\in G}\varphi(g)\overline{\psi(g)}. \] If \(V\) and \(W\) are irreducible \(\mathbb{C}G\)-modules with characters \(\chi_V,\chi_W\), then \[ \langle \chi_V,\chi_W\rangle \;=\; \begin{cases}1 & \text{if } V\cong W,\\[2pt] 0 & \text{if } V\not\cong W.\end{cases} \] Equivalently: the set of irreducible characters of \(G\) is an orthonormal set in the inner product space of class functions \(G\to\mathbb{C}\).

Why it matters

This single computation converts representation theory into linear algebra. Once the irreducible characters are known to be orthonormal, decomposing an arbitrary representation \(U\) into irreducibles reduces to computing inner products: the multiplicity of \(V_i\) in \(U\) is exactly \(\langle \chi_U,\chi_i\rangle\), no explicit change of basis required. Orthonormality also bounds the number of irreducible representations: since class functions on \(G\) form a vector space of dimension equal to the number of conjugacy classes, an orthonormal set inside it is automatically linearly independent, so \(G\) has at most that many irreducibles (equality is proved separately, but this theorem supplies the inequality for free).

The result is the finite-group ancestor of Fourier analysis: characters play the role of the exponentials \(e^{in\theta}\), the inner product above is the discrete analogue of \(\int_0^{2\pi}\), and decomposing a representation into irreducibles is literally computing a Fourier expansion in this basis.

Hypotheses

\(G\) is finite.The averaging \(\frac{1}{|G|}\sum_{g\in G}\) needs a finite, invariant, total-mass-one measure. Take \(G=\mathbb{Z}\): there is no such average over an infinite discrete group at all (no finitely additive translation-invariant probability measure on all of \(\mathbb{Z}\) exists), so the pairing is not even defined and the argument cannot start. Representations are finite-dimensional.Traces, and hence characters, are undefined for infinite-dimensional operators in general; the averaging operator \(f^{\circ}\) below is built from a trace argument that requires \(\dim V,\dim W\lt\infty\). The ground field is \(\mathbb{C}\) (or any field of characteristic \(0\) that is algebraically closed).Algebraic closure is used to guarantee every \(G\)-endomorphism of an irreducible module has an eigenvalue (Schur's Lemma, second form). Characteristic \(0\) with \(|G|\lt\infty\) is used to guarantee Maschke's Lemma (complete reducibility) and to guarantee \(\rho(g)\) is diagonalizable with root-of-unity eigenvalues. Over a field of characteristic \(p\) dividing \(|G|\) (modular representation theory) Maschke's Lemma fails, indecomposable modules need not be irreducible, and the orthogonality relations fail in this exact form. \(V,W\) are irreducible.For a reducible \(U\), \(\chi_U=\sum_i m_i\chi_i\) is a genuine sum, so \(\langle\chi_U,\chi_U\rangle=\sum_i m_i^2\), which exceeds \(1\) as soon as \(U\) is not irreducible (e.g. \(U=V\oplus V\) gives \(\langle\chi_U,\chi_U\rangle=4\)).

Proof

Fix bases of \(V,W\) and write \(\rho:G\to GL(V)\), \(\sigma:G\to GL(W)\) for the representations, with matrix entries \(R(g)_{ij}\), \(S(g)_{kl}\) (\(1\le i,j\le \dim V\), \(1\le k,l\le\dim W\)), so \(\chi_V(g)=\sum_i R(g)_{ii}\), \(\chi_W(g)=\sum_k S(g)_{kk}\).

Setup: a representation is a group homomorphism into \(GL(V)\); choosing a basis converts it to matrices. A

For any linear map \(T:W\to V\) define the averaged map \(T^{\circ}=\dfrac{1}{|G|}\sum_{g\in G}\rho(g)\,T\,\sigma(g)^{-1}:W\to V\). Then \(T^{\circ}\) is \(G\)-equivariant: \(\rho(h)T^{\circ}=T^{\circ}\sigma(h)\) for every \(h\in G\).

Direct check: \(\rho(h)T^{\circ}=\frac1{|G|}\sum_g \rho(hg)T\sigma(g)^{-1}\); substitute \(g'=hg\), use \(\sigma(g)^{-1}=\sigma(h)^{-1}\sigma(g')^{-1}\), and reindex the sum over \(g'\in G\) (a bijection of \(G\) with itself). This is the standard "averaging trick", the same idea that proves Maschke's Lemma. B

\(\textbf{Schur's Lemma.}\) If \(\varphi:V\to W\) is a homomorphism of \(\mathbb{C}G\)-modules and \(V,W\) are irreducible, then either \(\varphi=0\) or \(\varphi\) is an isomorphism. Consequently \(\mathrm{Hom}_G(V,W)=0\) if \(V\not\cong W\); and if \(V=W\) then every \(\varphi\in\mathrm{Hom}_G(V,V)\) is a scalar, \(\varphi=\lambda\,\mathrm{Id}_V\).

\(\ker\varphi\) and \(\mathrm{im}\,\varphi\) are submodules of \(V,W\) respectively (immediate from equivariance), so irreducibility forces \(\ker\varphi\in\{0,V\}\); if \(\ker\varphi=0\) then \(\varphi\) is injective, so \(\mathrm{im}\,\varphi\neq 0\) is a submodule of \(W\), hence \(\mathrm{im}\,\varphi=W\) and \(\varphi\) is an isomorphism. For the scalar form: \(\mathbb{C}\) is algebraically closed and \(\dim V\lt\infty\), so \(\varphi\) has an eigenvalue \(\lambda\); then \(\varphi-\lambda\,\mathrm{Id}_V\) is again a \(G\)-endomorphism of \(V\) with nonzero kernel, so by the first part it is the zero map, i.e. \(\varphi=\lambda\,\mathrm{Id}_V\). C

By Step 2, \(T^{\circ}\in\mathrm{Hom}_G(W,V)\) for every linear \(T:W\to V\). By Schur's Lemma (Step 3): if \(V\not\cong W\) then \(T^{\circ}=0\) for every \(T\); if \(V=W\) then \(T^{\circ}=\dfrac{\mathrm{tr}(T)}{\dim V}\,\mathrm{Id}_V\).

Direct application of Step 3 to the equivariant map produced in Step 2; the scalar in the second case is pinned down by taking traces of both sides, \(\mathrm{tr}(T^{\circ})=\frac1{|G|}\sum_g\mathrm{tr}(\rho(g)T\rho(g)^{-1})=\mathrm{tr}(T)\) since trace is conjugation-invariant, so \(\lambda\dim V=\mathrm{tr}(T)\). B

Take \(T=E_{pq}:W\to V\), the elementary map sending the \(q\)-th basis vector of \(W\) to the \(p\)-th basis vector of \(V\) and every other basis vector to \(0\). Its \((i,k)\) matrix entry after averaging is \[ (T^{\circ})_{ik}=\frac{1}{|G|}\sum_{g\in G} R(g)_{ip}\,S(g^{-1})_{qk}. \]

Matrix multiplication: \((\rho(g)E_{pq}\sigma(g)^{-1})_{ik}=\sum_{a,b}R(g)_{ia}(E_{pq})_{ab}S(g^{-1})_{bk}=R(g)_{ip}S(g^{-1})_{qk}\), since \((E_{pq})_{ab}=\delta_{ap}\delta_{bq}\); average over \(g\). C

Combining Steps 4 and 5: if \(V\not\cong W\), \(\dfrac{1}{|G|}\sum_g R(g)_{ip}S(g^{-1})_{qk}=0\) for all \(i,k,p,q\); if \(V=W\) (so \(R=S\)), \(\dfrac{1}{|G|}\sum_g R(g)_{ip}R(g^{-1})_{qk}=\dfrac{\delta_{pq}\delta_{ik}}{\dim V}\), using \(\mathrm{tr}(E_{pq})=\delta_{pq}\) and \((\mathrm{Id}_V)_{ik}=\delta_{ik}\).

Read off entrywise equality of the matrices \(T^{\circ}=0\), resp. \(T^{\circ}=\frac{\delta_{pq}}{\dim V}\mathrm{Id}_V\), from Step 4 against the formula of Step 5. A

Set \(p=i\), \(q=k\) in Step 6 and sum over \(1\le i\le\dim V\), \(1\le k\le\dim W\) (or \(\dim V\) in both index ranges when \(V=W\)): \[ \frac{1}{|G|}\sum_{g\in G}\chi_V(g)\,\chi_W(g^{-1}) \;=\; \begin{cases}0,& V\not\cong W,\\[2pt] 1,& V=W.\end{cases} \]

\(\sum_i R(g)_{ii}=\chi_V(g)\), \(\sum_k S(g^{-1})_{kk}=\chi_W(g^{-1})\), and in the equal case \(\sum_{i,k}\delta_{ik}\delta_{ik}=\sum_i 1=\dim V\), which cancels the \(1/\dim V\) from Step 6. Interchange of the finite sums over \(g\) and over \(i,k\) is legitimate (finite double sum). B

For every \(g\in G\), \(\chi_V(g^{-1})=\overline{\chi_V(g)}\).

\(G\) finite \(\Rightarrow\) \(g\) has finite order \(n\), so \(\rho(g)^n=\mathrm{Id}\); the minimal polynomial of \(\rho(g)\) divides \(x^n-1\), which has \(n\) distinct roots in \(\mathbb{C}\) (characteristic \(0\)), so \(\rho(g)\) is diagonalizable with eigenvalues that are \(n\)-th roots of unity, hence of modulus \(1\). If \(\lambda_1,\dots,\lambda_d\) are these eigenvalues then \(\rho(g^{-1})=\rho(g)^{-1}\) has eigenvalues \(\lambda_j^{-1}=\overline{\lambda_j}\), so \(\chi_V(g^{-1})=\sum_j\lambda_j^{-1}=\sum_j\overline{\lambda_j}=\overline{\chi_V(g)}\). B

Substituting Step 8 into Step 7 gives \(\dfrac{1}{|G|}\sum_g \chi_V(g)\overline{\chi_W(g)} = \langle\chi_V,\chi_W\rangle\), equal to \(0\) if \(V\not\cong W\) and \(1\) if \(V=W\). If \(V\cong W\) via any \(G\)-isomorphism, conjugate matrices have equal trace, so \(\chi_V=\chi_W\) as functions and this is the same as the case \(V=W\).

Direct substitution and the definition of \(\langle\cdot,\cdot\rangle\) from the Statement; the isomorphism-invariance of characters is immediate since \(\rho'(g)=A\rho(g)A^{-1}\) has the same trace as \(\rho(g)\) for any invertible \(A\). A

Result

\(\langle \chi_V,\chi_W\rangle=\delta_{VW}\) for irreducible \(\mathbb{C}G\)-modules \(V,W\), \(G\) finite

Reading. Two non-isomorphic irreducible representations have characters that are exactly orthogonal under the averaged inner product on class functions; each irreducible character has norm exactly \(1\). The irreducible characters form an orthonormal, in particular linearly independent, set.

Scope. Applies to finite-dimensional complex representations of a finite group \(G\). Extends verbatim to compact groups with the counting measure \(\frac1{|G|}\sum_g\) replaced by normalized Haar measure \(\int_G\); does not hold as stated for infinite discrete groups, for infinite-dimensional representations, or in modular characteristic (\(\mathrm{char}(F)\mid |G|\)).

Corollaries & converses

Irreducible characters are linearly independent as functions \(G\to\mathbb{C}\) (an orthonormal set is automatically independent), hence the number of irreducible representations of \(G\) is at most the number of conjugacy classes (equality is a separate theorem, proved using the group algebra decomposition).
For any representation \(U\) with character \(\chi_U=\sum_i m_i\chi_i\) (\(m_i\ge 0\) the multiplicity of the irreducible \(V_i\)), \(m_i=\langle\chi_U,\chi_i\rangle\) and \(\langle\chi_U,\chi_U\rangle=\sum_i m_i^2\); in particular \(U\) is irreducible if and only if \(\langle\chi_U,\chi_U\rangle=1\) — this converse direction does hold, since \(\sum m_i^2=1\) with \(m_i\in\mathbb{Z}_{\ge0}\) forces exactly one \(m_i=1\).
Converse of the bare orthogonality statement (does \(\langle\varphi,\varphi\rangle=1\) for a class function \(\varphi\) imply \(\varphi\) is an irreducible character?) is false in general: \(\varphi\) also needs to be an actual character (a non-negative integer combination of irreducibles arising from an honest representation), not just any class function of norm \(1\); e.g. a signed combination like \(\varphi=\frac{1}{\sqrt2}\chi_1-\frac1{\sqrt2}\chi_2\) can have norm \(1\) without being a character at all.

Fails without

Irreducibility dropped. Take \(V=U\oplus U\) for any nontrivial irreducible \(U\); then \(\chi_V=2\chi_U\) and \(\langle\chi_V,\chi_V\rangle=4\langle\chi_U,\chi_U\rangle=4\neq1\). More strikingly, for the regular representation \(\mathbb{C}G\) one has \(\chi_{\mathrm{reg}}(e)=|G|\) and \(\chi_{\mathrm{reg}}(g)=0\) for \(g\neq e\), so \(\langle\chi_{\mathrm{reg}},\chi_{\mathrm{reg}}\rangle=\frac1{|G|}\cdot|G|^2=|G|\), which is \(1\) only for the trivial group.
Finiteness of \(G\) dropped. For \(G=\mathbb{Z}\) there is no finitely additive translation-invariant probability measure on all subsets, so \(\frac{1}{|G|}\sum_{g\in G}\) does not make sense and the pairing \(\langle\cdot,\cdot\rangle\) is not even defined; the theorem has no analogue without first restricting to a class of "nice" (e.g. compact, or unitary and square-summable) representations and replacing the sum by Haar/Plancherel integration.
Characteristic-0 / algebraic-closure dropped. Working over a field of characteristic \(p\) with \(p\mid|G|\), Maschke's Lemma fails: not every submodule has a complement, so "irreducible" and "indecomposable" split apart, and the trace-based averaging argument in Step 4 (which used Schur's Lemma over an algebraically closed field of characteristic \(0\)) breaks down; the modular character table need not satisfy \(\langle\chi_i,\chi_j\rangle=\delta_{ij}\) under the same pairing.

Common errors

Writing \(\chi(g^{-1})=\chi(g)^{-1}\) (treating the character value as if it were the group element being inverted); the correct identity from Step 8 is \(\chi(g^{-1})=\overline{\chi(g)}\).
Dropping the \(\frac{1}{|G|}\) normalization and reporting \(\sum_g\chi_V(g)\overline{\chi_W(g)}=\delta_{VW}\) instead of \(\frac1{|G|}\sum_g(\cdots)=\delta_{VW}\).
Applying the formula to a reducible character and expecting \(1\), instead of computing \(\sum_i m_i^2\).
Computing the sum "by conjugacy class" as \(\sum_{c}\chi(c)\overline{\chi'(c)}\) and forgetting to weight each class representative by its class size \(|C_c|\), i.e. omitting the correct form \(\frac{1}{|G|}\sum_c|C_c|\,\chi(c)\overline{\chi'(c)}\).
Assuming \(\chi(g^{-1})=\chi(g)\) always (true only when \(\chi\) is real-valued, e.g. for real or self-dual representations, not in general).
Using orthogonality on a function that is not actually constant on conjugacy classes (checking a "character-like" formula on individual matrix entries rather than the trace).

Discussion

The proof above is essentially Frobenius's original 1896 argument, later streamlined by Schur into the averaging-operator form given here; the elementary matrices \(E_{pq}\) in Step 5 produce the finer "great orthogonality relations" for matrix coefficients, of which character orthogonality is the diagonal, basis-independent shadow. Historically this predates the general representation theory of compact groups; the Peter–Weyl theorem later reproduced the same orthonormality statement for compact (possibly infinite) groups with Haar measure replacing the counting average, showing the finite-group case is the prototype of a much more general phenomenon.

Conceptually, the theorem says the characters \(\{\chi_1,\dots,\chi_r\}\) form an orthonormal basis for the space of class functions once one also knows \(r\) equals the number of conjugacy classes (a fact proved independently via the structure of the group algebra \(\mathbb{C}G\cong\bigoplus_i \mathrm{End}(V_i)\)). Given that, any class function — in particular any character — has a unique expansion in this basis, and the coefficients are computed exactly as Fourier coefficients are: by taking the inner product with the basis vector.

The deeper reason Schur's Lemma produces scalars rather than merely "some endomorphism" is that \(\mathrm{End}_G(V)\) is a finite-dimensional division algebra over \(\mathbb{C}\) whenever \(V\) is irreducible (Step 3, first part, shows every nonzero equivariant endomorphism is invertible); since \(\mathbb{C}\) is algebraically closed, the only such division algebra is \(\mathbb{C}\) itself. Over \(\mathbb{R}\), by contrast, \(\mathrm{End}_G(V)\) can also be \(\mathbb{R}\) itself or the quaternions \(\mathbb{H}\), and the orthogonality relations acquire the well-known Frobenius–Schur correction factors — a genuinely different, though related, computation.

Common misconception. Students often think orthogonality of characters is a coincidence of the trace map; it is not — it is a direct print-out of Schur's Lemma applied to matrix units, and every non-diagonal instance of the "great orthogonality relations" for individual matrix entries (Step 6) is strictly stronger than the character-level statement, which is only the trace (diagonal sum) of that finer relation.

Worked examples

Let \(G=S_3\), with conjugacy classes \(\{e\}\), \(\{(12),(13),(23)\}\), \(\{(123),(132)\}\) of sizes \(1,3,2\). \(S_3\) has three irreducibles: trivial \(\chi_1\), sign \(\chi_{\mathrm{sgn}}\), and the \(2\)-dimensional standard representation \(\chi_2\), with values \[ \chi_2(e)=2,\quad \chi_2((12))=0,\quad \chi_2((123))=-1. \]

Standard character table of \(S_3\), obtainable e.g. from the permutation representation on \(\{1,2,3\}\) (character \(3,1,0\) on the three classes) minus the trivial character. A

Verify \(\langle\chi_2,\chi_2\rangle=1\) using class sizes: \(\displaystyle \langle\chi_2,\chi_2\rangle=\frac1{6}\Big(1\cdot 2^2+3\cdot 0^2+2\cdot(-1)^2\Big)=\frac1{6}(4+0+2)=1.\)

Grouping the sum over \(g\in G\) by conjugacy class and using that \(\chi_2\) is real (so \(\overline{\chi_2}=\chi_2\)), \(\langle\varphi,\psi\rangle=\frac1{|G|}\sum_c |C_c|\varphi(c)\overline{\psi(c)}\), an immediate reindexing of the defining sum. A

Verify \(\langle\chi_2,\chi_1\rangle=0\): \(\displaystyle \frac16\big(1\cdot2\cdot1+3\cdot0\cdot1+2\cdot(-1)\cdot1\big)=\frac16(2+0-2)=0.\)

Same reindexed formula with \(\chi_1\equiv1\); confirms the theorem's cross-term prediction \(\langle\chi_2,\chi_1\rangle=0\) since \(V_2\not\cong V_1\). A

\(\langle\chi_2,\chi_2\rangle=1,\ \langle\chi_2,\chi_1\rangle=0\) — orthonormality confirmed directly on \(S_3\)

Reading. The 2-dimensional irreducible of \(S_3\) has norm exactly \(1\), and is orthogonal to the trivial character, exactly as the theorem forces.

Let \(U=\mathbb{C}^3\) be the permutation representation of \(G=S_3\) acting on \(\{1,2,3\}\), with character \(\chi_U(e)=3,\ \chi_U((12))=1,\ \chi_U((123))=0\). Decompose \(U\) into irreducibles using the Corollary \(m_i=\langle\chi_U,\chi_i\rangle\).

\(\chi_U(g)=\#\{\text{fixed points of }g\}\), the standard formula for a permutation character. A

\(m_1=\langle\chi_U,\chi_1\rangle=\frac16(1\cdot3\cdot1+3\cdot1\cdot1+2\cdot0\cdot1)=\frac16(3+3+0)=1.\)

Same class-weighted inner product formula as Example 1, Step 2, now paired against the trivial character. A

\(m_{\mathrm{sgn}}=\langle\chi_U,\chi_{\mathrm{sgn}}\rangle=\frac16(1\cdot3\cdot1+3\cdot1\cdot(-1)+2\cdot0\cdot1)=\frac16(3-3+0)=0.\)

\(\chi_{\mathrm{sgn}}\) takes values \(1,-1,1\) on the three classes; substitute directly. A

\(m_2=\langle\chi_U,\chi_2\rangle=\frac16(1\cdot3\cdot2+3\cdot1\cdot0+2\cdot0\cdot(-1))=\frac16(6+0+0)=1.\) Check: \(\langle\chi_U,\chi_U\rangle=1^2+0^2+1^2=2\), matching \(\frac16(9+3+0)=2\) directly, and \(\chi_1+\chi_2\) evaluates to \((3,1,0)=\chi_U\).

Same formula against \(\chi_2\) from Example 1; the consistency check uses the Corollary \(\langle\chi_U,\chi_U\rangle=\sum m_i^2\), itself a direct consequence of orthonormality of \(\{\chi_1,\chi_{\mathrm{sgn}},\chi_2\}\). B

\(U\cong V_1\oplus V_2\) (trivial \(\oplus\) standard), multiplicities read off purely from inner products of characters

Reading. No explicit \(G\)-equivariant map or change of basis was ever constructed; orthonormality of the irreducible characters alone determined the decomposition of the permutation module.

Problems

Compute \(\langle\chi_1,\chi_1\rangle\) for the trivial character \(\chi_1\equiv1\) of any finite group \(G\), directly from the definition.
Solution
\(\langle\chi_1,\chi_1\rangle=\frac1{|G|}\sum_{g\in G}1\cdot\overline1=\frac1{|G|}\cdot|G|=1\), consistent with the theorem since the trivial representation is irreducible.
Let \(G\) be a finite group with exactly \(3\) conjugacy classes. Using only the theorem (not the separate fact that the number of irreducibles equals the number of classes), show \(G\) has at most \(3\) pairwise non-isomorphic irreducible representations.
Solution
Class functions \(G\to\mathbb{C}\) form a vector space of dimension equal to the number of conjugacy classes, here \(3\) (a class function is determined by, and freely specifiable on, one representative per class). By the theorem the irreducible characters \(\chi_1,\dots,\chi_r\) are pairwise orthogonal and each has norm \(1\), hence are linearly independent vectors in this \(3\)-dimensional space (an orthonormal set is always independent: if \(\sum a_i\chi_i=0\), pairing with \(\chi_j\) gives \(a_j=0\) for each \(j\)). Independent vectors in a \(3\)-dimensional space number at most \(3\), so \(r\le3\).
The cyclic group \(G=C_4=\langle x\mid x^4=1\rangle\) has four 1-dimensional irreducible characters \(\chi_k(x^j)=i^{kj}\) for \(k=0,1,2,3\) (\(i=\sqrt{-1}\)). Verify \(\langle\chi_1,\chi_3\rangle=0\) directly.
Solution
\(\chi_1(x^j)=i^j\), \(\chi_3(x^j)=i^{3j}=(-i)^j\), so \(\overline{\chi_3(x^j)}=\overline{(-i)^j}=i^j\) (since \(\overline{-i}=i\)). Then \(\langle\chi_1,\chi_3\rangle=\frac14\sum_{j=0}^3 i^j\cdot i^j=\frac14\sum_{j=0}^3 i^{2j}=\frac14(1+(-1)+1+(-1))=\frac14\cdot0=0\), confirming orthogonality of the two distinct irreducibles, consistent with \(\chi_3=\overline{\chi_1}\) and \(\chi_1\ne\chi_3\) as \(C_4\) is not isomorphic to its own dual class function trivially (they differ as functions since \(\chi_1(x)=i\neq -i=\chi_3(x)\)).
Explain precisely why \(\langle\chi,\chi\rangle=1\) for a character \(\chi\) of an honest (not necessarily irreducible a priori) representation \(U\) implies \(U\) is irreducible, using the Corollary stated on this page, and explain why the same conclusion would not follow from merely knowing \(\varphi\) is some class function with \(\langle\varphi,\varphi\rangle=1\).
Solution
Write \(\chi_U=\sum_i m_i\chi_i\) with \(m_i\in\mathbb{Z}_{\ge0}\) the multiplicities of the irreducibles \(V_i\) in \(U\) (Maschke's Lemma guarantees such a decomposition exists at all, since \(G\) is finite and the field has characteristic \(0\)). By orthonormality of the \(\chi_i\), \(\langle\chi_U,\chi_U\rangle=\sum_i m_i^2\). If this equals \(1\) with all \(m_i\) non-negative integers, exactly one \(m_i\) is \(1\) and the rest are \(0\), so \(U\cong V_i\) is irreducible. This argument used crucially that the \(m_i\) are non-negative integers coming from an actual module decomposition; an arbitrary class function \(\varphi\) with \(\langle\varphi,\varphi\rangle=1\) need not have integer (or even real) coefficients when expanded in the \(\chi_i\) basis, e.g. \(\varphi=\frac{1}{\sqrt2}\chi_1-\frac1{\sqrt2}\chi_2\) has norm \(1\) but is not any group's character.
\(G=S_3\) has irreducible characters \(\chi_1,\chi_{\mathrm{sgn}},\chi_2\) as in the Worked Examples. A representation \(U\) has character \(\chi_U(e)=8,\ \chi_U((12))=0,\ \chi_U((123))=-1\). Decompose \(U\) into irreducibles and verify \(\dim U\) is consistent.
Solution
Using \(\langle\chi_U,\chi_i\rangle=\frac16\sum_c|C_c|\chi_U(c)\overline{\chi_i(c)}\) with class sizes \(1,3,2\): \(m_1=\frac16(1\cdot8\cdot1+3\cdot0\cdot1+2\cdot(-1)\cdot1)=\frac16(8+0-2)=1.\) \(m_{\mathrm{sgn}}=\frac16(1\cdot8\cdot1+3\cdot0\cdot(-1)+2\cdot(-1)\cdot1)=\frac16(8+0-2)=1.\) \(m_2=\frac16(1\cdot8\cdot2+3\cdot0\cdot0+2\cdot(-1)\cdot(-1))=\frac16(16+0+2)=3.\) So \(U\cong V_1\oplus V_{\mathrm{sgn}}\oplus 3V_2\). Dimension check: \(1\cdot1+1\cdot1+3\cdot2=1+1+6=8=\chi_U(e)\), consistent. As a further check, \(\langle\chi_U,\chi_U\rangle\) should equal \(m_1^2+m_{\mathrm{sgn}}^2+m_2^2=1+1+9=11\); directly, \(\frac16(1\cdot64+3\cdot0+2\cdot1)=\frac16(64+0+2)=11\), matching.