An intertwining operator T between representations ρ and σ satisfies Tρ(g) = σ(g)T for all g ∈ G. If T is also invertible, then ρ and σ are equivalent. What happens if T is not invertible?
AT cannot exist unless it is invertible
BT is still called an intertwining operator (or G-map), but it establishes a morphism rather than an isomorphism between the representations
CThe relation Tρ(g) = σ(g)T cannot hold for a singular T
DT defines an equivalence only between subrepresentations
A linear map T: V → W satisfying Tρ(g) = σ(g)T for all g is an intertwining operator (or G-homomorphism) regardless of invertibility. Such maps form the vector space Hom_G(V, W). If T is invertible, it establishes an isomorphism (equivalence); if not, its kernel and image are subrepresentations — a fact that plays a key role in Schur's lemma and the general structure theory.
Question 2 True / False
If two representations of G have different dimensions, they cannot be equivalent.
TTrue
FFalse
Answer: True
Equivalence requires an invertible linear map T: V → W, which can only exist when dim(V) = dim(W). This is immediate from linear algebra: an invertible linear map is a bijection, and finite-dimensional spaces of different dimensions cannot be in bijection. So dimension is the crudest invariant of a representation — equivalent representations must have the same degree.
Question 3 Short Answer
Explain why two representations could have the same dimension and the same character (trace function) yet still fail to be equivalent over ℝ, while being equivalent over ℂ.
Think about your answer, then reveal below.
Model answer: Over ℂ, representations are determined up to equivalence by their characters (for finite groups), so same character implies equivalence. Over ℝ, this can fail because the intertwining matrix P may require complex entries. The real representations may be inequivalent because the necessary change-of-basis matrix does not exist within GL_n(ℝ).
This subtlety arises because algebraic closure matters. For example, two real representations might become equivalent after extending scalars to ℂ — each matrix can be conjugated to the same form over ℂ, but no real matrix achieves the conjugation. This is why representation theory over ℂ (where the field is algebraically closed) is cleaner than over ℝ.
Question 4 True / False
The set of all intertwining operators from ρ to σ forms a vector space.
TTrue
FFalse
Answer: True
If T₁ and T₂ both satisfy Tᵢρ(g) = σ(g)Tᵢ for all g, then (αT₁ + βT₂)ρ(g) = ασ(g)T₁ + βσ(g)T₂ = σ(g)(αT₁ + βT₂). So linear combinations of intertwining operators are intertwining operators. This space Hom_G(V, W) is a subspace of Hom(V, W), and its dimension carries important structural information — Schur's lemma tells us this dimension is 0 or 1 when both representations are irreducible.