Where do the real analytic Eisenstein series live? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T19:52:59Z http://mathoverflow.net/feeds/question/78994 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78994/where-do-the-real-analytic-eisenstein-series-live Where do the real analytic Eisenstein series live? Eren Mehmet Kiral 2011-10-24T16:26:35Z 2012-11-20T07:02:01Z <p>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.</p> <p>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 $s$?</p> <p>A similar question of course goes for $\mathbb{R}$, where does $e^{2\pi i x}$ live with respect to $(L^2(\mathbb{R}), \rho)$?</p> <p>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.</p> <p>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. </p> http://mathoverflow.net/questions/78994/where-do-the-real-analytic-eisenstein-series-live/79036#79036 Answer by BR for Where do the real analytic Eisenstein series live? BR 2011-10-24T23:07:01Z 2012-11-20T07:02:01Z <p>This is kind of a complicated question, since there isn't really a single good answer.</p> <p>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.</p> <p>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 <a href="http://en.wikipedia.org/wiki/Paley%E2%80%93Wiener_theorem" rel="nofollow">Paley-Wiener theorems</a>, 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.) <br> 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.</p> <p>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 <a href="http://en.wikipedia.org/wiki/Selberg%27s_conjecture" rel="nofollow">Selberg's Conjecture</a>.)</p> <p>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, <a href="http://www.math.ubc.ca/~cass/research/pdf/pw.pdf" rel="nofollow">here</a> and <a href="http://www.math.ubc.ca/~cass/research/pdf/schwartz.pdf" rel="nofollow">here</a>. In a different direction, there is Schmid and Miller's work on automorphic distributions, e.g. <a href="http://arxiv.org/pdf/math/0605783v1" rel="nofollow">here</a>.</p> http://mathoverflow.net/questions/78994/where-do-the-real-analytic-eisenstein-series-live/79037#79037 Answer by GH for Where do the real analytic Eisenstein series live? GH 2011-10-24T23:12:05Z 2011-10-24T23:12:05Z <p>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 $L^{2-\epsilon}(\Gamma\backslash G)$.</p> http://mathoverflow.net/questions/78994/where-do-the-real-analytic-eisenstein-series-live/79043#79043 Answer by paul garrett for Where do the real analytic Eisenstein series live? paul garrett 2011-10-25T01:26:59Z 2011-10-25T01:26:59Z <p>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".</p> <p>As in other answers, one interpretation of Eisenstein series is as being in the dual to "rapidly decreasing" functions. This has various weaknesses.</p> <p>"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]. </p> <p>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.</p> <p>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". </p> http://mathoverflow.net/questions/78994/where-do-the-real-analytic-eisenstein-series-live/91062#91062 Answer by Marc Palm for Where do the real analytic Eisenstein series live? Marc Palm 2012-03-13T08:04:31Z 2012-04-09T10:27:31Z <p>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.</p> <p>Let $G =PSL_2(\mathbb{R})$ with standard parabolic $B$ and $\Gamma$ a cofinite lattice.</p> <p>My intuition is that te analytic Eisenstein series on $\Gamma \backslash G$ are vectors of the induced representation:</p> <p>$$Ind_{\Gamma N}^G 1,$$</p> <p>but there is one major issue with this, namely that $\Gamma N$ is not a group.</p> <p>Rigorously seen they are given as $P$ series, i.e.</p> <p>$$E: C_c^\infty(N \backslash G) \rightarrow L^2(\Gamma \backslash G),$$</p> <p>by defining the $B$ series $$E(f) (g)= \sum\limits_{B \cap \Gamma \backslash \Gamma} f(\gamma g).$$</p> <p>The image of $E$ generates a dense subspace of the orthocomponent to the cuspidal forms.</p> <p>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</p> <p>$$Ind_N^G 1 \cong \int\limits_{\Re s = 0}^{\oplus} Ind_B^G | \cdotp |^s.$$</p> <p>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 $\Re s = 0$.</p> <p>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 $K= PSO(2)$.</p>