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The j-invariant as a modular function is typically defined $$j(\tau) = \frac{E_4(\tau)^3}{\Delta(\tau)}$$ since $E_4$ is a modular form of weight 4 and $\Delta$ has weight 12, it follows that $j$ is a modular function.

My first question: Is there a different derivation of the $j$-invariant? Perhaps one using the geometry of the modular curve $X(1) = \mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}^*$ or one more algebraic in nature? I.e., if I wanted to construct a function that is $\mathrm{SL}_2(\mathbb{Z})$-invariant, how should I start?

For my second question, There is the well known result that the field of all modular functions is equal to $\mathbb{C}(j)$. This is typically using the q-expansions of $j$ and a residue like theorem for modular forms.

However, $X(1) = \mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}^*$ has a model so that $X(1)(F)$ parametrizes $\bar{F}$-isomorphism classes of elliptic curves defined over $F$. So, one might expect that there is an algebraic proof that the $j$-invariant generates all modular functions that works over any field. Does such a proof exist?

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The first question comes naturally under Schwarz's theory of uniformization of hyperbolic triangles: let $\tau_0=(1+\sqrt{-3})/2$, $\tau_1=i$, and $\tau_\infty=i\infty$, and denote by $\Delta$ the hyperbolic triangle with vertices $\tau_z$, so one half of the usual fundamental domain for $X(1)$. If $H^*$ denotes the closure of the upper-half plane, Schwarz theory tells you that there exists a unique analytic map from $H^*$ to $\Delta$ such that $D(z)=\tau_z$ for $z=0$, $1$, $\infty$. Denote by $J$ the inverse map (so $J$ is from $\Delta$ to $H^*$ and satisfies $J(\tau_z)=z$). By Schwarz's reflection principle, coming from the tesselation by $\Delta$, one can extend $J$ into a meromorphic function from $H^*$ to $P^1(\mathbb C)$ which will be invariant under the subgroup of orientation preserving maps of the group generated by reflections along the sides of $\Delta$, here $SL_2(\mathbb Z)$, and of course $j(\tau)=1728 J(\tau)$. This can of course be done for any hyperbolic triangle.

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  • $\begingroup$ This is exactly the kind of answer I was looking for. Do you have a reference for some of the details? $\endgroup$
    – Rdrr
    Oct 4, 2021 at 21:13
  • $\begingroup$ This is classical (Schwarz is 19th century) but was explained to me by Frits Beukers. I am sure some reader of MO can give suitable references. $\endgroup$ Oct 5, 2021 at 17:45

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