## Where do the real analytic Eisenstein series live?

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In obtaining the spectral decomposition of $L^2(\Gamma \backslash G)$ where $G=SL_2(\mathbb{R})$, and $\Gamma$ is an arithmetic subgroup (I am satisfied with $\Gamma = SL (2,\mathbb{Z})$) we have a basis of eigenfunctions of the hyperbolic Laplacian, and orthogonal to that we have the space spanned by the incomplete Eisenstein series, $$E(z,\psi) = \sum_{\Gamma_\infty \backslash \Gamma} \psi (\Im(\gamma z)) = \frac{1}{2\pi i}\int_{(\sigma)} E(z,s)\tilde{\psi}(s)\mathrm{d}s$$ where $\psi \in C_c^\infty(\mathbb{R}^+)$, $\tilde{\psi}$ is its Mellin transform, and $E(z,s) = \sum_{\gamma \in \Gamma_\infty \backslash \Gamma} \Im(\gamma(z))^s$ is the usual Eisenstein series.

My question is, where does $E(z,s)$ itself live with respect to the vector space $V = L^2(\Gamma \backslash G)$ which can be considered as the vector space of the right regular representation of $G$, and what is this parameter $s$?

A similar question of course goes for $\mathbb{R}$, where does $e^{2\pi i x}$ live with respect to $(L^2(\mathbb{R}), \rho)$?

I would appreciate a representation theoretic flavored answer, that is why I mentioned representations, but any other answer would also be an addition to my understanding of this.

In general, is there an associated space to $(V,\pi)$, an automorphic representation, such that the elements of the vector space are of moderate or rapid growth, instead of decay.

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I know very little about this kind of representation theory, but it seems that the concept of "rigged Hilbert spaces" is relevant here. See here: en.wikipedia.org/wiki/Rigged_Hilbert_space – Mark Schwarzmann Apr 9 2012 at 13:27

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At their parents' place? – Emilio Pisanty Apr 9 2012 at 16:22

## 4 Answers

Surely there is not a single good answer, since the question is about how to legitimize "generalized eigenvectors", and there is no single-best notion of "legitimize".

As in other answers, one interpretation of Eisenstein series is as being in the dual to "rapidly decreasing" functions. This has various weaknesses.

"Continuous, moderate growth" is a better space for many purposes, but, note, it does not contain $L^2$ (!), but does contain suitable positively-indexed Sobolev spaces [sic].

There are interesting difficulties in understanding what "moderate growth" (of a given exponent) might mean, if/when one wants these spaces to be representation spaces for $G$ on $\Gamma\backslash G$. The most naive-and-appealing definitions of topological vector space structures are not $G$-stable, for elementary reasons, but sensible adaptations are easy, when one allows (by now 60-year-old) topological vector space notions.

In a different direction, note that the Plancherel theorem for afms does not depend upon knowing a space in which $E_{1/2+it}$ lies, any more than the usual Plancherel for Fourier transform on $\mathbb R$ depends on knowing "where $e^{i\xi x}$ lies".

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Thank you for your answer. I did not understand what you meant by "the most naive and appealing definitions of topological vector space structures". The way I know to incorporate that some function on G is moderately growing is that we take its NAK decomposition, our functions will be periodic and bounded with respect to N, and K is compact anyway, and we have some ordering on A to talk about growth. And given a representation and a vector (and another vector in the dual space) we may talk about the matrix coefficient to pass to a function on G. – Eren Mehmet Kiral Oct 29 2011 at 20:03
@Eren Mehmet Kiral, yes, as you say, there are several devices to topologize functions on $G$ (or on Siegel sets, or...) which have moderate growth. In my experience, one often must be alert to an implicit question, of whether one wants/needs the resulting space to be a (continuous!) representation space for $G$. Sometimes not, but sometimes yes. In brief, it is easy to tell needless falsehoods. This is already clear in the action of $G=\mathbb R$ on itself: the space of continuous, bounded functions on $\mathbb R$ is not a repn space, but continuous and going to 0 at $\infty$ is. Thus, ...  – paul garrett Oct 31 2011 at 13:32
... ascending unions (filtered colimits) or nested intersections (filtered limits) of spaces of various exponents of moderate growth (on $\mathbb R$ or on $SL(2,\mathbb)$) can readily be demonstrated to be (continuous!) repn spaces, while the "limitands" require slightly more delicate treatment. Often that delicacy is irrelevant to the problem at hand, of course.  Also, notions of moderate growth or rapid decay for afms really best should refer to Siegel sets. E.g., as $z=x+iy$ approaches the real axis in a fashion cutting across infinitely-many distince fundamental regions, ... – paul garrett Oct 31 2011 at 13:38
... waveforms' sups of absolute values do not go to zero: "nothing is of rapid decay"??? But this misunderstanding is needless, too.   – paul garrett Oct 31 2011 at 13:39

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This is kind of a complicated question, since there isn't really a single good answer.

We begin with a simple Lie group $G$ (for simplicity!). On the one hand, we hopefully have a description of the unitary representations of $G$. On the other hand, we may want to understand how spaces such as $L^2(H\backslash G)$, where $H$ may be trivial or discrete or maximal compact or etc), decompose into unitary representations of $G$ (that it will decompose is known on general (highly nontrivial) principles). At least two issues arise.

First, what does it mean for a representation to "appear" in the decomposition of $L^2(H\backslash G)$? We'd like it to mean that there exists an $f\in L^2(H\backslash G)$ such that $f$ generates the representation. This can't possibly work in general, and it already fails for $L^2(\mathbb R)$. Basically, whenever $H\backslash G$ is not compact, there will be a "continuous" part to the decomposition made up of unitary representations that can't be found as subrepresentations of $L^2(H\backslash G)$. Personally, a priori, it is surprising to me that you can integrate a bunch of stuff not in $L^2$ and wind up with something in $L^2$. But then, I think about Fourier inversion and Paley-Wiener theorems, and it's not so surprising. (In fact, if you believe your future will contain a nontrivial amount of harmonic analysis, you should try to become well-acquainted with Fourier theory.)
Now, there are functions on $H\backslash G$ that generate these representations and they usually aren't very far from being in $L^2(H\backslash G)$ (like $e^{ix}$ on $\mathbb R$ and Eisenstein series on $\Gamma\backslash\mathfrak H$), but there really is no way to force them in there. A person might wonder why a benevolent God would allow this to happen, but that is outside of my expertise.

Second, which representations will appear in $L^2(H\backslash G)$? For example, the trivial representation appears in $L^2(H\backslash G)$ if and only if $H\backslash G$ has finite volume. And complementary series representations don't seem to appear at all (usually)! (This is Selberg's Conjecture.)

On a hopefully more helpful note, with certain definitions of a Schwartz space on $H\backslash G$, you can realize these functions as tempered distributions (meaning continuous linear functionals on the Schwartz space). In fact, the space of functions with uniform moderate growth on $\Gamma\backslash \mathfrak H$ contains Eisenstein series and is contained in the dual of the Schwartz space for $\Gamma\backslash\mathfrak H$. See some of Casselman's work, here and here. In a different direction, there is Schmid and Miller's work on automorphic distributions, e.g. here.

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The dual of the Schwartz space is good. But what about functions that grow exponentially, such as the I-Bessel function. Do we need a space of functions which decay even faster, and look at their dual space to fit exponentially growing functions in. – Eren Mehmet Kiral Oct 29 2011 at 20:23
You could try something like that, though perhaps it would be simpler to consider the dual space to smooth functions with compact support. I'm not sure you will gain much from this perspective. – BR Oct 31 2011 at 20:03

For any $s\in\mathbb{C}$, the Eisenstein series $E(z,s)$ does not live in $L^2(\Gamma\backslash\mathcal{H})$, where $\mathcal{H}$ is the upper half-plane. It is "smallest" when $\Re(s)=1/2$ in which case it almost lives there, namely it is in $L^{2-\epsilon}(\Gamma\backslash\mathcal{H})$. In this case we can also regard $E(z,s)$ as a function $E(g,s)$ in $L^{2-\epsilon}(\Gamma\backslash G)$ which is right invariant by $K=SO_2(\mathbb{R})$. Applying the Maass raising and lowering operators on $E(g,s)$ results in functions in $L^{2-\epsilon}(\Gamma\backslash G)$ which transform by a character of $K$ under the right $K$-action. The vector space generated by these Maass shifts is the automorphic representation associated with $E(g,s)$: it is an infinite dimensional subspace $V\subset L^{2-\epsilon}(\Gamma\backslash G)$ consisting of $K$-finite functions. It is also useful to consider the closure of $V$ in $L^{2-\epsilon}(\Gamma\backslash G)$ or natural spaces in between these two extremes: e.g. the set of functions whose partial derivatives all lie in $L^{2-\epsilon}(\Gamma\backslash G)$.

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Thank you for your answer. Why would we take the smallest possible and take $\Re(s) = 1/2$ in order to form the representation you have described? It seems like any $s$ in the critical strip should produce a representation $V$ lying in some $L^p(\Gamma\backslash G)$ for $1<p<2$. Why would only $\Re(s) = 1/2$ show up in the decomposition of $L^2(\Gamma \backslash G)$ appear into unitary representations. Or maybe do the other Eisenstein series show up in the decomposition of non $L^2$ spaces?  – Eren Mehmet Kiral Oct 29 2011 at 20:41
I don't think $L^p(\Gamma\backslash G)$ for $1<p<2$ has a similar decomposition as for $p=2$. The proof for the case $p=2$ makes great use of the extra sructure available: $L^2(\Gamma\backslash G)$ is a Hilbert space and the Laplace-Beltrami operator is self-adjoint on this space. The decomposition we are talking about is really the spectral decomposition of a self-adjoint operator on a Hilbert space. This is by no means an easy result, I could not give a quick reason why it is true. (To be continued.) – GH Oct 29 2011 at 21:47
(Continued.) At any rate, self-adjointness and positivity implies (or at least indicates) that only Laplacian eigenvalues $\lambda>0$ are present. For $E(z,s)$ we have $\lambda=s(1-s)$ which forces $\Re(s)=1/2$ or $0<s<1$. The missing of values $0<s<1$ is not so obvious, it is part of the proof. In fact we conjecture that in the full decomposition we don't see Laplacian eigenvalues $0<\lambda<1/4$ at all (this is the Selberg conjecture), but we don't have a proof yet. – GH Oct 29 2011 at 21:53

I can give an additional point of view, which is coming from the theory of parabolic induction. Parabolic induction plays a prominent role in representation theory and gives you a better intuition for the higher rank situation. This point of view is often better stressed in the adelic theory than in the classical picture.

Let $G =PSL_2(\mathbb{R})$ with standard parabolic $B$ and $\Gamma$ a cofinite lattice.

My intuition is that te analytic Eisenstein series on $\Gamma \backslash G$ are vectors of the induced representation:

$$Ind_{\Gamma N}^G 1,$$

but there is one major issue with this, namely that $\Gamma N$ is not a group.

Rigorously seen they are given as $P$ series, i.e.

$$E: C_c^\infty(N \backslash G) \rightarrow L^2(\Gamma \backslash G),$$

by defining the $B$ series $$E(f) (g)= \sum\limits_{B \cap \Gamma \backslash \Gamma} f(\gamma g).$$

The image of $E$ generates a dense subspace of the orthocomponent to the cuspidal forms.

Now, we one notices that $C_c^\infty(N \backslash G)$ is a dense subspace of $Ind_N^G 1$. Induction by steps gives a decomposition $$Ind_N^G 1 \cong Ind_B^G Ind_N^B 1.$$ Now for $Ind_N^B 1 \cong L^2(B/N) = L^2(M)$, where $M$ are the diagonal matrices. Pontryagin duality gives you a direct integral decomposition of $L^2(M)$, and we have as a result

$$Ind_N^G 1 \cong \int\limits_{\Re s = 0}^{\oplus} Ind_B^G | \cdotp |^s.$$

Certainly one hopes that $E$ extends to $Ind_B^G | \cdotp |^s$, but convergence only happens $\Re s >1/2$, and the operators has to be defined by analytic continuation to make sense on $\Re s = 0$.

Perhaps it useful to give at least one definition here: functions $f \in Ind_B^G | \cdotp |^s$ are defined as $f(bg) = |b_{1,1} / b_{2,2}|^{s+1/2} f(g)$ for $b \in B$ with $f|_K\in L^2(K)$ for $K= PSO(2)$.

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