The $j$-ivariant has the following Fourier expansion $$j(\tau)=\frac 1q +\sum_{n=0}^{\infty}a_nq^n=\frac{1}{q}+744+196884q+21493760q^2+\cdots.$$ Here is $q=e^{2\pi i \tau}$.

Is there some simple effective bound on the coefficients $a_n$?


This question comes from On the “gap” in a theorem of Heegner. Let $D$ be a negative discriminant such that $h(D)=1$. We want to show that $J=j(\sqrt{D})$ generates a cubic extension of $\mathbf Q$. Since we have at our disposal a monic cubic polynomial with rational coefficients, the modular equation $\Phi_2(X,j)$, whose root is $J$, and the other two roots are non-real, it is sufficient to show that $J$ is not an integer.

In this case $j=j\left(\frac{-1+\sqrt D}{2} \right)$ is also an integer. Set

$$t=e^{2\pi i(-1+\sqrt D)/2}.$$





On the left there is an integer. However the right side tends to zero as $|D|$ gets large. Stark asserts that $|D|>60$ is enough for the RHS to be less than 1. Why is it enough?


2 Answers 2


Once you know that the coefficients are all positive (see postscript), it's easy to get an effective upper bound that grows as $\exp(4\pi \sqrt{n})$, which is within a factor $O(\sqrt n)$ of the correct order of growth. Start from the inequality $$ a_n = q^{-n} (a_n q^n) < q^{-n} \sum_{k=-1}^\infty a_k q^k = q^{-n} j(\tau) $$ for any purely imaginary $\tau = it$ (because $q = e^{-2\pi t} > 0$ so each term $a_n q^n$ is positive). If $t \geq 1$ then $$ j(it) = e^{2\pi t} + \sum_{n=0}^\infty a_n e^{-2\pi n t} \leq e^{2\pi t} + \sum_{n=0}^\infty a_n e^{-2\pi n} = e^{2\pi t} + j(i) - e^{2\pi} < e^{2\pi t} + 1728. $$ Since $j(i/t) = j(it)$ it follows that also $$ j(it) < e^{2\pi/t} + 1728 $$ for $t \leq 1$. Thus our inequality on $a_n$ yields $$ a_n < q^{-n} j(\tau) < e^{2\pi n t} (e^{2\pi/t} + 1728). $$ The main term $\exp(2\pi (nt+1/t))$ is minimized at $t = 1/\sqrt{n}$ where it equals $\exp(4\pi \sqrt{n})$. Choosing this value of $t$ yields $$ a_n < e^{4\pi \sqrt n} + 1728 e^{2\pi \sqrt n} $$ which is an effective bound of the desired kind.

Postscript: one easy proof of $a_n>0$ starts from the formula $j = E_4^3 / \Delta$: the coefficients of $E_4$ are all positive, so the same is true for $E_4^3$; and $1 / \Delta = q^{-1} \prod_{m=1}^\infty ((1-q^m)^{-1})^{24}$ where each factor has nonnegative coefficients because $(1-q^m)^{-1} = \sum_{k=0}^\infty q^{km}$. So the product $E_4^3 \cdot 1/\Delta$ also has positive coefficients.

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    $\begingroup$ I fixed what I think were typos in the first two displayed equations (missing $q^{-n}$ on RHS of first, spurious $t$ in exponent in second.) You might want to check that I didn't create new errors. $\endgroup$ Jun 14, 2018 at 14:29
  • $\begingroup$ Looks right to me; thanks for the corrections (and also for summing over $k$ rather than $n$ in the first display). $\endgroup$ Jun 14, 2018 at 14:36
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    $\begingroup$ Thank you for your answer. Is there a missing $q$ in the product for $\Delta$? $\endgroup$
    – Shimrod
    Jun 14, 2018 at 14:36
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    $\begingroup$ This is reminiscent of the famous asymptotic formula of Hardy and Ramanujan for the partition function (whose values are essentially the coefficients of the inverse of the Dedekind eta function). Maybe there is a way to derive an asymptotic formula in this case as well. $\endgroup$ Jun 14, 2018 at 14:54
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    $\begingroup$ Such a formula is known, and indeed GH from MO posted a link jstor.org/stable/2371313 to the relevant Rademacher paper "The Fourier Coefficients of the Modular Invariant $J(\tau)$" (Amer. J. Math. 60 #2 (1938), 501--512), before deleting his answer on the grounds that it's overkill for this purpose. $\endgroup$ Jun 14, 2018 at 15:20

By a variation of Elkies's answer we can even get $a_n<e^{4\pi\sqrt{n}}$ without using $j(i)=1728$.

For $n=1$ the claim is clear. Now let $0<t<1$ and use the identity $j(it)=j(i/t)$. After expanding and rearranging, we get $$\sum_{n=1}^\infty a_n(e^{-2\pi nt}-e^{-2\pi n/t})=e^{2\pi/t}-e^{2\pi t}.$$ It follows that $$a_n<\frac{e^{2\pi/t}-e^{2\pi t}}{e^{-2\pi nt}-e^{-2\pi n/t}},\qquad n\geq 1.$$ Putting $t:=1/\sqrt{n}$, we get $$a_n<\frac{e^{2\pi\sqrt{n}}-e^{2\pi/\sqrt{n}}}{e^{-2\pi\sqrt{n}}-e^{-2\pi n\sqrt{n}}},\qquad n\geq 2.$$ so it suffices to show that the RHS is less than $e^{4\pi\sqrt{n}}$. Equivalently, $$4\pi\sqrt{n}-2\pi n\sqrt{n}<2\pi/\sqrt{n},\qquad n\geq 2,$$ $$2<n+n^{-1},\qquad n\geq 2.$$ The last inequality is obvious, so we are done.


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