Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Representations of SU(2) are equivalent to their duals

  1. Aug 10, 2007 #1

    arg

    User Avatar

    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 [tex]f[/tex] to
    [tex]f(g^{-1} \left(
    \begin{array}{cc}
    x\\
    y \end{array}
    \right) ) [/tex]
    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.
     
  2. jcsd
  3. Aug 11, 2007 #2

    matt grime

    User Avatar
    Science Advisor
    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)?
     
  4. Aug 11, 2007 #3

    arg

    User Avatar

    Hi Matt. Thanks for your quick reply.

    So if the degree of the homogenous polynomials is n the basis is:
    [tex]x^n, x^{n-1}y, x^{n-2}y^2, ... , xy^{n-1}, y^n[/tex] 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 [tex]x^ky^{n-k}[/tex] 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 [tex]f(x,y)[/tex] is to take the matrix [tex]g^{-1}[/tex] and multiply it on the right by the column vector [tex]\left(\begin{array}(x\\y\end{array} \right)[/tex]. 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.
     
  5. Aug 11, 2007 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    The double dual of V is what you started with....
     
  6. Aug 11, 2007 #5

    arg

    User Avatar

    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.
     
    Last edited: Aug 11, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Representations of SU(2) are equivalent to their duals
Loading...