Actually I have a few related questions.

Here, by $Y(1)$ I mean the affine $j$-line $\text{SL}_2(\mathbb{Z})\backslash\mathcal{H}$.

I know $Y(1)$ is only a coarse moduli space, so there isn't a universal elliptic curve over it, but does there exist an elliptic surface over it such that the fiber above every point is the elliptic curve corresponding to that point?

Now, I know that above the open set $Y(1) \setminus \{i, e^{2\pi i/3}\}$, there is such an elliptic surface. My second question is: Does there exist an open cover of $Y(1)$ such that above each open set in the cover, there exists an elliptic surface with the above property? (I'm not asking for the surfaces to glue.)

Complex analytically, one can consider the space: $$\mathcal{H}\times\mathbb{C}$$ On this space we have an action (on the right) by $\mathbb{Z}\times\mathbb{Z}$, acting via $$(\tau,x)\cdot(a,b) = (\tau, x + a\tau + b),$$ and an action (on the left) by $\text{SL}_2(\mathbb{Z})$, acting via $$\gamma\cdot(\tau,x) = (\gamma\tau, x)$$ where $\gamma$ acts on $\mathcal{H}$ by fractional linear transformations. Since intuitively, $\mathbb{Z}\times\mathbb{Z}$ acts on $\mathcal{H}\times\mathbb{C}$ "discretely", so the quotient $\mathcal{H}\times\mathbb{C}/\mathbb{Z}\times\mathbb{Z}$ should be a complex manifold, and is essentially an elliptic surface over $\mathcal{H}$.

If $\Gamma \subset \text{SL}_2(\mathbb{Z})$, then we may also try to form the quotient $$\mathbb{E}(\Gamma) := \Gamma\backslash\mathcal{H}\times\mathbb{C}/\mathbb{Z}\times\mathbb{Z}.$$ Intuitively, if $\Gamma$ has no elliptic elements, then it should act "discretely" on $\mathcal{H}\times\mathbb{C}/\mathbb{Z}\times\mathbb{Z}$ and the quotient ought to be a manifold. Hence, if $\Gamma = \Gamma(2)$ (ie, matrices congruent to the identity mod 2), then since $\Gamma(2)$ has no elliptic elements, shouldn't $\mathbb{E}(\Gamma(2))$ be a (complex) manifold? If it is, then it's a complex manifold above $Y(2) := \Gamma(2)\backslash\mathcal{H}$, which is again only a coarse moduli scheme, and hence has no universal elliptic curve. In this case, my third question is: is $\mathbb{E}(\Gamma(2))$algebraic? and if it is, can someone describe heuristically how it's different from a universal elliptic curve over $Y(2)$, if one existed?

The context for these questions comes from me trying to understand why Katz's definition of modular forms for $\Gamma(N)$ (in his paper on $p$-adic modular forms) is properly a generalization of the analytic definition of modular forms. In particular, I'm trying to understand why they must give holomorphic functions on $\mathcal{H}$. I see why this must be the case when $N \ge 3$, since then you have a universal elliptic curve, and in this case holomorphicity is a result of the required compatibility with base change, but in the case of $N = 1$ and $N = 2$, I'm still a little confused. Relevant references would be appreciated as well.


  • will
  • $\begingroup$ Dear oxeimon, There is a paper of Igusa from the 50s (if I remember correctly) which shows that the jump in the size of the automorphism group at $j = 0$ and $j = 1728$ provides a local obstruction to constructing a family over a n.h. of either of these $j$-invariants whose $j$-invariant is equal to $j$. The argument is probably very similar to Will Sawin's below. Regards, $\endgroup$
    – Emerton
    Apr 18, 2013 at 2:04

1 Answer 1


1 The last exercise of Hartshorne chapter 4 section 4 proves that there is no elliptic surface over $\mathbb A^1$ with nonconstant $j$ invariant. Obviously this is a special case.

2 No as well. Put the fiber in Weirstrass form near $j=0$. Then one can write it as $y^2=x^3-g_2x-g_3$, where $g_2$ and $g_3$ are both functions of $j$ that are well-defined at $j=0$. Furthermore clearly $g_3$ is nonvanishing at $j=0$ and $g_2$ is vanishing at $j=0$. Then computing the $j$ invariant near $j=0$ as $g_2^3/(4g_2^3-27g_3^2)$ up to a constant, we see it vanishes to at least third order, an obvious falsehood.

3 Yes, you're thinking of the Legendre family $y^2=x(x-1)(x-\lambda)$. The Legendre family is universal up to a quadratic twist - every family with full level two structure is a pullback of Legendre, up to a quadratic twist. So it is as close an approximation to the universal elliptic curve over $Y(2)$ as you can get in the category of schemes.

  • 1
    $\begingroup$ I don't think your answer to (2) is quite correct. When counting orders of zeros on $Y(1)$, you have to take ramification into account. Since $j = 0$ at the triply-ramified point $e^{2\pi i/3}$, its order is actually its order as a holomorphic function on $\mathcal{H}$, divided by 3. $\endgroup$
    – Will Chen
    Apr 17, 2013 at 22:19
  • 1
    $\begingroup$ Will Sawin's answer is correct. $\endgroup$
    – Angelo
    Apr 18, 2013 at 2:33
  • 3
    $\begingroup$ Dear oxeimon, You are getting confused. The curve $Y(1)$ is simply the affine line, with coordinate $j$. You are asking if it is possible to find a family of elliptic curves parameterized by some n.h. around $j = 0$ whose $j$-invariant is equal to the coordinate $j$ itself. Clearly $j$ vanishes to first order at $j = 0$ (not 3rd order). On the other hand, if such a family existed, then (writing it in Weierstrass form, which is always possible in a suff. small n.h. of any point) one deduces that in fact $j$ vanishes to third order at $j = 0$. This is a contradiction, hence no such ... $\endgroup$
    – Emerton
    Apr 18, 2013 at 14:16
  • 2
    $\begingroup$ $Y(2)$ is a projective line missing three points. Your $\mathbb E(\Gamma(2))$ is just the Legendre family with the two singular fibers removed. $\endgroup$
    – Will Sawin
    Apr 18, 2013 at 23:08
  • 3
    $\begingroup$ Dear Will, a small point --- my name is Emerton, with a t, not an s. Cheers, Matt $\endgroup$
    – Emerton
    Apr 18, 2013 at 23:14

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