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We have the $j$-invariant defined as

I have that $$ j(\tau)=\frac{1}{q}+\sum_{k\geq 0}c_kq^k, $$ where $q=e^{-2\pi t}$ ($\tau=it$) and $c_k\sim e^{4\pi\sqrt{k}}/(k^{3/4}\sqrt{2})$.

The inversion formula for the $j$-invariant is $$ q=j^{-1}+\sum_{k\geq 2}d_kj^{-k}. $$

Thus, I would like to know some upper bound or asymptotic formula for $d_k$.

Any hint or reference?

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    $\begingroup$ Have you see this paper (arxiv.org/pdf/1708.02725.pdf)? You might be able to find what you are looking for there. $\endgroup$ Commented Dec 17, 2021 at 20:13
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    $\begingroup$ The OEIS sequence A091406 entry probably has the information you are looking for. The series is $q = 1/j + 744/j^2 + 750420/j^3 + 872769632/j^4 + \cdots$ $\endgroup$
    – Somos
    Commented Dec 17, 2021 at 20:46
  • $\begingroup$ @Somos Thanks for your answer. Actually, I would like to find an explicit upper bound for $\sum_{k\geq 1}\frac{kd_k}{(1728)^{k+1}}$ in terms of $1/(q-q_0)$, for $q_0=e^{-2\pi}$. Any suggestion? $\endgroup$
    – Jean
    Commented Dec 17, 2021 at 21:37
  • $\begingroup$ Please edit your question to include this vital piece of context for everyone to see. $\endgroup$
    – Somos
    Commented Dec 17, 2021 at 21:38
  • $\begingroup$ @Somos Thanks for suggestion! $\endgroup$
    – Jean
    Commented Dec 17, 2021 at 22:04

3 Answers 3

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It's in the OEIS: https://oeis.org/A066396

There's a formula there that gives an approximation of the form (in your notation) $$ d_k \sim A \cdot (-1)^{k+1}\cdot B^k / k^{3/2} $$ where $A\approx1943.54943\dots$ and $B\approx2311.3945621\dots\,$.

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    $\begingroup$ No, that is a different inverse where $t := q^2 j = q + 744 q^2 + 196884 q^3 + \cdots $ implies $ q = t - 744t^2 + 910188t^3 + \cdots.$ $\endgroup$
    – Somos
    Commented Dec 17, 2021 at 20:53
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Extended comment: Some analysis I've done recently in free probability theory gives the compositional inverse of a Laurent series of the form

$$f(z) = \frac{1}{z} + a_1 + a_2 z + a_3z^2 + \cdots$$

as

$$f^{(-1)}(z) =\frac{1}{z} + \frac{a_1}{z^2} + \frac{a_1^2+a_2}{z^3} + \frac{a_3+3a_1a_2+a_1^3}{z^4}+\cdots,$$

where the numerator polynomials are given in OEIS A134264 and are related to noncrossing partitions, Dyck lattice paths, and other combinatorial constructs.

This supports Somos' claim that OEIS A091406 gives the inverse of the j-invariant.

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  • $\begingroup$ I had forgotten that this result is also found on p. 11 of "Modular Matrix Models" by Yang-Hui He and Vishnu Jejjala (arxiv.org/pdf/hep-th/0307293.pdf). I arrived at it recently (once again after encountering it several years ago) through a different method while relating an identity connected to a Virasoro group action presented by A. Alexandov in "Enumerative geometry, tau-functions and Heisenberg–Virasoro algebra" to earlier notes of mine. $\endgroup$ Commented Dec 18, 2021 at 8:10
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$j(\tau)=J(e^{2i\pi \tau})$, since $J(q)\in q^{-1}+\Bbb{Z}[[q]]$ then in formal series $$f(s)=J^{-1}(1/s)=\sum_{k\ge 1} d_k s^k \text{ is in }\Bbb{Z}[[s]] \tag{1}$$

$f^{-1}(q)=1/J(q)$ is surjective from $|q|< 1$ to $\Bbb{C}$ and locally biholomorphic away from $J(q)\in 0,1728,i\infty$. Whence in the branch where $f(0)=0$ then $f$ is analytic for $|q|<1/1728$, and it will be given by the series $(1)$.

The radius of convergence is $1/1728$ and $f'(1/1728)=\sum_{k\ge 1} \frac{k\, d_k}{1728^k}$ doesn't converge, this follows from that $J'(i)=0\implies \lim_{s\to 1/1728^-}f'(s)=\infty$.

To get an upper bound on the $d_k$, let $m(y)=\sup_{\Im(\tau)>y} |1/j(\tau)|$. For $|s|<m(y)$ we'll have $J^{-1}(1/s)<e^{-2\pi y}$ so that $|d_k| \le m(y)^{-k-1} e^{-2\pi y}$ whenever $m(y)< 1/1728$. A numerical check shows that $m(2)< 10^{-5}$.

The monodromy group of $j^{-1}(s)$ and $f(s)$ is interesting:

  • in the branch where $j^{-1}(\infty)= i\infty$ when rotating around $\infty$ it transforms $j^{-1}(s)$ to $j^{-1}(s)-1=\pmatrix{1&-1\\0&1}j^{-1}(s)$.

  • In the branch where $j^{-1}(1728)=i$ when rotating around $1728$ it transforms $j^{-1}(s)$ to $\pmatrix{0&1\\-1&0} j^{-1}(s)$.

  • In the branch where $j^{-1}(0))=e^{2i\pi/3}$ when rotating around $0$ it transforms $j^{-1}(s)$ to $\pmatrix{0&-1\\1&1} j^{-1}(s)$.

  • In the branch where $j^{-1}(\infty)=\pmatrix{a&b\\c&d}i\infty=a/c$ it gets different, rotating around $\infty$ transforms $j^{-1}(s)$ to $\pmatrix{a&b\\c&d} \pmatrix{1&-1\\0&1} \pmatrix{a&b\\c&d}^{-1} j^{-1}(s)$. Whence in the branch such that $f(0)=\exp(-2i\pi a/d)$ then $f(s) = \exp(-2i\pi j^{-1}(s))$ isn't analytic at $0$ anymore.

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