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I want to show that for the representations $\pi_n$ of $SU(2)$ the characters are given by:

$$\chi_{\pi_n}=\frac{\sin(n+1)\phi}{\sin(\phi)}.$$

The representation $\pi_n$ is defined to be the restriction of the representation $\pi$ to the set of homogenious polynomials of degree $n$. The representation $\pi$ is defined by $\pi(g)p(z)=p(g^{-1}z)=p(\overline{\alpha}z_1+\overline{\beta}z_2,-\beta z_1 +\alpha z_2)$, where $g=\left(\begin{smallmatrix} \alpha & -\overline{\beta} \\ \beta & \overline{\alpha} \end{smallmatrix}\right)$.

I know that $\chi_\pi(x)=\text{tr}(\pi(x))$. But I'm not sure how to continue. I'm also a little confused on what the representation $\pi$ actually is doing. Normaly we work with a vector space $V$ and we have $\pi(g)\cdot v$. Are we taking $\mathbb C[z_1,z_2]$ to be our vector space here?

I also know that characters have the property that, $\chi(x)=\chi(gxg^{-1})$. And in $SU(2)$ every matrix $x$ is similar to a matrix $\left(\begin{smallmatrix} e^{i \phi_n} & 0 \\ 0 & e^{-i\phi_n} \end{smallmatrix}\right)$. So we should be able to use this to help calculate $\text{tr}(\pi(x))$?

I think we can use this by $\pi(g)p(z)=p(g^{-1}z)=p(\overline{\alpha}z_1+\overline{\beta}z_2,-\beta z_1 +\alpha z_2)$ gives

$p(e^{i\phi}z_1,e^{i\phi} z_2)=(e^{i\phi})^np(z_1,z_2)$, for a degree $n$ homogenious polynomial?

Emily
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1 Answers1

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You started out ok by making the observation that it suffices to consider the diagonalizable matrices.

The space $V_n$ of homogeneous polynomials of degree $n$ is the span of the monomials $p_k=z_1^kz_2^{n-k}$ with the integer parameter $k$ ranging over $0\le k\le n$. Depending on how you defined the representation, the monomial $p_k$ is an eigenvector of $g(\phi):=\mathrm{diag}(e^{i\phi},e^{-i\phi})$ belonging to the eigenvalue $\lambda_k=e^{i(2k-n)}$ or its complex conjugate (makes no difference in the end as the eigenvalues come in conjugate pairs anyway).

Your task is simply to calculate the trace $$ tr(\pi(g(\phi)))=\sum_{k=0}^n\lambda_k. $$ Notice that this is a geometric sum, so you have a useful sum formula.

Jyrki Lahtonen
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