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Over a splitting field for finite group representations of characteristic $p$, the irreducible Brauer characters form a complex basis of the class functions on the p-regular elements. Independence is the character form of the Brauer–Nesbitt theorem. To obtain spanning, extend such a class function by zero on the p-singular classes. Ordinary irreducible characters form a basis of all class functions by character orthogonality. Restricting them to p-regular elements yields Brauer characters of reductions of an integral form of a group representation in a compatible splitting p-modular system; if needed, first extend scalars, which does not change the simple-module list under the splitting hypothesis. Each restriction is a nonnegative integral sum of simple Brauer characters by exact-sequence additivity and the Jordan–Hölder theorem. Thus these restrictions span, proving the assertion. Consequently the number of simple modules equals the number of p-regular conjugacy classes, and evaluation identifies the complexified modular representation ring with the product of one copy of $\mathbb{C}$ for each such class.

The logarithm from the Euler product of an Artin L-function has first-prime-power term $\sum \chi (g_{\mathfrak{p}})(N\mathfrak{p}{)}^{-s}$. Higher powers are uniformly bounded as real $s\downarrow 1$: $|\chi (g)|\le \chi (1)$ and ${\sum }_{\mathfrak{p}}{\sum }_{m\ge 2}(N\mathfrak{p}{)}^{-ms}/m\le 2{\sum }_{\mathfrak{p}}(N\mathfrak{p}{)}^{-2}<\infty$. The finitely many ramified local factors also have bounded logarithms. The Artin L-function order at one therefore proves the displayed estimate. Expanding a conjugacy-class indicator by character orthogonality leaves only the trivial character's logarithmic term, with coefficient $|C|/|G|$. This proves Dirichlet density directly; a natural-density assertion needs additional analytic arguments.

For a finite group $G$ represented unitarily by $U(g)$, the central projector onto the isotypic subspace of an irreducible representation $\lambda$ is $P_{\lambda }=(d_{\lambda }/|G|){\sum }_g{\chi }_{\lambda }(g{)}^{\ast }U(g)$, where $d_{\lambda }$ is its dimension. Character orthogonality implies that the operator is identity on copies of $\lambda$ and zero on other irreducibles. For a one-dimensional irreducible, applying it to an atomic orbital gives either zero or a symmetry-adapted linear combination of that symmetry, to be normalized afterward. A multidimensional central projector selects an entire isotypic component rather than one uniquely chosen component function.

Write the Klein four-group as $\{1,a,b,ab\}$. Its four one-dimensional irreducible representations independently assign signs to $a,b$, so the displayed characters have $p,q\in \{0,1\}$. The character table isEvery conjugacy class is a singleton because the group is abelian. The four rows exhaust the complex irreducibles by the equality between their number and the number of classes. If $H$ is this table, $H{H}^T=4I$ proves character orthogonality. The diagonal three-dimensional representation by half-turns about the coordinate axes is the direct sum of the three nontrivial characters, with total character $(3,-1,-1,-1)$.

In the order of the conjugacy classes of the symmetry group of a square displayed there, its irreducible complex character table isThe four one-dimensional representations assign independent signs to $r$ and $s$. The remaining row is the character of the ordinary planar representation: a rotation has trace $2\cos \theta$ and a reflection has trace zero. Weighted by class sizes, these rows satisfy character orthogonality. Their squared dimensions sum to eight, so they exhaust the irreducible representations.

For a finite Galois extension $L/K$ and a conjugacy class $C$ in $G=\mathrm{Gal}(L/K)$, the unramified primes with Frobenius conjugacy class $C$ have Dirichlet density $|C|/|G|$. In particular,One proof uses Brauer induction, the Artin L-function order at one, and character orthogonality. The trivial Artin L-function is the Dedekind zeta function of $K$. The full theorem also gives the natural density, but this logarithmic estimate alone establishes the Dirichlet density assertion.

Write $S(\theta )={\sum }_{x\le n}e(\theta x^3)$ and $I_j={\int }_0^1|S{|}^j$. Character orthogonality gives $I_2=n$. For $I_4$, fix the difference $D=x^3-y^3$. If $D=0$, there are $n^2$ choices for the two equal pairs. If $D\ne 0$, any solution $z^3-w^3=D$ has $z-w\mid D$; fixing this divisor gives a quadratic equation for $w$, hence at most two choices. The subpower bound for the divisor function yields $I_4{\ll }_{\eta }{n}^{2+\eta }$.

The twice-differenced inequality in Weyl differencing reads $|S{|}^4\le 2n{\sum }_v{c}_{v}e(\theta v)$, with $c_v$ counting triples $(h,l,x)$ for which $v=3hl(2x+h+l)$ and all four shifted arguments are in $[1,n]$. Thus $c_0\ll n^2$, and $c_v{\ll }_{\eta }{n}^{\eta }$ for $v\ne 0$: the product $hl$ divides $v$, and the remaining equation determines $x$. The Fourier coefficients $r(v)$ of $|S{|}^4$ are nonnegative counts, with $r(0)=I_4$ and ${\sum }_{v}r(v)=n^4$. Multiply the inequality by $|S{|}^4$ and integrate. It follows that $I_8\ll n(n^2{I}_4+n^{\eta }{n}^4){\ll }_{\eta }{n}^{5+\eta }$, after choosing the divisor-bound exponents small enough.

For an irreducible complex character of a representation, the indicator is $1$ for real type, $-1$ for quaternionic type and $0$ for complex type, the latter meaning $\chi \ne \overline{\chi }$. It detects the invariant bilinear form on an irreducible representation. For an odd-order group, squaring permutes its elements, so every nontrivial irreducible character has indicator zero by character orthogonality.

For a finite abelian group $G$, $U(h)|g⟩=|g+h⟩$ defines the regular representation on the basis labelled by $G$. The linear characters give the common eigenbasisCharacter orthogonality proves orthonormality, and changing variables by the translation in a group proves the eigenvalue formula.

In a completely reducible group representation $V={⨁}_{\lambda }{m}_{\lambda }{V}_{\lambda }$, the nonnegative integer $m_{\lambda }$ counts the copies of the corresponding irreducible representation. For complex representations of a finite group, character orthogonality gives $m_{\lambda }=⟨{\chi }_{\lambda },{\chi }_V⟩$.

A nonprincipal Dirichlet character differs from the principal Dirichlet character. Equivalently, it is a nontrivial character on the unit group, so its sum over one full residue period is zero by character orthogonality. Its Dirichlet L-function is entire by Mellin continuation of a nonprincipal Dirichlet L-function.

For the tetrahedral symmetry classes $(E,8C_3,3C_2,6S_4,6{\sigma }_d)$, each set of four $s$ atomic orbitals hasThe counts are the fixed sites of each symmetry operation: a threefold rotation fixes one vertex of each tetrahedron, a twofold or improper fourfold operation fixes none, and a diagonal mirror fixes two. Character orthogonality reduces each set toThe dimensions $1+3=4$ provide a check. Thus the complete eight-orbital model contains two copies of each of $A_1$ and $T_2$, which can mix across the two species.

For a real-character irreducible molecular representation with character $\chi$, the multiplicity of the fully symmetric representation in its tensor square isby character orthogonality. The identity uses the real-character assumption appropriate to the usual real irreducible representations intended here. For a general complex irreducible representation, the trivial representation in a tensor square has multiplicity one precisely when it is self-dual, and zero otherwise, by Schur's lemma. The universally valid tensor product is $V\otimes V^{\ast }$.

Use the conjugacy class order of solution. The defining representation has characterIts character inner product with itself is $(4+4)/8=1$, proving that it is an irreducible representation. There is also a direct invariant subspace argument: $B$ has two distinct coordinate eigenspaces, and $A$ exchanges them, so no line is invariant under both.

Let $L_{st}$ denote the one-dimensional representation with $L_{st}(A)=s$ and $L_{st}(B)=t$. Then $L_{st}(-I)=1$ and $L_{st}(C)=st$. The complete character table isThe four scalar rows obey character orthogonality, and the dimensions exhaust the group order as established in solution.

In a one-dimensional representation, the scalar images of $Q$ and $P$ commute. The relation $PQP=Q^{-1}$ gives $R(Q{)}^2=1$, while $R(Q{)}^5=1$, so $R(Q)=1$ and $R(P)=\pm 1$. These give exactly $R_0$ and $R_1$.

For $k=1,2$, define the two-dimensional representationsThe same invariant subspace argument as in solution proves that each is irreducible. Set $R_2=D_1$ and $R_3=D_2$. Their characters differ at $Q$, so they are inequivalent. The sum of squares of irreducible representation dimensions is already $1^2+1^2+2^2+2^2=10$, proving completeness.

Put $s=2\cos (2\pi /5)=(\sqrt{5}-1)/2$ and $t=2\cos (4\pi /5)=-(\sqrt{5}+1)/2$. Taking traces yields the character tableThe identities $s+t=-1$, $st=-1$, and $s^2+t^2=3$ also verify character orthogonality directly.

Only the character ${\chi }_2$ of the two-dimensional irreducible representation remains unknown. Write it as $(2,a,b,c,d)$. The regular representation has character zero off the identity and decomposes with multiplicities equal to the irreducible representation dimensions. Thus, for a nonidentity element $g$,For the four nonidentity conjugacy classes this gives $a=0$, $b=2$, $c=-1$, $d=0$. The complete character table is consequentlyFor example, $⟨{\chi }_2,{\chi }_2⟩=(4+12+8)/24=1$, agreeing with character orthogonality.

A group representation on a vector space $V$ is a group homomorphism $R:G\to \mathrm{GL}(V)$: $R(gh)=R(g)R(h)$ and $R(e)=I$. A reducible representation has a nonzero proper invariant subspace $W\subset V$, meaning $R(g)W\subseteq W$ for every $g$. Over the complex numbers, a finite-dimensional representation of a finite group is a direct sum of irreducible representations, by Maschke's theorem.

The character of a representation is ${\chi }_R(g)=\mathrm{tr}R(g)$. The cyclic property of the trace makes it a class function:Thus a character is specified by one number per conjugacy class, much less information than all the representation matrices. Moreover,Here character orthogonality says that $m_r$ is the multiplicity of the irreducible representation $r$. Consequently characters determine finite-dimensional complex representations up to equivalence of group representations and compute tensor product of group representations decompositions without constructing a new basis explicitly.

All three displayed transposition matrices have trace zero. Multiplying two of them givessince ${\alpha }^3=1$. Their traces are $\alpha +{\alpha }^{-1}=-1$, whereas $\mathrm{tr}R(e)=2$. ThereforeBy character orthogonality, $R$ is the irreducible representation $2$ from c.

There is also a direct invariant subspace argument. The two distinct eigenvalues of $R({\tau }_1{\tau }_2)$ force any invariant line to be one of the two coordinate lines. But $R({\tau }_3)$ exchanges those lines. Neither is invariant under the whole group, so this two-dimensional representation is irreducible.

A group representation on a vector space $V$ is a group homomorphism $R:G\to \mathrm{GL}(V)$, so $R(gh)=R(g)R(h)$ and $R(e)=I$. A nonzero group representation is an irreducible representation if its only invariant subspaces are $0$ and $V$. Here the representations and characters are finite-dimensional over the complex numbers.

The character of a representation is ${\chi }_R(g)=\mathrm{tr}R(g)$. Three useful properties are:

In particular, ${\chi }_R(e)=\dim V$ and ${\chi }_R(g^{-1})={\chi }_R(g{)}^{\ast }$, the latter following from unitarization of a finite-group representation. The character inner product $⟨{\chi }_r,{\chi }_R⟩$ counts the multiplicity of an irreducible representation $r$ in $R$.

Write $1$ for the trivial representation, $\epsilon$ for the sign representation, and $2$ for the given two-dimensional group representation. Their matrix representations on the generators of a group arewith $\alpha =e^{2\pi i/3}$. The two eigenvalues of $r$ in $2$ are distinct, and $s$ exchanges its two eigenlines; no line is invariant under both. Thus $2$ is an irreducible representation. The dimensions already exhaust the sum of squares of irreducible representation dimensions, so this is the complete list.

Since $\alpha +{\alpha }^{-1}=-1$, the character table isMultiplying characters and using character orthogonality gives every tensor product of group representations:For example ${\chi }_2^2=(4,1,0)={\chi }_1+{\chi }_{\epsilon }+{\chi }_2$. Explicit Clebsch-Gordan coefficients can be read from two doublets $x=(x_1,x_2{)}^T$, $y=(y_1,y_2{)}^T$:The first two combinations are invariant under $r$ and respectively even and odd under $s$; the last transforms by the displayed doublet matrices.