Let $M$ be the coarse moduli space of smooth connected hypersurfaces of degree $d$ in some fixed smooth projective variety $X$.

If $X$ is projective space, it is easy to see that $M$ is connected. One simply notes that $tF +(1-t)G = 0$ connects the hypersurfaces $F=0$ and $G=0$, where $t$ is a coordinate function on $\mathbb A^1$.

Does this argument also work to show that $M$ is connected when $X$ is not necessarily projective space?

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    $\begingroup$ What if $X$ is a smooth quadric surface in $\mathbb{P}^3$, and you consider the parameter space for degree $1$ "hypersurfaces" (i.e., Cartier divisors) in $X$? $\endgroup$ – Jason Starr May 24 '14 at 17:23
  • $\begingroup$ Assuming that "degree $d$" stands for $\mathcal{O}_X(d)$, you are merely speaking about a linear system and the same argument works: the moduli space in question is just a projective space. $\endgroup$ – Alex Degtyarev May 24 '14 at 17:36

The curve you constructed $tF+(1-t)G$ is just the line through $F$ and $G$. Assuming that one could do a similar construction it would mean that $M$ is rationally connected.

Let say that your veriety $X$ is embedded in $\mathbb{P}^n$ and the degree $d$ is given by the hyperplane section of $\mathbb{P}^n$.

Take a smooth quadric $Q\subset\mathbb{P}^3$ and $d = 1$. There are two families of lines in $Q$. Each one is parametrized by a conic in the Grasmannian $\mathbb{G}(1,3)\subset\mathbb{P}^5$. There two conics lie in complementary planes in $\mathbb{P}^5$. In particular the two conics does not intersect and $M$ is not connected.

If $X\subset\mathbb{P}^3$ is a general smooth cubic and $d = 1$. Then $M$ is just a set of twenty-seven points. Therefore it is not connected.


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