Finding a unique projective subspace

Question

Yesterday I started studying projective geometry and today I was making some exercices on this. The following problem is were I got stuck, it goes as follows: In $P(\mathbb{R}^{n+1})$ Let $P^r$, $P^q$ and $P^s$ be projective subspaces which are pairwise crossing so there intersection is empty. s, r and q denote their respective dimensions. Assume that $P^q$ and $P^r$ span the whole space en that $2s \leq n-1$. We need to show that there is unique projective subspace $P^{2s+1}$ of dimension $2s+1$ for which $P^s$ is a subspace and also $dim(P^q \cap P^{2s+1}) = dim(P^r \cap P^{2s+1}) = s$. Now If $2s = n-1$ this is trivial since then $P^{2s+1} = P(\mathbb{R}^{n+1})$. However I'm struggling with the case when $2s < n-1$, I have deduced that $2s \leq q+r$ but I'm not able to go any further. Does anyone have a tip how I could continue with this problem? Any help would be grealty appreciated :))

Answer 1

Finding a candidate subspace:

The dimension theorem for projective subspaces tells us that $$ \dim P^q + \dim P^r = \dim (P^q + P^r) + \dim(P^q \cap P^r) $$ from which we find (since $P^q$ and $P^r$ span the space and are disjoint) that $q + r = n - 1$.

Now, applying the same dimension theorem for the disjoint subspace combinations $P^q, P^s$ and $P^r, P^s$ we get $$ \dim P^q + \dim P^s = q + s = \dim (P^q + P^s) + \dim(P^q \cap P^s) $$ and $$ \dim P^r + \dim P^s = r + s = \dim (P^r + P^s) + \dim(P^r \cap P^s). $$ Therefore it follow that $\dim (P^q + P^s) = q + s + 1$ and similarly $\dim (P^r + P^s) = r + s + 1$.

Look at the subspace $(P^q + P^s) \cap (P^r + P^s)$. Again applying the dimension theorem gives us $$ \dim(P^q + P^s) + \dim(P^r + P^s) = \dim ((P^q + P^s) + (P^r + P^s)) + \dim((P^q + P^s) \cap (P^r + P^s)). $$ Substituting all the dimensions we have already found we get $$ r + s + 1 + q + s + 1 = n + 2s + 1 = \dim((P^q + P^s) \cap (P^r + P^s)) + n$$ where we also used that $(P^q + P^s)$ and $(P^r + P^s)$ obviously must also span the space.

This tells us that $(P^q + P^s) \cap (P^r + P^s)$ is a space of dimension $2s + 1$ that clearly contains $P^s$ and satisfies the other conditions.

Uniqueness:

Suppose now that $P$ is a subspace that satisfies the given conditions.

Then using the dimension theorem we find $$ 2s + 1 + q = \dim(P + P^q) + \dim(P \cap P^q) = \dim(P + P^q) + s $$ So $\dim(P + P^q) = q + s + 1$. But we also know that $\dim(P^s + P^q) = q + s + 1$. Combining this with the fact that $(P^s + P^q) \subseteq (P + P^q)$ gives us $(P^s + P^q) = (P + P^q)$. But therefore $P \subseteq (P^s + P^q)$. Analogously we get $(P^s + P^r)$.

Combining the previous two inclusions results in $P \subseteq (P^s + P^q) \cap (P^s + P^r)$ but since both these sets have dimension $2s + 1$ we must have $P = (P^s + P^q) \cap (P^s + P^r)$ which proves the uniqueness.