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Let $Y\subseteq X\subsetneq\mathbb{P}^{N}$ be smooth projective varieties, and let $$ S_{X,Y}=\overline{\{(x,y,z)\in X\times Y\times \mathbb{P}^{N}:x\neq y, z\in\langle x,y\rangle\}}. $$ Can we deduce the smoothness of $S_{X,Y}$ from the smoothness of $X,Y$?

I don't know theorems about the behaviour of smoothness under morphisms, but it could be easy if we knew any, via considering the projection $$ S_{X,Y}\rightarrow X\times Y. $$ (Notice that the fiber of every $p=(x,y)\in X\times Y$ such that $x\neq y$ is a line, and the fiber of the points $p=(y,y)\in X\times Y$ is isomorphic to $T_{y}X$).

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  • $\begingroup$ Usually, it is not smooth. For the simplest example take $X$ to be a twisted cubic curve and $Y$ a point on it. Then $S_{X,Y}$ is a quadratic cone with vertex at this point. $\endgroup$
    – Sasha
    Feb 27, 2016 at 13:47
  • $\begingroup$ It looks like your $S_{X, Y}$ is a $\mathbb{P}^1$-bundle over $Y$ times the blowup of $X$ along $Y$, and hence is smooth. $\endgroup$ Feb 27, 2016 at 14:02
  • $\begingroup$ @Sasha I see that the image of $S_{X,Y}$ under the projection to $\mathbb{P}^{N}$ is a cone. How do we know that $S_{X,Y}$ is a cone too? $\endgroup$
    – JosuaJones
    Feb 27, 2016 at 14:07
  • $\begingroup$ @JosuaJones: Sorry, I misread your question. Your $S_{X,Y}$ is indeed, a $P^1$-bundle over the blowup of $Y \times X$ in the graph of the embedding $Y \to X$ (almost as Piotr Achinger said). In fact, you could first ask this question about $X = Y = P^n$. What you get will be a $P^1$-bundle over the blowup of $P^n \times P^n$ in the diagonal. And then you can base change to $X \times Y \subset P^n \times P^n$. $\endgroup$
    – Sasha
    Feb 27, 2016 at 15:12
  • $\begingroup$ @Sasha Ok, thank you! I'd appreciate if you could give me an idea how to show that it is a $\mathbb{P}^{1}$ bundle over that blow-up. Why does it imply that it is smooth? (I am not used to work with projective bundles, sorry) $\endgroup$
    – JosuaJones
    Feb 28, 2016 at 11:20

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