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Let $q=e^{2\pi i\tau}$ and $$E_2(\tau)=1-24\sum_{n=1}^\infty n\frac{q^n}{1-q^n},$$ $$E_4(\tau)=1+240\sum_{n=1}^\infty n^3\frac{q^n}{1-q^n},$$ $$E_6(\tau)=1-504\sum_{n=1}^\infty n^5\frac{q^n}{1-q^n},$$ $$J(\tau)=\frac{E_4(\tau)^3}{E_4(\tau)^3-E_6(\tau)^2},$$ $$s_2(\tau)=\frac{E_4(\tau)}{E_6(\tau)}\left(E_2(\tau)-\frac{3}{\pi \operatorname{Im}(\tau)}\right).$$

The paper https://arxiv.org/abs/1809.00533 proves that $$\frac{1}{2\pi\operatorname{Im}(\tau)}\sqrt{\frac{J(\tau)}{J(\tau)-1}}=\sum_{n=0}^\infty \left(\frac{1-s_2(\tau)}{6}+n\right)\frac{(6n)!}{(3n)!n!^3}\frac{1}{(1728J(\tau))^n}$$ for $\operatorname{Im}(\tau)\gt 1.25$ where $√$ is the principal branch of the square root, i.e. it is the square root function "using" the principal complex argument, the one ranging from $-\pi$ to $\pi$.

Consequently, this principal square root function has a branch cut at the negative real axis. Obviously one can choose another square root function, the branch cut can be elsewhere, for example it can be the ray emanating from $0$ and going in the direction of $i$ towards $i\infty$.

On p. 39 of the linked paper, the author argues that

"Thus we still have to prove for any $\tau$ with $\operatorname{Im}(\tau) \gt 1.25$ that our result doesn’t contain any hidden complex roots of unity: We choose $\tau_8=\frac{i\sqrt{8}}{2}$. Here we get that $q = e^{2\pi i\tau_8} = e^{-2π\sqrt{2}}$ is a real number. Thus (from Thm. 4.5) both $J(\tau_8)$ and $s_2(\tau_8)$ are also real-valued. The approximations together with the error estimates from Thm. 5.1 and 5.3 tell usthat both $J(\tau_8)$ and $\frac{1−s_2(\tau_8)}{6}$ are positive real numbers. This shows that all quantities in the equation of Thm. 9.7 are real-valued and positive at $\tau=\tau_8$. The region $\operatorname{Im}(\tau) \gt 1.25$ is connected, both sides of the equation depend continuously on $\tau$ and are not zero. This proves that the equation is exact..."

What does he mean by "depend continuously on $\tau$?" What theorem does the text in bold even refer to? When I saw "connected", "continuously", it reminded me of the identity theorem, but reuns commented that $\frac{1}{2\pi\operatorname{Im}(\tau)}\sqrt{\frac{J(\tau)}{J(\tau)-1}}$ isn't holomorphic anywhere.

It seems that the author proved the theorem only for some square root function, not necessarily the principal one. I can't see how choosing just one point suffices to determine the complex square root function.

I took a look at the original paper of the Chudnovsky brothers. They didn't consider the equality for arbitrary $\tau$ with $\operatorname{Im}(\tau)\gt 1.25$, but they considered $\tau$ at imaginary quadratic irrationals at which $J(\tau)$ is real, so they didn't have to take care about these subtleties.

L. Milla
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Vestoo
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1 Answers1

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First, let's denote the left-hand side of $$\frac{1}{2\pi\operatorname{Im}(\tau)}\sqrt{\frac{J(\tau)}{J(\tau)-1}}=\sum_{k=0}^\infty \left(\frac{1-s_2(\tau)}{6}+k\right)\frac{(6k)!}{(3k)!k!^3}\frac{1}{(1728J(\tau))^k}$$ with $l(\tau)$ and the right-hand side with $r(\tau)$.

Before the piece of proof you quoted (on pages 34-39; see Remark 8.2), many $n$-th roots of unity appear, for example twelfth roots. In this part, we didn't specify which branch of the root we use, so that the equations are only proven up to a $n$-th root of unity.

To be specific, we have only proven $l(\tau)=\varepsilon(\tau)r(\tau)$ so far, where $\varepsilon(\tau)$ is a complex $n$-th root of unity and $n$ doesn't depend on $\tau$.

Both sides $l(\tau)$ and $r(\tau)$ of the equation are continuous functions of $\tau$, which means that there are no abrupt changes in value.

Since both sides have their zeroes outside the considered region with $\operatorname{Im}(\tau)>1.25$, we can write $\varepsilon(\tau)=l(\tau)/r(\tau)$ and obtain another continuous function.

This is a continuous function from a connected subset of $\mathbb C$ to a discrete set of $n$-th roots of unity, thus it must be constant. This is why it suffices to check it at one point $\tau$.

As you pointed out, it is not the identity theorem but much more basic. It's a direct consequence of continuity with discrete values, I would say.

L. Milla
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