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Let $X$ be a connected projective scheme over $\mathbb{C}$ and $E$ a coherent sheaf on $X$. Consider the Quot scheme $\operatorname{Quot}_X(E,P)$ of quotients of $E$ of fixed Hilbert polynomial $P$. As far as I know, this scheme is connected at least in the following cases:

  1. for Hilbert schemes on $\mathbb{P}^n$ (by Hartshorne)
  2. for punctual Quot schemes on smooth surfaces (by Fogarty, Ellingsrud–Lehn, Baranovski, etc.).

Are there any other known cases for which $\operatorname{Quot}_X(E,P)$ is connected? Could you give some examples where the connectedness fails? I am particularly interested in higher dimensions or singular surfaces/varieties.

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The simplest example when connectedness fails is the Hilbert scheme of lines on a quadric surface: $$ \mathrm{Hilb}_{1+t}(\mathbb{P}^1 \times \mathbb{P}^1) = \mathbb{P}^1 \sqcup \mathbb{P}^1, $$ where the polarization is by the line bundle $\mathcal{O}(1,1)$.

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    $\begingroup$ You cannot “fix” this by using a “refined” Hilbert polynomial: let $Y$ in $\mathbb{P}^3\times\mathbb{P}^9$ be the universal quadric surface, and let $X$ be $Y\times\mathbb{P}^1$. The cone of effective curve classes on this Fano manifold is the cone of triples of nonnegative integers: the degrees of the projections to $\mathbb{P}^1$, to $\mathbb{P}^3$, and to $\mathbb{P}^9$. Consider the curve class where the triple is $(1,4,0)$. The actual curves of arithmetic genus $0$ project to curves in $Y$ in a fiber ,I.e., a quartic in a quadric surface. The curve class might be $(2,2)$ or $(3,1)$. $\endgroup$
    – Jason Starr
    Feb 22, 2022 at 7:07
  • $\begingroup$ Please note that this $X$ is a Fano manifold. This does not contradict Cohen-Jones-Segal, because they require the curve class to be far from the boundary of the cone of effective curve classes. $\endgroup$
    – Jason Starr
    Feb 22, 2022 at 7:18
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Let $X$ be an irreducible variety, $E$ a locally free sheaf on $X$, and $n \geq 0$ an integer. The Quot scheme $\textrm{Quot}_X(E,n)$ is connected by Thm. 1.4 here. To achieve disconnectedness, probably one has to look at nonconstant polynomials, as Sasha's example suggests.

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  • $\begingroup$ do you think it is necessary to assume that $X$ is irreducible (rather than connected). It seems to me that the result should be true if $X$ is just connected. $\endgroup$
    – Samuel
    Feb 28, 2022 at 11:29
  • $\begingroup$ you're probably right! $\endgroup$
    – Andrea Ricolfi
    Mar 1, 2022 at 20:33

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