DualRepresentation

In mathematics, if $G$ is a group and $ρ$ is a linear representation of it on the vector space $V$, then the dual representation $ρ*$ is defined over the dual vector space $V*$ as follows:

$ρ^{*}(g)$ is the transpose of $ρ(g^{-1})$, that is,

$$ρ^{*}(g) = ρ(g^{-1})^T \forall g∈G$$

The dual representation is also known as the contragredient representation.

If $g$ is a Lie algebra and $π$ is a representation of it on the vector space $V$, then the dual representation $π*$ is defined over the dual vector space $V*$ as follows:

$$π^{*}(X) = −π(X)^T \forall X ∈ g$$

The motivation for this definition is that Lie algebra representation associated to the dual of a Lie group representation is computed by the above formula. But the definition of the dual of a Lie algebra representation makes sense even if it does not come from a Lie group representation.

In both cases, the dual representation is a representation in the usual sense.

Properties

Irreducibility and second dual

If a (finite-dimensional) representation is irreducible, then the dual representation is also irreducible[^4^]—but not necessarily isomorphic to the original representation. On the other hand, the dual of the dual of any representation is isomorphic to the original representation.

Unitary representations

Consider a unitary representation $\rho$ of a group $G$, and let us work in an orthonormal basis. Thus, $\rho$ maps $G$ into the group of unitary matrices. Then the abstract transpose in the definition of the dual representation may be identified with the ordinary matrix transpose. Since the adjoint of a matrix is the complex conjugate of the transpose, the transpose is the conjugate of the adjoint. Thus, $\rho^*(g)$ is the complex conjugate of the adjoint of the inverse of $\rho(g)$. But since $\rho(g)$ is assumed to be unitary, the adjoint of the inverse of $\rho(g)$ is just $\rho(g)$.

The upshot of this discussion is that when working with unitary representations in an orthonormal basis, $\rho^*(g)$ is just the complex conjugate of $\rho(g)$.

The SU(2) and SU(3) cases

In the representation theory of SU(2), the dual of each irreducible representation does turn out to be isomorphic to the representation. But for the representations of SU(3), the dual of the irreducible representation with label $(m_1,m_2)$ is the irreducible representation with label $(m_2,m_1)$.[^5^] In particular, the standard three-dimensional representation of SU(3) (with highest weight $(1,0)$) is not isomorphic to its dual. In the theory of quarks in the physics literature, the standard representation and its dual are called "$3$" and "$\bar 3$."