Linear subspaces of projective space.

Question

I'm following a basic course in Algebraic Geometry where the lectures are based on the first chapter of Algebraic Geometry by Robin Hartshorne. Our lecturer gave an additional advanced exercise after the first three sections. I don't really have a clue how to start off with it.

Let $k<n$ be natural numbers. Suppose that $\Lambda$ and $\Gamma$ are linear subspaces of dimension $k$ and $n-k-1$ of $\mathbb{P}^n$ over an algebraically closed field. Moreover, suppose that $\Lambda \cap \Gamma = \emptyset$.

1. Show that for each $Q \in \mathbb{P}^n\setminus \Lambda$ the intersection of $<Q,\Lambda>$ and $\Gamma$ has only one point $\pi(Q)$. Here $<Q,\Lambda>$ is the union of all projective lines $\mathbb{P}^1\subset \mathbb{P}^n$ connecting points in $\Lambda$ with $Q$.
2. Show that $\pi$ is a morphism.
3. Let $\Theta$ be a linear subspace of $\mathbb{P}^n$ and $\Lambda \cap \Theta= \emptyset$. Is $\pi(\Theta)$ a linear subspace of $\Gamma$?
4. The same question as (3) but without the condition that $\Lambda \cap \Theta= \emptyset$.

Does anybody have some hints, references or answers that could help me? Thanks!

Answer 1

I think that the trick here should be to write down everything explicitly in coordinates.Let's call the field of definition $K$.

1. Let's fix coordinates $x_0,\dots x_n$ for $\mathbb{P}^n$ and let's suppose $\Lambda=\{x_{n-k}=\dots x_n=0 \}$ and $\Gamma=\{x_0=\dots x_{n-k-1}=0\}$. Now $Q \in \mathbb{P}^n-\Lambda$ so that this point has $x_{i} \neq 0$ with $n-k \leq i \leq n$. Now, we have $Q=[q_0 : \dots q_n]$. So we have $ Span(Q,\Lambda)=[\lambda q_0+ \mu x_0 \dots \lambda q_{n-k-1}+\mu x_{n-k-1} : \lambda q_{n-k} \dots \lambda q_n]$ with $x_i \in K$ and $\lambda \neq 0$.

So, when we do $Span(Q,\Lambda ) \cap \Gamma $ what we get is $[0 : \dots 0: q_{n-k} : \dots q_n]$. Note that this tells you also that the projection you have just defined is also a morphism, so that we get free 2.

3.Now, let's call $p: K^{n+1}-0 \to \mathbb{P}^n$ your standard projection. Let's call $W \subset K^{n+1}$ the vectorial subspace such that $p(W)=\Theta$ , $V $ the one such that $p(V)=\Lambda$ and $Z$ the one such that $p(Z)=\Gamma$. The condition on the intersection is nothing else that $V$ is in direct sum with $W$.We also have $V \oplus Z=K^{n+1}$. Let's call $\phi:K^{n+1} \to Z$ the projection induced by such a decomposition. As you can check, the morphism $\pi$ one defines is such that $\pi(p)=p(\phi)$ (where everything is defined !) so that one gets also $\pi(p(W))$ is $p(\phi(W))$ and so a vectorial subspace.

4. I'm assuming that this question is referred to $\pi(\Theta)$ as the image of the points where $\pi$ is defined. So now let's take $n=2$, $\Lambda=\{x_0=x_1=0\}$ and $\Gamma=\{x_2=0\}$. Let $\Theta=\{x_1=x_2=0\}$: we get $\pi(\Theta)=[x_0 : 0 : 0 ]$ with $x_0 \neq 0$ that is not a subspace.