Abelian surfaces can be defined abstractly. Using an ample divisor we can embed it into projective space.
Abelian surfaces could be product of elliptic curves.
Abelian surface can appear as Jacobian of a genus 2 curve, might look like
\[C: y^2 + g(x)y=f(x)\]
with \(\deg f\in \{5,6\}\), \(\deg g\le 3\).
Then the \(\mathrm{Jac}(C) = \mathrm{Div}^0(C)/\sim\) is an abelian surface.
(Flynn 98) When \(\mathrm{char}(K)\neq 2\), \(\mathrm{Jac}(C)\hookrightarrow \mathbb{P}^{15}\) defined by 72 quadrics.
How do we work with something like this? We don’t work with Jacobian directly but we work with a Kummer surface.
\[\tau:A\to A: a \mapsto -a\]
We can consider the quotient \(\Kum(A) := A/\tau\).
If \(A=\mathrm{Jac}(C)\), then we can embed \(\Kum(A) \hookrightarrow \mathbb{P}^3\) defined by \(1\) quartic.
We can
Add points (on an elliptic curve)
Compute isogenies
Compute the Mordell-Weil group
Compute the descent
Somehow, Kummer surfaces allow us to understand abelian surfaces. The goal of this talk is rather to go in the opposite direction. How abelian surfaces allow us to study K3 surfaces.
A K3 surface is a smooth surface \(X\) satisfying
\(\omega_X\cong \mathcal{O}_X\)
\(h^1(X,\mathcal{O}_X)=0\)
A very large class of K3 surfaces are quartics in \(\mathbb{P}^3\). Kummer surfaces are K3 surfaces (\(\mathrm{char}K\neq 2\)).
When is the quotient of an abelian surface a K3 surface?
Solved by Katsura in 1987 through the following theorem:
Theorem (Katsura 87)
Suppose \(A\) is an abelian surface and
\[G\subset \mathrm{End}(A)^\times = \{ \phi : A\to A \text{ is an invertible endomorphism} \}\]
Then \(A/G\) is a K3 surface if and only if the following 3 conditions hold:
\(G\) acts symplectically on \(A\), for all \(g\) we have \(g^* \omega_A = \omega_A\).
\(G\) acts rigidly, \(|A^g|<\infty\).
Singularities of \(A/G\) are nice (ADE singularities). (This ensures that when you blow up the singularity, the canonical divisor does not change.)
He provided a full list of \(G\) such that there exists an \(A\) with \(A/G\) a K3 surface.
\(G=C_n\) where \(n\in \{2,3,4,5,6,8,10,12\}\)
Other groups
(only true when \(\mathrm{char}K \not| |G|\))
If we have \(g\in G\) such that \(|g|=p^n\) then \(A^g \le A[p]\). In \(\mathrm{char}(K)\neq p\), the \(p\) torsions \(A[p]=(\mathbb{Z}/p\mathbb{Z})^4\). When \(\mathrm{char}(K)=p\) we have \(A[p]=(\mathbb{Z}/p\mathbb{Z})^f\) where \(0\le f\le 2\).
For the elliptic curve \(E_{1728}:y^2=x^3+x\), we have \(\mathrm{End}(E_{1728})=\mathbb{Z}[\sqrt{-1}]\).
Consider
\[\mathrm{End}(E_1\times E_2) =\begin{cases} \mathrm{End}(E_1)\times \mathrm{End}(E_2) & E_1\not\simeq E_2 \\ ? & E_1\simeq E_2 \\ M_2(\mathrm{End}(E)) & E_1 = E_2 \end{cases} \]
\(M_2\) acts symplectically and rigidly on \(E\times E\), equivalent to say the eigenvalues of \(M\) are \(\{\zeta_n^{\pm 1}\}\) for some \(n\).
For example for
\(M_2 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\)
we have \((P,Q)\mapsto (P,-Q)\).
and for
\(M_3 = \begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}\)
its eigenvalues are \(\zeta_3^{\pm 1}\).
Now consider \(E\times E\to E\times E\) given by \((P,Q)\mapsto (Q, -P-Q)\). If you observe what it means, note that \(P,Q,-P-Q\) are collinear. This is a symplectic action and it just permutes on three points.
Whose fixed points are of the form \(\{(P,P): P\in E[3]\}\).
When we consider \(M_4\), we have \(M_4 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\)
But on \(E_{1728}\) has complex multiplication by \(\sqrt{-1}\), we can use
\(M_4 = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}\)
and this seems very interesting.
The goal is to completely classify K3 quotients of the form \(A/G\).
Strategy is to find all possible group actions \(G\mathrel{\circlearrowright}A\) (complete)
Then construct explicit examples of \(A\mathrel{\circlearrowright}G\) such that \(A/G\) is a K3 surface.