I've seen the following sentence come up a few times in papers:

Let $E$ be the universal elliptic curve over the modular curve $Y_1(N)$. Then the localization of $E$ at any choice of cusp is isomorphic to the Tate curve with some suitable level structure.

Could somebody explain what exactly this sentence means? What does it mean to localize $E$ at a cusp? (Here $Y_1(N)$ is the open modular curve so it has no cusps.) And in what sense is this localization of $E$ at a cusp given by the Tate curve?

My only exposure to Tate curves has been from Silverman's Advanced Topics book, where he explains how the Tate Curve $E_q$ over $\mathbf{Q}_p$ can be $p$-adically uniformized. But I'm not so comfortable with how the Tate curve shows up when dealing with universal elliptic curves. Could someone shed some light on the connection between Tate curves and universal elliptic curves?


1 Answer 1


Let $E_1(N)$ denote the universal elliptic curve you are referring to. Note that we must assume $N\ge 4$, or else there is no such universal elliptic curve. Localizing $E_1(N)$ at a cusp means, roughly, to look at how it behaves near a cusp. As you rightly point out, $Y_1(N)$ has no cusps. In this case, the analytic picture that the sentence suggests you consider is the restriction of $E_1(N)$ to a small punctured neighborhood at a cusp.

The algebraic analog of a small neighborhood of a point $x$ is the spectrum of the completion of the local ring at $x$. In the case of a curve $X$ over a field $k$, the complete local ring at a smooth $k$-rational point is isomorphic to $k[[t]]$. Thus the algebraic analog of a small punctured neighborhood at $x$ is $\text{Spec }k((t))$. To be precise, for $x\in X$, there is a map $l_x : \text{Spec }k[[t]]\cong\text{Spec }\widehat{\mathcal{O}_{X,x}}\rightarrow \text{Spec }\mathcal{O}_{X,x}\rightarrow X$. If $Y\subset X$ is an open not containing $x$, then the pullback of $l_x$ by the inclusion $Y\hookrightarrow X$ gives a map $\text{Spec }k((t))\rightarrow Y$. The quoted sentence presumably asks you to consider the the restriction of $E_1(N)$ to $\text{Spec }k((t))$ via this map.

As for the relation with the Tate curve, the canonical reference is probably chapter 8 of Katz-Mazur. Below I give a rough sketch of the relationship:

In Silverman's advanced topics book, he gives an explicit equation for the Tate curve, where the coefficients are power series in $q$. In his book $q$ is taken to be a nonzero element of a $p$-adic field, but it's not hard to see that the same equations in fact defines an elliptic curve over $\mathbb{Z}((q)) := \mathbb{Z}[[q]][q^{-1}]$ (in fact it defines a stable 1-pointed curve of genus 1 over $\mathbb{Z}[[q]]$, whose fibers above $q = 0$ give "nodal cubics"). If you consider $Y_1(N)$ over a field $k$, then these equations define the Tate curve as an elliptic curve over $k((q))$, with $j$-invariant $$j(q) = \frac{1}{q} + 744 + 196884q + \cdots\in k((q)).$$ At this point, one can "simply" say that the Tate curve, viewed as a stable pointed curve over $k[[q]]$, is the universal deformation of its central fiber, and $k[[q]]$ is the universal deformation ring of a nodal cubic. This also implies that if $\overline{\mathcal{M}_{1,1}}$ is the moduli stack of stable pointed curves of genus 1 and $\text{Spec }R\rightarrow\overline{\mathcal{M}_{1,1}}$ is the completed etale local ring at the cusp, then $R$ is isomorphic to $k[[q]]$, and via this isomorphism the associated map $\text{Spec }k[[q]]\rightarrow \overline{\mathcal{M}_{1,1}}$ corresponds to the Tate curve. This covers the case $N = 1$. For general $N\ge 4$, one should note that the universal elliptic curve over $Y_1(N)$ is just the pullback of the universal elliptic curve over the moduli stack of elliptic curves. Thus, the complete local rings near the cusps are given by the Tate curve with scalars extended to $k[[q^{1/n}]]$, where $n$ denotes the width of the cusp.

Here is a more verbose but elementary illustration of the previous paragraph: For $N = 1$, $Y(1) = Y_1(N)$ is the coarse moduli scheme of elliptic curves. Explicitly, $Y(1)$ is an affine line, with coordinate given by the $j$-invariant; we will write $Y(1) = \text{Spec }k[j]$ (think of the case $k = \mathbb{C}$ and $Y(1) = \mathcal{H}/\text{SL}_2(\mathbb{Z})$). In this case a punctured neighborhood at $j = \infty$ is given by $\text{Spec }k((j^{-1}))$. Taking $q$-expansions formally (i.e., map $j^{-1}$ to $j(q)^{-1}$), we obtain an isomorphism $\text{Spec }k((j^{-1}))\cong\text{Spec }k((q))$, and hence have a map $$t : \text{Spec }k((q))\longrightarrow Y(1)$$ where we view $\text{Spec }k((q))$ as a punctured neighborhood of the cusp "$j = \infty$" of $Y(1)$. Note that when $k = \mathbb{C}$, and $Y(1)$ is the spectrum of the ring of modular functions of level 1 on the upper half plane which are meromorphic at the cusps, then the map $t$ induces precisely the analytic $q$-expansion map at the level of rings.

The map $t$ defines a $k((q))$-valued point of $Y(1)$. It is a defining property of coarse schemes that $t$ uniquely defines an equivalence class of elliptic curves over $k((q))$, where two elliptic curves are equivalent if and only if they are isomorphic over an algebraic closure (i.e., they are twists of each other). This equivalence class is exactly the data of a $j$-invariant, and the $j$-invariant corresponding to $t$ is by definition the image of $j\in\Gamma(Y(1),\mathcal{O}_{Y(1)})$ in $k((q))$. Since the image of $j$ is $j(q)$, $t$ corresponds to the twist-equivalence class of an elliptic curve with $j$-invariant $j(q)$. Since the Tate curve has the right $j$-invariant, $t$ must correspond to the class of the Tate curve.

This gives an algebraic picture of the Tate curve. There is also an analytic picture, starting from the universal elliptic curve over the upper half plane. I've worked out some of these details in section 5 of my notes here, though I hesitate to link them since I wrote them a long time ago for myself, and to my knowledge nobody has ever checked it for correctness.

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    $\begingroup$ This is an excellent answer. Incidentally, the OP's quote was written by me, so I can confirm that when you write "the quoted sentence presumably asks you", you are presuming correctly. :-). $\endgroup$ Aug 14, 2022 at 9:06
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    $\begingroup$ thanks so much! this was very illuminating. I'll take some time to go through your answer in detail but I really appreciate it. And thank you @DavidLoeffler for writing the quote in the question + suggesting to me this connection :-) $\endgroup$ Aug 14, 2022 at 13:48

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