Let W be a subvariety of some Hilbert scheme of $P^n$, $n \geq 3$. Assume the general member $C_w \subset P^n$ of W is pure positive dimensional, not necessarily reduced (e.g., defined by the square of the ideal of a (variable) smooth curve in $P^3$).

Let $V_d={ (w , F_d) | F_d \; \text{is a hypersurface of degree d containing} \; C_w}$.

Then for all $d>>0$ the map $(w,F_d) \rightarrow F_d$ is generically injective.

In other words, for all $d>>0$, if w is general in W and $F_d$ is general in the space of hypersurfaces of degree d containing $C_w$ then w is the only member of W such that $F_d$ contains $C_w$.

  • 4
    $\begingroup$ What is the question? $\endgroup$ – Sasha Apr 19 '18 at 13:20
  • $\begingroup$ @Sasha. The question is whether, for $d$ sufficiently positive, for a sufficiently general closed point $w$ of $W$ parameterizing a closed subscheme $\text{Zero}(\mathcal{I}_w)$ of $\mathbb{P}^n$, for a general member $F_d$ of the linear series $\mathcal{I}_w\cdot \mathcal{O}(d)$, is $w$ the unique point $v$ of $W$ such that $F_d$ is in the linear series $\mathcal{I}_v\cdot \mathcal{O}(d)$. This can fail for small values of $d$, e.g., if $d$ equals $1$ and $W$ parameterizes closed subschemes that do not span $\mathbb{P}^n$. $\endgroup$ – Jason Starr Apr 19 '18 at 15:21
  • $\begingroup$ Stated as above, the question is whether, for every sufficiently general $w$ in $W$, for every $v$ in $W\setminus\{w\}$, does the quotient sheaf $\mathcal{I}_w/(\mathcal{I}_v\cap \mathcal{I}_w)$ have positive-dimensional support. If so, then the Hilbert function of this quotient increases as a function of $d$. Together with the usual upper semicontinuity arguments, there exists an integer $d_0$ such that for every integer $d\geq d_0$, the Hilbert function in degree $d$ is strictly greater than the dimension of $W$, and the result follows. $\endgroup$ – Jason Starr Apr 19 '18 at 17:06

I believe that this is not true. For every integer $g$ that is sufficiently positive, I believe that Henry Pinkham has proved that projective cones in $\mathbb{P}^g$ over canonically embedded, genus $g$ curves in $\mathbb{P}^{g-1}$ are "rigid" in the sense that the only (embedded) deformations arise as cones over deformations of the curve in $\mathbb{P}^{g-1}$. (Ed. Actually, that might be due to Mumford, but I am having trouble tracking down a reference.) Assuming this, let $\mathcal{I}\subset \mathcal{O}_{\mathbb{P}^g}$ be the ideal sheaf of such a cone.

Denote by $\mathfrak{m}$ the maximal ideal sheaf in $\mathcal{O}_{\mathbb{P}^g}$ of the vertex of the cone. Consider the ideal sheaves $\mathcal{J}$ such that $$\mathfrak{m}\cdot \mathcal{I} \subset \mathcal{J} \subset \mathcal{I}$$ and such that the quotient $\mathcal{I}/\mathcal{J}$ has length $1$, i.e., the quotient equals the skyscraper sheaf of the vertex. By Max Noether's Theorem, the dimension of the locus of such ideals (for fixed ideal $\mathcal{I}$) equals $(g^2-5g+4)/2$. On the other hand, the locus parameterizing ideal sheaves $\mathcal{G}\subset \mathcal{I}$ of colength $1$ whose cokernel has support disjoint from the vertex has dimension $g$. If $g\geq 8$, it appears that the locus of ideal sheaves $\mathcal{J}$ as above forms an irreducible component of the Hilbert scheme (allowing also the cone to vary).

That is bad news. For each such ideal sheaf $\mathcal{I}$ and ideal sheaf $\mathcal{J}$, consider any other ideal sheaf $\mathcal{J}'\subset \mathcal{I}$ of colength $1$ containing $\mathfrak{m}\cdot \mathcal{I}$. Among polynomials $F_d$ of degree $d$ that are contained in $\mathcal{J}\cdot \mathcal{O}(d)$, it is only one more condition to be contained in $\mathcal{J}\cap \mathcal{J}'\cdot \mathcal{O}(d)$. Thus, varying $\mathcal{J}'$ in a $1$-parameter family, this collection of hyperplanes in $H^0(\mathbb{P}^g,\mathcal{J}\cdot \mathcal{O}(d))$ should sweep out the entire vector space. Therefore, it appears to me that the question above has a negative answer.


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.