# How can one understand the Eisenstein series E2 in terms of automorphic representation?

The weight 2, level 1 Eisenstein series $E_2(z)$ is a non-holomorphic automorphic form. It is defined as the analytic continuation to $s = 0$ of the series $$E_2(z, s) = \sum_{\substack{m, n \in \mathbf{Z} \\ (m, n) \ne (0,0)}} \frac{\operatorname{Im}(z)^s}{(mz + n)^2 |mz + n|^{2s}}$$ which is convergent for $\operatorname{Re}(s) > 0$ (but not for $s = 0$). Moreover for any prime p the function $E_2(z)-pE_2(pz)$ is holomorphic.

My question is: what's the Archimedean component of the automorphic representation corresponding to $E_2$? (If it it the holomorphic discrete series of weight two then seems the vector corresponding to $E_2$ should be holomorphic.)

• I think this is a really good question, so I've cleaned it up a little and retagged it to make it more visible. – David Loeffler Apr 20 '17 at 9:54

Excellent question indeed. The quick answer is that $E_2(z)$ is an almost holomorphic modular form of weight $2$ and level $1$, so the automorphic representation generated by it is not irreducible. For more details (and my thought process), read below.

Consider the Maass raising operator in weight $0$, $$R:=y\left(i\frac{\partial}{\partial x}+\frac{\partial}{\partial y}\right).$$ Let $(m,n)\in\mathbb{Z}^2$ be a nonzero pair of integers. Then a small calculation gives that, for $z=x+iy$ and $s\in\mathbb{C}$, $$R\left(\frac{y^s}{|mz+n|^{2s}}\right) =\frac{sy^s}{(mz+n)^2|mz+n|^{2s-2}}.$$

Now let us introduce the usual weight $0$ level $1$ (nonholomorphic) Eisenstein series $$E(z,s):=\sum_{\substack{m, n \in \mathbb{Z} \\ (m, n) \ne (0,0)}} \frac{\operatorname{Im}(z)^s}{|mz + n|^{2s}},\qquad \operatorname{Re}(s)>1$$ then we see that $$R\,E(z,s+1) = (s+1)\,yE_2(z,s),\qquad \operatorname{Re}(s)>0.\tag{1}$$ On the right hand side, $yE_2(z,s)$ is the canonical weight $2$ level $1$ (nonholomorphic) Eisenstein series, the one which transforms as a weight $2$ Maass form. It is worthwhile to recall here that weight $k$ holomorphic forms embed into the weight $k$ Maass spectrum by multiplying each weight $k$ holomorphic form by $y^{k/2}$. In our case $k=2$, which explains why we multiply by $y$.

So your Eisenstein series, after inserting the factor $y$ to make it into a canonical weight $2$ form, and also inserting the scaling factor $s+1$, equals the Maass raising shift of $E(z,s+1)$. It belongs to the same automorphic representation as $E(z,s+1)$, hence it has the same Langlands parameters as $E(z,s+1)$ at every place. In particular, the archimedean Langlands parameters are $$(s+1)-\frac{1}{2}=s+\frac{1}{2}\qquad\text{and}\qquad \frac{1}{2}-(s+1)=-s-\frac{1}{2}.$$

Added and revised. Well, we still need to specify all this to $s=0$, but in this case the above argument breaks down, because $E(z,s+1)$ has a pole at $s=0$. So we need to be more careful. Let us use the results and notation of Section 4 of Duke-Friedlander-Iwaniec: The subconvexity problem for Artin L-functions (Invent. Math. 149 (2002), 489-577). Then for $\operatorname{Re}(s)>1$ we have the Fourier decomposition \begin{align*} \frac{1}{2}E(z,s)&=\ \zeta(2s)y^s+\pi^{2s-1}\frac{\Gamma(1-s)}{\Gamma(s)}\zeta(2-2s)y^{1-s}\\&+\ \frac{\pi^s}{\Gamma(s)}\sum_{n=1}^\infty\frac{\sigma_{2s-1}(n)}{n^s}\bigl\{f_0^+(nz,s)+f_0^-(nz,s)\bigr\}.\end{align*} Let us replace $s$ by $s+1$ here, and then apply the raising operator $R$ along with $(1)$. Then for $\operatorname{Re}(s)>0$ we obtain the Fourier decomposition \begin{align*} \frac{1}{2}E_2(z,s)&=\ \zeta(2s+2)y^s+\pi^{2s+1}\frac{\Gamma(1-s)}{\Gamma(2+s)}\zeta(-2s)y^{-s-1}\\&-\ \frac{\pi^{s+1}}{y\Gamma(2+s)}\sum_{n=1}^\infty\frac{\sigma_{2s+1}(n)}{n^{s+1}}\bigl\{f_2^+(nz,s+1)+s(s+1)f_2^-(nz,s+1)\bigr\}.\end{align*} The right hand side is indeed holomorphic at $s=0$, and at this value it specifies to \begin{align} \frac{1}{2}E_2(z)&=\ \frac{\pi^2}{6}-\frac{\pi}{2y}- \frac{\pi}{y}\sum_{n=1}^\infty\frac{\sigma_1(n)}{n}f_2^+(nz,1)\\ &=\ \frac{\pi^2}{6}-\frac{\pi}{2y}-4\pi^2\sum_{n=1}^\infty\sigma_1(n)e(nz).\tag{2}\end{align} It is clear now that the $L$-function of $E_2(z)$ is $\zeta(s-1/2)\zeta(s+1/2)$, and $E_2(z)$ should belong to the holomorphic discrete series of weight $2$ even though its constant term is not holomorphic. I think this paradox arises from the fact that $E_2(z)$ is not a true automorphic form. (More precisely, $E_2(z)$ is not a vector from an irreducible automorphic representation, see the Added 3 section below.)

Added 2. Indeed, $E_2(z)$ is an almost holomorphic modular form of weight $2$ and level $1$: it equals $2G_2^*(z)$ in the notation of Section 2.3 of Bruinier-v.d.Geer-Harder-Zagier's book "The 1-2-3 of modular forms". In particular, (19) and (21) in this book reveal that $E_2(z)$ transforms precisely like a holomorphic modular form of weight $2$ and level $1$, even though it is not holomorphic. Of course, the same also follows from the fact that $E_2(z)=\lim_{s->0+}E_2(z,s)$, where $yE_2(z,s)$ for $\operatorname{Re}(s)>0$ transforms precisely like a Maass form of weight $2$ and level $1$. One can read more about almost holomorphic modular forms in Section 5.3 of the book.

Added 3. We can learn more by applying the Maass lowering operator in weight $2$, $$L:=1+y\left(i\frac{\partial}{\partial x}-\frac{\partial}{\partial y}\right).$$ Using the Fourier decomposition (2), we can see directly that $$L(yE_2(z))=-\pi,$$ which harmonizes with the facts that $L(yE_2(z,s)) = -s E(z,s+1)$ for $\operatorname{Re}(s)>0$ and $\operatorname{res}_{s=0}E(z,s+1)=\pi$. So we see that the automorphic representation generated by the weight $2$ vector $yE_2(z)$ is reducible: it contains the trivial representation $\mathbb{C}$ as a subrepresentation, while its quotient by $\mathbb{C}$ is irreducible and belongs to the holomorphic discrete series of weight $2$. See also Theorem 2.5.2 in Bump: Automorphic forms and representations, specified to $k=2$ and $\lambda=0$, which helped me understand what is going on.

• I can never remember the normalisations for these things; but isn't it true that when $s = 0$ this is exactly the case where the archimedean principal-series representation is reducible, with the weight 2 holomorphic discrete series occurring as the sub and the trivial representation as the quotient? If I've got that right, that might explain why $E_2$ is so "close" to being holomorphic. – David Loeffler Apr 20 '17 at 21:01
• Great explanation (and great question) – Venkataramana Apr 21 '17 at 0:33
• Sorry I still don't quite get the point. I guess you mean the E(z,s+1) and yE_2(z,s) at s=0 belong to the limit of discrete series. But shouldn't the weights in limit of discrete series be odd numbers? Which thing do you mean corresponds to David Loeffler's "case where the archimedean principal-series representation is reducible, with the weight 2 holomorphic discrete series occurring as the sub and the trivial representation as the quotient"? Also in Bump's book page 66 it is mentioned there the E(z,s) has a simple pole at s=1 with residue being constant function. This seems a little strange. – little dog Apr 21 '17 at 7:43
• @GHfromMO I share littledog's concerns -- are you sure this is correct? Bump's representations $D^{\pm}(1)$ correspond to holomorphic modular forms of weight 1, while the OP's question was ultimately about weight 2, so something doesn't fit. I think you want the case covered by Theorem 2.5.3(ii) in Bump's book with $k = 2$, where you have a principal-series representation with a unique invariant subspace (isomorphic to $\mathcal{D}^{\pm}(2)$) and the quotient is the 1-dimensional trivial representation. – David Loeffler Apr 21 '17 at 8:04
• @GHfromMO This time I think makes much more sense. – little dog Apr 23 '17 at 5:46

(I took @LSpice's inquiry as encouragement to add some remarks to @GH from MO's good answer... But/and one of the issues that helped me overcome my skepticism about the utility of representation theory long ago was its clarification of exactly these issues about "Hecke summation", Maass-Shimura operators, and such. In particular, that it is not necessary to write out Fourier expansions using special functions, etc.)

First, as in the comments that helped @GH...'s answer get on track, in a typical (naive) indexing scheme, the $2k$-th holomorphic discrete series is not a subrepresentation of the $2k$-th principal series (also in a naive normalization), but just of the $k$-th. Also, there are choices about when-and-how to append/renormalize the $y^k$ to weight-$2k$ automorphic forms, to make them left $\Gamma$-invariant and right $K$-equivariant. Lots of chances to make a mess...

A more interesting issue than normalization is the fact that (the map...) formation of Eisenstein series (manifestly) does not commute with meromorphic continuation. Such a thing already occurs in looking at the spectral decomposition of the space of pseudo-Eisenstein series in terms of Eisenstein series.

For generic principal series $I_s$, the raising and lowering operators $R,L$ change the $K$-type by $\pm 2$, with the lowering operator having a coefficient of $s-k$ (here the normalization matters, obviously). So, up to irrelevant constants, with $E_{s,2k}$ being the Eisenstein series hit by $I_s$ with $K$-type $2k$, $LE_{s,2k}=(s-k)E_{s,2k-2}$. Thus, at $s=1$, if $E_{s,2k-2}$ has no pole, then $E_{s,2k}$ is annihilated by the lowering operator.

It is partly some sort of crazy luck that the lowering operator $L$, in coordinates, is essentially the Cauchy-Riemann operator, so annihilation by it implies holomorphy.

But/and of course with $2k=2$, the Eisenstein series $E_{s,0}$ has a pole at $s=1$. The residue is just a constant, not very exciting, but not $0$. Thus, in whatever normalization one operates, the application of the lowering operator (Cauchy-Riemann) to (the meromorphically continued) $E_{s,2}$ gives that residue, not $0$.

For Hilbert modular forms over (totally real) number fields of degree $>1$ over $\mathbb Q$, the lowering operators attached to the various archimedean factors of the group do not map $E_{s,(2,2,2,2,...)}$ to $E_{s,0}$, where there is a pole, so they annihilate $E_{s,(2,2,...)}$.

The weight-one Eisenstein series are another story...

(Edit: some "$s-1$"'s should have been "$s-k$"'s, but I was thinking about $k=1$... maybe repaired...)

• I can't do more than just say thanks for this right now, because I barely understand Eisenstein series; but I consider my bugging you an investment on behalf of my future self, who is learning about Eisenstein series and encounters this or a related question. – LSpice Jul 8 '17 at 0:23
• @LSpice, :) ...... – paul garrett Jul 8 '17 at 1:13