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The standard definition for a half-integral weight meromorphic modular form is a meromorphic function that obeys that following functional equation for all matrices $\begin{bmatrix} a & b \\ c & d \end{bmatrix} \in \Gamma_0(4N)$:

$$ g(\dfrac{a z + b}{c z + d}) = \chi(d) (\dfrac{c}{d})^{2 \lambda + 1} \epsilon_d^{-1 - 2 \lambda} (c z +d)^{\lambda + 1/2} g(z).$$

My understanding is that this particular transformation law is inspired in part by analogy with Jacobi's theta function $\theta(\tau) = \sum_{n \in \mathbb{Z}} q^{n^2},$ which obeys the same transformation law as above for $\Gamma_0(4).$

What I do not understand is why in this definition for half-integral weight modular forms always restricts to congruence subgroups of level 4N. I understand that the Jacobi theta function is one of the fundamental examples and is of itself level 4, but I do not understand why then every definition of half-integral modular forms is then on level 4N. I have torn my hair out looking through references and the ones I have found simply start by defining half-integral modular forms on level 4N without explaining WHY the definition is this way or motivating it (past the Jacobi theta function).

Can anybody provide me with intuition for why half-integral modular forms are always defined on level $4N$? Or give insight into what would happen if one attempted to construct a half-integral weight modular form on a different congruence subgroup?

Thank you for your insights!

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This is all about how to make consistent choices of square roots. If you want to attach a meaning to half-integer weight modular forms of level $\Gamma$, then you need to pick, in some consistent fashion, a square root of $c\tau + d$, for every $\tau \in \mathcal{H}$ and every $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ in your group $\Gamma$. If $\Gamma \subseteq \Gamma_0(4)$, then the transformation law for the Jacobi theta function shows that there's a consistent set of choices, and that leads to the "usual" definitions.

So the question is: if $\Gamma$ isn't contained in $\Gamma_0(4)$, how do we choose the square roots? There's a gadget called the metaplectic group $M$ which is cooked up specifically to understand what this means: an element of $M$ consists of a pair $(\gamma, j_\gamma)$, where $\gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is an element of of $SL_2(\mathbb{R})$, and $j_\gamma$ is a holomorphic function $\mathcal{H} \to \mathbb{C}$ such that $j_\gamma(\tau)^2 = c\tau + d$. The group law in $M$ is a bit messy to write down, although it's very natural; it's rigged so that

  • Forgetting $j_\gamma$ gives a group homomorphism $M \to SL_2(\mathbb{R})$ with kernel $\pm 1$.
  • For any $k \in \tfrac{1}{2}\mathbb{Z}$, there's a "weight $k$" right action of $M$ on holomorphic functions $\mathcal{H} \to \mathbb{C}$, given by $$\left[f \mid (\gamma, j_\gamma)\right](\tau) = j_\gamma(\tau)^{-2k} f(\gamma \cdot \tau).$$ If $k \in \mathbb{Z}$ then this action factors through the map $M \to \operatorname{SL}_2(\mathbb{R})$ and it's just the usual weight $k$ action; if $k$ is a half-integer, then it really matters which $j_\gamma$ you pick.

So it's obvious how to define half-integer weight modular forms for every discrete subgroup $G$ of $M$.

(If you care about Lie groups, you might be interested to know that $\operatorname{SL}_2(\mathbb{R})$ is homotopy equivalent to the circle group $SO_2(\mathbb{R})$ so its fundamental group is $\mathbb{Z}$. Thus it has a unique degree $n$ covering for every $n \ge 1$. It turns out that these coverings are also Lie groups, and $M$ is the $n=2$ case of this construction. This $M$ is an example of a real Lie group having no finite-dimensional faithful representation.)

Now here's the point: the exact sequence $$ 0 \to \{ \pm 1\} \to M \to SL_2(\mathbb{R}) \to 0$$ splits over $\Gamma_0(4)$, i.e. there's a homomorphism $\Gamma_0(4) \to M$ whose composite with the projection $M \to SL_2(\mathbb{R})$ is the identity map of $\Gamma_0(4)$. (This homomorphism is exactly the choice of square root in the transformation law for $\theta$.) Hence any subgroup of $\Gamma_0(4)$ can be regarded as a subgroup of $M$ in a natural way. However, if you have some more general subgroup of $SL_2(\mathbb{Z})$, you have to choose how you want to lift it to $M$, and there might not even exist such a lifting (if I remember correctly there is an obstruction coming from some class in an $H^2$).

In some sense the "right" notion of level for a half-integer weight form is a subgroup of $M$, not of $\operatorname{SL}_2(\mathbb{R})$; the existence of the splitting over $\Gamma_0(4)$ is somehow an "accident", and trying to make the definitions work for more general subgroups of $\operatorname{SL}_2(\mathbb{Z})$ is a morally unsound question.

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