# Characterizing the real analytic Eisenstein series

Consider the classical real analytic Eisenstein series $$E(z,s)=\left(\pi^{-s}\Gamma(s)\frac{1}{2}\right)\sum_{(m,n)\neq(0,0)}\frac{y^s}{|mz+n|^{2s}},$$ where $z=x+iy$. We think of $E(z,s)$ as a function on $(z,s)\in\mathfrak{h}\times\mathbf{C}$. The function $E(z,s)$ satisfies the following properties

(1) For a fixed $z\in\mathfrak{h}$, $s\mapsto E(z,s)$ is holomorphic except with poles of order $1$ at $s=1$ and $s=0$ with residues $1/2$ and $-1/2$ respectively (the knowledge of one residue implies the knowledge of the other from the functional equation in $s$, see below).

(2) $E(z,s)$ is $SL_2(\mathbb{Z})$-invariant in $z$

(3) $\Delta_h E(z,s)=s(1-s)E(z,s)$ where $\Delta_h$ is the hyperbolic Laplacian.

(4) $E(z,s)=E(z,1-s)$

(5) For a fixed $s\in\mathbf{C}\backslash\{\frac{1}{2}\}$, we have $E(z,s)=O(y^{\sigma})$ as $y\rightarrow \infty$ where $\sigma=\max(\Re(s),1-\Re(s))$.

Q1 Do the properties (1), (2), (3), (4) and (5) characterize $E(z,s)$ ?

Q2 Is there some redundancy among properties (1), (2), (3), (4) and (5)?

Q3 What is a good way to characterize what $E(z,s)$ is ? (I guess that representation theorists should have something nice to say for Q3)

added Note that $E(z,\frac{1}{2})$ is not square integrable. Indeed, looking at the constant term of the Fourier series in $z$ of $E(z,1/2)$ we find that $E(z,1/2)\sim Cte\cdot\log(y)\sqrt{y}$. So if one integrates in the usual fundamental domain $\mathcal{D}_{T}$ of $SL_2(\mathbb{Z})$, up to height $T$, with respect to the Poincare volume, we find that $$\int_{\mathcal{D}_T}|E(z,1/2)|^2\frac{dxdy}{y^2}\sim \int_{1}^{T} \frac{\log(y)^{2}dy}{y}\sim \frac{1}{3}\log(T)^3.$$ So as $T\rightarrow \infty$ the integral diverges. Note though that it is "almost" square integrable in the sense that it diverges extremely slowly.

• Your (4) and (5) are not correct. To fix them, you should modify your definition of $E(z,s)$ by including the factor $\pi^{-s}\Gamma(s)$ in front of the $(m,n)$-sum. See for example Theorem 1.6.1 in Bump: Automorphic forms and representations, and note that $E(z,s)$ there denotes the Eisenstein series with the extra factors included (Bump also has a factor of $1/2$). – GH from MO Oct 22 '14 at 22:59
• But, $\Delta(E(z,s)+c)=s(1-s)E(z,s)\neq s(1-s)(E(z,s)+c)$ – Subhajit Jana Oct 22 '14 at 23:41
• From the spectral decomposition $$L^2(\Gamma\backslash G)=L^2_{cusp}\oplus \mathbb{C}\oplus L_{cont}^2,$$ Any function satisfying (2),(3) and (5) should be in the continuous spectrum. Therefore it can be described by the given Eisenstein series (as it has only one cusp at $\infty$. (2), (3) and (5) imply (1) and (4) with @GHfromMO's correction. – Subhajit Jana Oct 22 '14 at 23:56
• @Kunnysan: You have to be careful. Eisenstein series do not lie in the $L^2$-space, not even those which contribute to the spectral decomposition: $E(z,s)$ with $\Re(s)=1/2$. – GH from MO Oct 23 '14 at 0:43
• Dear GH, thanks for the comment you are perfectly. I'll add the Euler factor with the factor $1/2$ so that I at least get the right residues! – Hugo Chapdelaine Oct 23 '14 at 1:18

For $Re(s)>1$, the function $E(z,s)$ is smooth on $\mathbb{H}$ and satisfies $(2)$, $(3)$ and $$(*) \ E(z,s)-\xi(2s) \cdot y^s=O(y^{1-s}) \ \text { as } y \rightarrow +\infty.$$ (Here $\xi(s)=\pi^{-s/2}\Gamma(s/2)\zeta(s)$ is the completed Riemann zeta function.) One can prove this by computing the Fourier expansion of $E(z,s)$. These properties characterize $E(z,s)$ since the difference of any two functions satisfying them is square-integrable on $SL_2(\mathbb{Z})\backslash \mathbb{H}$ (think of the usual fundamental domain) and every eigenvalue $\lambda$ of $\Delta$ in $L^2(SL_2(\mathbb{Z})\backslash \mathbb{H})$ is $\lambda \geq 0$.

Together with having meromorphic continuation to $s \in \mathbb{C}$ for fixed $z$, this characterises $E(z,s)$.

(Of course, this sort of characterisation of $E(z,s)$ is well-known: see e.g. Lemma 2.5.1 in these notes or Lemma 1 in here.)

• Dear Luis, this is a nice characterization. Of course specifying partly what the constant term of the Fourier series is, is not as much conceptual as what I was hoping at first, but may be one cannot do better than that. – Hugo Chapdelaine Oct 23 '14 at 20:52
• Dear Hugo, I am a bit confused since I did not mention the constant term, just a growth estimate. If you can be a bit more precise about what you would consider more conceptual, then I'll try to think about it! – Luis Garcia Oct 23 '14 at 21:27
• Well, by subtracting $\xi(2s)$ to $E(z,s)$ combined with some growth estimate "seems to be close" to saying that $\xi(2s)y^s$ is part of the constant term of the Fourier series $E(z,s)$. For example, if $s=3/4$ it says that $E(z,s)$ behaves asymptotically exactly like $\xi(3/2)\cdot y^{3/4}$. – Hugo Chapdelaine Oct 23 '14 at 22:00
• I see. But any function satisfying (2) and (3) has a constant term which will be of the form $Ay^s + B y^{1-s}$ (by invariance of the Laplacian under translations). For $Re(s)>1$, you can't have $A=0$ for the reasons given in my answer. So you are just rescaling to $A=\xi(2s)$. – Luis Garcia Oct 23 '14 at 23:01
• I think I see what you're getting at now. Suppose $F(z,s)$ satisfies (2),(3), (5), then it has a Fourier expansion with constant term $A(s) y^s + B(s) y^{1-s}$ (e.g. see Thm 3.1 of Iwaniec's spectral methods book). Then $\xi(2s) F(z,s) -A(s) E(z,s)$ is $L^2$ for $\text{Re}(s) > 1$ but this means it vanishes at such an $s$. By meromorphic continuation, it vanishes everywhere. – Matt Young Oct 24 '14 at 15:13

The properties (1)-(5) do not characterize $E(z,s)$. The issue is that there's no enough control on it as a function of $s$. For an example, let $$F(z,s) = e^{s(1-s)} E(z,s).$$ Then $F(z,s)$ satisfies properties (1)-(5).

• Thanks Matt, this is good observation! Probably, one should put some explicit restrictions on the constant $C(s)$ which appears implicitly in the big O notation of property (5). – Hugo Chapdelaine Oct 23 '14 at 16:26
• Do you have any precise idea on how to fix it? – Hugo Chapdelaine Oct 23 '14 at 16:27
• I'm not sure what is the best way to fix it... – Matt Young Oct 23 '14 at 17:06