Counting dimensions in $\mathbb{C}P^n$?

Question

"The space of lines in $\mathbb{C}P^n$ that meet two generic subspaces of complex dimension $i$ and $j$ fills out a subspace of dimension $i + j + 1$." One does one prove this, or at least intuitively understand why it is true? Does one pull back to $\mathbb{C}^{n+1} \setminus \{0\}$ and translate this into a linear algebra problem?

Answer 1

You can pull back to $\mathbb C^{n+1}$, but you can also argue directly via geometry in $\mathbb CP^n$.

You have a line meeting each of the two subspaces (call them $V$ and $W$). Unless it is contained in either one of them, it meets them each in exactly one point. So the set of lines passing through a point of $V\setminus W$ and a point of $W\setminus V$ is dense in the set of all such lines. Now the idea is that there is an $i$-dimensional set of points on $V\setminus W$, and a $j$-dimensional set of points on $W\setminus V$, and for each such pair of points, there is a one dimensional set of points on the line joining them. Adding up all the dimensions gives $i + j + 1$ in total.

To make this precise, you have to think more carefully. The most subtle point is that a point on the union of all the lines may lie on more than one line (in fact on a positive dimensional collection of lines) and hence may be over-counted. If this happens for a general point on the union, then $i+j+1$ will be an over-count of the dimension.

You can see this in the case $n = 2$ and $i = j = 1$.

On the other hand, if $i + j < n,$ and $V$ and $W$ are generic, then $V \cap W = \emptyset$, and in fact you can see that the line joining $(v,w)$ and $(v',w')$ cannot meet unless $v = v'$ or $w = w'$ (and they meet just at $v$ or just at $w$, unless $(v,w) = (v',w')$, in which case of course they are same line). So in this case the dimension count is correct.

[To see this, suppose that these two lines meet in a point $x$ not lying on $V$ or $W$. Firstly if either $v = v'$ or $w = w'$ then these two lines have two points in common, and so they actually coincide (and $(v,w) = (v',w')$). Otherwise we have $v \neq v'$ and $w\neq w'$. Then the two lines span a $\mathbb CP^2$ in $\mathbb CP^n$, and any two lines in the projective plane meet. In particular, the line joining $v$ and $v'$ and the line joining $w$ and $w'$, both of which line in this plane, meet. But the resulting point lies on $V \cap W$, which we assumed was empty. Thus $x$ doesn't exist.]