0
$\begingroup$

Claim: Let $V \subset \mathbb{C}P^n$ be a non-singular projective algebraic variety of complex dimension $k$ and let $P \subset \mathbb{C}P^n$ be a hyperplane. Then $V \setminus (V \cap P)$ is a complex $k$-dimensional non-singular affine algebraic variety in $\mathbb{C}^n$.

This fact is used (but not proved) in J. Milnors book Morse Theory in the proof of the Lefschetz Hyperplane Theorem (Corollary 7.3). (Actually, there it would be enough to know that $V \setminus (V \cap P)$ is a complex manifold of dimension $k$, but still the above is stated.)

My thoughts so far:

I first proved that $\mathbb{C}P^n \setminus P \cong \mathbb{C}^n$. Therefore the embedding in $\mathbb{C}^n$ doesn't cause any problem.

Now assume that $p_1, \ldots, p_j : \mathbb{C}^{n+1} \rightarrow \mathbb{C}$ are complex homogeneous polynomials such that the variety $V$ is induced from the set $$\bigcap_{r=1}^j p_r^{-1}(0) \subset \mathbb{C}^{n+1}.$$ Furthermore let $l: \mathbb{C}^{n+1} \rightarrow \mathbb{C}$ be a linear polynomial such that $P = l^{-1}(0)$. Then I would like to write something like \begin{align} V \setminus (V \cap P) &= V \cap (V \cap P)^c = V \cap P^c \\ &= \left( \bigcap p_r^{-1}(0) \right) \cap \{ l \neq 0 \} \\ &\overset{?}{=} \bigcap \left( \frac{p_r}{l} \right)^{-1}(0) \end{align} where the $p_r$ now denote the induced functions $\mathbb{C}P^n \rightarrow \mathbb{C}P^n$. Then $V \setminus (V \cap P)$ would be cut out by the "polynomials" $\frac{p_r}{l}$ and I'd be done. But this doesn't seem to be formally correct...

Thanks in advance for any help!

$\endgroup$
  • 4
    $\begingroup$ This is almost correct. But instead of $\frac{p_r}{l}$ you need $\frac{p_r}{\lambda_r}$, where $\lambda_r$ is $l$ raised to the power $\operatorname{deg }p_r$. These $p_r/\lambda_r$ give actual polynomials on $\mathbf C^n$, cutting out your affine variety. $\endgroup$ – user64687 Jun 13 '14 at 10:16
  • $\begingroup$ Thank you for your answer. But why are te $p_r/\lambda_r$ actual polynomials? And why in $\mathbb{C}^n$ and not in $\mathbb{C}^{n+1}$? For example in the case $n=1$ consider $p, l: \mathbb{C}^2 \rightarrow \mathbb{C}$ with $p(z_1, z_2) = z_1 z_2$ and $l(z_1, z_2) = z_1 + z_2$. Then $\frac{p}{l^2} = \frac{z_1 z_2}{z_1^2 + 2z_1 z_2 + z_2^2}$ is not a polynomial in $\mathbb{C}$ right? $\endgroup$ – Tom Bombadil Jun 13 '14 at 11:39
  • 2
    $\begingroup$ They are homogeneous rational functions of degree 0 on C^{n+1} with poles only along $l^{-1}(0)$. So they descend to well-defined functions on $P^n \setminus P \sim C^n$. $\endgroup$ – user64687 Jun 13 '14 at 11:43
  • $\begingroup$ Okay, I understand that. Now I only still need to show the non-singularity of $V \setminus (V \cap P)$. It would make sense, if this simply follows from the non-singularity of $V$, since $V \setminus (V \cap P)$ is a subset of $V$. However I only really know the definition of non-singularity as the requirement that the derivative of the combined function $\bigoplus_{r=1}^j p_r$ is surjective at every point in the variety. How do I solve this problem? $\endgroup$ – Tom Bombadil Jun 13 '14 at 12:20
  • 2
    $\begingroup$ Dear Tom,for an open subset of a non-singular variety, this is true.I don't have time to write out the details now, but you can try to work through the definitions yourself. $\endgroup$ – user64687 Jun 13 '14 at 12:40
1
$\begingroup$

The algebraic projective variety $V\subset \mathbb{P}^n$ is given by the zero locus of homogeneous polynomials $f_i\in \mathbb{C}[x_0,\dots,x_n]$.

Take now the open subset $U$ of $\mathbb{P}^n$ where $x_0\not=0$, which is an affine space. You can assume that $x_0=1$ and obtain then coordinates $x_1,\dots,x_n$. This gives an isomorphism $$\begin{array}{rcl}\mathbb{C}^n&\to& U\\ (x_1,\dots,x_n)&\mapsto & [1:x_1:\dots:x_n]\end{array}$$

Then $U\cap V$ is the locus of points of the form $[x_0:\dots:x_n]$ such that $x_0=1$ and $f_i(x_0,\dots,x_n)=0$. This shows that you obtain an affine variety given by the polynomials $f_i(1,x_1,\dots,x_n)$.

This process is very classical and can be founded in any course of algebraic geometry.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.