# Dimension in $\mathbb{P}^4$ of $\langle L,M \rangle \cap N$ with $L,M,N$ pairwise non-intersecting and not in one hyperplane

Given three lines, $$L, M, N \in\mathbb{P}^4$$, not in one hyperplane and not pairwise intersecting, I need to calculate

$$\dim(\langle L,M\rangle\cap N).$$

By the dimension of intersection theorem for projective spaces we have

$$\dim(\langle L,M \rangle \cap N) = \dim\langle L,M \rangle + \dim N - \dim\langle \langle L,M \rangle N \rangle.$$

But I do not know how to interpret the angle bracket notation for two lines. I do know:

What I know: A projective line through two points $$P=(p_0:...:p_n), Q=(q_0:...:q_n)$$ is defined by first moving these points to $$\mathbb{R}^{n+1}$$, and then we have $$PQ=\langle P,Q\rangle=\{\lambda p_0 + \mu q_0 : ... : \lambda x_n + \mu q_n \mid (\lambda,\mu)\neq (0,0)\}.$$

I also think that $$\dim N$$ must be equal to 2, since a projective line is a plane in Euclidian space, so given two points of $$N$$, we have that $$N$$ is the span of two lines passing through these two points respectively.

Furthermore I know that $$\dim \mathbb{P}^4 = 4$$.

• @Berci well I don't understand. My book says that $H$, a hyperplane of $\mathbb{P}^n$, corresponds to an n-dimensional subspace $W\subset \mathbb{R}^{n+1}$, but then $\dim H = n$ right? But that is already the dimension of $\mathbb{P}^n$.. – The Coding Wombat Mar 27 at 23:22

$$\langle L,M\rangle$$ is itself a hyperplane if $$L$$ and $$M$$ don't intersect.
Since it doesn't contain $$N$$, it will intersect $$N$$ in a single point.
• Yes, it does. Turning everything into $\Bbb R^5$ gives $L, M, N$ as 2 dimensional linear subspaces. If they don't intersect, it means $L\cap M=\{0\}$, so their generated linear subspace $\langle L, M\rangle=L+M$ has dimension $4$ in $\Bbb R^5$ (so projective dimension is $3$). The plane $N$ is not contained in this hyperplane by hypothesis, so they must meet in a line in $\Bbb R^5$, i.e. a point in $\Bbb P^4$. – Berci Mar 28 at 9:09
• But L and M also do not intersect N, so then wouldn't the dimension of the entire space $L + M + N$ have to be six, which couldn't be possible since we're in $\mathbb{R}^5$? I cannot visualize any of this :(. And since we can turn $\mathbb{P}^4$ to $\mathbb{R}^5$, does this mean that $\dim(\mathbb{P}^4)=5$? (See also my comment in response to your now deleted comment under the original question) – The Coding Wombat Mar 28 at 14:07