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Recall that for a subgroup $\Gamma \subset SL_2(\mathbb{Z})$ a modular form $f$ of weight $k$ is a holomorphic function from the upper-half plane into the complex numbers such that for any

$\begin{pmatrix} a & b \\ c& d \end{pmatrix}\in \Gamma$

and any $z$ in the upper half-plane, we have $f(\frac{az+b}{cz+d}) = (cz+d)^k f(z)$ and $f$ is meromorphic (or holomorphic, depending on taste) at the cusps, which can be formulated as a growth condition, which I won't make precise here.

For many subgroups $\Gamma$, we have a modular interpretation of this. For example, for $\Gamma = SL_2(\mathbb{Z})$, a modular form corresponds to a section of $\omega^{\otimes k}$ on the (compactified, if we want holomorphic at the cusps) moduli stack of elliptic curves over $\mathbb{C}$. For $\Gamma = \Gamma_0(N) = \{\begin{pmatrix} a & b \\ c& d \end{pmatrix}\in SL_2(\mathbb{Z}) | c \equiv 0 \mod N\}$, we consider instead elliptic curves with a fixed subgroup of order $N$. As this all makes sense over $\mathbb{Z}$ (or, at least, $\mathbb{Z}[\frac{1}{N}]$) instead of $\mathbb{C}$, we also get integral versions of the rings of modular forms.

But in some sense, this seems to be just the tip of the iceberg. Many functions which show some "modular behaviour" are not modular forms in the sense above. In particular, this is true for many kinds of $\Theta$-functions, where both the aspects of a nebentypus and a half-integral weight come in.

Let $\chi: (\mathbb{Z}/N)^\times \to \mathbb{C}$ be a character. Then a modular form of weight $k$, level $N$ (or level group $\Gamma_0(N)$) with nebentypus $\chi$ is a holomorphic function $f$ on the upper half-plane such that for any

$\begin{pmatrix} a & b \\ c& d \end{pmatrix}\in \Gamma_0(n)$

and any $z$ in the upper half-plane, we have

$f(\frac{az+b}{cz+d}) = \chi(d)(cz+d)^k f(z)$

and $f$ is meromorphic (or holomorphic) at the cusps.

Is there any modular interpretation for modular forms with nebentypus? Are there integral versions of rings of modular forms with nebentypus? And how about the story when we have only half-integral weight?

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1 Answer 1

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Yes, there is such an interpretation, which is standard (cf. Delinge-Rapoport or Katz-Mazur). You begin by interpreting the modular forms of level $\Gamma_1(N)$ and weight $k \in \mathbb N$ as section of $\omega^k$ (plus condition at infinity) on the moduli space $Y_1(N)$ (over $\mathbb C$) of all elliptic curves $E$ together with what is called "a $\Gamma_1(N)$-level structure" and is simply a point $P$ on your elliptic curve of order exactly $N$. Then you define an action of $(\mathbb Z/N\mathbb Z) ^\ast$ on $Y_1(N)$ simply by $x \cdot (E,P)=(E,xP)$. Note that since $x$ is invertible modulo $N$, $xP$ is still of exact order $N$. This action induces an action on the sheaf $\omega$, hence on the sections of $\omega^k$, hence (putting under carpet the condition at infinity) hence on the space of modular forms. This actions is the same as the action of the diamond operators. Hence, to get your space with Nebentypus $\epsilon$, you just take the $\epsilon$-part for that action.

An other way to say the same thing is to say that a modular form with Nebentypus $\epsilon$ of weight $k$ and level $N$ is a law, which to any triple $(E/S,w,P)$ where $S$ is a scheme over $\mathbb C$, $E$ an elliptic curve over $S$, $w$ an invariant differential form on $E$, $P$ a point in $E(S)$ of order $N$ attaches an element of $\Gamma(S,O_S)$. The law being subject to those three conditions: (a) it commutes with base change $S' \rightarrow S$. (b) Multiplying $\omega$ by $\lambda$ multiplies the results by $\lambda^{-k}$, for $\lambda$ in $\Gamma(S,anneau_S)$. (c) Multiplying $P$ by $x \in (\mathbb Z/N\mathbb Z)^\ast$ multiplies the result by $\epsilon(x)$.

Of course all of this has an interpretation over $\mathbb Q(\zeta_N)$ and furnishes an algebraic version of your space of modular forms. You can also go to integers if you like.

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Thanks! That was easier than I thought. –  Lennart Meier Mar 21 '13 at 17:56
    
So, do I understand it correctly, that the action of $(\mathbb{Z}/n\mathbb{Z})^\times$ on $\omega^k$ defines an equivariant line bundle on $Y_1(N)$, hence a line bundle on $Y_0(N)$ and modular forms with the given Nebentypus are then just sections of this line bundle? –  Lennart Meier Mar 21 '13 at 21:22

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