How many regions does n non-concurrent lines divide a Projective Plane into? (probably a standard problem, but I am having conflicting answers)
Projective plane division by non-concurrent lines
3
$\begingroup$
projective-geometry
projective-space
-
0I guess it will become clearer if you lower the dimension and look at lines passing through the origin in $\mathbb{R}^2$. $2$ lines will divide it inso $4$ regions, but $3$ lines produce only $6$ regions instead of $8$. – 2012-11-08
1 Answers
3
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}. $