Counting zero-sum subspaces for the multiplicative inverse function

Abstract

Let $q=2^n$ with $n\ge 6$, and let ${\mathbb{F}}_q$ be the finite field with $q$ elements. For a subspace $E$ of ${\mathbb{F}}_q$, define \[ S(E)=\sum_{x\in E\setminus\{0\}}x^{-1}.\] Let $N_{n,k}$ denote the number of $k$-dimensional ${\mathbb{F}}_2$-subspaces $E$ for which $S(E)=0$. We prove that, if $k\ge 3$ and $n\ge 2k+1$, then \[ \left|\frac{2^nN_{n,k}}{\genfrac{[}{]}{0pt}{}{n}{k}_{2}}-1\right|<2^{r^2+r}2^{n(1-r/2)},\] where $r=\lfloor \frac{k-1}{2}\rfloor$, and ${\left[\genfrac{}{}{0ex}{}{n}{k}\right]}_2$ is the Gaussian binomial coefficient. In particular, \[ N_{n,k}\sim2^{-n}\genfrac{[}{]}{0pt}{}{n}{k}_{2}\quad (n\to\infty)\] uniformly for $n/3<k\le (n-1)/2$. Combined with the known low-dimensional cases and the symmetry between dimensions $k$ and $n-k$, and the elementary middle-dimensional subfield construction, this proves a conjecture of Carlet: for every $3\le k\le n-3$, there exists a $k$-dimensional ${\mathbb{F}}_2$-subspace $E$ such that $S(E)=0$.

arXiv comment17 pages