Let $\eta(z) = q^{1/24} \prod_{n=1}^\infty (1-q^n)$ be the usual Dedekind $\eta$-function, $q = e^{2\pi i z}$, and let an arbitrary $\eta$-quotient be $$f(q) = \prod_{\delta \in \mathbb{N}} \eta(\delta z)^{r_\delta}$$ with $r_\delta \in \mathbb{Z}$, all but finitely many entries of the sequence $(r_\delta)$ being zero.

A standard theorem of Gordon, Hughes, and Newman gives conditions on $(r_\delta)$ under which $f(q)$ is a weakly holomorphic modular form of weight $k$, level $N$, and character $\chi$, including giving $k$, $N$, and $\chi$.

After years of struggle (I don't know why the subject seems so impenetrable to me) I think I can claim some facility with using this theorem, along with the theorem of Ligozat (to bound the order at cusps), the standard fact that $M_k \cdot M_\ell = M_{k+\ell}$ for weakly holomorphic modular forms of integral weight (to get rid of cusps by appropriate multiplications), and Sturm's bounds on the resulting holomorphic forms (for congruences) to do the usual combinatorial work I need with these.

Currently, however, I find myself staring down some $\eta$-quotients which I believe are half-integral weight, which are of course an entirely different and much angrier kettle of fish. I have the following rather fundamental questions:

1.) Is it true that, for arbitrary $(r_\delta)$, $f(q)$ is a weakly holomorphic modular form of half-integral weight? I have heard this in non-rigorous settings but I'm not certain of it.

2.) If not, is there a half-integral weight equivalent to the theorem of Gordon, Hughes and Newman which will tell me conditions under which it is?

3.) Once it is, will that theorem, or another, tell me in what space $M_{k+1/2}(\Gamma_0(N),\chi)$ I might find $f(q)$?

These don't seem to be in the usual textbooks on modular forms which I have readily accessible to consult. Thanks for any help.

  • $\begingroup$ Let me edit to add: if the following things are all true, then my question is sufficiently answered: (1) $\eta$ is a weakly holomorphic half-integral weight modular form of weight 1/2 for $\Gamma_0(4)$; (2) $1/\eta$ is a whhiwmf of weight -1/2 for $\Gamma_0(576)$; (3) the map $V_m : \eta(z) \rightarrow \eta(mz)$ preserves modularity and raises level the same way as it does for integral weight; (4) the theorem $M_k \cdot m_\ell = M_{k + \ell}$ applies for whhiwmf as well as integral weight. $\endgroup$ Oct 3, 2019 at 16:12

1 Answer 1


Let $$f(z) = \prod_{j=1}^{2k} \eta(a_j z), a_j \in \Bbb{Z}_{\ge 1}$$

$\eta(z)^2 \in M_1(\Gamma_1(24))$ thus $\eta(a_j z)^2 $ is weight $1$ modular for $\Gamma_1(a_j 24)$

so that $g(z)=f(z)^2$ is weight $2k$ modular for $\Gamma_1(N),N=24\prod_{j=1}^{2k} a_j$ ie. $(cz+d)^{-2k}g(\gamma(z))=g|_{2k}\gamma(z) = g(z)$ so that

$(cz+d)^{-k} f(\gamma(z))=f|_k\gamma(z) = \psi(\gamma) f(z)$ with $\psi(\gamma)= \pm 1$

Then $\psi(\beta \gamma) f(z)=f|_k\beta\gamma(z) =(f|_k\gamma)|_k\beta(z)= \psi(\beta)\psi(\gamma) f(z)$

thus $\psi$ is a character $\Gamma_1(N) \to \pm 1$,

$\Gamma=\ker(\psi)$ is of index $1$ or $2$,

the main point is to show $\Gamma$ is a congruence subgroup (not sure how, from $[\Gamma_1(N):\Gamma]=2,\Gamma \supset [\Gamma_1(N),\Gamma_1(N)]$ and the metaplectic double cover coming from $\eta$, or directly from the weight $2$ Eisenstein series expression for $\log f$)

once it is done $\Gamma$ contains $\alpha \Gamma_1(M)\alpha^{-1}$ for some $\alpha \in SL_2(\Bbb{Z}),M$

and $f$ is weight $k$ modular for $\Gamma$, analytic (with no zeros) on the upper half-plane and with poles at the cusps.

There is a formula for the order of the poles at the cusps but it involves complicated multiplicative sums.

From there it is clear in what space you'll find the weight $k+1/2$ versions $f(z)\eta(a_{2k+1}z)$ and the quotients of such eta products.

  • $\begingroup$ I don't understand your response. To start with my question (1), are you saying, no, it is not true that, for arbitrary $(r_\delta)$, $f(q)$ is a weakly holomorphic modular form of half-integral weight? $\endgroup$ Oct 3, 2019 at 19:53
  • 1
    $\begingroup$ What do you not understand ? I am saying you need to look at the even $2k= \sum r_\delta$ weight $k$ version which is a modular form with poles at the cusps for some congruence subgroup and to understand the odd $\sum r_\delta$ half integral weight from it. To find precisely the congruence subgroup you need to understand the kernel of $\psi$. $\endgroup$
    – reuns
    Oct 4, 2019 at 11:46

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