Classification of order sixteen non-symplectic automorphisms on \(K3\) surfaces (Q2835286)

This paper classifies complex \(K3\) surfaces with non-symplectic automorphisms of order \(16\) in full generality.NEWLINENEWLINELet \(X\) be a \(K3\) surface, \(\omega_X\) a generator of \(H^{2,0}(X)\), \(\sigma\) an order \(16\) automorphism such that \(\sigma^*\omega_X=\zeta_{16}\omega_X\), where \(\zeta_{16}\) denotes a primitive \(16\)-th root of unity. It is assumed throughout the paper that \(\sigma^8\) acts as the identity on \(\mathrm{Pic}(X)\). Under this situation, the rank of \(\mathrm{Pic}(X)\) can only be \(6\) or \(14\).NEWLINENEWLINEThe first results are about the fixed locus of \(\sigma\) and that of \(\sigma^4\): if \(\mathrm{Fix}(\sigma)\) contains a curve, its genus is zero, and \(\mathrm{Fix}(\sigma^4)\) always contains at least a curve of genus \(0\) or \(1\).NEWLINENEWLINEIf \(\mathrm{rk}\, \mathrm{Pic}(X)=6\), there are two possibilities: the lattice structure of \(\mathrm{Pic}(X)\), the number \(N\) of \(\sigma\)-fixed points, and the number \(k\) of \(\sigma\)-fixed rational curves are classified as NEWLINE\[NEWLINE(\mathrm{Pic}(X),N,k)=(U\oplus D_4, 6,1),\quad\text{{or}}\quad (U(2)\oplus D_4, 4,0).NEWLINE\]NEWLINENEWLINENEWLINEIf \(\mathrm{rk}\, \mathrm{Pic}(X)=14\), there are five possibilities: the number \(N\) of isolated \(\sigma\)-fixed points and the number \(k\) of \(\sigma\)-fixed rational curves are \((N,k)=(8,1)\) or \((6,0)\), and if \(\mathrm{Fix}(\sigma^4)\) contains at least a curve and its genus is at most \(0\), there are three cases for \((\mathrm{Pic}(X),N,k)\): NEWLINE\[NEWLINE(\mathrm{Pic}(X),N,k)=(U\oplus D_4\oplus E_8, 12,1),(U(2) \oplus D_4\oplus E_8, 4,0), (U(2)\oplus D_4\oplus E_8,10,1).NEWLINE\]NEWLINENEWLINENEWLINEThese classification results imply the non-existence of a \(K3\) surface \(X\) with \mathrm{Pic}ard number \(14\) with a non-symplectic automorphism of order \(16\) acting trivially on \(\mathrm{Pic}(X)\). Also it is shown that if the action of the automorphism of order \(16\) is trivial on \(\mathrm{Pic}(X)\), then the \mathrm{Pic}ard number must be \(6\).