# construct the elliptic fibration of elliptic k3 surface

Hi all,

As we know, every elliptic k3 surface admits an elliptic fibration over $P^1$, but generally how do we construct this fibration? For example, how to get such a fibration for Fermat quartic?

Moreover, as we know all (elliptic) k3 surfaces are differential equivalent to each other, does this mean: topologically the elliptic fibration we get for each elliptic fibraion is the same, which is just the torus fibration over $S^2$ with 24 node singularities? Or, the totally space is the same, but different complex data(structure) provides different way or "direction" of projection onto $S^2$, thus induces different type of fibrations?

Thanks!

• One elliptic fibration of the Fermat quartic that was known (in all but name) to Euler is described on pages 12-13 of the lecture notes at math.harvard.edu/~elkies/euler_11c.pdf, which also describe a few other ways to work with elliptic fibrations of K3's. Commented Feb 6, 2012 at 5:41
• Certainly the topological fibrations of elliptically fibered K3 surfaces are not all topologically equivalent. The number of singular fibers, as well as the types of singularities, can vary. The "weighted sum" of the singularities of fibers always equals 24, but that leaves a lot of room for variation. Commented Feb 6, 2012 at 11:20
• Some more questions in this direction: -- @ Jason Starr: What is the weight you are mentioning? --So the number of singular fibers (which I think is the degree of j-map to $\mathbb{P}^1$ can be different in different examples? --Are there examples where the singular fiber is "not" normal-crossing" (or semi-stable)? Commented Mar 27, 2013 at 2:07
• When talking about j-map above, I am assuming there is a section. Commented Mar 27, 2013 at 2:52
• This looks fun: grdb.lboro.ac.uk/search/ellk3 Commented Mar 27, 2013 at 2:57

Let $S$ be a smooth projective $K3$ surface, say over the complex numbers, and suppose that $S$ admits a (non-constant) fibration $\pi\colon S\to C$ over a curve $C$.

By the universal property of the normalization, we can suppose that $C$ is normal, hence smooth. Now, this curve $C$ must be $\mathbb P^1$, since otherwise you would have (by pulling-back) some non-trivial global holomorphic $1$-form on $S$, contradicting $h^{1,0}(S)=0$.

Next, this fibration is clearly given by the linear system $|\pi^*H^0(\mathbb P^1,\mathcal O(1))|\subset |L|$ of the pull-back $L:=\pi^*\mathcal O(1)$, which is spanned by two independent sections, say $\sigma$ and $\tau$. Now, take a general fiber: it is a smooth curve $F\subset S$ which is a divisor in the above-mentioned linear system, of the form $\{\lambda s+\mu t=0\}$. In particular $\mathcal O_S(F)\simeq L$. Moreover, by definition, $\mathcal O_F(F)\simeq L|_F=\pi^*\mathcal O(1)|_F$ which is trivial.

By adjunction, $K_F\simeq (K_S\otimes\mathcal O_S(F))|_F\simeq\mathcal O_F$ is trivial, so that $F$ is an elliptic curve.

Thus, any fibration of a smooth projective $K3$ surface is an elliptic fibration over $\mathbb P^1$, obtained as above.

Now, let's consider the more specific case of the Fermat's quartic $S:\{x^4-y^4-z^4+t^4=0\}$ in $\mathbb P^3$ (it is the standard Fermat's quartic up to multiplying $y$ and $z$ by a $4$th root of $-1$). Then, we can factorize it in the following way $$(x^2+y^2)(x^2-y^2)-(z^2+t^2)(z^2-t^2)=0.$$ This shows that, for $[\lambda:\mu]\in\mathbb P^1$, the complete intersection given by $$C_{[\lambda:\mu]}:=\begin{cases} \lambda(x^2-y^2)=\mu(z^2+t^2) \\ \mu(x^2+y^2)=\lambda(z^2-t^2) \end{cases}$$ is contained in $S$. For generic $[\lambda:\mu]\in\mathbb P^1$, this is a smooth elliptic curve, since its tangent bundle fits in the following short exact sequence $$0\to T_{C_{[\lambda:\mu]}}\to T_{\mathbb P^3}|_{C_{[\lambda:\mu]}}\to\mathcal O_{C_{[\lambda:\mu]}}(2)\oplus\mathcal O_{C_{[\lambda:\mu]}}(2)\to 0.$$ The function $[\lambda:\mu]$ defines a map from $S$ onto $\mathbb P^1$, which is the elliptic pencil on $S$ you were looking for.

Note that, for $\lambda/\mu=0,\pm 1,\pm i,\infty$, $C_{[\lambda:\mu]}$ degenerates into a cycle of four lines. This gives you the 24 singularities.

• Good! So is this answer satisfactory or you wanted to know more specific things? Commented Feb 7, 2012 at 9:40
• that is great~ thanks! by the way, do you know anything about the existence of sections? maybe just a topological section...
– Jay
Commented Feb 8, 2012 at 5:11
• @Jay: regarding sections, look at my answer below. Commented Nov 13, 2013 at 18:54
• @diverietti Why is the curve not allowed to have genus $1$? The canonical bundle of an elliptic curve is trivial, so all the holomorphic forms on it are constant... Commented Aug 21, 2020 at 1:05
• I don't understand. The general fibre has genus one, indeed. Commented Aug 24, 2020 at 13:03

Here is an alternative way to construct elliptic fibrations of quartic surfaces containing a line.

Let $X\subset \mathbb P^3$ be a smooth quartic surface and assume that it contains a line $L\subset X$. Let $\mathfrak d$ denote the $1$-dimensional linear system of hyperplanes in $\mathbb P^3$ containing $L$ restricted to $X$. Further let $H\in \mathfrak d$ and $E=H-L$.

Observe that $E^2=H^2-2H\cdot L L^2= 4 - 2\times 1 -2=0$ and that by construction $\dim |E|\geq 1$. It follows that for any $E_1,E_2\in |E|$, $E_1\cap E_2=\emptyset$ and hence $|E|$ is basepoint-free and $\dim |E|=1$.

Therefore $|E|$ defines a morphism $f: X\to \mathbb P^1$ and it follows from the construction that the fibers of $f$ are cubic plane curves, so this is indeed an elliptic fibration.

Remark 1: One can get more data out of this construction.

1. Since each fiber is a plane cubic, this construction limits the possible singularities of the fibers quite a bit.
2. The line we started with is a triple section.
3. If $X$ contains another line, say $C$, such that $C\cap L=\emptyset$ then $C$ gives a section of $f$: Indeed if $C$ is a line, then $C\cdot H=1$ and if $C\cap L=\emptyset$, then $C\cdot E=1$.
4. If $X$ contains another line, say $C$, such that $C\cap L\neq\emptyset$, then $C$ is contained in a fiber intersecting $L$ exactly once.
5. You can keep playing with this to get more.

Remark 2: For an elliptic K3 the Picard number is at least $2$ and if it is exactly $2$, then it admits at most two different elliptic fibrations. If the Picard number is at least $3$, then an elliptic K3 may admit infinitely many elliptic fibrations, so obviously in many cases you don't get all the elliptic fibrations this way.

Remark 3: Finally, note that the Fermat quartic contains lines: Using diverietti's notation the 16 lines $$\ell_{\xi,\xi'}=\big\{[a:\xi a:b:\xi' b] \ \vert\ [a:b]\in \mathbb P^1\big\}$$ where $\xi,\xi'$ are 4$^\text{th}$ roots of unity all lie on that Fermat quartic, so one obtains 16 different elliptic fibrations this way.

Further note that if $\xi\neq \xi''$ and $\xi'\neq \xi'''$, then $\ell_{\xi,\xi'}\cap \ell_{\xi'',\xi'''}=\emptyset$, so this way we get a couple of sections for each of the elliptic fibrations as described above.