Consider $\mathfrak{sl}_2$ and its fundamental weight $\lambda_1$. The character of the simple representation $L(n\lambda_1)$ with highest weight $n\lambda_1$ is given by a polynomial in $x=\mathrm{ch}L(\lambda_1)$. Writing $P_n(x)=\mathrm{ch}L(n\lambda_1)$, we have a recurrence $P_n(x)=xP_{n-1}(x)-P_{n-2}$. The first few polynomials are $1,x,x^2-1,x^3-2x\dots$ These are actually the Chebyshev polynomials of the second kind on the interval $[-2,2]$.
Consider the same question for $\mathfrak{sl}_3$, with simple roots $\alpha_1$, $\alpha_2$ and corresponding fundamental weights $\lambda_1$, $\lambda_2$. Let $x=\mathrm{ch}L(\lambda_1)$ and $y=\mathrm{ch}L(\lambda_2)$. Let $P_{m,n}(x,y)=\mathrm{ch}L(m\lambda_1+n\lambda_2)$.
- What recurrence do the $P_{m,n}(x,y)$ satisfy?
Recall that the classical Chebyshev polynomials satisfy $\int_{-2}^2P_m(x)P_n(x)\sqrt{4-x^2}dx=\delta_{m,n}$.
- Are the $P_{m, n}$ orthogonal in the way that the Chebyshev polynomials are orthogonal?
Best Answer
Hint:
Try to prove that $$ P_{m+1,n}(x,y)=x P_{m,n}(x,y)-P_{m,n-1}(x,y)-P_{m-1,n+1}(x,y)\\ P_{m,n+1}(x,y)=y P_{m,n}(x,y)-P_{m-1,n}(x,y)-P_{m+1,n-1}(x,y) $$
and then try eliminate the terms $P_{m-1,n+1}(x,y)$ and $P_{m+1,n-1}(x,y).$