Curve through the 16 singular points of a Kummer surface

Question

Let $X$ be an abelian surface over $\mathbb{C}$. Consider the Kummer surface $K$ associated to $X$, that is the quotient of $X$ by the action of involution on $X$, $x\mapsto -x$. Kummer surface is a singular surface with 16 nodes. Does this mean that every curve through the 16 nodes is a singular curve? If we start with a smooth curve on a the abelian surface through the sixteen 2-torsion points, is it not isomorphic to its image? Thanks!

Answer 1

Regarding your first question, the answer is no.

Following J. Silverman's suggestion, let me provide an example of an abelian surface $A$ and a smooth curve $C$, fixed by the involution $(-1)_A$, and containing all the $2$-torsion points of $A$. Therefore, the image $C'$ of $C$ in $K := \textrm{Kum}(A)$ will be a smooth curve through the $16$ nodes of $K$.

Let $(A, \, \mathcal{L})$ be a $(1, \, 2)$-polarized abelian surface. We can assume with no loss of generality that $\mathcal{L}$ is a symmetric polarization, i.e. that $(-1)_A^* \mathcal{L} = \mathcal{L}$. Then it is possible to show what follows:

• $\dim H^0(A, \, \mathcal{L}^2)=8;$
• $\dim H^0(A, \, \mathcal{L}^2)^+=6\,$ and $\, \dim H^0(A, \, \mathcal{L}^2)^-=2,$ where "+" and "-" denote the space of $(-1)_A$-invariant and $(-1)_A$-anti-invariant sections, respectively;
• the pencil $|\mathcal{L}^2|^-$ of anti-invariant sections has $16$ base points, namely the $16$ points of 2-torsion of $A$, and the general element of this pencil is smooth at the base points, hence smooth everywhere by Bertini theorem.

Since the zero locus of every invariant or anti-invariant section is fixed by $(-1)_A$, we can take as $C$ a general element of $|\mathcal{L}^2|^-$.

Note that the involution on $A$ induces a double cover $C \to C'$ ramified at $16$ points. Since $C^2=16$, by the genus formula it follows $g(C)=9$; thus the Riemann-Hurwitz formula implies $g(C')=1$.

References.

W. Barth: Abelian surfaces with $(1,\, 2)$-polarization, in Algebraic Geometry, Sendai, 1985, Adv. Stud. Pure Math. 10, North-Holland, Amsterdam, 1987, 41–84.

See in particular Section 5.

Answer 2

To answer your second question, no, $C$ does not have to be isomorphic to its image. First, if the image is singular, you may have the common situation of a map $C\to C'$ that is bijective on points, but not an isomorphism, because $C'$ is singular. (Example: $\mathbb P^1$ mapping to the cuspidal cubic $y^2=x^3$.) Second, suppose that you take a smooth curve $C$ on $X$ with the property that the map $x\to-x$ maps $C$ to itself. Then the map $C\to C'$ will be a double cover. I think you can probably find such a curve $C$ that goes through the 16 points fixed by the involution, which will give you an example where the image curve $C'$ is smooth.