Projectively equivalent and non-equivalent configurations in $\mathbb{P}^n$

Question

I'm trying to show all the non-projectively equivalent configurations of linear subspaces.
Two subspaces $\Lambda_1,\Lambda_2$ are projectively equivalent if

$\exists\omega:\mathbb{P}^n\longrightarrow \mathbb{P}^n$ homography such that $\omega(\Lambda_1)=\Lambda_2$.

For example, in $\mathbb{P}^4$ let $l_1,l_2$ be two distinct lines and $P$ a point such that such that $P \notin l_1\cup l_2$.

What's the number of non-projectively equivalent configurations of the two lines and the point $P$?

My idea is to study the dimension of the intersection space of the two lines.
Infact, for Grassmann's formula we have: $$\mathrm{dim}(l_1 \cup l_2)=\mathrm{dim}(l_1)+\mathrm{dim}(l_2)-\mathrm{dim}(l_1 \cap l_2) \le4$$
We know that $\mathrm{dim}(l_1 \cap l_2)= \begin{cases} 0 \\ -1 \end{cases}$

If $\mathrm{dim}(l_1 \cap l_2)=0$ it means that $l_1 \cap l_2=\{Q\}$ (case $1$), while if $\mathrm{dim}(l_1 \cap l_2)=-1$ it means that $l_1 \cap l_2 = \emptyset$ (case $2$). We can now study all these cases.

$1$a) $<l_1,l_2>=\pi$, where $\pi$ is a plane, $P \in \pi$ and $\mathrm{dim}(<l_1,l_2,P>)=2$

$1$b) $<l_1,l_2>=\pi$, where $\pi$ is a plane, $P \notin \pi$ and $\mathrm{dim}(<l_1,l_2,P>)=3$

$2$a) $<l_1,l_2>=\Theta$, where $\Theta$ is a hyperplane, $P \in \Theta$ and $\mathrm{dim}(<l_1,l_2,P>)=3$

$2$b) $<l_1,l_2>=\Theta$, where $\Theta$ is a hyperplane, $P \notin \Theta$ and $\mathrm{dim}(<l_1,l_2,P>)=4$

All these are non-projectively equivalent because a homography $\omega$ doesn't change the dimension of linear subspaces in $\mathbb{P}^n$. Is that correct? Thank you in advance.

Answer 1

Yep, you are correct about why your four cases are all distinct. In fact, they are the only distinct cases: any two configurations in the same case are equivalent. To prove this, I find it convenient to pull everything back to $k^5$ (where $k$ is the field) and just think about linear subspaces of $k^5$ using linear algebra.

So, for instance, suppose we are in your case (2a) (similar arguments can be made in your other three cases). Then $\ell_1$ and $\ell_2$ correspond to two-dimensional subspaces $V_1,V_2\subseteq k^5$ with trivial intersection, and $P$ corresponds to a 1-dimensional subspace $W$ which is contained in $V_1+V_2$ but not contained in $V_1$ or $V_2$. Pick a nonzero vector $w\in W$, and write $w=w_1+w_2$ for $w_1\in V_1$ and $w_2\in V_2$ (since $w\in W\subseteq V_1+V_2$). Since $W$ is not contained in $V_1$ or $V_2$, both $w_1$ and $w_2$ are nonzero. So, we can pick $v_1\in V_1$ and $v_2\in V_2$ such that $\{v_1,w_1\}$ and $\{v_2,w_2\}$ are bases for $V_1$ and $V_2$. Finally, pick a nonzero vector $u\in k^5\setminus (V_1+V_2)$, so $\{v_1,w_1,v_2,w_2,u\}$ is a basis for $k^5$.

Now, given any other configuration $(\ell_1',\ell_2',P')$ which is also in case (2a), we can similarly pick a basis $\{v_1',w_1',v_2',w_2',u'\}$. There is then a linear isomorphism $T:k^5\to k^5$ mapping $v_1$ to $v_1'$, $w_1$ to $w_1'$, and so on. This $T$ will then map $V_1$ to $V_1'$, $V_2$ to $V_2'$, and $W$ to $W'$, so it induces a homography that maps $\ell_1$ to $\ell_1'$, $\ell_2$ to $\ell_2'$, and $P$ to $P'$.