I know that every weight zero modular function can be written as a rational polynomial in the J-invariant, but I'm not sure how to explicitly calculate the rational polynomial for a given weight zero modular function.
Specifically, how can I write, $$2/5 q^4 - 3/10q^5 - 3150143/10q^6 - 34094690q^7 + 768512401908/5q^8 + 168243374209287/5q^9 - 289421248307239262/5q^{10} - 98695108126255661076/5q^{11} + O(q^{12})$$ as a Polynomial in J?
Thank you in advance!
On
I think I figured it out. The idea is to use the fact that the J-invariant has as a first-term $q^{-1}$ in its $q$ expansion, and using that to simplify the terms in the modular form.
So from the modular function $$2/5q^4−3/10q^5−3150143/10q^6−34094690q^7+768512401908/5q^8+168243374209287/5q^9−289421248307239262/5q^{10}−98695108126255661076/5q^{11}+O(q^{12})$$ we subtract $2/(5J^4)$ to remove the $2/5 q^4$ term. The result is $$-3/10q^5 + 1/10q^6 + 295326q^7 + 160612548/5q^8 - 173151338853q^9 - 189873719201502/5q^{10}+O(q^{11})$$ Next we add $3/(10J^5)$ to get $$1/10*q^6 - 590652/5*q^8 - 12896256*q^9 + 404421849378/5*q^{10}+O(q^{11})$$. Finally subtracting $1/(10J^6)$ gives just $$O(q^{11})$$ Putting this all together gives $$f = \frac{(4J^3-3J^2+J)}{(10J^7)} $$ Up to a constant. (since I've normalized J be removing the constant term)
A general method is to let $\; j=1/q+744+196884q+\cdots,\;$ and define the inverse function $\;f(z):= z+744z^2+7540420z^3+\cdots,\;$ such that $ q=f(1/j).\;$ For any Laurent series in $q,$ $ s(q):=\frac25 q^4 -\frac3{10}q^5 -\frac{3150143}{10}q^6 +\cdots,\;$ let $ g(z):=s(f(z)) $$ = \frac25z^4+\frac{11901}{10}z^5 +\frac{22130281}{10}z^6+\cdots.$ Now $\;g(1/j) = s(q).\;$ All that is left is to determine whether $\;g(1/j)\;$ is a polynomial in $\;j.\;$ In your case it doesn't look like that is true. As a test, let us try the method on a known example.
Let $\;s(q):=q^{-2} +42987520q + 40491909396q^2 + O(q)^3,\;$ $g(z)=z^{-2}-1488z^{-1}+159768.$ This implies that $\;s(q) = j^2 - 1488j +159768\;$ which is correct.