3
$\begingroup$

At the beginning of Eichler and Zagier's book on Jacobi Forms, there is the following diagram, summarizing part of the special role played by Jacobi forms of index 1.

enter image description here

One of the things confusing me is that by "Jacobi form" I believe they mean holomorphic Jacobi form, which are those forms with Fourier coefficients $c(n,r)=0$ unless $\Delta := 4n-r^{2} \geq 0$, where I have used that the index is 1. If $\varphi_{k, 1}( \tau , z)$ is any Jacobi form (holomorphic, weak, or otherwise) of index 1, the Fourier coefficients depend only on $\Delta$. So in the case where $\varphi_{k, 1}( \tau , z)$ is holomorphic, the second correspondence above is given by

$$ \varphi_{k, 1}( \tau , z) \longrightarrow \sum_{\substack{\Delta = 0,3 \,mod\,4\\\Delta \geq 0}} c(\Delta) q^{\Delta}$$

Quite simply, I'm wondering if this sequences of correspondences, or a similar sequence of correspondences, continue to hold in the case where $\varphi_{k, 1}( \tau , z)$ is only a weak Jacobi form. On the one hand I feel like it should because in index 1, all that would change is that $\Delta$ may take finitely many negative values. So would you produce a weakly holomorphic modular form in $M_{k-\frac{1}{2}}\big(\Gamma_{0}(4)\big)$ instead?

Assuming this works in some sense, maybe some one could point out sources where this is applied to the weak Jacobi form of weight -2 and index 1,

$$\Theta^{2} = \frac{\vartheta^{2}(\tau, z)}{\eta^{6}(\tau)}.$$ In this case, $\Delta = 0,3 \, \text{mod} \,4$ and $\Delta \geq -1$, so I'm wondering if this gives rise to a weakly holomorphic modular form of weight -3/2 associated to some congruence subgroup.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

Section 4.2 of this paper https://arxiv.org/pdf/1208.4074.pdf by Dabholkar, Murthy and Zagier may be what you want. The theta coefficients of weak Jacobi forms of weight k and index m are 2m component, vector-valued weakly holomorphic forms of weight $k-1/2$ for $SL(2,\mathbb{Z})$. For index one there is also an isomorphism $J_{k,1} \cong M^!_{k-1/2}(\Gamma_0(4))$ (see equation 4.23). (Following the discussion below I think $J_{k,1}$ should be $\tilde J^!_{k,1}$).

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks a lot. Indeed, it was (4.23) of DMZ which got me thinking about this. But when they write $J_{k,1}$ I believe they mean holomorphic Jacobi forms. Can I clarify that you're claiming the weak Jacobi forms (what DMZ call $\tilde{J}_{k,1}$) are isomorphic to the weakly holomorphic modular forms of weight $k-1/2$ with respect to the congruence subgroup? If so, I assume it is still induced by sending $\varphi$ to $\sum_{\Delta} c(\Delta)q^{\Delta}$, where now $\Delta$ may take negative values. $\endgroup$
    – Benighted
    Commented Jan 17, 2018 at 2:30
  • 2
    $\begingroup$ I think the $J_{k,1}$ might be a typo as below eqn 4.11 they say the $h_\ell$ are weakly holomorphic, holomorphic, or cuspidal if $\phi$ is a weak Jacobi form, Jacobi form or Jacobi cusp form. My papers with Cheng and Duncan on Umbral Moonshine use an analog of this for mock Jacobi forms and the corresponding weak mock modular forms which have a finite number of negative powers of $q$ in their $q$ expansions. $\endgroup$
    – Jeff Harvey
    Commented Jan 17, 2018 at 3:20
  • $\begingroup$ That second isomorphism in the figure from Eichler and Zagier which I provide seems to be exactly equation (4.23) in DMZ. You expect the conclusion should be that whatever "kind" of Jacobi form you start with, you produce the corresponding kind of modular object under that sequence of isomorphisms? $\endgroup$
    – Benighted
    Commented Jan 17, 2018 at 4:12
  • $\begingroup$ Yes, that is correct. The modular properties of the theta coefficients $h_\ell$ follow from those of the Jacobi form $\phi$ and the theta functions $\theta_{k,m}$ and do not care whether there are polar terms or not. By simply tracking through the singular terms you can see that $h_\ell$ will in general have negative powers of $q$ if $\phi$ is weakly holomorphic rather than holomorphic. $\endgroup$
    – Jeff Harvey
    Commented Jan 17, 2018 at 15:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .