Maximum subsets of ${\mathbb{F}}_q^n$ containing no right angles

Recently, Croot, Lev, and Pach (Ann. of Math., 185:331–337, 2017.) and Ellenberg and Gijswijt (Ann. of Math., 185:339–443, 2017.) developed a new polynomial method and used it to prove upper bounds for three-term arithmetic progression free sets in ${\mathbb{Z}}_4^n$ and ${\mathbb{F}}_3^n$, respectively. Their approach was later summarized by Tao and is now known as the slice rank method. In this paper, we apply this method to obtain a new upper bound on the cardinality of subsets of ${\mathbb{F}}_q^n$ which contain no right angles. More precisely, let $q$ be a fixed odd prime power and $x\cdot y$ be the standard inner product of two vectors $x,y\in {\mathbb{F}}_q^n$, we prove that the maximum cardinality of a subset $A\subseteq {\mathbb{F}}_q^n$ without three distinct elements $x,y,z\in A$ satisfying $(z-x)\cdot (y-x)=0$ is at most $\left(\genfrac{}{}{0px}{}{n+q}{q-1}\right)+3$. For sufficiently large $n$, our result significantly improves the previous upper bound of Bennett (European J. Combin., 70:155–163, 2018.), who showed that $\mathrm{\mid }A\mathrm{\mid }=\mathcal{O}(q^{\frac{n+2}{3}})$.