New hemisystems of the Hermitian surface

Abstract

Constructing hemisystems of the Hermitian surface is a well known, apparently difficult, problem in Finite geometry. So far, a few infinite families and some sporadic examples have been constructed. One of the different approaches relies on the Fuhrmann-Torres maximal curve and provides a hemisystem in $PG(3,p^2)$ for every prime p of the form $p=1+4a^2$, a even. Here we show that this approach also works in $PG(3,p^2)$ for every prime $p=1+4a^2$, a odd. The resulting hemisystem gives rise to two weight linear codes and strongly regular graphs whose properties are also investigated.

Appendix A

We provide a proof of Proposition 6.7. Since our proof relies on cyclotomic fields from algebraic number theory, we present it in the form of an appendix.

Let ${\mathbb {Q}}(\zeta _m)$ the cyclotomic field of mth roots of unity with $\zeta _m=e^{2\pi i /m} \in {\mathbb {C}}$. In particular, the cyclotomic field ${\mathbb {Q}}(\zeta _{16})$ contains $\sqrt{2}$ as an integer. Let ${\mathfrak {b}}$ a prime ideal of ${\mathbb {Q}}(\zeta _{16})$ such that ${\mathfrak {b}}$ contains p (i.e. ${\mathfrak {b}} \mid p$). The extension ${\mathfrak {b}} \mid p$ is unramified and ${\mathbb {Z}}[\zeta _{16}]/{\mathfrak {b}} \cong {\mathbb {F}}_{p^4}$; see [12, Proposition 13.2.5] and [11, Section 4.5]. Note that $h=\pm \sqrt{2} \pmod {{\mathfrak {b}}}$. We may assume $h \equiv \sqrt{2} \pmod {{\mathfrak {b}}}$.

Proof of Proposition 6.7

We do the computation for $q \equiv 13 \pmod {16}$, the proofs for the other cases being analogous.

$$\begin{aligned} \begin{aligned} (1+h)^\frac{q+1}{2}h^\frac{q-1}{2}&\equiv (1+\sqrt{2})^\frac{q+1}{2} (\sqrt{2})^\frac{q-1}{2} \pmod {{\mathfrak {b}}} \\&=(\sqrt{2}+2)^\frac{q+1}{2} \frac{1}{\sqrt{2}}\\&=(\zeta _8+\zeta _8^{-1}+2)^\frac{q+1}{2}\frac{1}{\sqrt{2}}\\&=(\zeta _{16}+\zeta _{16}^{-1})^{q+1}\frac{1}{\sqrt{2}}\\&\equiv (\zeta _{16}+\zeta _{16}^{-1})(\zeta _{16}^{13}+\zeta _{16}^{-13}) \frac{1}{\sqrt{2}} \pmod {{\mathfrak {b}}}\\&\equiv (\zeta _{16}+\zeta _{16}^{-1})(\zeta _{16}^{-3}+\zeta _{16}^{3}) \frac{1}{\sqrt{2}} \pmod {{\mathfrak {b}}}\\&=(\zeta _{16}^4+\zeta _{16}^{-2}+\zeta _{16}^2+\zeta _{16}^{-4})\frac{1}{\sqrt{2}}\\&=(\zeta _8+\zeta _8^{-1})\frac{1}{\sqrt{2}}=1 \square \end{aligned} \end{aligned}$$