Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
1  
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
1  
@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
1  
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
2  
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
1  
@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 $$\frac{1}{2}+\sqrt{2q^4+2q^2+\tfrac{9}{4}}\leq\frac{1}{2}+\sqrt{2}(q^2+1).$$

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.