Let $Q \subset \mathbb{P}^n$ be a smooth quadric hypersurface. Where can I find a proof of/can anyone supply a proof of$$\text{Hilb}\_{2m + 1}(Q) \cong \text{Bl}\_{OG(3, n+1)}G(3, n+1)?$$Can we conclude that when $n > 5$, the Picard group of $\text{Hilb}\_{2m+1}(Q)$ has rank two?


Apologies in advance, for I am not an algebraic geometrer...



Smooth quadric hypersurface, Hilbert scheme is blowup of Grassmannian?
I am sure there are more direct references, but it is fairly easy to prove as well. First of all, the Hilbert scheme $\text{Hilb}\_{2m+1}(\mathbb{P}^n)$ is a $\mathbb{P}^5$-bundle over the Grassmannian $G(3,n+1)$. There must be other sources, but one source is Theorem 3.4.1 of Alex Lee's senior thesis.


Alex Lee  

The Hilbert Scheme of Curves in $\mathbb{P}^3$  

<http://www.uio.no/studier/emner/matnat/math/MAT4230/h10/undervisningsmateriale/ALee_Hilbertschemes.pdf>


Of course there is an obvious morphism from the total space of that $\mathbb{P}^5$-bundle to $\text{Hilb}\_{2m+1}(\mathbb{P}^n)$. It requires just a little work to prove that this morphism is an isomorphism.


Given this identification, it is easy to see that the underlying closed subset of $\text{Hilb}\_{2m+1}(Q^n)$ and the closed subset $\text{Bl}\_{OG(3,n+1)}G(3,n+1)$ are equal as closed subsets of $\text{Hilb}\_{2m+1}(\mathbb{P}^n)$. It only remains to prove that $\text{Hilb}\_{2m+1}(Q^n)$ is reduced as a scheme.


On $\text{Hilb}\_{2m+1}(\mathbb{P}^n)$ there is a rank $5$, locally free sheaf obtained by pulling back $\mathcal{O}\_{\mathbb{P}^n}(2)$ to the universal curve and then pushing forward to the Hilbert scheme. It requires some computation to prove that there are no higher direct images and that the pushforward is locally free, but that is straightforward because the fibers are all degree $2$ Cartier divisors in $2$-planes (so just use standard results about cohomology of invertible sheaves on $\mathbb{P}^2$). The defining equation of your quadric defines a global section of this locally free sheaf, and $\text{Hilb}\_{2m+1}(Q^n)$ is the zero scheme of this global section. Thus, wherever $\text{Hilb}\_{2m+1}(Q^n)$ has codimension $5$ in $\text{Hilb}\_{2m+1}(\mathbb{P}^n)$, it is a local complete intersection. By the set-theoretic argument, $\text{Hilb}\_{2m+1}(Q^n)$ is everywhere a local complete intersection. In particular, this scheme is Cohen-Macaulay.


Finally, on the dense open subset of $\text{Hilb}\_{2m+1}(Q^n)$ parameterizing smooth, genus $0$ curves, it is straightforward to compute that $\text{Hilb}\_{2m+1}(Q^n)$ is smooth by checking vanishing of $H^1$ of the normal sheaf of the curve in $Q^n$: the tangent sheaf of $Q^n$ is globally generated, hence the normal sheaf is globally generated, and every globally generated coherent sheaf on $\mathbb{P}^1$ has vanishing $H^1$. Since $\text{Hilb}\_{2m+1}(Q^n)$ is generically smooth, it is generically reduced. Every Cohen-Macaulay scheme that is generically reduced is everywhere reduced. Since it is everywhere reduced, $\text{Hilb}\_{2m+1}(Q^n)$ equals the reduced closed subscheme $\text{Bl}\_{OG(3,n+1)}G(3,n+1)$ of $\text{Hilb}\_{2m+1}(\mathbb{P}^n)$.


