MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question
I believe the group is $B_3$, the universal central extension. See… . – 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
up vote 6 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 $Mp_2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.