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Consider a finite field $\mathbb{F}_d$ of order $d$, and let the vector space $V=\mathbb{F}_d^n$. Let $\mathbf{Gr}(m,V)$ be the Grassmanian containing all subspaces in $V$ of dimension $m$. Suppose $S\subset \mathbf{Gr}(m,V)$ is such that the union of all subspaces in $S$ is all of $V$. I would like to determine the smallest possible size of $S$.

When $m=1$, the answer is $\frac{d^n-1}{d-1}$ because any two pairs of lines intersect only at the origin. In fact, I conjecture that when $m|n$, we will be able to find $\frac{d^n-1}{d^m-1}$ subspaces that pairwise only intersect at the origin. More generally, $\frac{d^n-1}{d^m-1}$ is not an integer, but it will be a lower bound on the answer. Any thoughts on how to proceed with this problem?

Damalone
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  • A similar simple counting argument shows that when $m=n-1$, we need $|S|\ge d+1$. That bound is achievable by a set consisting of all the subspaces of dimension $n-1$ containing a given subspace of dimension $n-2$. Basically we are then covering a projective line one point at a time. – Jyrki Lahtonen Jun 20 '24 at 06:36
  • Doesn't your conjecture follow from the fact that when $m\mid n$ we can think of $V$ as a $k$-dimensional space over the extension field $\Bbb{F}{d^m}$ ($k=n/m$)? Again, we are covering a projective space over $\Bbb{F}{d^m}$ one point at a time. – Jyrki Lahtonen Jun 20 '24 at 06:42

1 Answers1

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The question is investigated in the paper: A. Beutelspacher, "On t-covers in finite projective spaces", J.Geom. 12:10–16 (1979), DOI: 10.1007/BF01920229.

Beutelspracher proves that the number of $t$-dimensional projective spaces needed to cover a $d$-dimensional projective space over the field $GF(q)$ with $q$ elements is equal to $$q^{b + 1} \sum_{i = 0}^{a-1} q^{i(t+1)} + 1 = q^{b + 1} \cdot \frac{q^{a(t + 1)} - 1}{q^{t + 1} - 1} + 1$$ where $a = \lfloor\frac{d}{t + 1}\rfloor$ and $b = d - a(t + 1)$ are the quotient and the remainder of division of $d$ by $t + 1$.

Note that the dimensions are projective, so they differ by $1$ from dimensions in your question. In particular, if $d + 1$ is divisible by $t + 1$, then the number coincides with $\dfrac{q^{d + 1} - 1}{q^{t + 1} - 1}$. In this case there indeed exists a cover by pairwise non-intersecting subspaces, which can be constructed as follows: consider the $(d + 1)$-dimensional affine space as an affine space over $GF(q^{t + 1})$ and cover it in a trivial way with lines over $GF(q^{t + 1})$, which are $t + 1$-dimensional affine spaces over $GF(q)$.

Vladimir Lysikov
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