# How to partition a finite vector space into affine subspaces all of the same dimension

Given an $n$-dimension vector space $V$ over a finite field $\mathbb F_q$ and a natural number $d<n$, the goal is to write $V$ as disjoint union of $d$-dimensional affine subspaces $v_i+V_i$: $$V = \stackrel{.}{\cup}_i (v_i + V_i)\quad \mbox{ where } v_i\in V \mbox{ and } V_i\le V \mbox{ with } \dim_{\mathbb F_q} V_i = d$$

One way to obtain such a partition with not all $V_i$ being the same is to pick a hyperplane $H\le V$ (i.e., $\dim_{\mathbb F_q} H = n-1$), write $V = \stackrel{.}{\cup}_i (v_i + H)$, and refine this partition by partitioning each $v_i+H$ independently in a recursive way.

Are there partitions that cannot be derived in this way?

(If necessary you may assume $q=2$, as this is the case I'm mostly interested in.)

• What's that $\;q\;$ at the end? Perhaps $\;q=n\;$ ? Or else $\;q=d\;$ ? Nov 20, 2014 at 11:56
• @Timbuc: The order of the field.
– j.p.
Nov 20, 2014 at 12:56

## 1 Answer

No. With $q=2$ every two element subset $\{x,y\}$ is a coset of the subspace $\{0,x-y\}$. And you can partition a 3-dimensional space over $\Bbb{F}_2$ into four such doublets in such a way that no two of them form an affine 2-space. For example $$\{000,100\}, \{010,011\}, \{001,110\}, \{111,101\}.$$

• Thanks, this helps even for the case $d>1$ I'm normally considering: One just has to factor out a $(d-1)$-dimensional (common) subspace, and take the preimages of your solution.
– j.p.
Nov 20, 2014 at 13:01
• Exactly! I was prepared to give that recipe, if you had asked about it :-) Nov 20, 2014 at 13:04
• Do you happen to know something about the case $q>2$?
– j.p.
Nov 20, 2014 at 15:02
• Can you please elaborate more on your argument "One just has to factor out a (𝑑−1)-dimensional (common) subspace, and take the preimages of your solution"? Thanks in advance Nov 11, 2022 at 11:48