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Suppose my favorite way of thinking about modular forms is as functions on the space of (real, 2D) lattices. One can identify this space with $SL_2(\mathbb{Z}) \backslash GL_2(\mathbb{R})$, i.e. bases for the lattice up to reparameterization.

A function $f : SL_2(\mathbb{Z}) \backslash GL_2(\mathbb{R}) \rightarrow \mathbb{C}$ is a modular form of weight $k$ if it satisfies the "scaling relation":

$f(a\ R_\theta\ T) = a^{-k} e^{- i k \theta} f(T)$

where $R_\theta$ denotes the appropriate rotation matrix, and $a$ a positive real.

[More precisely $f$ only has to be defined on one connected component of the space.]

(I think) the standard definition is equivalent to this, by considering the value of $f$ on canonical lattices $\langle 1,\ z \rangle$ for $z$ in the complex upper half-plane. Note it's entirely clear in this language that e.g. $G_4$ is a modular form of weight 4.

So, my question is: is it possible to make sense of modular forms of half-integral weight (for concreteness, say $\vartheta$) in a similar way?

I'm aware that a necessary step is to pass to some kind of double cover such as $Mp_2$, to make the scaling relation make sense for $k=1/2$; but I am having trouble making this sufficient. In particular, as sub-questions:

  • What group plays the role of $SL_2(\mathbb{Z})$?
  • While I could extend $\vartheta$ by "brute force" to a function on e.g. $Mp_2$ (by callously applying the scaling relation), is there a natural way to define $\vartheta$ on the larger space, similar to the "obvious" definition:
    $G_4(\Lambda) = \sum_{w \in \Lambda \setminus 0} w^{-4}$

Apologies in advance if this is standard -- I've been unable to locate a satisfactory answer in the literature.


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I believe the group is $B_3$, the universal central extension. See mathoverflow.net/questions/20281/… . –  Qiaochu Yuan Jun 18 '12 at 11:22
Thanks, I will work through the answers to that question. I'm concerned though that I'm looking at the double cover rather than the universal (infinite cyclic) cover. Does the action of $B_3$ on $AdS_3$ project down nicely to an action of $B_3$ on the double cover $Mp_2$, or do we have to first pass to a subgroup of $B_3$? –  Freddie Manners Jun 18 '12 at 12:25
Minor nitpick: your $k$'s should be $-k$'s, I think. ($G_4$ is the sum of $z^{-4}$ over $z$ in your lattice, so it scales by $\lambda^{-4}$ if you enlarge the lattice by $\lambda$.) –  David Loeffler Jun 18 '12 at 14:02
@David Loeffler: fixed (I think), thanks. –  Freddie Manners Jun 20 '12 at 18:08

1 Answer 1

up vote 5 down vote accepted

You need to replace $SL_2(\mathbb{Z})$ with a discrete subgroup of $Mp_2$, and you want this subgroup not to contain the kernel of $Mp_2 \to SL_2$, since otherwise there are trivially no half-integer-weight forms. If you take a small enough finite-index subgroup of $SL_2(\mathbb{Z})$, then it will admit a lifting to $Mp_2$.

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