Schur's lemma

Schur's lemma is most naturally stated in the language of modules. Let $𝑉, 𝑊$ be simple modules over a ring $𝑅$ . Then any nonzero Module homomorphism $𝑓 : 𝑉 \to 𝑊$ is an isomorphism. #m/thm/module In particular, the endomorphism ring $E n d 𝑅 𝑉$ of a simple module is a division ring.

Very simple proof

Since $k e r 𝑓 \leq 𝑉$ and $i m 𝑓 \leq 𝑊$ are submodules of simple modules, they must either be trivial or equal to $𝑉$ and $𝑊$ respectively. If $𝑓 \neq 0$ then $k e r 𝑓 \neq 𝑉$ and $i m 𝑓 \neq 0$ , hence $𝑓$ is epic and monic and thus an $𝑅$ -module isomorphism.

$𝕂$ is an algebraically closed field and $𝑉$ is a module over a $𝕂$ -monoid $𝐴$ over $𝕂$ , there are a few cases where one can conclude $E n d 𝐴 𝑉 = 𝕂$ consists of homotheties, which is sometimes known as Schur's first lemma. Namely

If $d i m 𝕂 𝑉 < ℵ 0$ , by All elements of a finite-dimensional unital associative algebra are algebraic
If $d i m 𝕂 𝑉 < | 𝕂 |$ , by Dixmier's lemma
If $𝐴$ has a filtration like the universal enveloping algebra of a finite-dimensional Lie algebra, by Quillen's lemma

which also rely on the result from Division algebra with only algebraic elements over an algebraically closed field.

Schur's lemma for unitary group representations

Schur's lemma is a statement about linear maps which “commute” with an irrep.¹²

Schur's lemma, first form • Let $𝔛 : 𝐺 \to G L (𝑉)$ be a finite-dimensional (complex) Irrep and $𝐴 : 𝑉 \to 𝑉$ a linear endomorphism. If $𝐴$ commutes with $𝔛$ , i.e.

𝐴 𝔛 (𝑔) = 𝔛 (𝑔) 𝐴

for all $𝑔 \in 𝐺$ , then $𝐴 = 𝑐 𝐈$ for some $𝑐 \in ℂ$ . #m/thm/rep

Proof

Let $𝜆$ be an eigenvalue of $𝐴$ , and $𝑣 \in E 𝜆 (𝐴)$ . Then $𝐴 Γ (𝑔) 𝑣 = Γ (𝑔) 𝐴 𝑣 = 𝜆 Γ (𝑔)$ for all $𝑔 \in 𝐺$ . Therefore $Γ (𝑔) 𝑣 \in E 𝜆 (A)$ , meaning $E 𝜆 (𝐴)$ is $Γ$ -Invariant subspace. Since $Γ$ is irreducible and $E 𝜆 (𝐴) \neq {𝟎}$ , $𝐸 𝜆 (𝐴) = 𝑉$ . Therefore $𝐴 = 𝜆 𝟙$ .

Schur's lemma, second form • Let $𝔛 : 𝐺 \to G L (𝑉)$ and $𝔜 : 𝐺 \to G L (𝑊)$ be finite-dimensional unitary irreps and $𝑇 : 𝑉 \to 𝑊$ a linear map.

𝑇 𝔛 (𝑔) = 𝔜 (𝑔) 𝑇

for all $𝑔 \in 𝐺$ , then $𝑇 = 0$ or $𝔛$ and $𝔜$ are unitarily equivalent. #m/thm/rep $𝑇$ is thence called an intertwiner, which is unique up to scalar multiplication.

Proof

Taking the Hermitian conjugate of both sides gives $Γ (𝑔 - 1) 𝑇 † = 𝑇 † ˜ Γ (𝑔 - 1)$ for all $𝑔 \in 𝐺$ , i.e. $Γ (𝑔) 𝑇 † = 𝑇 † ˜ Γ (𝑔)$ . Hence

𝑇 † 𝑇 Γ (𝑔) = 𝑇 † ˜ Γ (𝑔) 𝑇 = Γ (𝑔) 𝑇 † 𝑇

thus $𝑇 † 𝑇$ commutes with $Γ$ , and by the first lemma, $𝑇 † 𝑇 = 𝑐 𝐈$ for some $𝑐 \in ℂ$ . Then $⟨ 𝑣 | 𝑇 † 𝑇 𝑣 ⟩ = ⟨ 𝑣 | 𝑐 𝑣 ⟩ = ⟨ 𝑇 𝑣 | 𝑇 𝑣 ⟩ = 𝑐 ‖ 𝑣 ‖ 2 = ‖ 𝑇 𝑣 ‖ 𝑣$ for all $𝑣 \in 𝑉$ and therefore $𝑐 = ‖ 𝑇 𝑣 ‖ 2 ‖ 𝑣 ‖ 2$ which is real and nonnegative. Therefore either $𝑐 = 0$ whence $𝑇 = 𝐎$ or $𝑐 > 0$ and $𝑈 = 1 \sqrt 𝑐 𝑇$ is unitary with the equivalence $˜ Γ (𝑔) = 𝑈 Γ (𝑔) 𝑈 †$ .

Corollaries

Irreps of abelian groups are 1-dimensional

Schur's lemma in abelian categories

Let $𝖢$ be an abelian category and $𝐴, 𝐵 \in 𝖢$ be simple objects. Then every nonzero morphism $𝑓 : 𝐴 \to 𝐵$ is an isomorphism. #m/thm/cat In particular, $E n d 𝖢 (𝐴)$ is a division ring.

Proof

Essentially, apply the Freyd-Mitchell theorem to the above proof for modules.

#state/tidy | #lang/en | #SemBr

2023, Groups and representations, p. 31 ↩
1996, Representations of finite and compact groups, §II.4, pp. 27–28. The proof offered here is virtually identical, but insists on using ∗-representations for reasons beyond me. ↩

Schur's lemma

Schur's lemma for unitary group representations

Corollaries

Schur's lemma in abelian categories

Footnotes