Finite Dimensional Representation of SU(2)

kakarukeys

Does anybody know whether the following irreducible representations of SU(2) are unitary?

g belongs to SU(2)
$$[U_j(g) f](v) = f(g^{-1} v)$$

f is an order-2j homogeneous complex polynomial of two complex variables v = (x, y)

e.g. for $$j = 1$$, $$f = 2x^2 + 3xy + 4y^2$$

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Careful

kakarukeys said:
Does anybody know whether the following irreducible representations of SU(2) are unitary?
g belongs to SU(2)
$$[U_j(g) f](v) = f(g^{-1} v)$$
f is an order-2j homogeneous complex polynomial of two complex variables v = (x, y)
e.g. for $$j = 1$$, $$f = 2x^2 + 3xy + 4y^2$$
Hi, they are. For example: lets look at the complex vector space V_2 of polynomials which are homogeneous of degree 2. This vectorspace is isomorphic to the vectorspace of 2 times 2 complex symmetric matrices through f <- -> A, f(v) = v^T A v. The action of the group corresponds then to A --> \bar{g} A g* where v^T is the transpose, g* g = 1 and \bar{g} is the complex conjugate of the matrix g. The scalar product on this space is < A,B> = tr(A*B) where tr is trace. Under the action of g, this becomes:
<U(g)A, U(g)B> = tr(g A* g^T \bar{g} B g* ) = tr(A*g^T \bar{g} B) = tr(A*B)
For higher j you can use the isomorphism with the symmetric tensors and generalized traces.

Cheers,

Careful

alexgs

I don't see how the trace corresponds to the inner product. Is this the map between the column vector f and the symmetric matrix A?
f=(f1,f2,f3)^T --> A = ( (f1, 1/2 f2), (1/2 f2, f3) )
since (x y) . A . (x y)^T = f1 x^2 + f2 xy + f3 y^2
If this is true then for two vectors f -> A and g -> B the diagonal elements of the product A*B are
A*B = ( (f1*g1 +f2*g2/4 , blah ) , (blah, f2*g2/4 + f3*g3) )
The trace of this matrix is Tr(A*B) = f1*g1 + f2*g2/2 + f3*g3
but this is not the inner product of f and g:
<f,g> = f1*g1 + f2*g2 + f3*g3

I'm sure your result is correct (that the representation is unitary) so how does one fix the proof?
Thanks
Alex

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