3
$\begingroup$

I wanted to read a bit about modular forms of half integral weight. There is the notion of a 'factor of automorphy' (as for example, R.Rankin uses it in the book 'Modular Forms and Functions'). Now I am confused about the following: In modern literature, people seem to follow Shimura, i.e. they define modular forms of half integral weight using some cover of GL$_2^+(\mathbb{R})$, that is the group $G$ consisting of all pairs $(\alpha, \phi)$ with $\alpha \in \text{GL}_2^+(\mathbb{R})$ and $\phi^2 = t \det(\alpha)^{-1/2} (c\tau + d)$ [where $t$ is some extra parameter in $S^1$] with the multiplication $$(\alpha, \phi)(\beta, \psi) = (\alpha \beta, \phi(\beta \tau) \psi(\tau))$$

My question is the following:

Why does one introduce this cover when there already exists a notion of modular forms of half-integral weight, namely in terms of these factors of automorphy?

Example: For the usual Theta Series $\sum_{n \in \mathbb{Z}} e^{2 \pi i n^2 \tau}$ one can define what it means that this is a modular form of weight $1/2$ without even mentioning the group $G$.

In the beginning I thought that this would help about the problem of 'choosing' branches of square roots in the following sense: I thought one takes a subgroup $\Delta \subset G$ and then defines the slash operator and demands that a modular form should satisfy $f|_\xi = f$ for all $\xi \in \Delta$. So all branches exist in 'friendly' coexistence and one does not have to choose branches anymore. What actually happens is the following: Shimura wants the natural projection $P : \Delta \to P(\Delta)$ to be a bijection but this is precisely choosing a smart branch of the square root for every matrix in question. So this does not seem to be the important point.

Of course, there is Hecke operators. If one just goes with the notion of a factor of automorphy, then this map is no more defined on matrices that are needed for defining Hecke operators. Furthermore, there is vector valued modular forms, say, for the Weil repn. Here it would be unnatural to say that the forms should have a translation behavior for some fixed lift of SL$_2(\mathbb{Z})$, so in view of these points I can understand why one uses this cover of GL$_2^+(\mathbb{R})$ but I have the feeling that I am missing out a much more basic point here...

Thanks in advance,

Fabian Werner

Improve this question
$\endgroup$
3
  • 3
    $\begingroup$ I am too sleepy for a proper answer, but basically the reason is this: a classical holomorphic modular form is a function on H=SL(2,R)/SO(2,R) that transforms in an appropriate way with respect to SL(2,Z) or a similar discrete subgroup. In particular, it is not SL(2,Z) invariant. Lifting to SL(2,R) means converting to a function that is left-invariant by SL(2,Z), the price to pay is that it is no longer not right-invariant by SO(2,R). For a half-integral weight holomorphic modular form you cannot lift to SL(2,R) this way, you need to consider a double cover of it. $\endgroup$
    – GH from MO
    Commented Oct 31, 2013 at 18:27
  • 1
    $\begingroup$ Some details on the lift mentioned in GH's comment can be found at mathoverflow.net/questions/20281/… . $\endgroup$
    – Qiaochu Yuan
    Commented Nov 1, 2013 at 2:21
  • $\begingroup$ Typo in my comment: "no longer not right-invariant" should read "no longer right-invariant". Thanks to Qiaochu for his valuable comment. $\endgroup$
    – GH from MO
    Commented Nov 1, 2013 at 13:28

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .