Projective line intersecting 3 projective subspaces

Question

I am trying to solve the following problem:

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Let $\mathbb{P}(U),$ $\mathbb{P}(V)$ and $\mathbb{P}(W)$ be projective subspaces of dimension $k,$ $l$ and $m$ respectively in $\mathbb{P}_K^n$.

Suppose $k+l+m\geq n-1$. Prove that there is a projective line intersecting all 3 subspaces.

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So if $\mathbb{P}(U)\cap \mathbb{P}(V)\neq \emptyset,$ then it's easy because we just take any (projective) point in their intersection and any (projective) point in $\mathbb{P}(W)$ and form the projective line between them.

If $\mathbb{P}(U)\cap \mathbb{P}(V)=\emptyset,$ then $U\cap V=\{0\}$ and so $\dim(U + V)=\dim(U)+\dim(V)$ and so $$\dim((U+V)\cap W)=k+l+m+3-\dim(U+V+W)\geq (k+l+m-(n-1))+1\geq 1$$ hence $\mathbb{P}(U+V)$ intersects $\mathbb{P}(U),$ $\mathbb{P}(V)$ and $\mathbb{P}(W)$ but it might not be a line... so now I'm stuck.

Any help greatly appreciated!

Answer 1

One possible solution to the problem is to use the Schubert calculus in the grassmannian $\mathbb G=\mathbb G(1,\mathbb P^n)$ of lines in $\mathbb P^n$.
That grassmannian has dimension $2n-2$ and the set of lines intersecting $\mathbb P(U)$ is a subvariety $\Omega _{n-1-k}$ (this is a standard notation) of codimension $ n-1-m$ in $\mathbb G$.
The lines intersecting $\mathbb P(V)$ and $\mathbb P(W)$ similarly form subvarieties $\Omega _{n-1-l}, \Omega _{n-1-m}$ of codimensions $n-1-l$ and $n-1-m$.
Pieri's formula implies that the intersection of those three $\Omega$ 's is non empty as soon as $$ (n-1-k)+(n-1-l)+(n-1-m) \leq \dim \mathbb G=2n-2 $$ and this is true according to your hypothesis.
Any point $\omega\in \Omega _{n-1-k}\cap\Omega _{n-1-l}\cap\Omega _{n-1-m}$ in this intersection then represents a line $L_\omega\subset \mathbb P^n$ cutting all three of $\mathbb{P}(U),\mathbb{P}(V)$ and $\mathbb{P}(W)$.

Edit
What I wrote applies to the context of the question but should not be unduly generalized: it is not true that arbitrary subvarieties of a grassmannian $\mathbb G$ whose dimensions sum to at most $\dim \mathbb G$ have a non-empty intersection.
For example in $\mathbb G=\mathbb G(1,\mathbb P^3)$ consider the subvariety $\Omega_2\subset \mathbb G$ of projective lines containing some point $q\in \mathbb P^3$ and the subvariety $\Omega_{(1,1)}\subset \mathbb G$ of projective lines contained in some projective plane $\mathbb P\subset \mathbb P^3$ such that $q\notin \mathbb P$.
Obviously $\Omega_2\cap \Omega_{(1,1)}=\emptyset$: indeed since $q\notin \mathbb P$ no projective line $L\subset \mathbb P$ can satisfy $L\supset q$.
Nevertheless $\dim \Omega_2+\dim \Omega_{(1,1)}=2+2=4=\dim \mathbb G$

To say it more forcefully: Bézout's theorem does not hold in Grassmannians

Answer 2

I've only got time to outline this, but here we go...

We have subspaces $U, V, W$ of $K^{n+1}$, of dimension $\tilde k = k+1, \tilde l = l+1$, and $\tilde m = m+1$ respectively, with $\tilde k + \tilde l + \tilde m \geq n + 2$.

First, note that if any two subspaces have a non-zero vector in common, we are done, because their projectivisations intersect, and we can just take a line from this intersection to any point of the third subspace. So assume that $U \oplus V \hookrightarrow K^{n+1}$. Now, $\dim (K^{n+1} / (U \oplus V)) = n+1 - \tilde k - \tilde l \leq \tilde m - 1 = \dim W - 1$. So there is at least one vector $w \in W \cap ( U \oplus V)$. Write $w = u + v$, with $u \in U, v \in V$. Assume $u \neq 0$ and $v\neq 0$, because otherwise $w \in V$ or $w \in U$, and we have already dealt with that case.

$w, u, v$ span a two-dimensional subspace, and upon projectivisation, this gives a line that intersects all three subspaces of interest.