Representations of SU(2) are equivalent to their duals

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SUMMARY

The irreducible representations of SU(2) are equivalent to their dual representations, specifically within the context of homogeneous polynomials in two complex variables of degree 2j, where j is a non-negative integer. The action of an element g from SU(2) on a polynomial f is defined by applying the inverse matrix g^{-1} to the vector (x, y) and substituting the resulting values back into f. The dual space consists of n+1 dimensional vector spaces, with the dual basis formed by 1-forms that evaluate the polynomials. The discussion highlights the need for a clear mapping between the polynomial space and its dual to establish the equivalence definitively.

PREREQUISITES
  • Understanding of SU(2) group theory
  • Familiarity with homogeneous polynomials in complex variables
  • Knowledge of dual spaces and dual bases in linear algebra
  • Experience with linear transformations and their properties
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  • Explore the concept of dual spaces in linear algebra
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Mathematicians, theoretical physicists, and graduate students specializing in representation theory, algebra, and complex analysis will benefit from this discussion.

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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(<br /> \begin{array}{cc}<br /> x\\<br /> y \end{array}<br /> \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.
 
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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)?
 
Hi Matt. Thanks for your quick reply.

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.
 
The double dual of V is what you started with...
 
I don't follow you. Can you be a bit more specific? Do you mean that the basis I listed is not the basis of the homogeneous polynomials of two complex variables? Thanks.
 
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