Is this some sort of modular form? $f(\tau)=\eta(a\tau)\eta(b\tau)$

Question

I recently found the following. Let $a,b\in\Bbb N$ with $24|(a+b)$ and consider the function $$f(\tau)=\eta(a\tau)\eta(b\tau).$$ We have the following symmetry relations: $$\begin{align} f(\tau+1)&=f(\tau),\tag1\\ f\left(-\tfrac{1}{ab\tau}\right)&=-i\tau\sqrt{ab}f(\tau),\tag2 \end{align}$$ which I will prove at the bottom of the post.

Considering that $\eta(\tau)$ is a modular form of weight $1/2$, is $f(\tau)$ some sort of modular form?

Just a word on notation. Throughout this question, I will use the notational conventions $\tau\in\Bbb H=\{z\in\Bbb C:\Im z>0\}$, $q:=e^{2i\pi \tau}$, and $\eta(\tau)=q^{1/24}\prod_{n\ge1}(1-q^n)=q^{1/24}(q;q)_\infty$.

Context: I was messing around with the $\eta$-function and I saw on Wikipedia that $$\eta(8\tau)\eta(16\tau)=\sum_{m,n\in\Bbb Z\\ m\le|3n|}(-1)^mq^{(2m+1)^2-32n^2},$$ which I found interesting, because the $q$-expansion $$\eta(\tau)=\sum_{n\in\Bbb Z}(-1)^nq^{(6n-1)^2/24}$$ has no integer-powers of $q$. After a little experimenting, I was able to show that the $q$-expansion for $\eta(a\tau)\eta(b\tau)$ has integer-powers of $q$ when $24|(a+b)$. So my next query was whether or not $f(\tau)$ was a modular form. Clearly the relation $(2)$ is not of the form $g(-1/\tau)=\tau^kg(\tau)$, but it is pretty similar, so I thought that maybe $f(\tau)$ was a modular form in some other sense, or for some subgroup of $\mathrm{SL}_2(\Bbb Z)$.

I also learned that $\eta$-quotients of the form $\prod_{d|n}\eta(d\tau)^{r_d}$ can be modular forms under certain restrictions on $r_d$ and $n$. Perhaps $f(\tau)$ satisfies these?

Forgive me if this question is obvious, I have only learned about this kind of thing from textbooks and Wikipedia, and I have yet to take any classes in number theory.

Proofs:

$(1)$: We can see that $$f(\tau)=\eta(a\tau)\eta(b\tau)=q^{(a+b)/24}(q^a;q^a)_\infty (q^b;q^b)_\infty=q^k(q^a;q^a)_\infty (q^b;q^b)_\infty,$$ for some $k\in\Bbb N$. Since $a,b\in\Bbb N$, clearly each product $(q^s;q^s)_\infty$, where $s=a,b$, will contribute only terms with integer-powers of $q$ to the $q$-expansion of $f$. And since $q^m$ is invariant under the transformation $\tau\mapsto\tau+1$ for integer $m$, we have that $f(\tau+1)=f(\tau)$.

$(2)$: From the relation $\eta(-\tfrac1{\tau})=\sqrt{-i\tau}\eta(\tau)$ it is easy to prove $(2)$.

Answer 1

Let $a,b\in\Bbb N$ such that $a+b=24$.

Theorem: The function $$f(\tau)=\eta(a\tau)\eta(b\tau)$$ is a cusp form of weight $1$ for the congruence subgroup $$\Gamma_1(ab)=\left\{\gamma\in\mathrm{SL}_2(\Bbb Z):\gamma\equiv\begin{pmatrix}1 & *\\ 0 & 1\end{pmatrix}\pmod{ab}\right\}.$$

Proof: Since $a+b=24$, we can write $$f(\tau)=q(q^a;q^a)_\infty(q^b;q^b)_\infty,$$ which is invariant under the map $\tau\mapsto \tau+1$, so $$f(\tau+1)=f(\tau).$$ Then, take any $$\gamma=\begin{pmatrix}p_1 & p_2\\ p_3 & p_4\end{pmatrix}\in\Gamma_1(ab),$$ and define $$\gamma (z):= \frac{p_1z+p_2}{p_3z+p_4},$$ as well as $$\gamma_M (z):=\frac{p_1z+Mp_2}{\frac{p_3}{M}z+p_4}$$ for any nonzero integer $M$. We have $$\begin{align} f(\gamma (z))=f\left(\frac{p_1z+p_2}{p_3z+p_4}\right)&=\eta\left(a\frac{p_1z+p_2}{p_3z+p_4}\right)\eta\left(b\frac{p_1z+p_2}{p_3z+p_4}\right)\\ &=\eta\left(\frac{ap_1z+ap_2}{p_3z+p_4}\right)\eta\left(\frac{bp_1z+bp_2}{p_3z+p_4}\right)\\ &=\eta\left(\frac{p_1(az)+ap_2}{\frac{p_3}{a}(az)+p_4}\right)\eta\left(\frac{p_1(bz)+bp_2}{\frac{p_3}{b}(bz)+p_4}\right)\\ &=\eta\left(\gamma_a(az)\right)\eta\left(\gamma_b(bz)\right). \end{align}$$ Then $$\det\gamma_M=\det\begin{pmatrix}p_1 & Mp_2\\ \tfrac{p_3}{M} & p_4\end{pmatrix}=p_1p_4-\left(Mp_2\right)\left(\tfrac{p_3}{M}\right)=p_1p_4-p_2p_3=\det\gamma=1,$$ and since $a$ and $b$ both divide $ab$ and by definition $p_3\equiv 0\pmod{ab}$, we have that $\gamma_a,\gamma_b\in\mathrm{SL}_2(\Bbb Z)$, and thus $$\eta(\gamma_M(Mz))=\varepsilon(\gamma_M)(\tfrac{p_3}{M}Mz+p_4)^{1/2}\eta(Mz)=(p_3z+p_4)^{1/2}\varepsilon_M\eta(Mz),$$ for $M=a,b$. Thus $$f\left(\frac{p_1z+p_2}{p_3z+p_4}\right)=\varepsilon_a\varepsilon_b (p_3z+p_4)f(z).\tag1$$ Here, $$\varepsilon\left[\begin{pmatrix}A & B\\ C & D\end{pmatrix}\right]=\begin{cases} e^{i\pi B/12} & C=0,\, D=1\\ \exp\left[i\pi\left(\tfrac{A+D}{12C}-\tfrac14-s(D,C)\right)\right] & C>0, \end{cases}$$ with $$s(D,C)=\sum_{n=1}^{C-1}\frac{n}{C}\left(\frac{Dn}{C}-\left\lfloor\frac{Dn}{C}\right\rfloor-\frac12\right),$$ and of course $\varepsilon_M=\varepsilon(\gamma_M)$.

From $(1)$ we then know that $f$ is a modular form of weight $1$ for $\Gamma_1(ab)$. As @reuns said, $f$ is a cusp form, because $f^{24}(\tau)=\Delta^{24}(a\tau)\Delta^{24}(b\tau)$, where $\Delta$ is the Modular Discriminant, which is a cusp form.

QED