\(U \oplus A_{2}(-2)\)

Let \(\Lambda = \mathrm{II}_{3,19}\) be the K3 lattice and \(G \subset O(\Lambda)\) a finite group of symplectic isometries. Row 48 of Table 10.3 of Hashimoto gives the invariant lattice \(\Lambda^G\) for the groups \(\mathfrak{G}_{48} \cong \mathfrak{S}_{3,3}\), \(\mathfrak{G}_{31} \cong C_3 \times D_6\) of Table 10.2, and this record is the lattice \(M\) with \(\Lambda^G = M(-3)\); the corpus records \(M\) and not its twist.

Roots: write \(r = a e_{1} + c e_{2} + w\) with \(w\) in the span \(W\) of \(e_{3}\) and \(e_{4}\). \(b(e_{1}, e_{2}) = -1\) is odd, \(W\) is orthogonal to \(e_{1}\) and \(e_{2}\), and \(b(w, w) \in 4\mathbb{Z}\) for \(w\) in \(W\), because the Gram tensor of \(W\) has its diagonal components in \(4\mathbb{Z}\) and the others in \(2\mathbb{Z}\). So \(b(r, e_{2}) = -a\), \(b(r, e_{1}) = -c\) and \(b(r, r) = -2 a c + b(w, w)\). If \(a\) is odd and \(c\) is even, then \(b(r, r) \equiv 0 \bmod 4\), and \(b(r, r)\) divides \(2 b(r, e_{2}) = -2 a \equiv 2 \bmod 4\), which is impossible; the same holds with \(a\) and \(c\) exchanged. Each root is therefore in \(L' = \{a \equiv c \bmod 2\}\), which has index 2 in \(L\). \(e_{1} + e_{2}\) has \(b(r, r) = -2\) and \(b(r, L) = \mathbb{Z}\), and \(-2\) divides \(2\). \(e_{1} - e_{2}\) has \(b(r, r) = 2\) and \(b(r, L) = \mathbb{Z}\), and \(2\) divides \(2\). \(e_{3}\) and \(e_{4}\) have \(b(r, r) = -4\) and \(b(r, L) = 2\mathbb{Z}\), and \(-4\) divides \(4\). These roots are in \(L'\), and their invariant factors in \(L\) are \(1, \dots, 1, 2\), so they generate \(L'\): \(\mathbb{Z}\Phi(L) = L'\), of index 2 in \(L\).