# There exists a subspace not intersecting the analytic variety.

Given an analytic subvariety $V$ of dimension $k$ in $\mathbb{P}^n$ and a point $p$ not in it, how do I show that there is an $n-k-1$ plane $\mathbb{P}^{n-k-1}$ containing $p$ and not intersecting $V$?

Is this some shady intersection theory stuff or something else?

I came across this while trying to understand the proof of Chow's Theorem in Griffiths-Harris.

$\DeclareMathOperator{\Span}{Span}\newcommand{\Proj}{\mathbf{P}}$If $\dim V = k$ and $p \not\in V$, fix a hyperplane $H$ not containing $p$; projection away from $p$ defines a map $V \to H$ whose image $V'$ is an analytic subvariety of dimension (at most) $k$. (The image may be interpreted as the intersection of $H$ with the chordal variety $$\Span(p, V) = \bigcup_{v \in V} \overline{pv},$$ comprising all lines joining $p$ to a point of $V$.)

If $k < n - 1$, then $\dim V' < \dim H$, so there exists a point $p'$ of $H$ not lying in $V'$; that is, there exists a line $L$ through $p$ and disjoint from $V$.

Assume inductively that for some integer $1 \leq \ell$, we have constructed an $\ell$-dimensional projective space $L$ through $p$ and disjoint from $V$. Fix an $(n - \ell - 1)$-dimensional projective space $H$ disjoint from (i.e., skew to) $L$. Projection away from $L$ defines a map $V \to H$ whose image is a subvariety $V'$ of dimension (at most) $k$. If $k < n - \ell - 1$, i.e., if $\ell < n - k - 1$, then $\dim V' < \dim H$, so there exists a point $p'$ of $H$ not lying in $V'$. The $(\ell + 1)$-dimensional projective space $\Span(p', L)$ is disjoint from $V$.

This induction guarantees existence of an $(n - k - 1)$-dimensional projective space through $p$ and disjoint from $V$.

Appendix: Since you're reading Griffiths and Harris, these ideas are probably familar, but for passersby and posterity:

If $L$ and $H$ are skew subspaces of maximal dimension in $\Proj^{n}$ (i.e., $L$ and $H$ are linear subspaces such that $L \cap H = \varnothing$ and $\dim L + \dim H + 1 = n$), then projection away from $L$ is the mapping $\Pi_{L}:\Proj^{n} \setminus L \to H$ that sends a point $q$ to the intersection of $\Span(q, L)$ with $H$.

Algebraically, if $z = (z_{0}, \dots, z_{\ell}, z_{\ell + 1}, \dots, z_{n})$ are affine coordinates on $K^{n+1}$ (the construction makes sense over an arbitrary field), and if $$L = \{z : z_{0} = \dots = z_{\ell} = 0\} \simeq \Proj^{\ell},\qquad H = \{z : z_{\ell + 1} = \dots = z_{n} = 0\} \simeq \Proj^{n - \ell - 1},$$ then $\Pi_{L}$ is ordinary orthogonal projection to the last $(n - \ell)$ coordinates: $$\Pi_{L}(z) = (0, \dots, 0, z_{\ell + 1}, \dots, z_{n}),$$ which is undefined in projective space precisely on $L$.

• Thanks so much; I understand your argument! – P. Brown Nov 26 '16 at 22:48