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The degree 3 modular equation for the Jacobi modular invariant $$ \lambda(q)=\biggl(\frac{\sum_{n\in\mathbb Z}q^{(n+1/2)^2}}{\sum_{n\in\mathbb Z}q^{n^2}}\biggr)^4 $$ is given by $$ (\alpha^2+\beta^2+6\alpha\beta)^2-16\alpha\beta\bigl(4(1+\alpha\beta)-3(\alpha+\beta)\bigr)^2=0, $$ where $\alpha=\lambda(q)$ and $\beta=\lambda(q^3)$. This has a very simple rational parametrization $$ \alpha=\frac{p(2+p)^3}{(1+2p)^3}, \qquad \beta=\frac{p^3(2+p)}{1+2p} $$ (with $p$ ranging from 0 to 1 as as $q$ changes in the range).

By certain heuristical reasons (which are hidden behind analysis in my recent joint work), I expect that modularity can occur in some other similar parameterizations. In particular, the expansions $$ \mu(q) = 4096q - 294912q^2 + 12238848q^3 - 379846656q^4 $$ $$ + 9737920512q^5 - 217011585024q^6 + 4333573472256q^7 - 79091807551488q^8 $$ $$ + 1337378422542336q^9 - 21157503871942656q^{10} + 315428695901356032q^{11} $$ $$ - 4455786006742302720q^{12} + 59885350975571779584q^{13} + O(q^{14}) $$ and $$ p = 4q + 12q^2 - 48q^3 + 156q^4 - 12q^5 - 6576q^6 + 78144q^7 - 607812q^8 $$ $$ + 3017364q^9 + 156q^{10} - 208502832q^{11} + 2876189520q^{12} - 24837585384q^{13} + O(q^{14}) $$ (which, of course, can be further extended) satisfy $$ \mu(q)=\frac{p(4+p)^5}{(1+4p)^5} \quad\text{and}\quad \mu(q^5)=\frac{p^5(4+p)}{1+4p}. $$ (These relations define $p$ and $\mu$ in a unique way.)

Is there any way to identify $\mu(q)$ with a known modular function, or to show that $\mu(q)$ is not modular at all?

Thanks! Best wishes already from 2011.

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If $\mu$ were modular then presumably there would be an algebraic relation between $\mu(q)$ and $\mu(q^n)$ for all positive integer values of $n$, the relation being of degree something like the index of $\Gamma_0(n)$ in $SL(2,Z)$ (but perhaps this isn't precisely right---the exact degree would depend on the level of $\mu$). So you could expand $\mu$ out to $O(q^1000)$ and then it would be easy to search for these relations. If it doesn't work out, i.e. if $n=5$ is OK but the others don't seem to come out, then this is evidence to suggest that $\mu$ isn't modular. – Kevin Buzzard Jan 1 '11 at 13:24
@Kevin, thanks for this hint. Because I have no guess about the index of the underlying group in $\Gamma(1)$, I am not sure that $O(q^{1000})$ would be enough. I didn't try to expand so far (the coefficients grow extremely fast), but what I did (up to $O(q^{50})$) was verifying a possible algebraic relation between $\mu(q)$ and the classical $j$-invariant. None was found... – Wadim Zudilin Jan 1 '11 at 13:52

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