Let $\mathcal{D} \approx \mathbb{P}^{\delta_d}$ be the space of homogeneous degree $d$ polynomials in three variables (up to scaling), where $\delta_d = \frac{d(d+3)}{2}$. Define $\mathcal{A}$ to be space of degree $d$ curves with a strict node at the point $[1,0,0]$, ie

$$ \mathcal{A} := \{ f \in \mathcal{D}: f([1,0,0]) =0, ~~ \nabla f |_{[1,0,0]} =0, \quad det \nabla^2 f ([1,0,0]) \neq 0 \}$$

Define $H_p$ to be the space of curves through the point $p$ and define

$$ H_p^* := \{ f \in \mathcal{D}: f(p) =0, ~~ \nabla f |_{p} \neq 0 \}$$

It can be shown that if $d$ is sufficiently large, then $\mathcal{A}$ is a smooth complex manifold of the expected dimension $k = \delta_d -3$. Also note that a generic element of $\mathcal{A}$ has only one node as pointed out in this discussion

Does a generic curve inside the space of curves with a node at a specific point have only finitely many nodes?

My question is the following........does it now follow that for generic choices of $k$ points, the following is true

$$ \overline{\mathcal{A}} \cap H_{p_1} \ldots \cap H_{p_k} = \mathcal{A} \cap H_{p_1}^* \ldots \cap H_{p_k}^* $$

Assume $d$ is sufficiently large.

Is there some reference for this fact?

$\textbf{EDIT:}$ I want to add one further question to what I had asked some time ago. As Jason has shown, for generic choice of points $(p_1, \ldots,p_k)$ the equality
$$ \overline{\mathcal{A}} \cap H_{p_1} \ldots \cap H_{p_k} = \mathcal{A} \cap H_{p_1}^* \ldots \cap H_{p_k}^* $$
holds as sets. My question is: Is it also true that the intersection $$ \mathcal{A} \cap H_{p_1}^* \ldots \cap H_{p_k}^*$$ is $\textit{transverse}$ for generic choices of $k$ points?


1 Answer 1


Dear Ritwik -- This is not a precise reference for the result you state. However, standard dimension results, i.e., Exercise II.3.22, p. 95 of Hartshorne's Algebraic Geometry, lead to a proof very quickly (and I would consider this as a standard application of dimension theory). Let $\overline{\mathcal{C}}\subset \overline{\mathcal{A}}\times \mathbb{P}^2$ be the universal curve over $\overline{\mathcal{A}}$, i.e., a (geometric) point of $\overline{\mathcal{C}}$ parameterizes a pair $([f],p)$ of a curve $[C]=[Z(f)]$ in $\overline{\mathcal{A}}$ and a point $p\in C$, possibly $p=[1,0,0]$. Note that the projection $$\pi:\overline{\mathcal{C}}\to \overline{\mathcal{A}}$$ is a flat morphism of relative dimension $1$. By the way, the other projection, $$\text{ev}:\overline{\mathcal{C}}\to \mathbb{P}^2$$ is certainly not flat, since the fiber over $[1,0,0]$ has larger dimension than the fiber over any other point $p$.

Also define $\mathcal{C}$ to be the open subset of $\overline{\mathcal{C}}$ which is the intersection of the open subsets $\pi^{-1}(\mathcal{A})$ and the smooth locus of the morphism $\pi$, i.e., in your terminology the open locus where $\nabla f|_p\neq 0$. Concretely, $\mathcal{C}$ parameterizes pairs $([C],p)$ where $C$ has a strict node at $[1,0,0]$ and where $p$ is a smooth point of $C$.

Now consider the $k$-fold fiber product $$\overline{\mathcal{C}}^k := \overline{\mathcal{C}}\times_{\overline{\mathcal{A}}} \dots \times_{\overline{\mathcal{A}}} \overline{\mathcal{C}}.$$ This parameterizes data $([C],p_1,\dots,p_k)$ of a curve $[C]=[Z(f)]$ in $\overline{\mathcal{A}}$ and $k$ points $p_1,\dots,p_k$ in $C$, not necessarily smooth and not necessarily distinct. Similarly, define $\mathcal{C}^k$ to be the $k$-fold fiber product $\mathcal{C}\times_{\mathcal{A}}\times \dots \times_{\mathcal{A}} \mathcal{C}$. This is an open subscheme of $\overline{\mathcal{C}}^k$. It parameterizes data $([C],p_1,\dots,p_k)$ such that $[C]$ is in $\mathcal{A}$ and such that $p_1,\dots,p_k$ are smooth points of $C$ (but possibly coincident -- this will be irrelevant soon).

There are a couple of observations. First, as a $k$-fold fiber product of flat morphisms of relative dimension $1$, the projection morphism, $$\pi_k:\overline{\mathcal{C}}^k \to \overline{\mathcal{A}},$$ is also flat of relative dimension $k$; thus $\overline{\mathcal{C}}^k$ has pure dimension $2k$. Similarly, since the projection $\mathcal{C}\to \mathcal{A}$ is smooth of relative dimension $1$, also the projection, $$\pi:\mathcal{C}^k\to \mathcal{A},$$ is smooth of relative dimension $k$; thus $\mathcal{C}^k$ is smooth of pure dimension $2k$. Second, the open subscheme $\mathcal{C}^k$ is dense in $\overline{\mathcal{C}}^k$, i.e., this open subscheme intersects every irreducible component (in fact there is only one irreducible component). To prove this, first note that every irreducible component dominates $\overline{\mathcal{A}}$ since $\pi$ is flat. Thus it suffices to prove that for a general point $[C]$ of $\overline{\mathcal{A}}$, the fiber of $\pi_k$ over $[C]$ intersects $\mathcal{C}^k$ in a dense open. As discussed in the previous post, $C$ is smooth away from $[1,0,0]$. The fiber of $\pi_k$ over $[C]$ is canonically isomorphic to the $k$-fold product $C^k := C\times \dots \times C$ (of course all absolute fiber products are relative to the Spec of our ground field). And the intersection with $\mathcal{C}^k$ is the fiber product $(C\setminus \{[1,0,0]\})^k$, which is clearly dense since the complement is the closed subset of strictly smaller dimension $(k-1)$ which is the union of the $k$ subvarieties of the form $$C\times \dots \times C \times \{[1,0,0]\} \times C \times \dots \times C.$$ Because $\mathcal{C}^k$ is dense in $\overline{\mathcal{C}}^k$, the complement $D$ is a proper subvariety of dimension $\leq 2k-1$.

Finally we are ready to set up the application of dimension theory. Consider the $k$-fold product of the morphism $\text{ev}$, $$\text{ev}:\overline{\mathcal{C}}^k \to (\mathbb{P}^2)^k, \ ([C],p_1,\dots,p_k)\mapsto (p_1,\dots,p_k).$$ For the smooth open $\mathcal{C}^k$ of pure dimension $2k$, Exercise II.3.22(e) states that there exists a dense open subset $W\subset (\mathbb{P}^2)^k$ such that for every point $(p_1,\dots,p_k)$ of $W$, the fiber $g^{-1}(p_1,\dots,p_k)\cap \mathcal{C}^k$ has the expected dimension $0$ (or it is empty). Next, we can write the closed complement $D$ of $\mathcal{C}^k$ as a union of locally closed subsets, $$D=D_1\sqcup \dots \sqcup D_r,$$ such that each $D_i$ is smooth and irreducible. As a locally closed subset of $D$, $\text{dim}(D_i) \leq \text{dim}(D) < 2k$. Since $(\mathbb{P}^2)^k$ has dimension $2k$, none of the morphisms $D_i \to (\mathbb{P}^2)^k$ are dominant, i.e., the image of each is contained in a proper, closed subset $E_i\subset (\mathbb{P}^2)^k$. Thus, up to replacing $W$ by the dense open subset $W\setminus (E_1\cup \dots E_r$, we conclude that for every point $(p_1,\dots,p_k)$ of $W$, the fiber $f^{-1}(p_1,\dots,p_k)\cap D_i$ is either empty or has the expected dimension $-1$. Since a variety cannot have negative dimension, Kleiman-Bertini precisely says that the fiber is empty. In other words, the fiber $f^{-1}(p_1,\dots,p_k)$ is disjoint from the closed subset $D$. Stated yet again, the fiber $f^{-1}(p_1,\dots,p_k)$ is completely contained in $\mathcal{C}^k$. Since $\pi(f^{-1}(p_1,\dots,p_k))$ equals $\overline{A}\cap H_{p_1}\cap \dots \cap H_{p_k}$, this gives your result.

In a manuscript, I believe that most algebraic geometry readers would be able to supply the argument above if you simply state that your result follows from dimension arguments.

Edit. In the first version of the answer, I used the Kleiman-Bertini theorem. This is valid (and is my own first instinct as an algebraic geometer). However, since there are issues with Kleiman-Bertini in positive characteristic, I feel it is better to phrase the argument in terms of dimension theory instead.

  • $\begingroup$ Thank you very much for your detailed reply. One further question as last time. I assume the argument will work if I change the question slightly, with A denoting the space of curves with an A_k singularity at [1,0,0]. $\endgroup$
    – Ritwik
    Commented Aug 4, 2012 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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