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Within each fixed unitary irreducible representation, choose the selected right singular vectors to be orthonormal. Their corresponding left singular vectors are also orthonormal, even when a singular value is repeated: choose an orthonormal basis in each eigenspace of $T_{\rho }^{\ast }{T}_{\rho }$ and put $B_p=T_{\rho }{A}_p/{\lambda }_p$. Therefore, when ${\rho }_p={\rho }_q$,This is (iv), with the unnormalized Hilbert-Schmidt inner product and the Kronecker delta. No orthogonality of differently shaped matrices is being asserted.

For the unheaded continuation, partition $b\in {\mathbb{C}}^m$ into blocks $b_p\in {\mathbb{C}}^{n_p}$. ThenThe Schur averaging of rectangular matrices formula isfor any $n_p\times n_q$ matrix $M$. The average is an intertwiner; the Schur lemma makes it zero for inequivalent irreducible representations, and a scalar multiple of the identity for the same representative. The trace determines that scalar in the latter case.

Apply this with $M=U(p{)}^{\ast }U(q)$. The Hilbert-Schmidt inner product normalization just proved givesExpanding the squared Euclidean norm now yieldsThus the final identity follows from averaging, although $U^{\ast }U$ itself need not be the identity. The whole construction is the spectral inverse theorem for the matrix-valued U2 quantity.

For the finite-dimensional Lie algebra representation specified in the PDF, a Lie-invariant bilinear form satisfiesThe dual Lie algebra representation has action $(x\varphi )(w)=-\varphi (xw)$, so the map $T_B:V\to V^{\ast }$ defined by $T_B(v)=B(v,-)$ is an intertwining operator. If $V$ is irreducible and $B\ne 0$, its kernel is zero. Since $V,V^{\ast }$ have equal finite dimension, $T_B$ is an isomorphism, and $B$ is a nondegenerate bilinear form.

For $B_1\ne 0$, the composition $T_{B_1}^{-1}{T}_{B_2}$ is a representation endomorphism. Over an algebraically closed field, the Schur lemma makes it scalar, provingTransposing $B$ gives another Lie-invariant bilinear form. For $B\ne 0$, write $B^T=\mu B$; transposing twice gives ${\mu }^2=1$. In field characteristic different from two, this means $\mu =1$ or $\mu =-1$. Thus the form is a symmetric bilinear form or an alternating bilinear form, respectively. The zero form has both properties. In the second case $B(v,v)=-B(v,v)$ forces $B(v,v)=0$, using the same characteristic assumption.

The local spectral interchanges preserve an irreducible representation, and standard Young tableaux of a fixed shape are connected by admissible adjacent interchanges. Different shapes have different multisets of Young-diagram cell contents. Their elementary symmetric functions in the Young–Jucys–Murphy elements are central: the identityproves this coefficient by coefficient. By the Schur lemma, one irreducible representation cannot therefore contain two different shape classes. The multiset determines a diagram because the counts on its positive and negative diagonals determine its arm and leg lengths at the diagonal cells.

Different irreducible representations cannot share a joint eigenvalue vector: the Gelfand–Tsetlin algebra is generated by the Young–Jucys–Murphy elements, so all its elements would act identically on that vector in both representations, whereas a central primitive idempotent distinguishes their two blocks. Thus distinct irreducibles give distinct shape classes.

The number of irreducible representations is the number $p(n)$ of conjugacy classes, each indexed by a partition of an integer. Every spectral vector gives a tableau by the preceding construction. Closure under admissible interchanges makes every shape class that occurs occur in full. The two counts $p(n)$ then force every shape to occur, with exactly one irreducible representation per shape. This supplies $\mathrm{Spec}(n)=\mathrm{Cont}(n)$.

On restricting to $S_{n-1}$, delete the last coordinate. For tableaux of shape $\lambda$, the entry $n$ occupies one Removable node of a Young diagram. Grouping by that node gives one copy of the corresponding module; the path decomposition has simple multiplicities. Hence the restriction branching rule for a symmetric group isThe different removable nodes produce distinct partitions. The multiplicity-free structural input defining the Gelfand–Tsetlin basis is distinct from identifying this graph with the graph of Young diagrams.

In an irreducible complex group representation $\rho$, every $\rho (z)$ for $z$ in the center of a group commutes with the entire representation. Schur lemma therefore gives $\rho (z)={\lambda }_{z}I$. Faithfulness makes $z\mapsto {\lambda }_z$ an injective homomorphism $Z(G)\to {\mathbb{C}}^{\times }$.

Its image is a finite subgroup. If its order is $m$, Lagrange's theorem for finite groups shows that every element is an $m$th root of unity. These roots form a cyclic group, and every subgroup of a cyclic group is cyclic. HenceBoth irreducibility over the algebraically closed field and faithfulness are used: scalar action alone would not embed the centre.

First the product is also split. Write $A_i=k{G}_i$ and $J_i=\mathrm{rad}{A}_i$. The idealis nilpotent: if $J_1^r=J_2^s=0$, every product of $r+s-1$ of its factors vanishes. The quotient isa product of full matrix algebras over $k$, because each $A_i$ is split. A nilpotent ideal lies in the Jacobson radical, and a semisimple quotient forces the reverse inclusion, so $I$ is exactly the radical. Thus $k$ is a splitting field for finite group representations of the product.

For simple modules $S_i$, Schur lemma and splitting give ${\mathrm{End}}_{A_i}(S_i)=k$. The Jacobson density theorem therefore makes the image of $A_i$ on $S_i$ the full ${\mathrm{End}}_k(S_i)$. Tensoring these maps shows that the product algebra acts on $S_1\otimes S_2$ through its full endomorphism algebra. Hence this tensor product of group representations is simple.

On restriction to $G_1$, it is a direct sum of $\dim S_2$ copies of $S_1$. If two external tensor products are isomorphic, their restrictions and the Jordan–Hölder theorem force $S_1\cong S_1^{\prime }$; restricting to $G_2$ similarly forces $S_2\cong S_2^{\prime }$. The converse follows by tensoring the isomorphisms.

Finally, the p-regular conjugacy classes of $G_1\times G_2$ are precisely pairs of such classes in the factors. The corrected result in part (b) counts as many simples for the product as pairs of simples for the two factors. Our pairwise nonisomorphic tensor products already attain that count, so they exhaust all simples. Thus $\boxed{{\mathrm{Irr}}_k(G_1\times G_2)=\{S_1\otimes S_2\}}$, uniquely indexed by pairs of simple isomorphism classes.

Completeness permits idempotent lifting from $kG$ to $\mathcal{O}G$. In particular, each projective cover $P_j$ lifts to a finite projective $\mathcal{O}G$-lattice ${\tilde{P}}_j$; one can lift an idempotent presenting it as a summand of a finite free module. Set $Q_j=K{\otimes }_{\mathcal{O}}{\tilde{P}}_j$.

For a simple $S_{\ell }$, every map $P_j\to S_{\ell }$ factors through its simple head $S_j$. Splitting and Schur lemma give ${\dim }_k{\mathrm{Hom}}_{kG}(P_j,S_{\ell })={\delta }_{j\ell }$. Exactness of this Hom functor along a composition series therefore givesPart (iii) identifies this with ${\dim }_K{\mathrm{Hom}}_{KG}(Q_j,V_i)$. By Maschke's theorem and splitting, $KG$ is split semisimple, soThe reductions of any two integral forms of the same ordinary module have identical Brauer characters, hence identical composition multiplicities. We may thus reduce a direct-sum form for the displayed decomposition instead of ${\tilde{P}}_j$. In the modular representation ring this givesComparing the simple basis coefficients proves $c_{\ell j}={\sum }_i{d}_{i\ell }{d}_{ij}$, or $\boxed{C=D^{T}D}$.

The row orthogonality relations for a character table state that for irreducible characters $\chi ,\psi$,Equivalently, if $c$ runs through representatives of the conjugacy classes $C$,

Fix an irreducible $\chi$. The sum of the elements in a conjugacy class $C$ is central in $\mathbb{C}G$, so Schur lemma says that it acts in the representation affording $\chi$ by the scalarThis central character value of a conjugacy-class sum is an algebraic integer: the class sum acts by a matrix with integer entries on the regular representation, and ${\omega }_{\chi }(C)$ is one of its eigenvalues. Also $\overline{\chi (c)}$ is an algebraic integer because character values are sums of roots of unity.

Row orthogonality with $\psi =\chi$ now givesThe right-hand side is an algebraic integer. The left-hand side is rational, and every rational algebraic integer is an integer. Therefore

Apply part (b) with $P=P_i$ and $U=S_j$. A homomorphism $P_i\to S_j$ kills $J(P_i)$ and therefore factors through the head $P_i/J(P_i)\cong S_i$. By Schur lemma over the splitting field,Consequently the two stated bases satisfy the Duality of simple and projective Brauer characters:

The form is nondegenerate directly: if $\varphi \ne 0$, thenEquivalently, in coordinates indexed by the p-regular conjugacy classes it is a positive diagonal Hermitian form with weights $1/|C_G(x)|$.

If $P$ is a prime ideal of the commutative Artinian ring $R$, then $R/P$ is an Artinian domain. For $0\ne x\in R/P$, the descending chain $(x)\supseteq (x^2)\supseteq \cdots$ stabilizes, so $x^n=x^{n+1}a$ for some $n$; cancellation gives $xa=1$. Thus $R/P$ is a field and $P$ is maximal.

There are only finitely many maximal ideals. Otherwise, distinct maximal ideals $P_i$ are pairwise comaximal and their finite products give the strictly descending chaincontradicting the Artinian condition. This proves the statement about the prime ideals of a commutative Artinian ring.

The nilradical $N$ is the ideal of all nilpotent elements, equivalently the intersection of all prime ideals. Its powers stabilize, say $N^r=N^{r+1}=I$. If $I\ne 0$, choose an ideal $J$ minimal subject to $IJ\ne 0$, then choose $x\in J$ with $Ix\ne 0$. Since $I^2=I$, minimality gives $Rx=Ix$, so $x=ax$ for some $a\in I$. But $a$ is nilpotent, hence $1-a$ is a unit, contradicting $(1-a)x=0$. Therefore the nilradical of a commutative Artinian ring is nilpotent.

If the maximal ideals are $P_1,\dots ,P_s$, the Chinese remainder theorem givesa finite product of fields. Each layer $N^j/N^{j+1}$ is an Artinian module over this product and therefore finite-dimensional. Since $N$ is nilpotent, these finitely many layers show that every ideal of $R$ is finitely generated. This proves the Artinian commutative ring is Noetherian theorem.

Now assume $R$ is local with maximal ideal $P$ and ${\dim }_{R/P}P/P^2=1$. The Nakayama lemma gives $P=(x)$. Since $P$ is nilpotent, its powers form a finite chain ending in zero. For a nonzero ideal $J$, choose the largest $r$ with $J\subseteq (x^r)$ and take $y\in J\setminus (x^{r+1})$. Write $y=x^{r}u$; then $u\notin P$, so $u$ is a unit. Henceand every ideal of $R$ is principal.

Finally, the Artin–Wedderburn theorem says that a finite-dimensional semisimple $k$-algebra has the formwhere each $D_i$ is a finite-dimensional division ring over $k$; the factors are unique up to permutation and isomorphism. By the assumed complete reducibility, write the right regular module aswith pairwise nonisomorphic simple right modules $S_i$. The Schur lemma says that $D_i={\mathrm{End}}_T(S_i)$ is a division ring and that homomorphisms between distinct $S_i$ vanish. Left multiplication and the endomorphism ring of the displayed direct sum therefore giveConversely, the regular module of $M_n(D)$ is a direct sum of $n$ simple column modules, proving that every algebra on the right is semisimple. Uniqueness follows from uniqueness of the simple summands and their multiplicities.

Schur lemma over $\mathbb{C}$ states that an intertwining operator $T:V\to W$ between finite-dimensional irreducible complex representations is either zero or an isomorphism. In particular, every intertwining endomorphism of $V$ is a scalar multiple of the identity.

Indeed, $\ker T\subseteq V$ and $\mathrm{im}T\subseteq W$ are invariant subspaces. If $T\ne 0$, irreducibility gives $\ker T=0$ and $\mathrm{im}T=W$, so $T$ is an isomorphism. For an endomorphism $T:V\to V$, the fundamental theorem of algebra gives an eigenvalue $\lambda \in \mathbb{C}$. Then $T-\lambda I$ is an intertwiner with nonzero kernel, so the first part forces $T-\lambda I=0$. Hence $T=\lambda I$, completing the Proof of Schur lemma.

The Jacobson radical $J=J(R)$ is the intersection of all maximal right ideals, equivalently the largest ideal annihilating every simple right module. The Artin–Wedderburn theorem givesbecause $k$ is algebraically closed.

The descending chain $J\supseteq J^2\supseteq \cdots$ stabilizes since $R$ is finite-dimensional. If $J^m=J^{m+1}$, Nakayama lemma applied to the finite right module $J^m$ gives $J^m=0$. Thus $J$ is nilpotent.

For $f\in {\mathrm{End}}_R(P_i)$, the Fitting lemma givesfor large $n$. Indecomposability makes one summand zero, so $f$ is either invertible or nilpotent. In the latter case $1-f$ is invertible. This is the criterion that ${\mathrm{End}}_R(P_i)$ is a local ring.

Let $M={⨁}_i{P}_i$. Reduction modulo $MJ$ definesSince $M/MJ={⨁}_i{S}_i$ and the $S_i$ are pairwise nonisomorphic simples,by Schur lemma. Arbitrary scalars on the direct summands lift to scalar identity maps on the $P_i$, so $\theta$ is surjective.

If $f\in \ker \theta$, then $f(M)\subseteq MJ$. A product of $r$ such maps sends $M$ into $M{J}^r$, so $\ker \theta$ is nilpotent. A nilpotent ideal lies in the Jacobson radical, while the semisimplicity of the quotient $k^n$ gives the reverse inclusion. HenceThis is exactly the definition of a basic algebra.

A representation is completely reducible, or semisimple, when it is a direct sum of irreducible representation. Equivalently, every invariant subspace has an invariant complement.

Maschke's theorem says that every finite-dimensional representation of a finite group $G$ over a field of characteristic zero is completely reducible. More generally, it is enough that the field characteristic does not divide $|G|$.

Schur lemma states the following.

Indeed, $\ker T\subseteq V$ and $\mathrm{im}T\subseteq W$ are invariant subspaces. If $T\ne 0$, irreducibility forcesso $T$ is an isomorphism. For an endomorphism of a finite-dimensional complex representation, choose an eigenvalue $\lambda$ of $T$. The intertwiner $T-\lambda I$ has nonzero kernel, so the first part forcesThis proves the Proof of Schur lemma.

Now let $\rho :G\to \mathrm{GL}(V)$ be faithful and irreducible over $\mathbb{C}$. For every $z\in Z(G)$, the operator $\rho (z)$ commutes with every $\rho (g)$. Schur's lemma therefore givesfor some ${\lambda }_z\in {\mathbb{C}}^{\times }$. Faithfulness makes $z\mapsto {\lambda }_z$ an embeddingSince every finite subgroup of the multiplicative complex numbers is cyclic,This is the cyclic-center obstruction to a faithful irreducible representation.

Because $M$ is a finite-dimensional semisimple module, it has an isotypic decompositionwhere the $S_i$ are pairwise nonisomorphic simple right $R$-modules. By Schur lemma,is a division ring, while ${\mathrm{Hom}}_R(S_i,S_j)=0$ for $i\ne j$. Consequently every endomorphism preserves the isotypic summands and is a matrix of entries from $D_i$ on each one. ThereforeIf the ground field is algebraically closed and the $S_i$ are finite-dimensional over it, Schur lemma gives $D_i=k$.

Maschke's theorem says that every invariant subspace of a finite-dimensional complex representation of a finite group has an invariant complement. Average any projection over the group to make it equivariant. Induction on dimension therefore decomposes every representation into irreducibles.

The character of a representation is ${\chi }_V(g)=\mathrm{tr}(g{|}_V)$. For a $G$-set $X$, the permutation representation $\mathbb{C}X$ has basis $X$, and its character is the number of fixed points:For the regular action,The multiplicity of an irreducible $V_i$ is the character inner productso

If $V\cong {⨁}_i{n}_i{V}_i$, Schur lemma givesFor $V=\mathbb{C}G$, this is ${\sum }_i(\dim V_i{)}^2=|G|$.

Now suppose $\chi (g)=0$ off the identity. The multiplicity of $V_i$ in $V$ isThus $\dim V/|G|$ is a nonnegative integer and $V$ is that many copies of the regular representation. Finally,so

Use the standard basisof ${\mathfrak{sl}}_2$. For the invariant trace form $B(x,y)=\mathrm{tr}(xy)$, the dual basis is $(f,e,h/2)$, soOn a highest-weight vector $v$ in the $(n+1)$-dimensional irreducible module, $ev=0$ and $hv=nv$. Since $efv=[e,f]v=nv$ and $fev=0$,Centrality and Schur lemma make this the eigenvalue on the whole module. Thus $\lambda =n(n+2)/2$, the Casimir eigenvalue for sl2. If the form is instead the Killing form, which is four times the trace form on ${\mathfrak{sl}}_2$, the corresponding Casimir and eigenvalue are divided by four.

Conjugation by a permutation merely permutes the transpositions. Their sum $C$ is therefore a conjugacy class sum and belongs to the center of an associative algebra $\mathbb{C}{S}_n$. Since the complex Specht module is an irreducible representation, Schur lemma says that $C$ acts on it as a scalar, say $c(\lambda )$.

Schur lemma says that an intertwining linear map between irreducible finite-dimensional complex representations is either zero or an isomorphism. In particular, every endomorphism of an irreducible complex representation is a scalar multiple of the identity.

A continuous representation of a topological group is a continuous homomorphismfor a finite-dimensional complex vector space $V$. It is a unitary representation if $V$ has a positive-definite Hermitian inner product for whichfor every $g\in G$ and $v,w\in V$.

For $G=S^1$, start with any positive-definite Hermitian form and average it using normalized Haar measure:Translation invariance makes this form $S^1$-invariant, and positivity is preserved, proving unitarity. Since $S^1$ is abelian, the operators $\rho (z)$ commute; since they are unitary, they are normal. Simultaneous diagonalization therefore decomposes $V$ into common one-dimensional eigenspaces. Thus every representation of the circle group is a direct sum of one-dimensional representations.

WriteThe group law and inverse areA calculation givesEvery element of the centre $Z$ occurs by taking, for example, $x^{\prime }=y=0$ and $y^{\prime }=1$. Hence the commutator subgroup is exactly $Z$. The image of a one-dimensional representation is abelian, so the one-dimensional representation kills the commutator subgroup and its kernel contains $Z$.

Now let $(\rho ,V)$ be a complex representation of $G/Z_0$. Its restriction to the central subgroupis a representation of the circle group. Decompose it into its distinct weight spaces:Because $Z/Z_0$ is central, every $\rho (g)$ commutes with its action and preserves every common eigenspace. Thus the $V_i$ are $G/Z_0$-subrepresentations, as asserted by the central circle weight-space decomposition.

Let $n_i=\dim V_i$. The mapis a one-dimensional representation of $G/Z_0$; after composition with $G\to G/Z_0$, its kernel contains $Z$. On $z\in Z/Z_0$ its value isso ${\theta }_i^{n_i}=1$. A continuous character of $S^1$ has the form $z\mapsto z^m$; the displayed identity forces $m=0$. Therefore every ${\theta }_i$ is trivial. Since the ${\theta }_i$ were distinct, $d=1$ and ${\theta }_1=1$.

It follows that every finite-dimensional complex representation of $G/Z_0$ kills the entire nontrivial central circle $Z/Z_0$. Its kernel is therefore nontrivial, so no such representation is faithful. This is the real Heisenberg quotient has no faithful finite-dimensional representation.

Write $e,h,f$ for the standard generators of the sl2 Lie algebra. For the representation $\varphi$, use the normalizationThis is the quadratic Casimir element, and by assumption it commutes with every $\varphi (x)$.

Schur lemma says that an endomorphism of a finite-dimensional irreducible complex representation which commutes with the representation is a scalar. Hence $\Omega =c{I}_V$ when $V$ is irreducible.

Let $v$ be a highest-weight vector of highest weight $m$, so $ev=0$ and $hv=mv$. Since $[e,f]=h$,ThereforeIt follows from scalarity thatThis is the Casimir eigenvalue for sl2 in the chosen normalization.

The Killing form of a complex simple Lie algebra is nondegenerate. Since $B$ is also nondegenerate, there is a unique endomorphism $T$ of $\mathfrak{g}$ satisfyingInvariance of both forms givesso $T$ intertwines the adjoint representation. That representation is irreducible because its invariant subspaces are ideals. The Schur lemma therefore gives $T=cI$. Since $\kappa$ is nondegenerate, $c\ne 0$, andThis is the uniqueness of an invariant bilinear form on a simple Lie algebra.