# Projective plane division by non-concurrent lines

How many regions does n non-concurrent lines divide a Projective Plane into? (probably a standard problem, but I am having conflicting answers)

• I would say $1+\frac{n(n-1)}{2}$. What are your answers? – Dan Shved Nov 8 '12 at 16:12
• Mine is $2^{n-1}$. How do you arrive at that? – Mandal Nov 8 '12 at 16:17
• I have posted an answer. This is actually my second solution. The first one was to use the similar problem for a usual Euclidean plane. For the usual plane $\mathbb{R}^2$, the answer is $1+n(n+1)/2$, and you you can turn $\mathbb{R}P^1$ into $\mathbb{R}^2$ by removing one of the lines, so the answer is $1+n(n-1)/2$. – Dan Shved Nov 8 '12 at 16:30
• @ Dan: thanks for the answer. I forgot to clarify, but by projective plane I wanted to mean $\mathbb{R}P^{2}$. But I guess we are talking of the same thing under different notations, right? – Mandal Nov 8 '12 at 16:52
• Also, it would be helpful if you find me the error in my argument. I was trying to go directly through dividing $\mathbb{R}{}^{3}$ by planes passing through origin. Each such plane $P$ creates regions $P^{+}$ and $P^{-}$. n such planes, no two of which intersect in a line, creates $2^{n}$ regions. (for 2 planes, they would be like $P_{1}^{+}P_{2}^{+},P_{1}^{+}P_{2}^{-},P_{1}^{-}P_{2}^{+},P_{1}^{-}P_{2}^{-}$). – Mandal Nov 8 '12 at 16:59

Well, $n$ non-concurrent lines turn the projective plane $\mathbb{R}P^2$ into a $CW$-complex. Denote by $k_i$, $i=0, 1, 2$, the number of $i$-dimensional cells. What we need to find is $k_2$. Since the the Euler characteristic of $\mathbb{R}P^2$ is $1$, we have $$k_0 - k_1 + k_2 = 1.$$ So we will know $k_2$ if we find $k_0$ and $k_1$. $k_0$ is the number of vertices. Since all the lines are not concurrent, it is clear that $k_0 = n(n-1)/2$. To find $k_1$, let's look at one individual line $l$. Topologically, $l$ is a circle $S^1$. It is split by the other $n-1$ lines into $n-1$ segments. So the total number of segments (= 1-dimensional cells) is $k_1 = n(n-1)$. And so we have $$k_2 = 1 + k_1 - k_0 = 1 + \frac{n(n-1)}{2}.$$