Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|cite|improve this question
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 $$ k_2 = 1 + k_1 - k_0 = 1 + \frac{n(n-1)}{2}. $$

share|cite|improve this answer

Your Answer


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.