Question

Let $X\subset\mathbb{P}^N$ be a complex irreducible projective variety. I recall that a line $l\subset\mathbb{P}^N$ is a secant of $X$ if the lenght of the scheme $l\cap X$ is $\geq2$. The union of all secant lines of $X$ is the secant variety of $X$, denoted by $\mathrm{Sec}(X)$. If $q\in\mathrm{Sec}(X)$, $L_q(X)$ denotes the union of secant lines of $X$ passing through $q$, and $$\Sigma_q(X):=L_q(X)\cap X=\{x\in X: \langle x, q\rangle\mbox{ is a secant line of $X$}\}.$$ Let us assume that exists a nonempty open set $V\subset\mathrm{Sec}(X)\setminus X$ such that for every $q\in V$ we have $L_q(X)=\mathbb{P}^{\delta+1}$ and $\Sigma_q(X)\subset L_q(X)$ is a quadric hypersurface, where $\delta:=2\dim(X)+1-\dim(\mathrm{Sec}(X))$. Now, for an open set $W\subseteq\mathrm{Sec}(X)$, let us consider the subset $$ \mathcal{A}(W):=\{ (x,q)\in X\times W: x\in\Sigma_q(X) \}\subseteq X\times W. $$

Is it true that $\mathcal{A}(V)$ (resp. $\mathcal{A}(\mathrm{Sec}(X)\setminus X)$) is a closed set in $X\times V$ (resp. in $X\times(\mathrm{Sec}(X)\setminus X)$)?

Let us notice that if $\mathrm{Hilb}^{Q}_{\delta}(X)$ denotes the Hilbert scheme of $\delta$-dimensional quadrics contained in $X$, then the fact that $\mathcal{A}(V)$ is closed implies the existence of a map of schemes $f_V:V\to \mathrm{Hilb}^{Q}_{\delta}(X)$ which sends $q\in V$ to $\Sigma_q(X)$.

Thanks.

Answer 1

I answer by itself. The projection $\pi:X\times \left(X\times\mathbb{P}^N \right)\longrightarrow X\times\mathbb{P}^N$ is a projective morphism and hence is closed. Moreover, if $W\subseteq\mathrm{Sec}(X)\setminus X$, we have \begin{eqnarray*} \mathcal{A}(W)&=&\{ (x,q)\in X\times W : x\in \Sigma_q(X) \}\\ &=&\pi\left(\overline{\{(z,x,q)\in X\times X\times \mathbb{P}^N: z\neq x\mbox{ and } q\in\langle z,x\rangle\}}\right)\cap X\times W, \end{eqnarray*} and then $\mathcal{A}(W)$ is a closed set of $X\times W$.