4
$\begingroup$

Assume we are given a simple abelian surface $A$ which has 2 non-equivalent principal polarizations $D_1$ and $D_2$ in $NS(A)$ (up to isomorphism), thus giving rise to two non-isomorphic smooth projective curves of genus 2 (we work over $\mathbb{C}$).

Each $D_i$ gives a map $|2D_i|: A \rightarrow \mathbb{P}^3$ which factors via an embedding $\iota_i: A/(-1) \hookrightarrow \mathbb{P}^3$.

Thus we can embed the Kummer surface $Km(A):=A/(-1)$ in two different ways into $\mathbb{P}^3$ giving us two quartic surfaces $S_1$ and $S_2$ in $\mathbb{P}^3$ with sixteen nodes.

$\textbf{Question:}$ What is the relationship between these two surfaces? Are they Cremona equivalent / isomorphic or are they even projectively equivalent?

Can we find $f\in Bir(\mathbb{P}^3)$ such that $f$ induces a birational map $\tilde{f}: S_1 \rightarrow S_2$ (Cremona equivalent) or even an isomorphism $\hat{f}: S_1\rightarrow S_2$ (Cremona isomorphic). Or is it possible to find $g\in Aut(\mathbb{P}^3)$ such that $g$ restricts to an isomorphism $\tilde{g}: S_1 \rightarrow S_2$ (projectively equivalent)?

If nothing of the above is possible, is there any other interesting relation between these surfaces in $\mathbb{P}^3$?

$\endgroup$
1
  • $\begingroup$ $S_1$ and $S_2$ are certainly not projectively equivalent since the linear systerms on $A$ are not the same. Whether or not there exists an $f$ is an interesting question; I had thought about this some years ago but was not able to reach any conclusion. $\endgroup$
    – naf
    Sep 22, 2017 at 5:18

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy