Hyper-reguli in PG(5,q)
classification
🧮 math.CO
keywords
exactlymathbbplanessetsswitchingandrargumentcounting
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A simple counting argument is used to show that for all $q$, an Andr\'e hyper-regulus $\mathbb X$ in $PG(5,q)$ has exactly two switching sets. Moreover, there are exactly $2(q^2+q+1)$ planes in $PG(5,q)$ that meet every plane of $\mathbb X$ in a point, namely the planes in the switching sets.
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