Ground field $\Bbb{C}$. Algebraic category. Elliptic surfaces are those surfaces endowed with a morphism onto some smooth curve, with generic fiber an elliptic curve.

Suppose $E$ is an elliptic curve and consider the ruled surface $$ S=\frac{E\times\Bbb{P}^1}{G} $$ where $G$ is a group of translations of $E$, acting on $\Bbb{P}^1$. Then $S$ is an elliptic surface, for the projection on the second factor induces a morphism $S\rightarrow\Bbb{P}^1$ whose fibers are elliptic curves ($F=E/G$, in fact).

Now, let $p\colon S\rightarrow C$ be an elliptic surface and suppose $S$ is ruled over an elliptic curve $E$.

Is $S$ isomorphic to the above example? Any hint for attacking this? Thank you.

  • $\begingroup$ An answer to this question is in fact contained in this article (see Theorem 5). $\endgroup$
    – HYL
    Apr 17 '19 at 20:08

EDIT We show that the answer to the OP's question is yes. Thanks to Will Sawin for his comments.

I use the notation of [Hartshorne, Algebraic Geometry, Chapter V Section 2].

Since $S$ is a ruled surface, there exists a section $C_0$ of minimal self-intersection; set $C_0^2 = -e$. If we write $S=\mathbb{P}(\mathcal{E})$, with $\mathcal{E}$ normalized, then $e= - \deg \mathcal{E}$.

Moreover, since the base of the ruling is an elliptic curve $E$, then $e \in \{-1, 0, 1, 2, 3, 4, \ldots \}$.

We can exclude the case $e >0$. Indeed, any divisor $D \in \textrm{Pic}(S)$ can be written as $a C_0 + b f$, where $f$ is the fibre of the ruling. Now the elliptic fibre of the morphism over $\mathbb{P}^1$ must satisfy $(aC_0+bf)^2=0$, that is $b = \frac{1}{2}ae$. But if $e >0$ then an effective divisor must also satisfy $b \geq ae$ (Hartshorne, Proposition 2.20 p. 382) and this is a contradiction.

Then the only possibility is $e \in \{-1, 0 \}$.

The case $e=-1$ corresponds to the second symmetric product $\textrm{Sym}^2(E)$; in this case $\mathcal{E}$ is the unique indecomposable rank two vector bundle on $E$ with $\deg \mathcal{E} =1$. Then the linear system $|4C_0-2f|$ is a pencil of elliptic curves. One can also write $S$ in the desired for (see Wil Sawin's first comment).

The case $\mathcal{E}=0$ corresponds either to the trivial bundle (so $S$ is a product), or to $\mathcal{O}_E \oplus \mathcal{L}$, with $\deg \mathcal{L}=0$ or to $\mathbb{P}(\mathcal{E})$, where $\mathcal{E}$ is the unique indecomposible rank two vector bundle on $E$ such that $\deg \mathcal{E}=0$.

Let us consider first the case $\mathcal{O}_E \oplus \mathcal{L}$, with $\deg \mathcal{L}=0$. Then the curves in the elliptic pencil should correspond to elements of $h^0(aC_0)$, that is to sections of $\textrm{Sym}^a (\mathcal{O}_E \oplus \mathcal{L})$. Since we must have two independent sections, we obtain that $\mathcal{L}$ is a torsion bundle. In this case the elliptic fibration do exist, and one can easily write $S$ in the desired form (see Will Sawin's second comment).

Finally, let us exclude the case $\mathbb{P}(\mathcal{E})$, with $\mathcal{E}$ indecomposable of degree $0$. Again, the curves in the elliptic pencil must correspond to elements of $h^0(aC_0)$, that is to sections of $\textrm{Sym}^a \mathcal{E}$. But from [Atiyah, Vector bundles on an elliptic curve, Theorem 9] one sees that $\textrm{Sym}^a \mathcal{E}$ is again indecomposable of degree $0$, and that $h^0(\textrm{Sym}^a \mathcal{E})=1$. Thus one does not have two independent sections, and the elliptic pencil cannot exist.

Summing up, we obtain

Let $p \colon S \to \mathbb{P}^1$ be an elliptic surface and assume that $S$ is ruled over an elliptic curve $E$. Then $S=\mathbb{P}(\mathcal{E})$, where either

$\bullet$ $\mathcal{E}$ is the unique indecomposable rank $2$ vector bundle of degree $1$ on $E$, or

$\bullet$ $\mathcal{E}= \mathcal{O}_E \oplus \mathcal{L}$, where $\mathcal{L}$ is a (possibly trivial) torsion line bundle.

In both cases, we can also write $S$ in the desired form $S=(E \times \mathbb{P}^1)/G$.

  • 1
    $\begingroup$ For $Sym^2 E$, you can. View $\mathbb P^1$ as $E / \{\pm 1\}$. Then $(x,\pm y)\to \{x+y,x-y\}$ is a map $E \times \mathbb P^1 \to Sym^2 E$. The map is clearly the same as the quotient by the group $G$ of $2$-torsion points, which acts on $E/\{\pm 1\}$ in the obvious way. For $\mathbb P(\mathcal E)$ for $\mathcal E$ indecomposable, I am worried that you can't. There doesn't seem to be much group-actiony about it. $\endgroup$
    – Will Sawin
    May 20 '13 at 16:38
  • 2
    $\begingroup$ For $e=0$, what about the cases of Hartshorne Example 2.11.2, where $\mathcal E = \mathcal O_C \oplus \mathcal L$? For $\mathcal L$ torsion, these are also elliptic surfaces, corresponding to $E \times \mathb P^1 / \left(\mathbb Z/n\right)$, with $\mathbb Z/n$ acting faithfully on both factors. For the $e=0$ indecomposable case, I'm pretty sure it's not an elliptic surface - the divisors that would correspond to pencils of elliptics are $nC_0$, but these have at most one-dimensional spaces of sections. $\endgroup$
    – Will Sawin
    May 21 '13 at 0:39
  • $\begingroup$ @Will: thank you, this is precisely the part I was missing :-) $\endgroup$ May 21 '13 at 7:03

One more attempt.

All the fibers of $p\colon S\to C$ dominate $E$, so they are all isogenous to $E$ and by Kodaira's classification of elliptic fibers they all have smooth support. Hence $p$ is a so-called "quasi-bundle'' and by a result of Serrano [F.Serrano, Isotrivial fibred surfaces Ann. Mat. Pura Appl. (4) 171 (1996), 63–81] is of the form $(F\times B)/G$, where:

  • $F$ is the elliptic fiber
  • $B$ is a smooth curve
  • $G$ is a finite group that acts faithfully on $F$, on $B$ and diagonally on the product $F\times B$
  • the action of $G$ on $F\times B$ is free.

The surface $S$ has exactly two morphims to a curve, $p_1\colon S\to F/G$ and $p_2\colon S\to B/G$. The general fiber of $p_1$ is $B$ and the general fiber of $p_2$ is $F$.

Since $S$ is ruled over an elliptic curve $E$, we have $B=\mathbb P^1$ and $F/G=E$. So $G$ acts on $F$ by translation and $S$ is as required.

A final remark: since $G$ acts faithfully also on $\mathbb P^1$, the possibilities for $G$ are $G=\mathbb Z_2\times \mathbb Z_2$ and $G=\mathbb Z_m$, $m\ge 2$.


Here is an alternative approach.

Since $S$ is an elliptic fibration, the rational curves of the ruling must cover $C$ and hence $C$ is also rational, i.e., $C\simeq \mathbb P^1$.

Next do a base change by the induced map from a (general) fiber of $p$, say $F$, to $E$. We get a new ellipticly ruled surface: $\rho: T=S\times_EF\to F$.

Now consider the Stein factorization of the composition $T\to S\to C\simeq \mathbb P^1$. Since $T$ is still a ruled surface, there are $\mathbb P^1$'s in $T$ that cover the Stein cover of $\mathbb P^1$, so that has to be a $\mathbb P^1$ again, so we get another elliptic fibration $q:T\to\mathbb P^1$ and by construction the fibers of $\rho$ and $q$ are transversal and meeting in a single point.

In other words the fibers of one morphism are sections of the other. It is easy to see that in this case this means that $T\simeq F\times\mathbb P^1$. The map $F\to E$ is an isogeny, so we may pick identities that it is a group homomorphism. Denoting the kernel by $G$ we get that $E\simeq F/G$ and $S\simeq (F\times \mathbb P^1)/G$.

  • $\begingroup$ I think the fibers of $q$ may be disjoint unions of elliptic curves. This leads to the action of $G$ on $\mathbb P^1$, because $q$ is an elliptic surface composed with the map from $\mathbb P^1$ to $\mathbb P^1$. $\endgroup$
    – Will Sawin
    May 20 '13 at 22:31
  • 2
    $\begingroup$ right. I had that in the first version, but then I convinced myself that it is not needed... I'll add that back in a bit. $\endgroup$ May 21 '13 at 0:14
  • $\begingroup$ Added Stein factorization back. Thanks, Will. $\endgroup$ May 21 '13 at 0:48

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