Is there some standard technique or approach to determine when a (irreducible) subvariety of a rationally connected variety is again rationally connected? Any reference/text dealing with this kind of question will be most welcome.

The example that I have in mind is the following: Let $P$ be the Hilbert polynomial of a complete intersection curve in $\mathbb{P}^3$ and $L$ an irreducible component of the corresponding Hilbert scheme parametrizing complete intersection curves in $\mathbb{P}^3$ with Hilbert polynomial $P$. Denote by $V \subset L$ the sublocus parametrizing curves which are not smooth. As far as I understand $L$ is rationally connected. Am I right? Then, I want to understand when is a connected component of $V$ again rationally connected.

  • 3
    $\begingroup$ Every projective variety embeds in projective space, and projective space is rationally connected. So, in complete generality, it cannot be easier to determine whether a general projective subvariety of a rationally connected variety is again rationally connected than it is to determine rational connectedness for an abstract projective variety. Having said this, let me be more honest: for specific pairs, of course you can try to use what you know about that pair to simplify the computation . . . $\endgroup$ – Jason Starr Sep 23 '15 at 10:29
  • 1
    $\begingroup$ For instance, every singular complete intersection curve is singular at some point with a given 2-dimensional subspace of the Zariski tangent space. So you can form the incidence variety of a complete intersection curve together with a singular point and a 2-dimensional tangent space. There is a forgetful morphism from the incidence correspondence to the dual projectivized tangent bundle of $\mathbb{P}^3$ that only remembers the singular point and tangent space (a Cheshire cat leaving behind its grin) . . . $\endgroup$ – Jason Starr Sep 23 '15 at 10:32
  • 1
    $\begingroup$ By homogeneity, the forgetful morphism is flat and surjective. Thus, by the rationally connected fibration theorem, it suffices to prove that the fibers are rationally connected. For a given point and tangent space, it is a linear condition on hypersurfaces to contain that point and to contain the given 2-dimensional subspace in its tangent space. Thus, the parameter space for tuples of hypersurfaces containing the given point and 2-dimensional subspace (i.e., non-reduced local Artin scheme) is unirational for the same reason that all of $L$ is rationally connected. $\endgroup$ – Jason Starr Sep 23 '15 at 10:37
  • $\begingroup$ @JasonStarr Thank you very much for the answer. This is very helpful. May be you could put this as an answer. $\endgroup$ – Ron Sep 23 '15 at 11:00

Let $[x,y,z,w]$ be homogeneous coordinates on $\mathbb{P}^3$ so that $\Gamma_*(\mathcal{O}_{\mathbb{P}^3})$ equals $k[x,y,z,w]$. Let $p$ be the point $[0,0,0,1]$ in these coordinates, whose associated homogeneous ideal $\Gamma_*(\mathcal{I}_{p/\mathbb{P}^3})$ is $\langle x,y,z \rangle$. Let $A\subset \mathbb{P}^3$ be the nonreduced, local Artin scheme supported at $p$ whose associated homogeneous ideal $I=\Gamma_*(\mathcal{I}_{A/\mathbb{P}^3})$ is $\langle x,y^2,yz,z^2 \rangle$.

For a given pair of positive integers, $(d,e)$, one parameter space (not the Hilbert scheme) for complete intersection curves is the open subscheme $L'$ of $$M':=\mathbb{P}k[x,y,z,w]_d \times_{\text{Spec}(k)} \mathbb{P}k[x,y,z,w]_e,$$ parameterizing pairs $([E(x,y,z,w)],[F(x,y,z,w)])$ of homogeneous polynomials of degree $d$, resp. $e$, such that the zero scheme $C:=\text{Zero}(E,F)\subset \mathbb{P}^3$ is a complete intersection curve. There is a dominant morphism, $$ \pi: L'\to L, \ ([E(x,y,z,w)],[F(x,y,z,w)]) \mapsto [C].$$ The fiber over a point $[C]$ is a dense open subset of the product of projective spaces $$\mathbb{P} \Gamma(\mathcal{I}_{C/\mathbb{P}^3}(d))\times_{\text{Spec}(k)}\mathbb{P}\Gamma(\mathcal{I}_{C/\mathbb{P}^3}(e)).$$

With this description, the subscheme of $K'$ parameterizing curves $C$ that contain $A$ is the intersection of the open subset $L'$ with the subvariety $$ N' = \mathbb{P}I_d \times_{\text{Spec}(k)}\mathbb{P}I_e.$$ In particular, $K'$ is a dense open subset of a product of projective spaces, hence $K'$ is rational.

Finally, there is a natural action of the automorphism scheme $\text{Aut}(\mathbb{P}^3,\mathcal{O}(1)) = \textbf{GL}_{4,k}$ on $\Gamma_*(\mathcal{O}_{\mathbb{P}^3})$ (in the sense of GIT, $G$-linearizations, etc.) That action induces an action on every projective space $\mathbb{P}k[x,y,z,w]_m$, and thus also an action on $M'$. The open subset $L'$ of $M'$ is invariant for this action. Denote the action morphism as follows, $$m:\textbf{GL}_{4,k} \times_{\text{Spec}(k)} L' \to L'.$$ There is an induced composition morphism, $$ \textbf{GL}_{4,k}\times_{\text{Spec}(k)} K' \xrightarrow{m} L' \xrightarrow{\pi} L.$$ I claim that the closure of the image is $V$. Thus, since $V$ is dominated by a rational variety, $V$ is unirational.

  • $\begingroup$ I am a little confused. Could you please clarify what is the role of the specific description of $A$? Is it related to the tangent space to the complete intersection curves at the singular point or the incidence correspondence you mention in your comments? In particular, I am not able to see the claim you make towards the end that "the closure of the image is $V$." $\endgroup$ – Ron Sep 27 '15 at 12:24
  • $\begingroup$ By construction, the image of $\textbf{GL}_{4,k}\times K'$ parameterizes those complete intersection curves that contain the image of $A$ under some projective automorphism of $\mathbb{P}^3$, i.e., they contain a point $p$ and a $2$-dimensional subspace of the Zariski tangent space of $\mathbb{P}^3$ at $p$. This is precisely to say, the image parameterizes complete intersection curve that are singular at some point $p$. $\endgroup$ – Jason Starr Sep 27 '15 at 13:30
  • $\begingroup$ To check if I am understanding correctly, are you saying that a complete intersection curve in $\mathbb{P}^3$ is singular if and only if after applying a projective automorphism the curve transforms into another curve whose ideal contains $A$? If so, is this fact obvious? $\endgroup$ – Ron Sep 27 '15 at 13:58
  • $\begingroup$ I am saying that a curve $C$ in $\mathbb{P}^3$, whether or not it is a complete intersection, is singular if and only if it contains a closed subscheme that is the image of $A$ under a projective automorphism of $\mathbb{P}^3$. Indeed, being singular at $p$ precisely means that the curve contains a $2$-dimensional subspace of the Zariski tangent space of $\mathbb{P}^3$ at $p$. $\endgroup$ – Jason Starr Sep 27 '15 at 14:01
  • $\begingroup$ Thank you. I had understood the last line ("Indeed, being singular at p ....") when you mentioned it in the comment itself, before the answer. What I was not sure was that having a two dimensional subspace of the Zariski tangent space is equivalent to $C$ containing $A$, upto projective automorphism. $\endgroup$ – Ron Sep 27 '15 at 14:41

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.