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

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I would add emphasize that main difference on "free-Sphere" from "just Sphere" is that x_i do NOT COMMUTE (just emphasize for better reading). At the moment it is not clear for me how to define "d" is "free" setup ? And also not clear for me definition of $\perp$, both free and non-free. Can we define "d" for "free-space" I mean if we do not impose condition $\sum x_i^2 =1 $ ? What will be the answer in this case ? – Alexander Chervov Jan 6 '13 at 14:51
please read second sentence as: At the moment it is not clear for me how to define "d" IN "free" setup ? – Alexander Chervov Jan 6 '13 at 14:53
Also, the metric enters in a more subtle way in the definition of $d^*$. The notion of adjoint uses some metric. – Liviu Nicolaescu Jan 6 '13 at 15:22
for usual sphere we can do everything with algebra and NO analysis - sl(n) will act on sphere and Laplacian (=dd^*) = Casimir (center of U(sl)), and hence representation theory of sl(n) applies. Do expect something like this for "free-sphere" ? At least do you expect that non-commutative polynoms of degree less than "k" will be preserved by hypothetical Laplacian dd^* ? – Alexander Chervov Jan 6 '13 at 20:40
Actually I do not see correct analogs of "Casimirs" in free setup... that is why it is somewhat surprising for me what you write... I may be quite wrong... Just feelings... – Alexander Chervov Jan 6 '13 at 22:14

1 Answer 1

up vote 9 down vote accepted

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 (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.

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+1, but somehow it seems the definition of "d" in question was not actually given, seems OP wrote that no one known, or I am not correct ? So is it a "theorem" or "guess" ? – Alexander Chervov Jan 8 '13 at 13:26
We are asking the same question, but in different terminology. "what is the Laplace operator on ...?" becomes "what is Brownian motion on ...?" For quantum groups we have Schürmann's theory of Lévy processes, the question becomes "which of the many Lévy processes on given quantum group deserves to be called a Brownian motion?" On $O^+$ we discovered that invariance under the adjoint action is a nice condition that leads to a subclass that we where able to classify. This is not the only condition one could imagine, but it is one that also works for compact simple connected Lie groups. – Uwe Franz Jan 8 '13 at 16:42
The generator of the Brownian motion is then our candidate for a Laplace operator. From the Laplace operator we get a Dirichlet form, under certain conditions and after fixing a reference state. If the reference state is tracial, then we can apply the construction by Cipriani and Sauvageot to get a derivation that implements the Dirichlet form via $\mathcal{E}[a]=||\partial a||^2$ and the Laplace operator as $\partial^*\partial$. The free sphere does have a tracial state that is invariant for the co-action of $O_n^+$, yes? – Uwe Franz Jan 8 '13 at 16:53
Thank you for your comments ! – Alexander Chervov Jan 9 '13 at 7:49

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