TITLE: Eigenvalues of the free sphere
QUESTION [8 upvotes]: Consider the usual sphere $S^{n-1}\subset\mathbb R^n$. By Stone-Weierstrass $C(S^{n-1})$ is generated by the standard coordinates $x_1,\ldots,x_n:\mathbb R^n\to\mathbb R$, and in fact we have the presentation result $C(S^{n-1})=C^*_{comm}(x_1,\ldots,x_n|x_i=x_i^*,\sum x_i^2=1)$. 
The Riemannian structure of $S^{n-1}$, or at least part of it, can be recaptured from this formula. Indeed, the eigenspaces of $D=\sqrt{d^*d}$ are $E_k=H_k\cap H_{k-1}^\perp$, where $H_k=span(x_{i_1}\ldots x_{i_r}|r\leq k)$, and the corresponding eigenvalues are $\lambda_k=k(k+n-2)$. This leads to the following question:

What is the free analogue of $\lambda_k$?

More precisely, consider the algebra $A=C^*(x_1,\ldots,x_n|x_i=x_i^*,\sum x_i^2=1)$, corresponding to the NCG-theoretic "free sphere". One can construct spaces $H_k,E_k$ as above, so this free sphere has indeed a spectral triple structure, and the question is to find the correct eigenvalues for $D=\sqrt{d^*d}$.

REPLY [10 votes]: How about
$\lambda_k = \frac{U'_k(n)}{U_k(n)}$
where the $U_k$ denote the Chebyshev polynomials of the second kind, $U_0(x)=1$, $U_1(x)=x$, and $U_k(x)=xU_{k-1}(x)-U_{k-2}(x)$ for $k\ge 2$.
In Section 10 of http://arxiv.org/abs/1210.6768 (See in particular Remark 10.4) we try to classify "Brownian motions" on $O_n^+$. The formula above follows, if you use the co-action of the free orthogonal quantum group on the free sphere to define an action of generator of "$O_n^+$-BM" on the free sphere.