# Duals

## Definition

Definition:

Given a representation R from G to G L V, the dual representation R star from G to G L of V dual is defined as follows.

• The dual vector space V dual or V star is the space hom from V to C of linear maps from V to C. Equivalently, if we think of V as a space of column vectors by picking a basis of V then we can think of V dual as the space of row vectors: given an element alpha in hom V, C, we get a row vector underline alpha as follows. Pick a basis of V so that elements can be written as column vectors v equals v_1 down to v_n. Then alpha of v equals alpha_1 v_1 plus dot dot dot plus alpha_n v_n for some coefficients alpha_i. Define underline alpha to be the row vector alpha_1 up to alpha_n (so alpha of v equals underline alpha matrix product with v).

Conversely, if you have a row vector w equals w_1 up to w_n then we get the linear map overline w in hom V, C, defined by overline w over v equals w_1 v_1 plus dot dot dot plus w_n v_n.

• The dual representation is defined as follows. For each g in G, define R star g from V dual to V dual to be the map which sends a row vector underline alpha to underline alpha matrix product with (R of g) inverse. In other words, underline (R star of g applied to alpha) equals alpha underline matrix product with (R of g) inverse.

The messing with underlines is just to distinguish between applying a linear map and multiplying a matrix, because we want to multiply on the right.

Lemma:

This defines a representation.

Proof:

We have the row vector corresponding to R star (g_1 g_2) applied to alpha equals alpha underline R of (g_1 times g_2) all inverse, which equals alpha underline, times R of g_2 inverse, times R of g_1 inverse. But alpha underline R of g_2 inverse equals the row vector corresponding to R star of g_2 applied to alpha, and so alpha underline, times R of g_2 inverse, times R of g_1 inverse equals the row vector corresponding to R star g_1 times R star g_2 applied to alpha Therefore R star (g_1 g_2) equals R star g_1 R star g_2 as required.

## Examples

### SU(2) representations

Lemma:

If R from SU(2) to G L V is a complex representation with a given weight diagram D then R star has the weight diagram D star equals D.

Remark:

We will prove that the weights of D star are given by minus the weights of D. Since weight diagrams of SU(2) representations are symmetric about the origin, this means D equals D star. By our classification theorem for SU(2) representations, we deduce that R is isomorphic to R star for any SU(2)-representation.

Proof:

Let W_n be the weight space with weight n. Recall that v in W_n then R of the diagonal matrix with diagonal entries e to the i theta, e to the minus i theta applied to v equals e to the i n theta times v. If we apply dagger (conjugate transpose) to both sides of the equation, we get: v dagger times (R of e to the i theta, e to the minus i theta) dagger equals e to the minus i n theta times v dagger.

Now v dagger is the row vector v_1 bar up to v_n bar.

We have an invariant Hermitian inner product on our representation. If we pick a basis which is orthonormal with respect to that then the matrices R of e to the i theta, e to the minus i theta of our representation (with respect to this basis) will be unitary, that is (R of e to the i theta, e to the minus i theta) dagger equals (R of e to the i theta, e to the minus i theta) inverse. In other words, this is the matrix which, when acting on the right on row vectors, defines R star of e to the i theta, e to the minus i theta.

This tells us that v dagger in V dual is a weight vector of R star with weight minus n, so the weights of R star are minus the weights of R. Since the weights of R are symmetric around the origin, this implies D equals D star and R star is isomorphic to R.

### SU(3) representations

For SU(3), the same argument tells us that the weights of R star are minus the weights of R.

Example:

If V is the standard representation then its weight diagram is

so its dual has the following weight diagram:

Since these diagrams are different, V is not isomorphic to V dual.

## Mesons

The quark model we discussed earlier can be extended to allow both quarks and antiquarks: antiquarks live in the dual of the standard representation. A meson is then a particle obtained by combining one quark and one antiquark, so lives in C 3 tensor C 3 dual. It's an exercise to see that this decomposes as Gamma_{1,1} direct sum C. This corresponds to the classification of mesons:

There are 8 particles in an octet (shown) and one particle (eta_1) in a singlet (with both strangeness and charge zero).

## Pre-class exercise

Exercise:

Find the weight diagram of C 3 tensor (C 3 dual) and verify that it decomposes as Gamma_{1, 1} direct sum C as claimed.