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It seems that

The $J$ invariant is the unique modular function of weight zero for $\operatorname{SL}(2,\mathbb{Z})$ which is holomorphic away from a simple pole at the cusp such that $$J(e^{2\pi i/3})=0,\, J(i)=1.$$

Is the above statement true? If so, can you refer me to a book where this is proved?

To make sure we're on the same page: $$J(\tau)=\frac{E_4(\tau)^3}{E_4(\tau)^3-E_6(\tau)^2}$$ where $E_4$ and $E_6$ are the normalized Eisenstein series, $$E_4(\tau)=1+240\sum_{n=1}^\infty \frac{n^3 e^{2\pi i\tau n}}{1-e^{2\pi i\tau n}},$$ $$E_6(\tau)=1-504\sum_{n=1}^\infty \frac{n^5 e^{2\pi i \tau n}}{1-e^{2\pi i\tau n}}.$$

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  • $\begingroup$ Isn't $j^2$ another one? $\endgroup$
    – Dave Benson
    Commented Nov 16, 2023 at 13:44
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    $\begingroup$ @DaveBenson No, $j^2$ doesn't have a simple pole at the cusp. $\endgroup$
    – Nomas2
    Commented Nov 16, 2023 at 13:59
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    $\begingroup$ Another such function, divided by j, would have no poles anywhere, hence be constant, and take the value 1 at one point, hence be 1. $\endgroup$
    – Will Sawin
    Commented Nov 16, 2023 at 14:06
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    $\begingroup$ A small variant on the strategy @WillSawin proposes: if $f$ is a modular function of weight $0$, holomorphic away from the cusp and with a simple pole at the cusp, then there is some complex number $a$ such that $f-aj$ is holomorphic at the cusp (cancel out the leading term of the pole). The same Liouville argument shows that $f-aj$ is constant, i.e. $f=aj+b$. If you want your normalisation conditions $f(e^{2\pi i/3})=0$ and $f(i)=1$, then this forces $a=1$ and $b=0$, so $f=j$. $\endgroup$
    – Alexander Betts
    Commented Nov 16, 2023 at 15:53
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    $\begingroup$ Serre, Cours d'arithmétique, chapter VII, §3, proposition 6(iii), proves that any modular function of weight 0 is a rational function of $j$. Translating your conditions on the rational function in question, it has a simple pole at infinity and none elsewhere, and the conditions at $0$ and $1$ imply that it is the identity. $\endgroup$
    – Gro-Tsen
    Commented Nov 16, 2023 at 20:08

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Any meromorphic modular function of weight $0$ for $\mathrm{SL}(2,\Bbb Z)$ is a rational function of $j$, say $P(j)$. Since your function is holomorphic, $P$ is a polynomial. Since your function has a simple pole at infinity, $P$ has degree one. But $P$ fixes $0$ and $1$, so it is the identity.

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