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Every complex character of a representation of a finite group $G$ is a rational linear combination of characters of induced representations from one-dimensional group representations of cyclic subgroups $C_j$. One-dimensionality follows because the irreducible representations of a finite cyclic group over the complex numbers are one-dimensional. The rational coefficients distinguish this statement from Brauer induction, which permits integer coefficients by allowing more general subgroups. For a rational-valued character, another form of Artin induction uses rational combinations of induced representations of trivial group representations of cyclic subgroups.

Every complex character of a representation of a finite group $G$ is an integer linear combination of characters of induced representations from one-dimensional group representations of subgroups $H_j\le G$. Negative coefficients are allowed. Unlike Artin induction, the subgroups need not be cyclic groups. Applied to an Artin L-function, compatibility with induced representations turns this character identity into a product of one-dimensional Artin L-functions with integer exponents, reducing the behavior at $s=1$ to abelian extensions.

For a finite group, the regular representation contains $d_{\rho }$ copies of each irreducible representation $\rho$. Its character of a representation is $|G|$ at the identity and zero elsewhere. Thus ${\sum }_{\rho }{d}_{\rho }{\chi }_{\rho }(x)=|G|1_{\{x=e\}}$, including the trivial representation. This formula converts independent uniform expectations into constrained configuration counts.

For a conjugacy class $C$ of a finite group, $z_C={\sum }_{g\in C}g$ is central in its group algebra. In a complex irreducible representation of degree $d$ and character of a representation $\chi$, it acts by $|C|\chi (g)/d$; its trace gives this formula and Schur lemma gives scalarity.

A continuous unitary character is a continuous group homomorphism from a topological group to the unit circle with multiplication. For abelian groups, multiplication of continuous unitary characters is pointwise. Such continuous unitary characters must be distinguished from the traces of higher-dimensional group representations, called characters of a representation, and from characters of an algebra.

For every integer $n$, the map ${\chi }_n(z)=z^n$ is a continuous unitary character of the circle group: ${\chi }_n(zw)={\chi }_n(z){\chi }_n(w)$ and $|{\chi }_n(z)|=1$. By contrast, the two-dimensional group representation $\rho (z)=\mathrm{diag}(z,z^{-1})$ has character of a representation $\mathrm{tr}\rho (e^{i\theta })=2\cos \theta$. Its value at the identity element is $2$, so this trace is not a continuous unitary character.

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.

Subgroups $H_1,H_2$ of a finite group $G$ are Gassmann equivalent if $|H_1\cap C|=|H_2\cap C|$ for every conjugacy class $C$. Equivalently their coset permutation representations have the same character of a representation. They need not be conjugate subgroups. The Sunada theorem turns this equality into equality of Laplacian eigenfunction multiplicities.

For a character $\psi$ of $H\le G$, the induced character isThis is the character of a representation of $\mathbb{C}[G]{\otimes }_{\mathbb{C}[H]}{V}_{\psi }$. Substituting this expression into the character inner product and putting $g=xh{x}^{-1}$ proves Frobenius reciprocity.

A one-dimensional character is a character of a representation arising from a one-dimensional representation, equivalently a homomorphism from a group to the multiplicative group of its scalar field.

A one-dimensional complex group representation acts on a one-dimensional complex vector space, and is equivalently a group homomorphism into the multiplicative group ${\mathbb{C}}^{\times }$. It is automatically an irreducible representation. Its character of a representation is its scalar value, a one-dimensional character. A change of basis cannot change those scalar values, so two such group representations are equivalent exactly when their homomorphisms agree. For a symmetric group $S_n$ with $n\ge 2$, all transpositions are conjugate, square to the identity and generate the group. Their common scalar image is therefore $1$ or $-1$, giving exactly the trivial representation and the sign representation.

A complex character of a representation of a finite group $G$ is the functionwhere $\rho :G\to \mathrm{GL}(V)$ is a finite-dimensional complex group representation. The character is a class function because trace is unchanged by conjugation. Its degree is $\chi (1)=\dim V$.

Use the classes $(E,2C_3,3{\sigma }_v)$ of the C3v point group. For the three hydrogen $1s$ atomic orbitals, the character of a representation counts orbitals fixed by each symmetry operation. Identity fixes three, a nontrivial threefold rotation fixes none, and a vertical mirror fixes one. ThereforeThe character inner product gives multiplicitiesThus ${\Gamma }_{\mathrm{H}}=A_1\oplus E$, with the $E$ contribution accounting for two independent symmetry orbitals.

A symmetry-closed orbital basis, often called a complete basis in this context, is a linearly independent set whose span is carried into itself by every group operation. No transformed basis function requires a function outside that span. This is completeness for the chosen group representation, not a claim that a small atomic-orbital set spans the full electronic Hilbert space.

A representation matrix $D(g)$ specifies the linear transformation produced by a molecular symmetry operation $g$. With the column convention,Its character of a representation is the matrix trace,The character is unchanged by a simultaneous change of basis and is constant on a conjugacy class, which is why character tables have one column per class.

A $180^{\circ }$ rotation leaves both metal centres fixed but reverses each transverse p orbital: $p_x\mapsto -p_x$ and $p_y\mapsto -p_y$. Its matrix is $-I_4$, so its character of a representation is $\chi (C_6^3)=-4$, also obtained from $4\cos \pi$.

Choose transverse directions parallel and perpendicular to the vertical mirror plane. At each metal the parallel p orbital is unchanged, whereas the perpendicular p orbital changes sign. Thus each two-dimensional block has trace $1-1=0$. The total character of a representation is $\chi ({\sigma }_d)=0$. A mirror plane passing through an orbital centre does not imply that every orbital on that centre contributes $+1$.

Use the axes actually printed: the two hydrogens lie in the $xz$ plane, $z$ is the principal twofold axis, and $y$ is perpendicular to that plane. For a Cartesian-displacement group representation, only atoms left at their own sites contribute to the trace; their vector components must also be transformed. In the order $(E,C_2(z),\sigma (xz),\sigma (yz))$, identity fixes six components, the twofold rotation exchanges the hydrogens, the $xz$ reflection fixes both hydrogens with vector matrix $\mathrm{diag}(1,-1,1)$, and the $yz$ reflection exchanges them. ThereforeThe nonzero mirror character of a representation is $2(1-1+1)=2$, not six.

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.

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 $R$.

Let $\mathrm{Irr}(G)$ denote the irreducible characters of finite-dimensional complex group representations of $G$. A generalized character, or virtual character, is an integral linear combination of these irreducible characters. A class function is a function constant on conjugacy classes. For class functions use the character inner productA p-elementary group is a direct product of groups $P\times C$, where $P$ is a p-group and $C$ is a cyclic group of order coprime to $p$. An elementary subgroup is a subgroup of this form for some prime $p$. Either factor may be trivial. This notion is different from an elementary abelian group.

Brauer's characterization of characters says that a complex class function $f$ on $G$ is a generalized character exactly when its restriction to every elementary subgroup is a generalized character there. Equivalently,The character orthogonality relations make these inner products the coefficients of the restrictions in their irreducible bases. The theorem detects the integral character lattice. An ordinary character of a representation additionally requires every global irreducible coefficient $⟨f,\chi {⟩}_G$ to be nonnegative; integrality alone characterizes virtual characters, not necessarily actual representations.

The Sunada theorem is the following. Let $X\to M_0$ be a finite normal Riemannian covering of a connected closed Riemannian manifold, with finite deck transformation group $T$. If $U_1,U_2\le T$ satisfy Gassmann equivalence, meaningthen $U_1\backslash X$ and $U_2\backslash X$ have the same scalar Laplacian spectrum, with multiplicities. Equivalently, an epimorphism ${\pi }_1(M_0)\to T$ yields isospectral covers corresponding to the inverse images of $U_1,U_2$.

Here is the representation-theoretic proof. Every finite-dimensional Laplacian eigenspace $E_{\lambda }$ of $X$ is a group representation of $T$, since isometries commute with the Laplace-Beltrami operator. Eigenfunctions on $U_i\backslash X$ identify with $E_{\lambda }^{U_i}$. Averaging gives the projection $P_i=|U_i{|}^{-1}{\sum }_{u\in U_i}u$, soThe two subgroup orders agree, and the character of a representation ${\chi }_{\lambda }$ is constant on conjugacy classes. The displayed multiplicities are therefore equal for every $\lambda$. This proof also applies when $T$ has fixed points on $X$, provided the two subgroups $U_i$ act freely, so their quotients are smooth manifolds. We shall use this extension for an orbifold base in Question 4.

The precise geometric test for an isometry uses the full isometry group of a common universal cover. Equip the simply connected cover $\tilde{X}$ of $M_0$ with the lifted metric and let ${\Gamma }_i$ be the subgroups of its deck transformation group corresponding to the two covers. ThenIndeed, such an $F$ descends to an isometry of the quotients. Conversely an isometry of the quotients lifts to their simply connected covers; lift its inverse with compatible basepoints to see that the lift is bijective. Since it preserves the lifted metrics locally, it is a global isometry, and equivariance of the covering maps shows that it conjugates the deck subgroups. This proves the criterion, rather than assuming that an isometry must preserve the map to $M_0$. In particular, nonconjugacy inside $T$ alone does not establish nonisometry; the full covering space may have additional isometries. Nonisomorphic fundamental groups are a sufficient topological obstruction, and length-identified geometric configurations provide another.

For a concrete nonisomorphic Gassmann equivalent regular subgroups example, fix an odd prime $p$ and take $T=S_{p^3}$. Embed $E=(C_p{)}^3$ and $B={\mathrm{UT}}_3({\mathbb{F}}_p)$ through their regular permutation actions. Both have order $p^3$. Every nonidentity element of $E$ has order $p$. For $B$, write its elements as $I+N$ with $N^3=0$; thenover ${\mathbb{F}}_p$, including $p=3$, so its nonidentity elements also have order $p$. In a regular action, each nonidentity order-$p$ element has cycle type $p^{p^2}$. Thus each subgroup meets the identity class in one element, the class of that cycle type in $p^3-1$ elements, and every other conjugacy class in zero elements. They are Gassmann equivalent. They are not isomorphic: $E$ is abelian, whereas the Heisenberg group $B$ has $[I+E_{12},I+E_{23}]=I+E_{13}\ne I$.

To obtain arbitrarily many examples in dimension five, choose $r$ distinct odd primes $p_1,\dots ,p_r$ with $2^r\ge n$. For each prime use the preceding pair $E_j,B_j\le S_{p_j^3}$. SetA conjugacy class in a direct product is a product of conjugacy classes, and its intersection count with $U_{\epsilon }$ is the product of the individual counts. Thus all $2^r$ subgroups are Gassmann equivalent. They are pairwise nonisomorphic: the unique Sylow $p_j$-subgroup of $U_{\epsilon }$ is abelian precisely when ${\epsilon }_j=0$, and an abstract group isomorphism preserves this property at each prime.

We use the closed-manifold realization of a finitely presented fundamental group in dimension five. Its construction here is short. From a finite presentation of $T$ with $s$ generators, start with the connected sum of $s$ copies of $S^1\times S^4$. Its fundamental group is free on those generators. Represent the finitely many relators by disjoint embedded circles, which is possible by general position for curves in a manifold in dimension five. Their oriented normal bundles are trivial. Perform framed circle surgery, removing $S^1\times D^4$ and inserting $D^2\times S^3$. The Seifert-van Kampen theorem shows that each surgery kills exactly the normal closure of the corresponding relator: the attaching circle bounds the new disk, and the new $S^3$ factor contributes no fundamental group. The resulting closed connected five-manifold $M_0$ has ${\pi }_1(M_0)=T$.

Choose any smooth Riemannian metric on $M_0$ and lift it to its universal cover $X$. Since $T$ is finite, $X$ is a compact simply connected five-manifold on which $T$ acts freely and isometrically. The Sunada theorem makes all $U_{\epsilon }\backslash X$ isospectral, whileshows that they are pairwise nonhomeomorphic, hence nonisometric. Taking any $n$ members proves for every positive integer $n$, there are $n$ pairwise nonisometric closed five-manifolds with the same Laplacian spectrum.