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In Alex Lee's undergraduate thesis (2000), it was said that the Hilbert scheme $H_{2m+2}(\mathbb{P}^3)$ has two components $\mathcal{H}',\mathcal{H}''$, where a general point of $\mathcal{H}'$ corresponds to a pair of skew lines and a general point of $\mathcal{H}''$ corresponds to a conic union a point. At the time of the thesis, the smoothness of neither of the two components were known.

Since then, it was shown in the paper Hilbert scheme of a pair of codimension two subspaces (2011) that $\mathcal{H}'$ is smooth. Is it known by now whether $\mathcal{H}''$ is smooth?

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2 Answers 2

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I think this paper by Chen and Nollet is relevant. Theorem 1.9 (proved as Theorem 4.3 on p. 16) shows the following about plane curves in $\mathbb P^3$.

The component $H_d\subset \textrm{Hilb}^{dz+2-g}(\mathbb P^3)$ whose general point is a degree $d$ plane curve union an isolated point is smooth for all $d\geq 1$.

Since there is a bijective morphism $f:\textrm{Bl}_\Sigma(\textrm{Hilb}^{dz+1-g}(\mathbb P^3)\times \mathbb P^3)\to H_d$, where $\Sigma$ is the incidence correspondence, $f$ is an isomorphism by Zariski's main theorem.

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I think this is correct. Moreover, I think one can describe explicitly $H''$ as follows.

First, consider $H_0$ the Hilbert scheme of conics. It is a $P^5$-bundle over $P^2$, in particular is smooth. Further, let $Z_0 \subset H_0 \times P^3$ be the universal conic. The natural map $Z_0 \to P^3$ is smooth (it is a fibration with fiber being $P^4$-bundle over $P^2$), hence $Z_0$ is smooth. And the claim is that $H''$ is isomorphic to the blowup of $H_0 \times P^3$ with center in $Z_0$, hence is smooth.

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  • $\begingroup$ I'd only be worried about whether the hilbert scheme is smooth at a point parameterizing a conic with an embedded point... $\endgroup$
    – DCT
    Sep 30, 2016 at 19:52
  • $\begingroup$ That is the exceptional divisor of the blowup. $\endgroup$
    – Sasha
    Sep 30, 2016 at 20:13
  • $\begingroup$ Is it clear the Hilbert scheme is reduced? $\endgroup$
    – DCT
    Sep 30, 2016 at 20:27
  • $\begingroup$ This particular component is isomorphic to the blowup of a smooth subvariety in a smooth variety. So it is smooth, and in particular reduced. If you want to check the reducedness of the whole Hilbert scheme, I think you should look carefully at the proof of the decomposition result mentioned in your question. $\endgroup$
    – Sasha
    Oct 1, 2016 at 5:32
  • $\begingroup$ Sorry for belabouring the point. I just think that your argument is set-theoretic. Without a calculation, it might be that there are obstructed infinitesimal deformations of a conic with an embedded point. There is still something to prove (which I think is done in the paper cited in the other answer as the title "detaching embedded points" suggests), since it might be that the relevant component of the Hilbert scheme is set-theoretically as you claimed, but it has some nonreduced "fuzz" supported along the locus of conics with and embedded point and is not smooth. $\endgroup$
    – DCT
    Oct 1, 2016 at 13:34

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