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 $D=\sqrt{d^\*d}$.



Eigenvalues of the free sphere
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.


