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I have a question and I am stuck. I was wondering if anyone has a thought, before I start a brute-force search.

For $q$ a prime number and $n =6$, let $\mathbb {F}_{q}^{n}$ be an $n$-dimensional vector space over $\mathbb{F}_{q}$. Furthermore, let $ U_1, \dots, U_m$ be a family of $2$-dimensional subspaces of $\mathbb{F}_{q}^n$ such that $U_i \cap U_j = \{0\}$ and $\langle U_i, U_j \rangle \cap U_k = \{0\}$, for all $i, j, k \in \{1,\dots, n\}$, $i\neq j \neq k$. What is the biggest possible $m$?

Thanks in advance.

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Each $1$-dimensional subspace is contained in at most one $U_i$. This gives the upper bound $m \leq q^3 + 1$. – azimut Jul 4 '13 at 18:10
@JyrkiLahtonen: Thanks for your comment. You are right, $(q^6 - 1)/(q^2 - 1) = q^4 + q^2 + 1$ is the upper bound we get in this way. I did the wrong division $(q^6 - 1)/(q^3 - 1)$. What a shame that I cannot edit the other comment any more. – azimut Jul 4 '13 at 18:44
Don't worry about the old comment. The idea is fine. But I don't think it is sharp, because once we have picked two 2-d subspaces, their direct sum contains other 1-d subspaces (quite a few them actually), and these will then be off limits. But then it gets messy. – Jyrki Lahtonen Jul 4 '13 at 18:54
There is an easy construction of $q^2+1$ subspaces: The 1-d spaces $U_\alpha$ over $K=\mathbb{F}_{q^2}$ generated by a vector of the form $(1,\alpha,\alpha^2)$ with $\alpha\in K$, together with $U_\infty$ generated by $(0,0,1)$. Doesn't feel optimal. Viewing $\mathbb{F}_q^6$ as $K^3$ here obviously. – Jyrki Lahtonen Jul 4 '13 at 18:58
@JyrkiLahtonen: Yeah, I just had the same idea :) For $q$ even you can improve the lower bound to $q^2 + 2$, by taking a hyperoval. – azimut Jul 4 '13 at 19:02

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