Schur's lemma is this:

If (ρ1,V1) and (ρ2,V2) are irreducible representations of a group G, then any nonzero homomorphism ϕ:V1↦V2 is an isomorphism.


Schur's Lemma.
a. A unitary representation ρ of G is irreducible if and only if C(ρ) contains only scalar multiples of the identity. b. Suppose ρ1 and ρ2 are irreducible unitary representations of G. If ρ1 and ρ2 are equivalent then C( ρ1, ρ2) is one-dimensional; otherwise, C(ρI,ρ2) = {0}.

Whether these two are equivalent? If the answer is yes ,how isomorphism of ϕ causes that C( ρ1, ρ2) is one-dimensional?

  • $\begingroup$ Are we talking about finite groups and finite-dimensional spaces? $\endgroup$ – blue May 31 '14 at 22:29
  • $\begingroup$ G is a locally compact group. $\endgroup$ – user147972 Jun 1 '14 at 6:17

The first version of Schur's lemma is more general than the second (indeed it applies e.g. to simple modules over arbitrary rings, not just group algebras). It implies that if $V$ is an irreducible representation then $\text{End}(V)$ is a division ring. Now if $V$ is finite-dimensional over the base field $k$ then so is $\text{End}(V)$, but now it's an exercise to show that if $k$ is algebraically closed then the only finite-dimensional division ring over $k$ is $k$ itself. Hence in particular if $k = \mathbb{C}$ then $\text{End}(V) \cong \mathbb{C}$ consists of only scalar multiples of the identity.

(The more general version of Schur's lemma is important even in representation theory. For example, over $\mathbb{R}$ it implies that $\text{End}(V)$ must be one of $\mathbb{R}, \mathbb{C}$, or $\mathbb{H}$, and the distinction is important for understanding how real and complex representations relate to each other. See Frobenius-Schur indicator for more details.)

  • $\begingroup$ @ Qiaochu Yuan : thank you.I want to prove the second.Can I use the isomorphism the first one? $\endgroup$ – user147972 Jun 2 '14 at 5:59

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