Let T: V → W be a G-map between irreducible representations. Schur's lemma says T must be either zero or an isomorphism. Why can't T be a non-zero, non-invertible map?
Think about your answer, then reveal below.
Model answer: The kernel ker(T) is a G-invariant subspace of V, and the image im(T) is a G-invariant subspace of W. Since V is irreducible, ker(T) is either {0} or V. If T ≠ 0, then ker(T) ≠ V, so ker(T) = {0} (T is injective). Since W is irreducible, im(T) is either {0} or W. Since T ≠ 0, im(T) = W (T is surjective). So T is bijective.
This proof is a model of how irreducibility arguments work: G-invariance of kernels and images (which follows from the intertwining condition) constrains them to be trivial or the whole space, leaving no middle ground. The proof uses nothing beyond the definition of irreducibility and basic linear algebra.
Question 2 Multiple Choice
Over ℂ, if ρ: G → GL(V) is irreducible and T: V → V is a G-map, then T = λI for some scalar λ. Why does this fail over ℝ?
AThe real numbers are not a field
BT may have no real eigenvalues, so the argument 'T − λI has nontrivial kernel' cannot be started
CSchur's lemma does not apply over ℝ
DReal matrices cannot be scalar multiples of the identity
Over ℂ (algebraically closed), every linear operator has at least one eigenvalue λ. Then T − λI is a G-map with nontrivial kernel, so by Schur's lemma it must be zero, giving T = λI. Over ℝ, a linear operator need not have real eigenvalues (e.g., a rotation by 90° on ℝ²), so the argument breaks down at the first step. Schur's lemma (first part) still holds over ℝ — it is only the scalar conclusion that requires algebraic closure.
Question 3 True / False
Schur's lemma implies that irreducible representations of abelian groups over ℂ are one-dimensional.
TTrue
FFalse
Answer: True
For an abelian group, every ρ(g) commutes with every ρ(h). So each ρ(g) is a G-map from the representation to itself. By Schur's lemma over ℂ, each ρ(g) = λ_g · I. But then every subspace is G-invariant (since scalar matrices preserve all subspaces). For the representation to be irreducible, V must have no proper nontrivial subspaces, which forces dim(V) = 1. This is a powerful structural result derived from a few lines of reasoning.
Question 4 True / False
If V and W are non-isomorphic irreducible representations, then Hom_G(V, W) = {0}.
TTrue
FFalse
Answer: True
By Schur's lemma, any G-map T: V → W is either zero or an isomorphism. If V and W are not isomorphic, the isomorphism case is excluded, so every G-map must be zero. This means Hom_G(V, W) contains only the zero map. Combined with the scalar result for Hom_G(V, V) ≅ ℂ (over algebraically closed fields), this completely determines the structure of Hom spaces between irreducibles.