Isotrivial but non-trivial family of elliptic curves

Question

Consider the family $f\colon \mathfrak X \to \mathbb C^*$ of elliptic curves over $\mathbb C^*$ with coordinate $t$ given by the affine equation $y^2 = x^3 - t$. Then all the fibres have $j$-invariant $j = 0$, hence they are all isomorphic. In fact, it is easy to see that the family trivializes after the finite étale base change $s \mapsto s^6$.

On the other hand, it is claimed (e.g. in Exercise 1.6 of "Moduli of Curves" by Harris and Morrison) that the family $f$ is not trivial. How to show this?

One idea would be to show that the family does not have an algebraic/holomorphic section. However I don't know how to do this.

Or one could show that $R^1 f_* \mathbb Z_{\mathfrak X}$ has non-trivial monodromy using the Picard-Lefschetz formula. But this requires doing a semistable reduction first, which boils down to taking the 6-to-1 cover mentioned above and then the monodromy will be trivial because the family is trivial.

Answer 1

One idea is to show that the total space of the family is rational. In other words, we want to show that the surface $$ \operatorname{Spec} \frac{k[x,y,t]}{(x^3-y^2-t)} $$ is rational. This can easily be seen algebraically. At the level of rings, $$ \frac{k[x,y,t]}{(x^3-y^2-t)} \cong k[x,y] $$ where we used the fact that $R[t]/(t-a) = R$ for $a\in R$. In this case, $R=k[x,y]$. Consequently, our surface is isomorphic to $\operatorname{Spec} k[x,y]=\mathbb{A}^2$, so it is rational. (If you prefer a geometric argument, you can project away from a point.)

This proves that the given family is not trivial: Indeed, the total space of the trivial family is isomorphic to the surface $\mathbb{C}^* \times E$ where $E$ is the elliptic curve $y^2=x^3-1$ (since all the fibers are isomorphic elliptic curves). This surface is not rational, because a rational map $\mathbb{P}^2\dashrightarrow \mathbb{C}^*\times E$ composed with the projection $\mathbb{C}^*\times E\to E$ would give a dominant map $\mathbb{P}^2\dashrightarrow E$ which would mean that $E$ is unirational. Any unirational curve is rational (by Luroth's theorem), which is a contradiction (because an elliptic curve have genus 1, so it is not rational).

Answer 2

Whenever you have a family of (generically smooth, but possibly singular) elliptic curves over a smooth 1-dimensional base $B$, the local monodromy of the family around a point $b\in B$ is completely determined by the fiber at that point. Roughly speaking, this monodromy tells you how the homology groups of the fibers changes as you move along a loop in the base $B$. In this case, since the fibers are elliptic curves, their first homology is $\mathbb{Z}^2$, and thus the monodromy associated to a loop in $B$ will be a matrix in $GL_2(\mathbb{Z})$. The point here is that if the family is trivial, then the monodromy will be trivial (ie "the identity" $[[1,0],[0,1]]$).

When $B$ is a curve over $\mathbb{C}$ (or a number field), the relationship between monodromy and the singular fiber is completely classified by Kodaira in one of his papers "On the Structure of Compact Complex Analytic Surfaces" (he wrote a number of papers with this title). (I believe the case of elliptic surfaces over rings of integers in number fields was done by Neron/Tate)

Thankfully, one of the key results can be found here:

https://en.wikipedia.org/wiki/Elliptic_surface#Monodromy

In your case, your singular fiber above $t = 0$ is a cuspidal cubic (Kodaira type II), which according to the table has monodromy matrix $[[1,1],[-1,0]]$. Since this monodromy matrix isn't the identity, you can deduce that your family is not a trivial family. (Note that the only way to get trivial monodromy is for the "singular fiber" to actually be smooth, ie Kodaira type $I_0$).

Answer 3

As Ariyan commented, it can be done by hand, but it's not ugly at all. Let $k = \mathbb C$. Any change of variables preserving the short Weierstrass form $y^2 = x^3+Ax+B$ is $(x,y) = (u^2x', u^3y')$ for $u \neq 0$ (cf. Silverman III.1). (This needs that the characteristic is $\neq 2,3$). The coefficients transform as $B = u^6 B'$. Thus, an isomorphism between $\{y^2=x^3-t\}$ and $\{y'^2=x'^3-1\}$ over $k(t)$ would give $u \in k(t)^\times$ with $u^6=t$.

So we really need to make the étale base change $k(t) \to k(t^{1/6})$ to trivialise the family.

Answer 4

You can do this by computing the Picard-Fuchs equation.

Take the form $\omega(t)=dx/y$, and consider $$\pi(t) \ =\ \int_{\gamma(t)} \omega(t)$$ where for $U$ a small open set, $\gamma(t)\in H_1(X_t;\mathbf{Z})$ is a locally constant choice of loop for each $t\in U$. Then $$\pi(t) \ = \ \int_{\gamma(t)} \frac{dx}{\sqrt{x^3+t}} \ \ \ \text{ and } \ \ \ \pi'(t) \ =\ \int_{\gamma(t)}\frac{-dx}{2\sqrt{x^3+t}^3} $$ However, note that $$ df\ =\ 6t\omega'(t) \ - \omega(t)\ \ \ \text{ for }f \ =\ \frac{x}{\sqrt{x^3+t}}$$ Since the integral of an exact form is zero, $6t\pi'(t)+\pi(t)=0$, which implies $$\pi(t) \ =\ \pi(1)\exp\left( -\int_1^t\frac{d\zeta}{6\zeta}\right)$$ Thus looping around $t=0$ changes $\pi(t)$ by a primitive $6$th root of unity, i.e. we have detected that the initial and final values of $\gamma(t)$ are different. Of course the monodromy of a period in a trivial family is trivial, hence this family isn't trivial.

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Note the same $6$ appearing in marlu's answer. Also, the fact that one forms/periods appear is perhaps not surprising: to construct the moduli stack of elliptic curves $E$ you first construct the moduli stack of elliptic curves with a given form $(E,\omega)$, then take the quotient. See Nitsure's ''Theory of descent and algebraic stacks'', chapters $20-22$ for this.