Dual representation

1. Apr 23, 2006

Pietjuh

I've been starting to study some things about representation theory. I've come to the point where they introduced the dual of a representation.

Suppose that $\rho$ is a representation on a vector space V.
They then define the dual representation $\rho^*$ as:

$$\rho^*(g) = \rho(g^{-1})^t: V^* \to V^*$$

But the thing is that I don't see why they use $g^{-1}$ in this definition instead of just g?

2. Apr 23, 2006

matt grime

Because otherwise it would not be a representation (the map would not be a homomorphism, but an anti-homomorphism, that is denoting your notional maps as f, f(gh)=f(h)f(g))

Taking duals interchanges the order of composition, ie it makes a left representation into a right representation, fortunately groups possess this anti-involution that makes you able to correct it and turn it into a left representation again. Left means make the matrix act on the left, right means make the matrix act on the right.

Notice they *do* actually use g to define the representation p*, ie they do tell you how to work out p*(g), and it is p(g) 'inverse transpose'.

Given some representation p, there are many things that we can do to get another representation. This is just one of them, and it so happens the vector space of the representation is V*.

You shuold check that you do indeed find that the representation

f(g)=p(g) transpose

is not generically (which means 'usually', or 'except for certain cases' such as G being abelian, or p(G) being abelian) a (left) representation, ie the map f is not a group homomorphism.

Last edited: Apr 23, 2006
3. Apr 23, 2006

Pietjuh

Is this the definition of a dual map? Or is it derivable from something else?

4. Apr 23, 2006

matt grime

Given a representation, p,V, how can you make G act on the dual space? if f is in the dual space then the *only* obvious action of G on f is to define

g.f(v)=f(g.v)

for all v in V.: remember it suffices to define a linear map by how it acts on vectors, so we defince g.f to be the linear map that sends v go f(g.v).

All this is saying is that End(V) maps to End(V*) by taking transposes. And and (AB)^T = (B^T)(A^T) which we all learnt in our first lecture on dual spaces. So it's derivable just from elementary linear/quadratic maths.

Let's prove it reverses the order (this is just revision but in a different notation, probably): if you do this (gh).f(v) = f((gh).v)=f(g(h.v))=g.f(h.v)=h.(g.f(v), so it naturally changes the order.

But there is a way to correct this for *group representations*, by making

g.f(v)=f(g^{-1}(v))

for all v.

The composition of inverting matrices and then transposing them swaps the composition over twice.

Last edited: Apr 23, 2006