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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

Let $\mathcal{U}$ be a set of planes in $\Bbb{F}_q^6$ such that for any three planes $U_1,U_2,U_3\in\mathcal{U}$ we have $$U_i\cap U_j=\{0\}\qquad\text{ and }\qquad\langle U_i,U_j\rangle\cap U_k=\{0\}.$$ Then $\langle U_i,U_j,U_k\rangle=\Bbb{F}_q^6$, so no three planes in $\mathcal{U}$ are contained in a single $5$-dimensional subspace. Every pair $U_i,U_j\in\mathcal{U}$ spans a $4$-dimensional subspace, and there are $q+1$ different $5$-dimensional subspaces containing it. Then no other plane in $\mathcal{U}$ is contained in any of these $q+1$ subspaces. This holds for any pair of planes in $\mathcal{U}$. The total number of $5$-dimensional subspaces of $\Bbb{F}_q^6$ is $$\frac{q^6-1}{q-1}=q^5+q^4+q^3+q^2+q+1,$$ so the number of pairs of planes in $\mathcal{U}$ is at most $$\frac{q^5+q^4+q^3+q^2+q+1}{q+1}=q^4+q^2+1.$$ Let $m:=|\mathcal{U}|$ be the number of planes in $\mathcal{U}$. Then the above says that $$\binom{m}{2}\leq q^4+q^2+1,$$ where $\tbinom{m}{2}=\tfrac{1}{2}m(m-1)$. The quadratic formula then tells us that $$m\leq\frac{1}{2}+\frac{1}{2}\sqrt{1+8(q^4+q^2+1)}=\frac{1}{2}+\sqrt{2q^4+2q^2+\tfrac{9}{4}}.$$ This gives us an upper bound for the biggest possible $m$, which is close to the lower bound of $q^2+1\leq m$ given in the comments in the sense that $$\frac{1}{2}+\sqrt{2q^4+2q^2+\tfrac{9}{4}}\leq\frac{1}{2}+\sqrt{2}(q^2+1).$$

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