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Letwhere $dg$ is Haar measure. Left invariance gives $hK{h}^{-1}=K$ for every $h\in SU(2)$. Since the spin-$S$ representation is irreducible, Schur lemma impliesInserting this coherent-state resolution between $N$ short time steps and taking the continuum limit gives, for a noninteracting spin with zero Hamiltonian,up to the normalization generated by $c$. A Hamiltonian would add $-i\int ⟨g|H|g⟩dt$ in the exponent.

Fordirect differentiation and $S_3|\uparrow ⟩=S|\uparrow ⟩$ giveThe $\dot{\psi }$ term is a total derivative and records only the arbitrary phase used to represent a ray. It cancels against the endpoint-state phases, so the physical Spin coherent-state path integral depends only on the path on the two-sphere. For example, choosing the section $\psi =-\varphi$ gives the Spin coherent-state Berry phase proportional toChanging the surface used to fill a closed path changes its solid angle by $4\pi$. Single-valuedness of the path-integral phase requiresThus the Wess-Zumino coefficient is quantized in integer or half-integer units.

For the Affleck--Kennedy--Lieb--Tasaki state, take the dihedral subgroup of spin rotations generated by $\pi$ rotations about the $x$ and $z$ axes. On each physical spin-one site these generators commute and give an ordinary representation of ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$. On the virtual spin-one-half space they may be represented, up to phases, byThey anticommute,so the virtual action is the nontrivial projective representation of ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$.

The reduced density operator across a cut commutes with this irreducible projective action. By Schur lemma it is proportional to the identity within each virtual doublet. Consequently every level in the entanglement spectrum of a matrix product state is at least twofold degenerate, the characteristic protected edge-spin degeneracy of the AKLT phase.

An invariant bilinear form $B$ defines a linear mapInvariance givesso ${\Phi }_B$ is an $G$-homomorphism. Its kernel is a $G$-subrepresentation of irreducible $V$. Hence either ${\Phi }_B=0$, so $B=0$, or ${\Phi }_B$ is injective and therefore a linear isomorphism $V\to V^{\ast }$. In the latter case $B$ is a nondegenerate.

If $B$ and $C$ are two nonzero invariant forms, thenBy Schur lemma, this endomorphism is scalar, so $B=\lambda C$. Cases involving the zero form are immediate. Thus any two invariant bilinear forms are proportional.

For an irreducible $V$, Schur lemma makes the central element $-I$ act by a scalar $\epsilon$. Since $(-I{)}^2=I$, $\epsilon =\pm 1$, and on $V\otimes V$ it acts by ${\epsilon }^2=1$. Thus $\{\pm I\}$ lies in the kernel and the action factors through $S{U}_2/\{\pm I\}$, as in central parity on SU2 tensor products.

This fails for reducible representations. Take $V=V_0\oplus V_1$. On the summand $V_0\otimes V_1\subset V\otimes V$, the element $-I$ acts as $-1$, so the tensor-square action does not factor through the quotient.

Schur lemma says that a homomorphism between simple modules is either zero or an isomorphism; consequently the endomorphism ring of a simple module is a division ring. Indeed, the kernel and image of a module homomorphism are submodules. Simplicity makes each either zero or the whole module, proving both assertions.

The Trace form of a Lie algebra representation is invariant becauseChoose a basis orthonormal for the positive-definite form $-\kappa$ of the compact simple algebra. Any invariant bilinear form determines an endomorphism commuting with the irreducible adjoint action; Schur lemma makes it scalar. Thus $H(T_a,T_b)=c{\delta }_{ab}$. Since $d(T_a)$ is anti-Hermitian,The inequality is strict because the kernel of the nontrivial irreducible representation is an ideal and hence zero. Therefore $c=-\mu$ with $\mu >0$.

By Schur lemma, $z\in Z$ acts on irreducible $V$ by a scalar. Induced representation shows that $V$ occurs in ${\mathrm{Ind}}_Z^G\Theta$ iff that scalar character is $\Theta$, and then with multiplicity $\dim V$.

For chosen inequivalent unitary irreducible representations $\rho ,\sigma$ of a finite group and a $d_{\rho }\times d_{\sigma }$ matrix $M$, the average ${\mathbb{E}}_x\rho (x)M\sigma (x{)}^{\ast }$ is zero when $\rho \ne \sigma$. For $\rho =\sigma$, it is $(\mathrm{tr}M/d_{\rho })I$. This follows from the Schur lemma, since the average intertwines the two group representations. Equivalent representations in different bases require the corresponding intertwiner; the identity-matrix formula assumes literally the same representative.

For inequivalent chosen unitary irreducible representations of a finite group, uniform expectation givesThe Kronecker deltas in the first case refer to entries in the same chosen basis. Averaging an intertwiner and applying the Schur lemma proves the formula; summing diagonal entries gives character orthogonality.

For a complex semisimple Lie algebra $\mathfrak{g}\subseteq \mathrm{End}(V)$ and $X\in \mathfrak{g}$, both parts of its Additive Jordan decomposition lie in $\mathfrak{g}$. The polynomial semisimple part of $\mathrm{ad}X$ shows $X_s$ normalizes $\mathfrak{g}$. Split $\mathrm{End}(V)=\mathfrak{g}\oplus M$ by the Weyl complete reducibility theorem; the $M$-component of $X_s$ centralizes $\mathfrak{g}$. On each irreducible summand of $V$ it is scalar by the Schur lemma. Its trace is zero because $\mathfrak{g}$ is a perfect Lie algebra and the nilpotent part has zero trace, so those scalars vanish. Nonsemisimple subalgebras may fail this property: $\mathbb{C}(I+N)$ with $N\ne 0$ nilpotent contains neither Jordan part of $I+N$.

If $C{\gamma }^{\mu T}{C}^{-1}=-{\gamma }^{\mu }$, transposition shows that $C^T{C}^{-1}$ commutes with every gamma matrix. Irreducibility of the complex Dirac representation and the Schur lemma make it scalar. Transposing $C^T=zC$ again yields $z^2=1$, so $C^T=\pm C$.

Every invariant bilinear form on a finite-dimensional complex simple Lie algebra is a scalar multiple of its Killing form. A nondegenerate invariant form identifies the algebra with its dual; comparing this identification with the Killing form gives an endomorphism of the irreducible adjoint representation, so Schur lemma makes it scalar.