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.

  • 1
    $\begingroup$ 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 $1/2$). $\endgroup$
    – GH from MO
    Oct 22, 2014 at 22:59
  • $\begingroup$ But, $\Delta(E(z,s)+c)=s(1-s)E(z,s)\neq s(1-s)(E(z,s)+c)$ $\endgroup$ Oct 22, 2014 at 23:41
  • $\begingroup$ 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. $\endgroup$ Oct 22, 2014 at 23:56
  • 1
    $\begingroup$ @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 $\Re(s)=1/2$. $\endgroup$
    – GH from MO
    Oct 23, 2014 at 0:43
  • $\begingroup$ 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! $\endgroup$ Oct 23, 2014 at 1:18

2 Answers 2


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.)

  • $\begingroup$ 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. $\endgroup$ Oct 23, 2014 at 20:52
  • $\begingroup$ 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! $\endgroup$ Oct 23, 2014 at 21:27
  • $\begingroup$ 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 $\xi(3/2)\cdot y^{3/4}$. $\endgroup$ Oct 23, 2014 at 22:00
  • $\begingroup$ 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 $A=\xi(2s)$. $\endgroup$ Oct 23, 2014 at 23:01
  • 1
    $\begingroup$ 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. $\endgroup$
    – Matt Young
    Oct 24, 2014 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).

  • $\begingroup$ 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). $\endgroup$ Oct 23, 2014 at 16:26
  • $\begingroup$ Do you have any precise idea on how to fix it? $\endgroup$ Oct 23, 2014 at 16:27
  • $\begingroup$ I'm not sure what is the best way to fix it... $\endgroup$
    – Matt Young
    Oct 23, 2014 at 17:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.