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Let $p$ be an odd prime and let $\mathbb{F}_{p}^{n}$ be the $n$-dimensional vector space over the field of $p$ elements.

Consider a subset $A\subseteq \mathbb{F}_{p}^{n}$ with density at least $\epsilon$, i.e. $\frac{|A|}{p^{n}}\geq \epsilon$. An affine subspace $V\subseteq\mathbb{F}_{p}^{n}$ is called full (resp. $A$) if the density of $A$ in $V$ is at least $\epsilon/2$, i.e. $\frac{|A\cap V|}{|V|}\geq \epsilon/2$.

I want to prove that the density of full $k$-dimensional affine subspaces in the $k$-dimensional affine subspaces is at least $\epsilon/2$ (An affine subspace of $\mathbb{F}_{p}^{n}$ is a coset of a subspace of $\mathbb{F}_{p}^{n}$). Formally, consider the classes $\mathcal{A}_{k}$ and $\mathcal{F}_{k}$, of the $k$-dimensional affine subspaces and of the full $k$-dimensional subspaces of $\mathbb{F}_{p}^{n}$, respectively. We want to prove that $\frac{|\mathcal{F}_{k}|}{|\mathcal{A}_{k}|}\geq\epsilon/2$.

$|\mathcal{A}_{k}|$ is computed as follows: \begin{equation} |\mathcal{A}_{k}|=p^{n-k}\binom{n}{k}_{p} \end{equation} where $\binom{n}{k}_{p}$ is the Gaussian coefficient, which is defined as \begin{equation} \binom{n}{k}_{p}=\frac{(p^{n}-1)(p^{n}-p)\cdots(p^{n}-p^{k-1})}{(p^{k}-1)(p^{k}-p)\cdots (p^{k}-p^{k-1})} \end{equation} and it holds that it is equal to the number of $k$-dimensional subspaces of $\mathbb{F}_{p}^{n}$ (it is proved, for example, in this question).

I have no clue how to compute $|\mathcal{F}_{k}|$. Any ideas?

richarddedekind
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2 Answers2

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For a given $k$-dimensional subspace of $\mathbb F_p^n$, consider its $p^{n-k}$ cosets, whose union is $\mathbb F_p^n$. In the worst case, $m$ of them have density $1$ and $p^{n-k}-m$ have density just below $\frac\epsilon2$. Then

$$ \left(m+\left(p^{n-k}-m\right)\frac\epsilon2\right)p^k\gt\epsilon p^n $$

and thus

$$ m\gt p^{n-k}\frac{\epsilon-\frac\epsilon2}{1-\frac\epsilon2}=p^{n-k}\frac{\epsilon}{2-\epsilon}\gt p^{n-k}\frac\epsilon2\;. $$

Thus, at least a fraction $\frac\epsilon2$ of the cosets of each subspace are full, and hence also a fraction $\frac\epsilon2$ of all $k$-dimensonal affine subspaces.

joriki
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Define $\mathcal{A}_k=\{V\subseteq\mathbb{F}_p^n \mid V \ \text{is a $k$-dimensional affine subspace of }\mathbb{F}_p^n \}$.

Note that $|V|=p^k$ if $V\in \mathcal{A}_k$. Call $x$ the fraction of $\mathcal{A}_k$ of $k$-subspaces which are not full, so that $1-x$ is the fraction of full $k$-subspaces. Consider now all the partitions of $\mathbb{F}_p^n$, made of distinct elements of $\mathcal{A}_k$. Each of these partitions has exactly $p^{n-k}$ elements, so we have exactly $|\mathcal{A}_k|/p^{n-k}$ of these partitions. We can then write $\sum_{V\in\mathcal{A}_k}|A\cap V|=\frac{|A||{\mathcal{A}_k}|}{p^{n-k}}$ since if $(P_i)_{i=1}^{p^{n-k}}$ is a partition of $\mathbb{F}_p^n$, then $(A\cap P_i)_{i=1}^{p^{n-k}}$ is a partition of $A$. Using the density of $A$ and by the definition of $x$ we obtain inequalities for both members

\begin{gather*} \sum_{\substack{V\in\mathcal{A}_k \\ V \ \text{is full}}}|A\cap V|+\sum_{\substack{V\in\mathcal{A}_k \\ V \ \text{is not full}}}|A\cap V|=\sum_{V\in\mathcal{A}_k}|A\cap V|=\frac{|A||\mathcal{A}_k|}{p^{n-k}}\ge\frac{\epsilon p^n|\mathcal{A}_k|}{p^{n-k}}\\ \implies (1-x)|\mathcal{A}_k| p^k+x|\mathcal{A}_k|(\epsilon/2)p^k\ge\epsilon |\mathcal{A}_k| p^k\\ \implies (1-x)+x(\epsilon/2)\ge\epsilon \end{gather*}

from which follows that $x\le\frac{1-\epsilon}{1-(\epsilon/2)}$ and hence $1-x\ge1-\frac{1-\epsilon}{1-(\epsilon/2)}=\frac{\epsilon/2}{1-\epsilon/2}\ge\frac{\epsilon}{2}$, since $\epsilon<1$.

richarddedekind
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mazz096
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