## Eisenstein series as sections of line bundles on moduli spaces

It is well known that a modular form of weight k and level \Gamma is a global section of k-power of a Hodge line bundle over some modular curve. e.g. H^0(X,E^k).

My question is

How to characterize Eisenstein series among such sections using geometric datas?

For example, we know cusp forms are just sections of H^0(X,E^k(-cusps)).But how about Eisenstein series?

Actually in his Introduction to "Abelian Varieties" 1970, Mumford writes:

"It is interesting to ask whether further ties between the analytic and algebraic theories exist: e.g. an algebraic defintion of the Eisenstein series as a section of a line bundle on the moduli space. ..."

Could somebody explain the analytic-algebraic-representation aspects of Eisenstein series in some detail?

Thank you!

-
I wish an algebraic definition of Eisenstein series, as special section of line bundles , under some condition on cusps , hecke action, etc. Not need to mention cusp forms. – unknown (google) Aug 5 2010 at 14:58
Well "in the space spanned by non-cuspidal eigenforms" would work, right? – Kevin Buzzard Aug 5 2010 at 16:06
@Kevin This is exactly what I think. But I couldn't convince myself. Could you explain it with more detail? e.g. compare with the classical or representatio-theoric defintions? Thank you! – unknown (google) Aug 5 2010 at 16:11
@unknown: use Hecke operators T_p only for p prime to the level. Then classical theory says that these are all simultaneously diagonalisable. The cusp forms are a Hecke-stable subspace, spanned by eigenforms, and the Eisenstein series are a Hecke-stable subspace, spanned by eigenforms. Any eigenform is either a cusp form, so it lives in the cuspidal subspace, or an Eisenstein eigenform, so it lives in the Eisenstein subspace. My comment is really a rather trivial consequence, and Emerton's comment has the same idea at its heart. I don't really see what there is to prove :-/ – Kevin Buzzard Aug 5 2010 at 19:10

## 1 Answer

Here is one construction:

We have the exact sequence $$0 \to H^0(\omega^{\otimes k}(-\text{cusps})) \to H^0(\omega^{\otimes k}) \to H^0(\text{cusps}, \omega^{\otimes k}_{| \text{cusps}}).$$ (Here I am using $\omega$ for what you called $E$; this is the traditional notation for modular forms people.) It is easy to define a Hecke action on the third $H^0$ so that this exact sequence is Hecke equivariant.

The right hand map is surjective if $k > 2$, and its image has codimension one when $k = 2$. In any event, write $\mathcal I$ to denote the image, so that $$0 \to H^0(\omega^{\otimes k}(-\text{cusps})) \to H^0(\omega^{\otimes k}) \to \mathcal I \to 0$$ is short exact. One then shows that this short exact sequence has a unique Hecke equivariant splitting; i.e. there is a uniquely determined Hecke equivariant subspace $\mathcal E \subset H^0(\omega^{\otimes k})$ such that $\mathcal E$ projects isomorphically onto $\mathcal I$. This space $\mathcal E$ is the space of weight $k$ Eisenstein series (for whatever level we are working at).

-
@Emerton: Thanks, is this equiv.to "non-cuspidal+Hecke-eigen"? I still have some difficult to compare it with the classical definitions. – unknown (google) Aug 5 2010 at 16:54
And, could you get some reference? – unknown (google) Aug 5 2010 at 17:02
Dear Unknown, I don't know any references for this; hopefully someone who reads this will. This is saying that the Eisenstein series are determined as Hecke eigenforms by the Hecke action on their constant terms (which is one way of thinking about the classical definition). – Emerton Aug 5 2010 at 17:04
Sction 3 of: G. Faltings, B. Jordan, Crystalline cohomology and ${\rm GL}(2,Q)$. Israel J. Math. 90 (1995), no. 1-3, 1--66. – Felipe Voloch Aug 5 2010 at 18:52
I tried looking in Faltings-Jordan's paper. They indeed define Eisenstein series as eigenforms for which the Hecke action is determined by that on the coefficients. But still, this seems (to me) to be done in a very "analytic" way, by embedding suitable rings/fields in $\mathbb{C}$, ecc... May be I am wrong, and glad to be corrected, but I am still looking for a neat (detailed) construction in the spirit of Emerton's answer. – Filippo Alberto Edoardo Oct 22 at 3:19