# Representations of SU(2) are equivalent to their duals

Hi.
I am having trouble proving that the irreducible representations of SU(2) are equivalent to their dual representations.
The reps I am looking at are the spaces of homogenous polynomials in 2 complex variables of degree 2j (where j is 0, 1/2, 1,...). If f is such a polynomial the action of an element g of SU(2) is to take $$f$$ to
$$f(g^{-1} \left( \begin{array}{cc} x\\ y \end{array} \right) )$$
What is the dual space of this set of polynomials and how do you combine an element of the dual space with the original space to get a number?
I can find no proof of the equivalence of a representation with its dual. If anyone has any insight please let me know. Please let me know if I need to clarify anything.
Thanks very much.

matt grime
Homework Helper
That set of polynomials is _just_ a vector space. So write down the obvious basis, hence te dual basis, and now what is the action of SU(2)?

So if the degree of the homogenous polynomials is n the basis is:
$$x^n, x^{n-1}y, x^{n-2}y^2, ... , xy^{n-1}, y^n$$ so it is an n+1 dimensional vector space.
I guess the dual basis are the n+1 1-forms, the jth of which eats $$x^ky^{n-k}$$ and spits out 1 if k=j and 0 otherwise.
The way I understand the action of an element g of SU(2) on the polynomial $$f(x,y)$$ is to take the matrix $$g^{-1}$$ and multiply it on the right by the column vector $$\left(\begin{array}(x\\y\end{array} \right)$$. Then you get another column vector. Take the top element of this vector and plug it into the x-slot in f(x,y) and plug the bottom element of the vector into the y-slot. Now if you multiply everything out and regroup the terms you have another homogenous polynomial of degree n.
At this point I have several questions. Is there a way to work with these new polynomials without multiplying everything out by hand? How does one show that the action of g on this space of polynomials is 1-1 and onto? How can I come up with a good map from this space of polynomials to the dual space?

Thanks a lot.

matt grime