Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\mathcal{D} \approx \mathbb{P}^{\delta_d} $ be the space of nonzero homogeneous degree $d$ polynomials in three variables upto scaling, where $\delta_d = \frac{d(d+3)}{2} $ (basically degree $d$ curves in $\mathbb{P}^2$). Suppose $\mathcal{V}$ is a smooth sub variety of $\mathcal{D}$, not necessarily closed that has the following property: a generic element $[f] \in \mathcal{V}$ has a singularity only at the point $[1,0,0]$. Suppose $k$ is the dimension of $\mathcal{V}$. Define the following fibered products: $$ \overline{\mathcal{C}}(k):= \{ ([f], p_1, \ldots, p_k) \in \overline{\mathcal{V}} \times (\mathbb{P}^2)^k: f(p_1) =0, \ldots, f(p_k) =0 \} $$

$$ \mathcal{C}(k):= \{ ([f], p_1, \ldots, p_k) \in \mathcal{V} \times (\mathbb{P}^2)^k: f(p_1) =0, ~\nabla f|_{p_1} \neq 0, \ldots, f(p_k) =0, ~\nabla f|_{p_k} \neq 0 \} $$

Here $\overline{\mathcal{V}}$ denotes the closure of $\mathcal{V}$ in $\mathcal{D}$. I have two questions:

1) Is it true that $\mathcal{C}(k)$ is dense in $ \overline{\mathcal{C}}(k) $ (I apologize for the notation). And why?

2) Consider the projection map $$ \pi: \overline{\mathcal{C}}(k) \rightarrow (\mathbb{P}^2)^k $$ Is it true that the image $$ \pi( \overline{\mathcal{C}}(k) - \mathcal{C}(k) ) \subset (\mathbb{P}^2)^k$$ is "small" inside $(\mathbb{P}^2)^k$, ie it is contained in a union of subavrieties of dimension $2k-1$ or less?

The specific example I have in mind is when $\mathcal{V} = \mathcal{A}$, where $$\mathcal{A}:= \{ [f] \in \mathcal{D}: f([1,0,0]) =0, ~~\nabla f|_{[1,0,0]} =0, ~~det \nabla^2 f|_{[1,0,0]} \neq 0 \} $$

ie the space of curves having a simple node at $[1,0,0]$. It can be shown that a generic element of $\mathcal{A}$ has only one singular point (namely $[1,0,0]$).

share|improve this question
add comment

1 Answer

1) It is dense because, given a curve with a finite set of points, one can modify the curve's equation slightly so that it is smooth except for $[1,0,0]$, and then modify each of the points slightly so they still lie on the curve.

More precisely, given a point which might not lie in the closure, since the space of curves smooth away from $[1:0:0]$ is dense in the space of all curves, we can view the curve with the marked points as a special fiber to a family of curves over a DVR whose generic fiber is smooth. Then we lift the marked points, possibly having to make an integral extension to the DVR, just by fixing one coordinate and then adjoining a solution to the equation. Then we look at the image of the generic point in $C$ and deduce that the questionable point lies in the closure of that point.

2) Clearly not if $d$ is sufficiently large relative to $k$. Clearly so if $k$ is sufficiently large relative to $d$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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