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.


$(1)$: 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 $(2)$.

  • $\begingroup$ Yes, this is a typical modular form. $\endgroup$
    – WhatsUp
    Jan 29, 2021 at 22:38
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    $\begingroup$ A typical example is $\,a=1,b=23\,$ where the function $\,f(\tau)\,$ is the generating function of OEIS sequence A030199 which has much information. $\endgroup$
    – Somos
    Jan 30, 2021 at 3:42
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    $\begingroup$ Yes, that is exactly what it means. With weight one. $\endgroup$
    – Somos
    Feb 1, 2021 at 19:45
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    $\begingroup$ Yes, that is correct. You can check yourself with the Magma calculator in all cases. Some of this is already in the OEIS entry for A030199, A030200, A002655, A030202, A030213, A030203, A030214, A030204, A030215, A030216, A030217, A002107. $\endgroup$
    – Somos
    Feb 1, 2021 at 23:21
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    $\begingroup$ The functions you wrote down are examples of eta products and eta quotients, whose modular properties have been established by Gordon. Up to conjugation, they form modular forms for the congruence subgroup $\Gamma_0(N)$. $\endgroup$
    – El Rafu
    Feb 4, 2021 at 15:33

1 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.


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    $\begingroup$ :-) Vanishing at $i\infty$ isn't enough to be a cusp form. It is a cusp form because its $24$-th power is ($\Delta(az)\Delta(bz)$) thus it has to vanish at every cusp. $\endgroup$
    – reuns
    Feb 2, 2021 at 22:57

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