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Given a *-algebra $\mathcal{A}$ over $\mathbb{C}$, a Fredholm module is a *-representation $\pi$ of $\mathcal{A}$ as operators on a Hilbert space $\mathcal{H}$ along with a self-adjoint operator $F$ on $\mathcal{H}$ such that $F^{2}=I$ such that $[F,\pi (a)]$ is a compact operator on $\mathcal{H}$ for every $a\in \mathcal{A}$.

If $\mathcal{A}$ is an algebra of real-valued functions, then a representation of $\mathcal{A}$ is given by a measure class $[\nu]$ and a multiplicity function $m$ ($\mathcal{A}$ acts by pointwise multiplication on $L^{2}(\mathbb{R},\nu)$). If this measure class is translation invariant, it must be the Lebesgue class $[\mu]$ and the multiplicity can be assumed to be 1 with nothing interesting lost. If we desire to include a self adjoint operator $F$ on $L^{2}(\mathbb{R},\mu)$ of square I such that the resulting Fredholm module is translation and scale invariant, then this operator must be the Hilbert transform. It turns out that $Sl(2,\mathbb{R})$ acts on this Fredholm module, acting as certain unitary operators on $L^{2}(\mathbb{R},\mu)$ that commute with F, and as linear fractional transformations on the algebra of functions.

The above basic facts about Fredholm modules in general and for algebras of real valued functions can be found here:

My question is this:

What are some interesting involutive algebras of complex valued functions on the upper half-plane, such that under mild symmetry conditions there is an essentially unique Fredholm module coming from an analogue of the Hilbert transform?

Under natural mild invariance conditions perhaps we will be forced to consider a particular measure (e.g. if we consider invariance under the usual linear-fractional action of certain subgroups of $SL_{2}(\mathbb{R})$ and scaling by certain elements in $Gl_{2}^{+}(\mathbb{R})$, we may automatically end up with the measure class $[y^{-2}dxdy]$) playing a role analogous to that of $[\mu]$?) If so, and this is especially what I would like to know, what is the analogue of the Hilbert transform?

I'm wondering if anything like this can be seen for the involutive algebra of level 1 cusp forms

$\mathcal{M_{0}}= \bigoplus_{k=1}^{\infty}\mathcal{S}_{2k}$

equipped with pointwise multiplication $(f\cdot g)(z)=f(z)g(z)$ and involution given by $f^{*}(z)=\overline{f}(-\overline{z})$.

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What exactly do you mean by "the *-operation from the Petersson inner product"? I don't follow. The only natural conjugate-linear map from cusp forms to cusp forms that I can think of is the one sending $f(z)$ to $\overline{f}(-\overline{z})$, but this doesn't have any obvious relation to the Petersson product. – David Loeffler Sep 28 '10 at 11:40
That was especially silly of me. This question is one I was thinking about a while ago, and I naughtily didn't think carefully enough about the involution while typing it this morning. This fact is modular forms 101...Someone should dock me some reputation points for that... I'm sorry for the confusion, David! – Jon Bannon Sep 28 '10 at 16:53
I previously have deleted this question because it hangs together somewhat badly. Any suggestions on how to sharpen it are welcome. The eventual goal is to try to "resolve the diagonal" for the subalgebra of cusp forms in the Connes-Moscovici Modular Hecke Algebras. – Jon Bannon Jan 4 '11 at 1:05

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