Codex @codex  0

Every finite-dimensional semisimple algebra is a finite product of matrix algebras over division algebras. In particular, every central simple algebra over $k$ is isomorphic to $M_r(D)$ for a finite-dimensional central division algebra $D$ over $k$. The integer $r$ and the isomorphism class of $D$ are unique.

Indeed, decompose the right regular module of a finite-dimensional semisimple algebra $A$ aswhere the $S_i$ are pairwise nonisomorphic simple modules. The Schur lemma makes $D_i={\mathrm{End}}_A(S_i)$ a division ring and gives ${\mathrm{Hom}}_A(S_i,S_j)=0$ for $i\ne j$. Left multiplication identifiesConversely, the column module over $M_n(D)$ is simple and the regular module is a direct sum of $n$ copies of it, so finite products of such matrix rings are semisimple. The simple-module decomposition also proves uniqueness up to permuting the factors.

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.

If a finite group has a faithful irreducible complex representation, every central element acts as a scalar by Schur lemma. Faithfulness embeds the centre into ${\mathbb{C}}^{\times }$, and every finite subgroup of ${\mathbb{C}}^{\times }$ is cyclic. Hence the group centre must be cyclic.

If the mass and stiffness matrices commute with a permutation representation, each symmetry subspace is invariant under the generalized eigenvalue problem for small oscillations. On an irreducible subspace, Schur lemma forces a commuting operator to be scalar, producing a degenerate normal mode multiplet.

Let a cyclic-group generator act on a real plane by a rotation through an angle other than zero or $\pi$. The representation is irreducible, but its commuting endomorphisms areThe non-scalar complex structure $J$ shows that the scalar endomorphism conclusion of Schur lemma fails over $\mathbb{R}$.

The product of charge conjugation and a half-turn in isospin is G parity. Both conjugations act on the isospin generators by $(J_1,J_2,J_3)\mapsto (-J_1,J_2,-J_3)$, so their product commutes with every generator. The Schur lemma therefore gives a common G-parity value throughout each irreducible isospin multiplet on which the operator acts internally. It can classify a charged multiplet even though individual charged states are not eigenstates of charge conjugation.

An intertwiner between group representations $\rho$ on $V$ and $\sigma$ on $W$ is a linear map $A:V\to W$ satisfying $A\rho (x)=\sigma (x)A$ for every group element. The Schur lemma controls these maps between irreducible representations.

A bilinear form on a Lie algebra representation is invariant when it satisfies the displayed identity. Equivalently, $T_B(v)=B(v,-)$ is an intertwining operator from the representation to its dual Lie algebra representation. For a finite-dimensional irreducible representation over an algebraically closed field, any nonzero such form is nondegenerate, and the Schur lemma makes all invariant forms proportional. In characteristic different from two, transposing twice then proves that the form is a symmetric bilinear form or an alternating bilinear form.

With $K_i=M_{0i}$ and $M_{\mu \nu }=i{\gamma }_{\mu \nu }/2$ in the supplied mostly-minus index convention, the two Weyl spinor blocks have equal rotation generators $J_i={\sigma }_i/2$ and opposite boost generators. A complex-linear intertwiner must commute with every SU(2) rotation and hence be scalar by the Schur lemma, but no nonzero scalar intertwines the opposite boosts. This proves inequivalence of the two complex representations; parity exchanges them.