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I am trying to understand the decomposition $$L^2(SL_2(\mathbb{C})/\Gamma)=\oplus_i C_i \oplus V_{temp}$$ where $C_i$ are complementary series representations corresponding to eigenfunctions of the laplacian on $L^2(\mathbb{H}^3/\Gamma)$ and $V_{temp}$ a tempered representation.
I read that one can decompose it into a spherical and a non-spherical part. Why is the non-spherical part tempered ? How is the decomposition of the spherical part related to the spectral decomposition of $L^2(\mathbb{H}^3/\Gamma)$ with respect to $\Delta$ ?

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  • $\begingroup$ Do you have some assumptions over the Hausdorff dimension of the limit set? Those thing make dramatic impact on say $\lambda_{0}$. About your second question, the double coset space $K\backslash G / \Gamma$ is isomorphic to $\mathbb{H}^{3}/ \Gamma$ by the isomorphism of $SL_{2}(\mathbb{C})$ to $SO(3,1)(\mathbb{R})$. $\endgroup$
    – Asaf
    Dec 14, 2013 at 16:28
  • $\begingroup$ @Asaf The isomorphism can also be deduced from the Iwasawa decomposition, right? $\endgroup$
    – Marc Palm
    Dec 14, 2013 at 17:42
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    $\begingroup$ I think it's one of the exceptional isomorphism (I'm not an expert on alg. grps, but the fact that $SL_{2}(\mathbb{C})/SL_{2}(\mathbb{Z}[i])$ is isomorphic to $SO(3,1)(\mathbb{R})/SO(3,1)(\mathbb{R})$ is a bit exceptional, just like the $SL_2(\mathbb{R})$ vs $SO(2,1)(\mathbb{Z})$ case. $\endgroup$
    – Asaf
    Dec 14, 2013 at 18:51
  • $\begingroup$ One way to get it should be by taking universal covers, complexifiying and taking real form... One should probably move to connected components as well... $\endgroup$
    – Asaf
    Dec 14, 2013 at 18:52
  • $\begingroup$ Indeed, I assume the dimension of the limit set is greater than $1$ $\endgroup$
    – user7894
    Dec 14, 2013 at 19:17

2 Answers 2

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Because only infinite dimensional, unitary representation of $SL(2,F)$ for a local field $F$ can fail to be tempered, if they are spherical. This follows from the classification. This is true for $GL(2)$ as well modulo twisting by one-dimensional representations, which are trivial for $SL(2)$. I think Wallach "Real reductive groups vol 1+2" and Knapp "Representation theory of semisimple groups" covers this for $\mathbb{R}$ and $\mathbb{C}$.

As Asaf points out, looking at the $K=SU(2)$-invariant vectors will do the job. Be careful, you could think taking representations of $SU(2)$ to get something non-spherical is sufficient, but you need to take care that they are not contained in the Restriction of a spherical one;)

Getting to the finite-volume setting: You will obtain Eisenstein series and cusp forms and constant functions. There you actually have non-tempered representations besides the trivial representation. There are atmost finitely many. For $\Gamma$ a congruence subgroup of an imaginary quadratic field $k$, it is an important conjecture that there are non (the analogue of the Selberg eigenvalue conjecture). Then you actually have a much bigger group acting, i.e., $SL_2(A_k)$ via strong approximation, i.e. there exists an open subgroup$K_\Gamma$ of $SL_2(A_{k,f})$ (finite adeles) $$ \Gamma \backslash SL_2(\mathbb{C}) = SL_2(k) \backslash SL_2(A_k) / K_\Gamma.$$

If you consider an irreducible representation in there, it factors into representations of $SL_2(k_v)$ for each place as a tensor product, all but finitely many are spherical. It is assumed that they should all be tempered (besides the one-dimensional representations). This is know as the Ramanujan Peterson conjecture. Non trivial bounds are known due to Blomer and Brumley (they actually work with $GL(2)$). Here is survey: Blomer + Brumley -The role of the Ramanujan conjecture in analytic number theory, Bulletin AMS 50 (2013), 267-320

Usually the decomposition is $V_{cusp} + V_{const} + V_{cont}$, where $V_{cusp}$ are the cuspidal representations, $V_{const}$ the one-dimensional representation and $V_{cont}$ is the continuous representation. $V_{cont}$ is known to be tempered and can be explicitly given. $V_{cusp}$ is fairly unkown and only very few are conjectured to be related to Galoisrepresentations, which would imply temperedness automatically.

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  • $\begingroup$ @PM, I'm given the impression that the OP asked about infinite-volume spectral theory. $\endgroup$
    – Asaf
    Dec 14, 2013 at 17:37
  • $\begingroup$ The non-spherical part has no reason to be finite dimensional, has it ? $\endgroup$
    – user7894
    Dec 14, 2013 at 19:19
  • $\begingroup$ Yes, the one-dimensional representation is the trivial one and is spherical. $\endgroup$
    – Marc Palm
    Dec 14, 2013 at 19:25
  • $\begingroup$ Just to be clear, the spherical part is infinite dimensional because $L^2(\Gamma \backslash G /K)$ is and the one-dimensional only a subspace of the spherical part. We don't know in general if the cuspidal part is infinite-dimensional, but only in the congruence setting (Weyl law). $\endgroup$
    – Marc Palm
    Dec 14, 2013 at 19:28
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You can look at M.Cowling Almost $L^2$ matrix coefficients and Lax The Asymptotic Distribution of Lattice Points in Euclidean and Non-Euclidean Spaces.

I know the case of $SL_2(\mathbb R)$, but I am not familar with $SL_2(\mathbb C)$. For $SL_2(\mathbb R)$, we have a decomposition $L^2(\Gamma\backslash\mathbb H^2)=\mathbb Cv_0\oplus\cdots\oplus\mathbb Cv_N\oplus H$, where $v_j$ are all the eigenfunctions of $\Delta$ with eigenvalue in $(-1/4,0]$. In the paper of Lax, we have an estimation for a functions $\xi$ in $H$ $$(\pi(x)\xi,\xi)_{L^2(\Gamma\backslash SL_2(\mathbb R))}\in L^{2+\epsilon}(SL_2(\mathbb R)) $$
where $\pi$ is the right regular representation in $L^2(\Gamma\backslash SL_2(\mathbb R)$, $\xi$ regarded as a right $K$-invariant function. Then the theorem 1 in the paper of Cowling tells us the represention space generated by $\xi$ is weakly contained in the regular representaion of $L^2(SL_2(\mathbb R))$, so it is a tempered representation.

Now suppose $L^2(\Gamma\backslash SL_2(\mathbb R))=V_{v_0}\oplus\cdots\oplus V_{v_N}\oplus V_{temp}$. Using the argument above and Zorn's lemma, we can find a subspace of $V_{temp}$ which is tempered, and the rest space has no nonzero right $K$-invariant element. In the case of $SL_2(\mathbb R)$ we have to consider discret series, but in $SL_2(\mathbb C)$ we have done.

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