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Let $X$ be a very general hypersurface of degree $6$ in $\mathbb{P}^3$. Fix an integer $d$. Define $Y:= \{ C \in \mathbb{P}(H^0(\mathcal{O}(3))) \text{ such that } \text{dim}(\text{ Hilb}^d(X \cap C)) > d \}$, where $\text{Hilb}^d(X \cap C)$ denotes the Hilbert scheme of zero-dimensional subschemes of length $d$. Note that if the intersection $X \cap C$ is smooth then $\text{Hilb}^d(X \cap C)$ has dimension $d$. My question is the following: What is the dimension of $Y$ ? Can we give an effective bound on the dimension of $Y$ ?

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  • $\begingroup$ Out of curiosity: Why is $Y$ non-empty? Do you have an example? $\endgroup$ – red_trumpet Feb 14 at 9:56
  • $\begingroup$ I do not have any particular example. But if the intersection has some bad singularity then it is possible to have large dimensional Hilbert scheme. $\endgroup$ – user130022 Feb 14 at 9:59
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I believe that in characteristic zero for smooth $X$ we always have $\dim \text{Hilb}^d(X\cap C) = d$.

Sketch of proof: Let $Y = X\cap C$. For a point $p\in Y$ and any $e\in \mathbb{N}$ denote by $\text{Hilb}^e(Y, p) \subset \text{Hilb}^e(Y)$ the locus consisting of subschemes supported only at $p$. Since $X\subset \mathbb{P}^3$ is smooth, the singularities of $Y$ are planar, so that for every $p\in Y$ the locus $\text{Hilb}^e(Y, p)$ can be identified with a subscheme of $\text{Hilb}^e(\mathbb{A}^2, 0)$. Briançon proved that $\dim \text{Hilb}^e(\mathbb{A}^2, 0) = e-1$, so $\text{Hilb}^e(Y, p)\leq e-1$. (here we need characteristic zero)

Now fix a partition $\lambda = (\lambda_1,\ldots, \lambda_m)$ of $d$ and consider the locus $S_{\lambda}$ in $\text{Sym}^d(Y)$ consisting of cycles of the form $\sum \lambda_i p_i$ for distinct $p_i\in Y$. By the above estimate, the fiber of Hilbert-Chow morphism over any element of $S_{\lambda}$ has dimension at most $d-m$. The locus $S_{\lambda}$ has codimension $d-m$ in $\text{Sym}^d(Y)$, so we get that its preimage has dimension at most $d$, summing over all $\lambda$ we get the claim.

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  • $\begingroup$ thank you very much for the answer. $\endgroup$ – user130022 Feb 15 at 4:45

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