Dual and Contragredient Representations — Mathematics Knowledge System

Overview

Given a representation ρ: G → GL(V), the dual (or contragredient) representation ρ*: G → GL(V*) acts on the dual space V* = Hom(V, k) by the transpose-inverse formula. Dual representations appear naturally whenever one studies invariant bilinear forms, harmonic analysis, and the relationship between a representation and its complex conjugate. For unitary representations the contragredient coincides with the complex conjugate representation.

Intuition

If ρ acts on column vectors by left-multiplication, then ρ* acts on row vectors (linear functionals) so that the natural pairing ⟨f, v⟩ = f(v) is preserved: the group acts on the left on vectors and on the right on functionals. The transpose-inverse formula ensures this compatibility.

Formal Definition

Definition

Let ρ: G → GL(V) be a representation over a field k. The contragredient (dual) representation ρ*: G → GL(V*) is defined by

\rho^*(g)(f) = f \circ \rho(g^{-1}), \quad f \in V^*,\; g \in G

Contragredient action on dual space

\langle \rho^*(g)f,\, v \rangle = \langle f,\, \rho(g^{-1})v \rangle \quad \forall f \in V^*,\; v \in V

Compatibility with natural pairing

\text{In matrix form: } \rho^*(g) = \bigl(\rho(g^{-1})\bigr)^T = \bigl(\rho(g)^{-1}\bigr)^T

Matrix formula for contragredient

V \cong V^* \iff \exists \text{ non-degenerate } G\text{-invariant bilinear form on } V

Self-duality criterion

Notation

Notation	Meaning
$V^*$	Dual space Hom(V, k)
$\rho^*$	Contragredient representation on V*
$\langle f, v \rangle$	Natural pairing f(v)
$\rho^\vee$	Alternative notation for contragredient

Properties

Double dual

There is a natural G-isomorphism V \cong (V^*)^* for finite-dimensional V.

Tensor dual

\mathrm{Hom}_k(V, W) \cong V^* \otimes W as G-representations.

Irreducibility preserved

If ρ is irreducible, then ρ* is also irreducible.

Theorems

Theorem 1: Character of Dual Representation

For a finite-dimensional complex representation, the character of the contragredient satisfies \chi_{\rho^*}(g) = \overline{\chi_\rho(g)}.

Theorem 2: Self-Duality over ℝ

Every real representation is self-dual: ρ ≅ ρ* as real representations, via any non-degenerate invariant symmetric bilinear form (which exists after averaging).

Worked Examples

1
Let g be the generator. Then ρ(g) is the rotation matrix R = [[cos(2π/3), -sin(2π/3)], [sin(2π/3), cos(2π/3)]].
$\rho(g) = \begin{pmatrix} -1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & -1/2 \end{pmatrix}$
2
ρ*(g) = (ρ(g⁻¹))ᵀ = (ρ(g)⁻¹)ᵀ = (ρ(g²))ᵀ. Since ρ(g) is a rotation, ρ(g)⁻¹ = ρ(g)ᵀ (rotation by -2π/3), so ρ*(g) = (ρ(g)ᵀ)ᵀ = ρ(g).
$\rho^*(g) = \rho(g)$
3
The standard representation of ℤ/3ℤ by rotations is self-dual because rotations are orthogonal matrices.
$\rho \cong \rho^*$

✓ Answer

The contragredient equals the original representation; the rotation representation is self-dual.

Practice Problems

Mediumfree response

Let G = S₃ and ρ the sign representation (1-dimensional, g ↦ sgn(g) ∈ {±1}). What is ρ*?

Mediumproof writing

Prove that χ_{ρ*}(g) = χ_ρ(g⁻¹) for any finite-dimensional representation ρ.

Quiz

Summary

The contragredient ρ* of a representation ρ: G → GL(V) acts on V* by ρ*(g)(f) = f ∘ ρ(g⁻¹).
In matrix form, ρ*(g) = (ρ(g⁻¹))ᵀ = (ρ(g)⁻¹)ᵀ.
The character satisfies χ_{ρ*}(g) = χ_ρ(g⁻¹); over ℂ this equals χ̄_ρ(g) for unitary ρ.
ρ is self-dual iff there is a non-degenerate G-invariant bilinear form on V.
Irreducibility is preserved under taking the contragredient.

References

BookSerre, J.-P. — Linear Representations of Finite Groups (1977), §2.1
BookFulton, W. & Harris, J. — Representation Theory: A First Course (1991), §2