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For a complex irreducible representation $D$ of a finite group $G$, the sum $A={\sum }_{g}D(g^2)$ commutes with every $D(h)$: conjugation sends its summands to $D((hg{h}^{-1}{)}^2)$, merely permuting the summation index. Schur's lemma gives $A=cI$. Taking the trace yields $c\dim D={\sum }_g\chi (g^2)=|G|\nu (D)$, proving the identity. Here $\nu (D)$ is the Frobenius-Schur indicator. For a reducible representation the sum is still central in its image, but need not be one scalar on all irreducible constituents.

For a finite-dimensional complex semisimple representation $V={⨁}_{\lambda }({\mathbb{C}}^{m_{\lambda }}\otimes V_{\lambda })$ with pairwise inequivalent irreducible representations $V_{\lambda }$, its commuting endomorphisms are exactly ${⨁}_{\lambda }(A_{\lambda }\otimes I_{V_{\lambda }})$. To prove it, write each block as an intertwiner and apply Schur's lemma: blocks between inequivalent constituents vanish, and blocks between fixed copies of the same constituent are scalar. The dimension is ${\sum }_{\lambda }{m}_{\lambda }^2$.

A simple module is a nonzero module with no proper nonzero submodule. For a finite-dimensional simple complex module $S$, any endomorphism $\phi$ has a complex eigenvalue $\lambda$. The nonzero kernel of $\phi -\lambda I$ is an $A$-submodule, hence all of $S$. Thus $\phi =\lambda I$, proving Schur's lemma.

More generally a nonzero map between two simple modules is an isomorphism, because its kernel is zero and its image is the whole target. The endomorphism algebra of $S_1\oplus S_2$ is the block algebra with entries ${\mathrm{Hom}}_A(S_j,S_i)$. For nonisomorphic summands its off-diagonal entries vanish and its diagonal entries are scalars, giving $\mathbb{C}\times \mathbb{C}$. For isomorphic summands, choose one fixed isomorphism to identify them; every block is scalar and block composition is matrix multiplication, giving $M_2(\mathbb{C})$.

A finite-dimensional algebra is a semisimple algebra when its left regular module is a direct sum of simple modules. The complex Artin–Wedderburn theorem gives $A\cong {\prod }_i{M}_{d_i}(\mathbb{C})$, with the multiset of positive integers $d_i$ uniquely determined. Its dimension is ${\sum }_i{d}_i^2$. For dimension eight only sizes one and two can occur; the number of size-two blocks is zero, one, or two. Thus there are exactly three isomorphism types:They are distinct by the uniqueness clause, or by their numbers of one-dimensional simple modules. If $G$ is nonabelian, $\mathbb{C}G$ is noncommutative because its group elements form a basis, so the first type is excluded. The augmentation $\sum a_{g}g\mapsto \sum a_g$ provides a one-dimensional simple module, excluding the third type. ConsequentlyThere is one simple module for each matrix block, of dimension the block size, so exactly four are one-dimensional. Every nonabelian group of order eight therefore has this same complex group algebra.

In the finite-dimensional setting of the preceding part, let $S$ be a simple module. Each action map $s\mapsto as$ is an $A$-module endomorphism because $A$ is commutative. The complex Schur's lemma makes every such action a scalar. Therefore every linear subspace of $S$ is an submodule; simplicity forcesEquivalently, a simple module over a commutative ring is a quotient by a maximal ideal, and a finite-dimensional field extension of $\mathbb{C}$ is just $\mathbb{C}$.

The finite-dimensional context matters. If the clause is read as allowing an arbitrary commutative complex algebra, the unrestricted assertion is false: $A=\mathbb{C}(t)$ is simple as a module over itself and is infinite-dimensional over $\mathbb{C}$. This is simple module of a finite-dimensional commutative complex algebra. Thus the one-dimensional conclusion uses the finite-dimensional algebra context, or an alternative hypothesis such as finite dimensionality of the simple module.

For a finite-dimensional algebra, every simple module $S$ is finite-dimensional: choose $s\ne 0$; simplicity makes $A\to S$, $a\mapsto as$, surjective, so $\dim S\le \dim A$.

The general Schur's lemma says that a homomorphism $f:M\to N$ between simple modules over any unital ring is either zero or an isomorphism. Indeed, $\ker f$ is a submodule of $M$ and $\mathrm{im}f$ is a submodule of $N$. If $f\ne 0$, simplicity gives $\ker f=0$ and $\mathrm{im}f=N$. In particular, ${\mathrm{End}}_R(M)$ is a division ring.

For a complex algebra and a finite-dimensional simple module $M$, the stronger conclusion isFor $f\in {\mathrm{End}}_R(M)$, the fundamental theorem of algebra gives an eigenvalue $\lambda \in \mathbb{C}$. The endomorphism $f-\lambda I$ has nonzero kernel, so the general Schur's lemma forces it to be zero. Consequently a homomorphism between two such simple modules is zero if they are nonisomorphic, and their homomorphism space is one-dimensional if they are isomorphic.

For equal irreducible group representations, the intertwining identity and the complex Schur's lemma give $T=c{I}_n$. Conjugation preserves trace, soSince $\mathrm{tr}(c{I}_n)=nc$, we obtain

The preceding intertwining identity makes $T$ a homomorphism between the corresponding simple modules. By Schur's lemma, a nonzero such homomorphism would be an isomorphism. The representations are inequivalent, so this is impossible. Therefore

The general Schur's lemma says that a homomorphism $f:M\to N$ between simple modules over any unital ring is either zero or an isomorphism. Indeed, $\ker f$ is a submodule of $M$ and $\mathrm{im}f$ is a submodule of $N$. If $f\ne 0$, simplicity gives $\ker f=0$ and $\mathrm{im}f=N$. In particular, ${\mathrm{End}}_R(M)$ is a division ring.

For a complex algebra and a finite-dimensional simple module $M$, the stronger conclusion isFor $f\in {\mathrm{End}}_R(M)$, the fundamental theorem of algebra gives an eigenvalue $\lambda \in \mathbb{C}$. The endomorphism $f-\lambda I$ has nonzero kernel, so the general Schur's lemma forces it to be zero. Consequently a homomorphism between two such simple modules is zero if they are nonisomorphic, and their homomorphism space is one-dimensional if they are isomorphic.

The finite-dimensional qualification is essential for this scalar conclusion. For example, the field $\mathbb{C}(t)$ is a complex algebra, simple as a module over itself, and its endomorphisms include multiplication by $t$, which is not a complex scalar.

A symmetry orbital is a molecular orbital when it is also an eigenfunction of the effective one-electron Hamiltonian operator. If an irreducible representation occurs once in a complete invariant model space, a symmetry-invariant Hamiltonian acts as a scalar on that copy by Schur's lemma. Its basis functions are then degenerate eigenfunctions. When the same symmetry occurs in several copies, those copies can mix; one must diagonalize the corresponding block. Merely having a one-dimensional symmetry label is insufficient if other orbitals have that same label.

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 $V\otimes V^{\ast }$.

A group is a nonempty set $G$ with an operation $G\times G\to G$, $(g,h)\mapsto gh$, satisfying associativity, an identity element $e$ with $eg=ge=g$, and an inverse element $g^{-1}$ with $g{g}^{-1}=g^{-1}g=e$ for every $g$. These are the group axioms. An Abelian group additionally satisfies $gh=hg$ for every pair. A subgroup is a subset that is itself a group under the restricted operation; equivalently, it is nonempty and closed under $h{k}^{-1}$.

Let $R:G\to \mathrm{GL}(V)$ be a nonzero, finite-dimensional complex irreducible representation of an Abelian group. For each fixed $g$, commutativity givesThus $R(g)$ is an intertwining operator from $R$ to itself. By Schur's lemma over the complex numbers, $R(g)={\lambda }_g{I}_V$. Every line in $V$ is therefore an invariant subspace. Since $R$ is irreducible, this forcesThis is the finite-dimensional convention used throughout the paper. Without it, the literal assertion for arbitrary Abelian groups needs qualification. A concrete counterexample is $G=\mathbb{C}(t{)}^{\times }$ acting by multiplication on the complex vector space $V=\mathbb{C}(t)$ of rational functions. Any nonzero invariant complex vector subspace containing $f$ contains $gf$ for every nonzero rational $g$, hence contains every rational function. This is an infinite-dimensional irreducible representation. The scalar conclusion of Schur's lemma used above depends on finite dimensionality.

Schur's lemma states that an intertwiner $T:V\to W$ between irreducible representations is either zero or an isomorphism. In particular, inequivalent irreducible representations have no nonzero intertwiners. For a finite-dimensional irreducible representation over the complex numbers, every commuting endomorphism is scalar:Indeed, the kernel and image of an intertwiner are invariant subspaces, proving the first statement. For the second, choose an eigenvalue $\lambda$ of $T$. The intertwiner $T-\lambda I$ has nonzero kernel, so it must vanish. The scalar conclusion uses an algebraically closed coefficient field: over the real numbers, planar rotation through $\pi /2$ gives an irreducible real representation of ${\mathbb{Z}}_4$ and commutes with itself without being a real scalar matrix.

Partition $M$ into the same five-by-five blocks as $R$. The equation for block $(i,j)$ is $M_{ij}{R}_j(g)=R_i(g)M_{ij}$, so each block is an intertwiner. By Schur's lemma, blocks between the two inequivalent irreducible representations vanish, whereas blocks between identical copies are arbitrary scalar multiples of the relevant identity matrix. Thus the commutant of a semisimple representation isEquivalently, $M=(A\otimes I_{n_1})\oplus (B\otimes I_{n_2})$, where $A\in M_3(\mathbb{C})$ and $B\in M_2(\mathbb{C})$. Substitution proves every such matrix commutes with every $R(g)$. There is no invertibility requirement on $M$, and its solution space has complex dimension $3^2+2^2=13$, even if $n_1=n_2$.

On a monomial $X^a{Y}^b$, direct differentiation givesFor example $(EF-FE)(X^a{Y}^b)=[a(b+1)-b(a+1)]X^a{Y}^b=(a-b)X^a{Y}^b$. The same calculation with $H$ givesThese are exactly the brackets of $h,e,f$, so the linear map $\rho$ respects every bracket and is a Lie algebra representation.

For $m\ge 0$, let $W_m$ be the space of homogeneous polynomials of total degree $m$. Each operator preserves total degree. Every polynomial is a finite sum of homogeneous polynomials, soIn $W_m$, the monomials $v_j=X^{m-j}{Y}^j$ have distinct weights $m-2j$ and exactly the actions described in part (i). To see simplicity directly, a nonzero invariant subspace is invariant under $H$ and hence contains one of its weight components: polynomial interpolation in $H$ projects onto each of its distinct eigenspaces. Starting with a nonzero multiple of $v_j$, repeated application of $E$ reaches $v_0$, and repeated application of $F$ then reaches all $v_j$. All coefficients used are nonzero in characteristic zero. Therefore $W_m$ is simple and isomorphic to $L_m$.

ThusThis is an algebraic direct sum, not an infinite product or completion. Its simple submodules are precisely the $W_m$: they are pairwise nonisomorphic and each occurs once, so projection onto homogeneous degrees and Schur's lemma exclude diagonal copies between distinct degrees. Complete reducibility here follows from the explicit decomposition, rather than applying a finite-dimensional theorem directly to the infinite-dimensional $W$.

For a compact Lie group $G$, the complex representation ring of a compact group $R(G)$ is the Grothendieck group of finite-dimensional continuous complex group representations, with the relations $[M\oplus N]=[M]+[N]$. Multiplication is $[M][N]=[M\otimes N]$, and the unit is the trivial one-dimensional representation. The ring of class functions iswith pointwise addition and multiplication. The character of $M$ is ${\chi }_M(g)=\mathrm{tr}{\rho }_M(g)$. The trace identities for direct sums and tensor products makea well-defined ring homomorphism $\chi :R(G)\to c\ell (G)$.

Normalize Haar measure by ${\int }_{G}dg=1$. Averaging a positive definite Hermitian inner product giveswhich is positive definite and $G$-invariant. This proves unitarization of a compact-group representation. In a unitary representation the orthogonal complement of an invariant subspace is invariant. Induction on dimension therefore gives complete reducibility of compact-group representations. Hence every element of $R(G)$ is a finite integral linear combination of irreducible classes.

We establish the needed character orthogonality for compact groups carefully. For irreducible unitary representations $V,W$, let $G$ act on ${\mathrm{Hom}}_{\mathbb{C}}(W,V)$ byIts average $P={\int }_G(g\cdot )dg$ is a projection onto ${\mathrm{Hom}}_G(W,V)$: averaging makes every image invariant, and it fixes every intertwiner. The trace of this action is ${\chi }_V(g)\overline{{\chi }_W(g)}$, since the inverse of a unitary matrix has conjugate trace. Taking the trace of the projection givesFor completeness, Schur's lemma follows here from invariance of the kernel and image of an intertwiner. A nonzero intertwiner between irreducibles is an isomorphism. An endomorphism of an irreducible complex representation has an eigenvalue $\lambda$; the noninvertible intertwiner $A-\lambda I$ must vanish. Thus the last dimension is one for isomorphic irreducibles and zero for inequivalent ones.

If a virtual character ${\sum }_V{n}_V{\chi }_V$ vanishes, integrating it against $\overline{{\chi }_W}$ gives $n_W=0$ for every irreducible $W$. The irreducible multiplicities therefore distinguish the classes, andThis also proves that two finite-dimensional representations with the same character are isomorphic.

For SU(2), the irreducibles are $V_k={\mathrm{Sym}}^k({\mathbb{C}}^2)$, $k\ge 0$, of dimension $k+1$. One can see this classification through the usual ${\mathfrak{sl}}_2$ operators $H,E,F$: a highest-weight vector $v$ has $Hv=kv$, $Ev=0$, and the relations $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$ give, by induction,The torus weights are integers. If $F^{r}v\ne 0$ but $F^{r+1}v=0$, the identity at $j=r+1$ forces $k=r\ge 0$. The vectors $v,Fv,\dots ,F^{k}v$ span an invariant irreducible module, which in an irreducible representation is the whole space. This is the symmetric-power module. Conversely its successive monomial weight vectors are linked by $E,F$ with nonzero coefficients, proving its irreducibility. Thus the classification of finite-dimensional representations of SU2 gives every irreducible, rather than just a list of examples.

On the maximal torus $T=\{\mathrm{diag}(z,z^{-1}):|z|=1\}$, their characters areWith $x={\chi }_1=z+z^{-1}$, multiplication gives ${\chi }_{k+1}=x{\chi }_k-{\chi }_{k-1}$, starting with ${\chi }_0=1$. Hence each ${\chi }_k$ is a monic integral polynomial of degree $k$ in $x$. For example ${\chi }_2=x^2-1$ and ${\chi }_3=x^3-2x$. They form a basis over $\mathbb{Z}$, soThe character map realizes this ring as the finite integral polynomials in $z+z^{-1}$, interpreted as continuous class functions. Its image is not the whole infinite-dimensional ring of continuous class functions; injectivity is the assertion being proved.

Frobenius reciprocity is the adjointness of character induction and restriction: for class functions $f$ on $H$ and $u$ on $G$,For characters this says that the multiplicity of an irreducible $G$-character in an induced character equals the corresponding multiplicity after restriction. To prove the identity directly, insert the induced class function formula and interchange the two finite sums. For each $x$, substitute $g=xh{x}^{-1}$ with $h\in H$. Since $u$ is constant on conjugacy classes,This proves the theorem for all class functions, not just irreducible characters.

For the group-theoretic conclusion, suppose that $p$ divides $|G^{\prime }\cap Z(G)|$. A central element of order $p$ lies in every Sylow p-subgroup: adjoining it to a Sylow subgroup gives a p-subgroup, so maximality forces it into that subgroup. Hence there is a subgroup $Q\le P\cap G^{\prime }\cap Z(G)$ of order $p$. Choose a nontrivial linear character $\lambda$ of $Q$.

Since $P$ is abelian, $\lambda$ extends to a linear character $\mu$ of $P$. One direct justification is to extend from an abelian subgroup $A$ to $⟨A,x⟩$ one generator at a time: if $d$ is the least positive exponent with $x^d\in A$, choose a complex $d$th root of $\lambda (x^d)$ as the value at $x$. The formula on $a{x}^j$ is then well defined and multiplicative. Iteration yields the claimed extension to $P$.

The induced character ${\mu }^G$ has degree $[G:P]$, which is prime to $p$. In its decomposition into irreducible characters, at least one constituent $\chi$ must have degree prime to $p$; otherwise the total degree would be divisible by $p$. Frobenius reciprocity also gives $⟨{\chi }_P,\mu {⟩}_P>0$ for this constituent. Restriction to the central subgroup $Q$ forcesIndeed, the induced representation itself restricts to $Q$ as copies of $\lambda$, because conjugation by every coset representative is trivial on $Q$; alternatively this follows from Schur's lemma and the positive multiplicity after restriction.

Each $q\in Q$ consequently acts in the representation affording $\chi$ as the scalar $\lambda (q)$. Taking its determinant character givesThe left side is trivial since $Q\le G^{\prime }$ and every linear character kills $G^{\prime }$. The right side is nontrivial since $\lambda$ has order $p$ and $p\nmid \chi (1)$. This contradiction provesThe argument is the central commutator has no torsion from an abelian Sylow subgroup determinant obstruction.

For the symmetric group representation in part (a), the identity and three transpositions contribute $4I$ to $A={\sum }_{g}D(g^2)$. The two 3-cycles contribute $D(r)+D(r^2)=-I$. ThusMore generally, the central average of representation matrices at squares commutes with the representation, because conjugation permutes the summands. For a complex irreducible representation, Schur's lemma therefore gives $A=cI$. Taking the trace yields $c\dim D=|G|\nu$, proving the claimed formula under its correct hypotheses.

For the Klein four-group there is no two-dimensional irreducible representation, as established in part (b). Its genuine one-dimensional irreducibles have $A=4$ and $\nu =1$, so $A=(4\cdot 1/1)I$.

If the intended alternative is the real quarter-turn representation of the cyclic group of order four, its squared matrices sum to $I-I+I-I=0$. Thus $A=0=(4\cdot 0/2)I$ in that example. This equality is a direct calculation; the complex version of Schur's lemma cannot be applied to its reducible complexification.

Put $B=R/I$. By the definition of a left-primitive ideal, there is a faithful module $V$ that is a simple module on the left over $B$. The assumed relation passes to the quotient ring $B$, since any pair of elements of $B$ can be lifted to $R$.

By Schur's lemma, the endomorphism ring ${\mathrm{End}}_B(V)$ is a division ring: the kernel and image of a nonzero module endomorphism are submodules, so simplicity makes that module endomorphism both injective and surjective. SetEvaluation makes $V$ a right vector space over a division ring $\Delta$, and the $B$-action is $\Delta$-linear. The opposite ring in the definition ensures the correct order for this right action.

The Jacobson density theorem says that, for any finite $\Delta$-linearly independent family $v_1,\dots ,v_m$ and any $w_1,\dots ,w_m\in V$, some $b\in B$ satisfies $b{v}_i=w_i$ for every $i$.

Suppose that $V$ contains two $\Delta$-linearly independent vectors $v,w$. Applying the Jacobson density theorem twice, choose $a,b\in B$ satisfyingThe inherited relation giveswith integers $c_k$. Acting on $w$, the left side gives $baw=bv=w$, but every term on the right gives $a^{k}bw=0$. This contradicts $w\ne 0$. Thus ${\dim }_{\Delta }V=1$.

The ring homomorphismgiven by the action is injective because $V$ is a faithful module. It is surjective because the Jacobson density theorem, applied to a single nonzero basis vector, realizes every $\Delta$-linear endomorphism of this one-dimensional vector space over a division ring. Every nonzero such endomorphism is invertible. HenceThe use of the Jacobson density theorem here only prescribes actions on finitely many vectors; it does not assume that an arbitrary primitive ring is already a full endomorphism ring.

For a finite-dimensional commutative complex algebra, a simple module is a cyclic quotient of the algebra and hence finite-dimensional. Every action map is a module endomorphism, since the algebra is commutative. The complex Schur's lemma makes each action scalar, so every vector subspace is a submodule; simplicity forces dimension one. The finite-dimensional hypothesis cannot simply be deleted: $\mathbb{C}(t)$ acting on itself is a simple module of infinite complex dimension.

For a finite-dimensional irreducible complex representation $V$, invariant tensors in $V\otimes V$ correspond to equivariant maps $V^{\ast }\to V$. Schur's lemma makes their dimension one if $V$ is self-dual and zero otherwise. For a finite group, in particular, a real-character irreducible representation has precisely one trivial copy in its tensor square. For arbitrary complex irreducibles, the unconditional statement is that $V\otimes V^{\ast }$ contains the trivial representation once.

For an arbitrary group, a real-valued character need not imply self-duality: the one-dimensional group representation $n\mapsto 2^n$ of $\mathbb{Z}$ has real character but its tensor square has action $n\mapsto 4^n$, so contains no trivial representation.