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 $Y$ ?



complete intersection curves with large Hilbert scheme of points
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](https://link.springer.com/article/10.1007%2FBF01390164) 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.


