Index and congruence subgroup from scaling variables of Jacobi form

Question

Let $J_{k,m}(N)$ be the space of Jacobi forms of weight $k$, index $m$, and congruence subgroup $\Gamma_{0}(N) \rtimes \mathbb{Z}^{2}$. I do not believe it is relevant here to specify what type of Jacobi form (weak, holomorphic, etc). I can't find hardly any references on how the index, and the congruence subgroup change when scaling the variables. So my question is simply:

Given $\varphi(\tau, z) \in J_{k,m}(1)$, what is the index and congruence subgroup of the Jacobi form $$\psi_{d_{1}, d_{2}}(\tau, z) = \varphi(d_{1} \tau, d_{2} z)$$ for general $d_{1}, d_{2} \in \mathbb{Z}_{>0}$?

There are a few relevant things I know. For an ordinary modular form $f(\tau)$ of weight $k$ on $SL_{2}(\mathbb{Z})$, it's straightforward to use the "slash operator" to show that $f(d \tau)$ is a weight $k$ form on $\Gamma_{0}(d)$. (See Exercise 1.2.11 of Diamond and Shurman)

I also know that in the case of $d_{1}=1$, $\varphi(\tau, dz) \in J_{k, d^{2}m}(1)$. This is easy to show from the transformations, and moreover it coincides with a well-known Hecke-like operator

$$U_{d} : J_{k,m}(1) \to J_{k, d^{2}m}(1).$$

I'm unsure what the right statement is for general $d_{1}, d_{2}$ specifically perhaps $\varphi(d \tau, z)$ and $\varphi(d \tau, d z)$.

Part of what's confusing me is that if $F(\tau, z, \sigma)$ is a weight $k$, degree 2 Siegel modular form, then it has a Fourier-Jacobi expansion in $Q = e^{2 \pi i \sigma}$:

$$F(\tau, z, \sigma) = \sum_{m=0}^{\infty} Q^{m} \varphi_{k,m}(\tau, z), \,\,\,\,\,\,\,\,\,\,\,\,\,\, \varphi_{k,m} \in J_{k,m}(1)$$

I believe it's true that $F(d\tau, dz, d\sigma)$ is a weight $k$ form for the degree 2 congruence subgroup $\Gamma^{(2)}_{0}(d) \subset Sp_{4}(\mathbb{Z})$. So we have a Fourier-Jacobi expansion:

$$F(d\tau, dz, d\sigma) = \sum_{m=0}^{\infty} Q^{dm} \varphi_{k,m}(d\tau, dz).$$

But I think we need the index of the Jacobi forms to match the exponent of $Q$. So somehow, $\varphi_{k,m}(d\tau, dz) \in J_{k, dm}(d)$? So the index is multiplied by $d$, not $d^{2}$ in this case.

Answer 1

$\varphi(d_1\tau,d_2 z)$ will generally be a Jacobi form for a congruence subgroup of this kind only when $d_1 | d_2$. Otherwise it will not transform correctly under $(\tau,z) \mapsto (\tau,z+\tau)$.

In this case its congruence group is $\Gamma_0(d_1) \rtimes \mathbb{Z}^2$ and its index is $m\frac{d_2^2}{d_1}$, as you can see using the Fourier-Jacobi expansions (the index indeed matches the exponent) or by checking the transformations directly.