# Modularity of Euler $q$-series.

The Dedekind $$\eta$$ function is defined as a function on the upper half space $$\mathbb{H}$$ as $$\eta(\tau) = e^{\frac{\pi i \tau}{12}}\prod_{n>0}(1-e^{2\pi i n\tau})$$ or, using the circular variable $$q=e^{2\pi i \tau}$$, as the following $$q$$-series $$\eta(q) = q^{\frac{1}{24}}\prod_{n>0}(1-q^n).$$ It is well know that the Dedekind $$\eta$$ function satisfies a modular property very similar to the one satisfied by a weight $$1/2$$ modular form, i.e. $$\eta\left(\dfrac{a\tau+b}{c\tau+d}\right) = c(a,b,c,d)(c\tau+d)^{1/2}\eta(\tau)$$ essentially given by the fact the the $$24^{\mathrm{th}}$$-power of $$\eta$$ is a modular form of weight $$12$$. What I am interested in is the function $$\phi(q):=q^{-\frac{1}{24}}\eta(q) = \prod_{n>0}(1-q^n)$$ known as the Euler function or Euler $$q$$-series. The question is: does $$\phi$$ satisfies any kind of modular property?

If we had $$\phi \in M_k(\Gamma, \chi)$$ for some half-integer $$k \in \frac{1}{2} \Bbb Z$$ and congruence subgroup $$\Gamma$$ and a character $$\chi$$ of finite order on it, then $$q(z)^{-1 / 24}$$ would be modular as well (in $$M_{k - 1/2}(\Gamma, \chi / c)$$), which cannot happen. Indeed, since $$\chi / c$$ has finite order ($$\chi$$ has order 24), then by raising to a sufficiently large power (some multiple of 24), we get that $$f : z \mapsto e^{2 \pi i d z} \in M_w(\Gamma)$$ for some integers $$w, d \in \Bbb Z$$.
Take some $$N > 1$$ so that $$\gamma := \begin{pmatrix} 0 & -1 \\ N & 0 \end{pmatrix} \in \Gamma$$. Then $$f(\gamma z) = f\left( \frac{-1}{N z} \right) = z^w f(z) \iff \exp(-2 \pi i d / (N z)) = z^w \exp( 2 \pi i d z )$$ for all $$z \in \Bbb H$$. But then raising to the power $$z$$ would give that $$z \mapsto \exp( z w \log(z) + 2 \pi i d z)$$ is constant, which it isn't.
• Note: $\chi$ is only a character for the metaplectic cover, apparently, but anyway, it takes values in the group of 24-th roots of unity which is enough for my argument to run. Jun 22, 2021 at 20:08
• Also at the very end, it should read $2 \pi i d \cdot z^2$, not $2 \pi i d z$. Jun 23, 2021 at 6:00