On the place where $\mathrm{Hilb}_{lines}^{x}(X)$ is smooth.

2011-06-28T06:48:03Z

Let $X\subset \mathbb{P}_{\mathbb{C}}^N$ be irreducible generically smooth closed subscheme and let $\mathrm{Hilb}_{lines}^{x}(X)$ denote the Hilbert scheme of lines contained in $X$ and passing through the point $x\in X$. Is it true that the set $$ \{ x\in X : \mathrm{Hilb}_{lines}^{x}(X) \mbox{ is smooth } \} $$ is constructible?

Thanks in advance.

Answer by ulrich for On the place where $\mathrm{Hilb}_{lines}^{x}(X)$ is smooth.

2011-06-28T14:48:29Z

Yes. This can be seen as follows:

Let $Hilb_{lines}(X)$ denote the Hilbert scheme of lines in $X$ and let $\Gamma \subset X \times Hilb_{lines}(X)$ be the correponding universal family. Then $Hilb_{lines}^x(X) = p^{-1}(x)$ where $p:\Gamma \to X$ is induced by the first projection. So we are reduced to the folowing (well-known) statement:

Let $f:Y \to X$ be a proper morphism of finite type schemes over a field. Then the set $\{x \in X | f^{-1}(x) \mbox{ is smooth } \}$ is constructible.

To prove this, by replacing $X$ by $X_{red}$ and $Y$ by $Y \times_X X_{red}$ we may assume $X$ is reduced (since we only care about the fibres). By generic flatness, we may find a finite stratification of $X$ by locally closed reduced subschemes $X_i$ so that the induced morphisms $Y \times_X X_i \to X_i$ are all flat. For a flat proper morphism the locus of points in the base so that the fibres are smooth is open. It follows that the set we are interested in is a finite union of open subsets of closed subsets of $X$, so is constructible.

On the place where $\mathrm{Hilb}_{lines}^{x}(X)$ is smooth. - MathOverflow

On the place where $\mathrm{Hilb}_{lines}^{x}(X)$ is smooth.

Answer by ulrich for On the place where $\mathrm{Hilb}_{lines}^{x}(X)$ is smooth.