# Reducibility of a Hilbert scheme in projective space

My question concerns the computation of the Hilbert scheme $\mathsf{Hilb}_{3}^{2x+1}$, which parametrizes all curves of degree $2$ and genus $0$ in $\mathbb{P}^{3}_{k}$, with $k$ algebraically closed. I would like to prove that $\mathsf{hilb}_{3}^{2x+1}$ is reduced. To do that, I proceed in the following way:

(1) I want to show that $\text{dim}(\mathsf{Hilb}^{2x+1}_{3})=8$;

(2) I want to show that $\text{dim}\ \text{T}_{p}(\mathsf{Hilb}^{2x+1}_{3})=8$, for any closed point $p\in \mathsf{Hilb}^{2x+1}_{3}$.

I cannot seem to be able to compute part (1). I thought about simply counting the parameters of the generic quadric in $\mathbb{P}^{3}$, but it clearly doesn't work. With part (2) I made a little progress. I found on "Curves in Projective Space", Joe Harris, Les Presses de L'université de Montréal, 1982, the following equivalence

$\qquad \qquad \qquad \qquad \qquad$ $h^{0}(C,\mathcal{N}_{C/\mathbb{P}^{3}})=h^{0}(C,\mathcal{O}(1)\oplus \mathcal{O}(2))=3+5$

I thought this could be traced back to some split exact sequence, but I don't seem to be able to find which one. I also don't understand why $h^{0}(C,\mathcal{O}(1))=3$ and $h^{0}(C,\mathcal{O}(2))=5$.

• Dear Fra, I think your title should read "reducedness" rather than "reducibility". Regards, – Matt E Feb 5 '14 at 1:58

(1) A naive count of parameters could go as follows: we have that the space of polynomials of degree $2$ in two variables (the homogeneous coordinates of $\mathbb P^1$) has dimension $h^0(\mathbb P^1,\mathscr O_{\mathbb P^1}(2))=3$. You need $4$ such polynomials to get a degree $2$ curve in $\mathbb P^3$. But you certainly want to identify conics given by polynomials who are multiples of each other (so we remove $1$ parameter), and you might also want to forget automorphisms of $\mathbb P^1$ (which form a $3$-dimensional group).

Putting all this together we get $$\#\,\textrm{parameters}=(4\cdot 3)-1-3=8.$$

This is not quite a proof that the Hilbert scheme has dimension $8$, because there might be extraneous components in the Hilbert scheme, like for twisted cubics for instance. But this is not the case, as (roughly) conics are much less rigid than twisted cubics.

(2) Let us embed $C$ as a conic hyperplane section inside a smooth quadric surface $S\subset \mathbb P^3$. Then we have the exact sequence $$0\to \mathcal N_{C/S}\to \mathcal N_{C/\mathbb P^3}\to \mathcal N_{S/\mathbb P^3}|_C\to 0.$$ This sequence splits because $C$ is a complete intersection. We have $$\mathcal N_{C/S}\cong\mathscr O_C(1)\cong\mathscr O_{\mathbb P^1}(2)$$ and $$\mathcal N_{S/\mathbb P^3}|_C\cong\mathscr O_C(2)\cong \mathscr O_{\mathbb P^1}(4).$$

Thus, the splitting condition says $\mathcal N_{C/\mathbb P^3}\cong\mathscr O_{\mathbb P^1}(2)\oplus\mathscr O_{\mathbb P^1}(4)$, and we have $$h^0(C,\mathcal N_{C/\mathbb P^3})=h^0(\mathscr O_{\mathbb P^1}(2))+h^0(\mathscr O_{\mathbb P^1}(4))=3+5=8.$$

• I'm a little skeptical about the part where you choose four degree $2$ polynomials from a $3$ dimensional space to get an embedding. I think instead one should argue that any degree $2$ curve in $\mathbb P^3$ is a plane conic, so the parameter count gives the number of planes ($3$ dimensional) $+$ the number of plane conics ($5$ dimensional). This gives a $3+5=8$ dimensional family of points of the Hilbert scheme. – Andrew Feb 4 '14 at 19:47
• Dear @Andrew, I wanted to count maps $\mathbb P^1\to\mathbb P^3$ of degree $2$, which are not necessarily embeddings. The image of any such map is what I call a rational curve of degree $2$ in $\mathbb P^3$. To give one such map is to give (up to constants) $4$ binary polynomials of degree $2$ which do not vanish at any point of $\mathbb P^1$. – Brenin Feb 4 '14 at 21:58
• But how do you control the arithmetic genus for the non-embeddings? – Andrew Feb 4 '14 at 22:09
• In any case, maybe it's not wrong, it's just a little less clear to me than the other argument. Btw, I meant smooth & irreducible degree $2$ curves. I'd have to think more about how to handle the singular curves, though they shouldn't matter for the purposes of a dimension count. – Andrew Feb 5 '14 at 0:38
• Yes I am also thinking how to deal with that. But what can actually happen? Are not all the singular ones also of arithmetic genus $0$? – Brenin Feb 5 '14 at 9:06