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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.

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    $\begingroup$ 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
    Commented Sep 23, 2015 at 10:29
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    $\begingroup$ 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
    Commented Sep 23, 2015 at 10:32
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    $\begingroup$ 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
    Commented Sep 23, 2015 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
    Commented Sep 23, 2015 at 11:00

1 Answer 1

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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.

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  • $\begingroup$ 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
    Commented Sep 27, 2015 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
    Commented Sep 27, 2015 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
    Commented Sep 27, 2015 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
    Commented Sep 27, 2015 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
    Commented Sep 27, 2015 at 14:41

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