Universal covers of dimensional dual hyperovals (Q488291)

A set \(\mathcal S\) of \(d\)-spaces of \(\mathrm{PG}(n,q)\) is a \(d\)-dimensional dual hyperoval of \(\mathrm{PG}(n,q)\) if every point of \(\mathrm{PG}(n,q)\) belongs to either \(0\) or \(2\) members of \(\mathcal{S}\), any two members of \(\mathcal{S}\) have precisely a point in common and the set of points belonging to the members of \(\mathcal{S}\) spans \(\mathrm{PG}(n,q)\). When \(n = 2\) and \(d=1\), we get the classical definition of dual hyperoval in a projective plane. Let \(\mathcal{S}_i\) be a \(d\)-dimensional dual hyperoval of \(\mathcal{P}_i\), \(i = 1,2\). If there exists a surjective map \(\pi: \mathcal{P}_1 \rightarrow \mathcal{P}_2\) such that \(\pi(\mathcal{S}_1) = \mathcal{S}_2\), then \(\mathcal{S}_2\) is a quotient of \(\mathcal{S}_1\), or \(\mathcal{S}_1\) is a cover of \(\mathcal{S}_2\). A cover is said to be proper if \(\pi\) is not an isomorphism. A \(d\)-dimensional dual hyperoval is said to be universal if it has no proper covers. In [Geom. Dedicata 79, No. 2, 157--178 (2009; Zbl 0948.51008)], \textit{A. Del Fra} observes that the dimension of the projective space containing a \(d\)-dimensional dual hyperoval is upperly bounded by a function of \(d\) and \(q\). It follows that, if \(\mathcal{S}\) is a \(d\)-dimensional dual hyperoval, then \(\mathcal{S}\) is the quotient of a universal \(d\)-dimensional dual hyperoval. In this note, the author proves that this universal example is uniquely determined by \(\mathcal{S}\).