# complete intersection curves with large Hilbert scheme of points

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 $$Y$$ ?

• Out of curiosity: Why is $Y$ non-empty? Do you have an example? – red_trumpet Feb 14 at 9:56
• 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. – user130022 Feb 14 at 9:59

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.