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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 $\dim \mathbb{P}^4 = 4$.

The Coding Wombat
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  • @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 '19 at 23:22

1 Answers1

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$\langle L,M\rangle$ 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.

Berci
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  • But can you tell me why? Does it follow from the definition of the angle brackets I gave in the quote in my question? – The Coding Wombat Mar 27 '19 at 23:23
  • 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 '19 at 09:09
  • 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 '19 at 14:07
  • My question should actually be: How can the intersection of <L,M> and N contain any points if L,M and N don't intersect? – The Coding Wombat Mar 28 '19 at 16:28
  • Projective dimension is always one less than the real dimension. A hyperplane always intersects any line in a projective space. – Berci Mar 28 '19 at 17:10
  • Choosing bases $l_1,l_2$ of $L\subseteq\Bbb R^5$ and $m_1,m_2$ of $M$, the condition $L\cap M={0} $ implies that $l_1,l_2,m_1,m_2$ are linearly independent, hence real dimension of $\langle L,M\rangle$ is $4$. Its projective dimension is $3$. It's a hyperplane. Its generatum with the 2 dimensional $N$ can't have dimension $6$, so their intersection has dimension $1$ or $2$, but $2$ would mean that the hyperplane contains $N$. – Berci Mar 28 '19 at 18:03
  • I'm sorry, but I don't understand why that would be the case. If you had two planes that ran parallel "on top of each other", so one plane being $\Pi=(x, y, 0, 0, 0)$ and one plane $\Pi'=(x, y, 1, 0, 0)$, then their intersection would be empty, but $\langle \Pi', \Pi \rangle$ would be three right? Since if you have two planes on top of each other in three dimensions, it would only require those three dimensions? – The Coding Wombat Mar 28 '19 at 18:14
  • In projective geometry, things that were parallel in the Euclidean part, will intersect. The problem with your argument is that $\Pi'$ is not a linear subspace, only an affine plane: it simply doesn't contain $0$. – Berci Mar 28 '19 at 18:51
  • Let us continue this discussion in chat. – The Coding Wombat Mar 28 '19 at 19:23