Connectedness of Quot schemes

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.

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 $$\mathcal{O}(1,1)$$.
• 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 $(3,1)$. Feb 22, 2022 at 7:07
• 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. Feb 22, 2022 at 7:18
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.
• 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. Feb 28, 2022 at 11:29