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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?

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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.

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    $\begingroup$ 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. $\endgroup$
    – Watson
    Jun 22, 2021 at 20:08
  • $\begingroup$ Also at the very end, it should read $2 \pi i d \cdot z^2$, not $2 \pi i d z$. $\endgroup$
    – Watson
    Jun 23, 2021 at 6:00

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