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Let $X$ be a complex projective scheme (resp. complex analytic space). The Hilbert scheme (resp. Douady space) parameterizes closed subschemes (resp. complex analytic subspaces) of $X$. More precisely, it represents the functor $F$ from schemes (resp. complex analytic spaces) to sets given by $$F(S)=\{Y\subset X\times S|\, Y \mbox{- closed subscheme (resp. anal. subspace), flat proper over } S\}.$$

Question 1. Let $X_1\subset X$ be a closed subscheme. Is it true that the Hilbert scheme of $X_1$ is a closed subscheme of the Hilbert scheme of $X$? The same question for the Douady spaces.

Question 2. I would like to have a modification of the Hilbert scheme/ Douady space which parameterizes closed subschemes of $X$ with a marked point. I guess one should consider the functor $$G(S)=\{Y\subset X\times S \mbox{ as above and a morphism } f\colon S\to Y|\, p\circ f=Id_S\},$$ where $p\colon Y\to S$ is the natural projection. Is the functor $G$ representable?

Question 3. Assume that Questoin 2 has a positive answer, namely the functor $G$ is representable by a scheme (resp. complex analytic space) $\mathcal{F}$. Obviously we have the morphisms $\mathcal{F}\to X$ and $\mathcal{F}\to Hilb(X)$ which forget either the subscheme or the marked point. Is it true that the induced morphism $\mathcal{F}\to X\times Hilb(X)$ is a closed imbedding? The same question for Douady space.

I believe answers to all these questions are well known to experts. Is there a standard reference to this material? The complex analytic case is more important for me, but the algebraic case is also of interest.

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Let me give an answer for Hilbert schemes.

Q. 1. Yes, the reason is that the functor $F_{X_1}$ is a closed subfunctor of the functor $F_X$ with respect to the natural embedding. In other words, for each $S$ and each $Y \in F_X(S)$ the maximal subscheme $S_1 \subset S$ such that $Y\times_S S_1 \in F_{X_1}(S_1)$ is closed in $S$. In fact subscheme $S_1$ can be defined as the zero locus of the canonical morphism $\pi_*I_{S\times X_1}(n) \to \pi_*O_Y(n)$ obtained by pushing forward the composition $I_{S\times X_1} \subset O_{S\times X_1} \to O_Y$ twisted by the sheaf $O_X(n)$ with $n$ sufficiently large (so that $I_{X_1}(n)$ is globally generated) via the projection $\pi:S\times X \to S$. The reference here is Grothendieck's FGA (or more recent "FGA explained").

Q. 2. I think it is more natural to consider so-called flag Hilbert scheme which parameterizes flags of subschemes $Y' \subset Y \subset S\times X$. If you specify the Hilbert polynomial of (fibers of) $Y'$ to be constant 1, this will give you precisely what you want. Representability then is known. I do not know a reference, but I think you can find one by googling for "flag Hilbert scheme".

Q. 3. Yes, this is a general fact for flag Hilbert schemes. Alternatively you can argue in the same way as in Q. 1.

I am completely sure that the same arguments should work for Douady spaces.

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    $\begingroup$ "I am completely sure that the same arguments should work for Douady spaces." I, also, am completely sure that the same basic results are true. However, the same arguments cannot work when there is no ample invertible sheaf $\mathcal{O}(1)$. Of course in the algebraic case, Artin's techniques apply. $\endgroup$ – Jason Starr Jul 21 '14 at 14:26
  • $\begingroup$ @Sasha: Many thanks. Could you say more precisely what is "FGA explained"? $\endgroup$ – makt Jul 22 '14 at 7:12
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    $\begingroup$ @semyon alesker: ams.org/bookstore-getitem/item=SURV-123-S $\endgroup$ – Sasha Jul 22 '14 at 7:37
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To your second and third question:

Let $Z \subset Hilb(X) \times X$ be the universal family over $Hilb(X)$. Then, immediately, $Z$ represents the functor you are asking for.

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  • $\begingroup$ How the universal family is defined in your approach? $\endgroup$ – makt Jul 22 '14 at 18:39
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    $\begingroup$ @semyonalesker: The universal family is part of the definition of the Hilbert functor / Douady functor. So your question is equivalent to asking: how is a representing object for the Hilbert functor defined? This is explained in the references that sasha listed. Of course the construction of the Douady space, representing the Douady functor, requires serious analysis, not just algebraic geometry. $\endgroup$ – Jason Starr Jul 23 '14 at 11:49

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