4

Let $P$ be a projective space of dimension $n$ and $Q$ a linear subspace of it. If the complement of $Q$ is affine, why must $Q$ be of dimension $n - 1$?

The following is my thought:

Take the homogeneous coordinate system $[T_0:\cdots:T_n]$. (Suppose that $Q$ is not a hyperplane.) Let $Q$ be of dimension $d(d < n - 1)$, satisfying $T_{i} = 0(i\in\{d+1,\cdots, n\})$. Then what's left is to show that $\{[T_0:\cdots:T_n]:\exists i\in\{d+1,\cdots, n\}, s.t. T_i\neq 0\}$ is not affine. How to prove this assertion?

For the other direction, I know it's obvious. Many thanks!

Stupid Math Lover
  • 87

2 Answers2

1

This is my first thought. It doesn't use much of what you've written and I regret that. Hopefully it works and I haven't said too much.

What's the fundamental fact about a pair of linear subspaces of $\mathbb{P}^n$? This is just a souped-up version of a fact from linear algebra. You should be able to use this idea to find something in the complement of $Q$ that isn't allowed inside of affine space.

Hoot
  • 3,501
  • Could you please explain it more explicitly? What do you mean by a pair of linear subspaces of $\mathbb{P}^n$? I thinks there is only a linear subspace $Q$ which matters here. – Stupid Math Lover Jan 06 '15 at 07:58
  • @StupidMathLover I'm just trying to not blurt out the answer. If I have two linear subspaces $\Lambda, \Lambda'$ of $\mathbb{P}^n$, what conditions on their dimensions forces them to intersect? I promise this is going somewhere. – Hoot Jan 06 '15 at 08:06
  • I think the necessary condition for them two to intersect is that $dim(\Lambda) + \dim(\Lambda') > n - 1$. – Stupid Math Lover Jan 06 '15 at 08:12
  • Okay, good. So, using that, a hyperplane meets every linear subspace that isn't just a point. But if the codimension is larger? [Maybe we can use your notation and explicit coordinates after all, but this is how I thought about it and it's a good fact.] – Hoot Jan 06 '15 at 08:16
  • [Yeah, I did mean "sufficient condition"XD] – Stupid Math Lover Jan 06 '15 at 08:20
  • Well, for the higher dimension: every linear subspace, which is not a hyperplane, meets every linear subspace, which is not just a line. But I can't see the connection with affine space. (I feel rather studpid TAT) – Stupid Math Lover Jan 06 '15 at 08:23
  • Okay, good, so maybe that will help. Actually, I claim that you can write down a copy of $\mathbb{P}^1$ inside the complement that you've specified above. – Hoot Jan 06 '15 at 08:24
  • This is just what I've claimed. – Stupid Math Lover Jan 06 '15 at 08:25
  • This implies that the complement contains $\mathbb{P}^1$. Does this mean the complement cannot be affine? – Stupid Math Lover Jan 06 '15 at 08:27
  • I don't know what facts you have available. An affine variety can't contain a copy of $\mathbb{P}^1$. To argue this you could use, for example, the fact that the only functions defined on all of $\mathbb{P}^1$ are constants. – Hoot Jan 06 '15 at 08:27
  • The only regular function defined over a projective variety is contant. I learnt this fact in my AG class and this seems to be an evidence for "affine variery can't contain a projective variety"? – Stupid Math Lover Jan 06 '15 at 08:33
0

Obviously it's not needed for this question (Hoot's answer gives a very nice direct solution), but it seems worth recording: it's a general fact that the complement of an affine open $U$ in a variety $V$ has codimension $1$ in $V$ (if it is non-empty).

Consequently, the complement of a closed subvariety $W$ of codimension $> 1$ in a variety $V$ cannot be affine.

tracing
  • 5,500