# General position argument for reasonable spaces

Let $\mathcal{D} \approx \mathbb{P}^{\delta_d}$ be the space of non zero homogeneous degree $d$ polynomials in three variables up to scaling, where $\delta_d:= \frac{d(d+3)}{2}$. Given a point $p \in \mathbb{P}^2$ define $H_p \subset \mathcal{D}$ to be the hyperplane $$H_p := \{ [f] \in \mathcal{D}: f(p) =0 \}.$$ Suppose $\mathcal{V}$ and $\mathcal{W}$ are two smooth algebraic varieties inside $\mathcal{D}$ of dimension $k$ and $k-1$ respectively. I have three questions:

1) Suppose that for a "generic" choice of $p \in \mathbb{P}^2$, $H_p$ intersects both $\mathcal{V}$ and $\mathcal{W}$ transversally (ie the set of $p$ for which the above statement is false is contained inside a finite union of subvarieties of $\mathbb{P}^2$ of strictly smaller dimension.)

Is the following statement necessarily true: For generic choices of $k$ points $p_1, \ldots, p_k$ the intersection $$\mathcal{V} \cap H_{p_1} \cap \ldots \cap H_{p_k}$$ is transverse and $$\mathcal{W} \cap H_{p_1} \cap \ldots \cap H_{p_k} = \emptyset$$ ?

2) Same hypothesis as in 1). Is it true that for generic choice of points $p_1$ and $p_2$, the intersection of $H_{p_2}$ with $\mathcal{V} \cap H_{p_1}$ is transverse? Note that by assumption all the three pairwise intersections are transverse if $p_1$ and $p_2$ are generic: $H_{p_1}$ and $\mathcal{V}$; $H_{p_2}$ and $\mathcal{V}$; $H_{p_1}$ and $H_{p_2}$. But that doesn't mean that $H_{p_2}$ will intersect transversally with $\mathcal{V} \cap H_{p_1}$. The claim/hope is that I can delete a further "small" set from $\mathbb{P}^2 \times \mathbb{P}^2$ to ensure that the desired triple intersection is transverse.

3) Let us make a much stronger assumption: For generic choices of $k$ points $p_1 \ldots p_k$ the intersection $$\mathcal{V} \cap H_{p_1} \cap \ldots \cap H_{p_k}$$ is transverse. Does it imply for generic choices of $k+1$ points, the following intersection is empty, ie $$\mathcal{V} \cap H_{p_1} \cap \ldots \cap H_{p_{k+1}} = \emptyset$$ ? Note that any $k+1$ fold intersection in the above expression is transverse (ie the intersection will be a finite bunch of points). That does not mean the $k+2$ fold intersection will be transverse (hence empty). But does it mean for generic choices of $k+1$ points the $k+2$ fold intersection will be empty? Forget about $\mathcal{W}$ for this question.

• This question is a bit more involved than it might look, since the varieties $\mathcal V$ and $\mathcal W$ are in the space of degree $d$ plane curves. Note that the space of all the $H_p$ is precisely the Veronese surface in $\mathcal D^*$; in particular, it is not a linear system. – Charles Staats Jul 1 '13 at 17:26
• For (3), a finite set of curves cannot cover all of $\mathbb{P}^2$. – Jack Huizenga Jul 1 '13 at 17:51
• @Jack Huizenga: Assuming that's true, how does it answer question 3)? – Ritwik Jul 1 '13 at 18:02
• @Charles Staats: I have added two further questions. Maybe question 3) is a little less involved, since I have a much stronger hypothesis. But I am not sure. – Ritwik Jul 1 '13 at 18:03
• The points in the (k+1)-fold intersection correspond to a finite set of curves. – Jack Huizenga Jul 1 '13 at 18:06

At the very least the intersections have the correct dimension. I'm not sure about actual transversality, but it wouldn't surprise me if the answer is no. For many applications, this fact is good enough; intersection theory will work, for instance, so long as you work with schemes.

In fact, suppose $X \subset \mathbb{P}^{\delta_d}$ is any closed subvariety of pure dimension $k$. Consider the universal degree $d$ curve

$\Sigma = \{(f,p):f(p)=0\} \subset \mathbb{P}^{\delta_d} \times \mathbb{P}^2$

with projections $\alpha,\beta$. Put $Y = \alpha^{-1}(X)$. The intersections $X \cap H_p$ are isomorphic to the fibers of the second projection $Y\to \mathbb{P}^2$.

Write $Y$ as a union of irreducible components. By the flatness of the universal degree $d$ curve and flatness of base change, every component of $Y$ has dimension $k+1$. Under the second projection $Y\to \mathbb{P}^2$, some of these components dominate $\mathbb{P}^2$ and some do not. The ones which don't contribute nothing to the general intersection $X\cap H_p$. On the other hand, if a component does dominate $\mathbb{P}^2$ then the general fiber of this projection has dimension $k-1$. We conclude that the general fiber of $Y\to \mathbb{P}^2$ is pure of dimension $k-1$, so $X\cap H_p$ is as well for general $p\in \mathbb{P}^2$.

Of course it can happen that the general intersection $X\cap H_p$ is singular or even everywhere nonreduced, as the example where $X$ is the Veronese surface parameterizing double lines in the space of conics shows. But at least the dimension will work out.

In particular, for any $X$ of dimension $k$ it is the case that $X \cap H_{p_1}\cap \cdots \cap H_{p_{k+1}}$ is empty.