Let $X$ and $Y$ be two proper varieties of dimension $n$ over a field $k$. I'm looking for "reasonable" conditions, under which, there exists a proper and dominant morphism $f:V\to \mathbb{P}^1_k$, with $V$ a variety of dimension $n+1$, such that $X=f^{-1}(0)$ and $Y=f^{-1}(\infty)$.
6
-
1$\begingroup$ Without some extra conditions, this question is trivial--do you want $V$ to be pure-dimensional? Do you want the morphism to be flat? $\endgroup$– Daniel LittCommented May 8, 2012 at 5:48
-
1$\begingroup$ I very much doubt that there are any reasonable general conditions. For example, if $X$ and $Y$ are curves of the same sufficiently large genus it is not always true (by results of Harris--Mumford and Eisenbud--Harris) that they can be put into a family over $\mathbb{P}^1$ but there do exist non-constant families over $\mathbb {P}^1$. $\endgroup$– nafCommented May 8, 2012 at 6:18
-
$\begingroup$ Does the OP want the fibers over $0$ and $\infty$ to be reduced / smooth? If not, then I think this could be interesting, already for curves. $\endgroup$– Jason StarrCommented May 8, 2012 at 12:56
-
$\begingroup$ In this questions, variety means integral scheme of finite type (so irreducible and reduced). This implies that the dominant map from $V$ to $\mathbb{P}^1_k$ is flat. $\endgroup$– hadimathCommented May 8, 2012 at 16:40
-
$\begingroup$ @hadimath: Do you want the fibers over $0$ and $\infty$ to be equal to $X$ and $Y$ as schemes, or do you want that the underlying reduced varieties of these fibers equal $X$ and $Y$? If the first, then, as pointed out, then this is extremely rare, already for curves. However, if you want the second condition, then this may be much more likely, cf. work of Bogomolov and Tschinkel over finite fields. $\endgroup$– Jason StarrCommented May 8, 2012 at 17:38
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
2
I will assume that by "variety" you mean (at least) "irreducible", otherwise as Daniel Litt points out the question is trivial.
Further, let's assume that $V$ is projective (and hence necessarily $X$ and $Y$), as ulrich points out, it will already be difficult enough.
Consider a common polarization of $X$ and $Y$ given by an $f$-ample line bundle. The first necessary condition is that the corresponding Hilbert polynomials on $X$ and $Y$ are equal. This follows from $f$ being flat.
So, if $X$ and $Y$ have the same Hilbert polynomial, then they are fibers of some huge family, which however may be disconnected and $X$ and $Y$ would have to be in the same irreducible component to satisfy your desired condition. This is pretty hard to force other than having a family over an irreducible base that admits both $X$ and $Y$ as fibers.
Then, if $X$ and $Y$ are in the same irreducible component of the Hilbert scheme, then it is yet another pretty difficult question whether their Hilbert points can be connected by rational curves. It turns out that this is a rare condition. As ulrich mentions, for instance for a general curve of high enough genus, there does not exist a non-trivial deformation over $\mathbb P^1$. In other words, already for $n=1$ and for most $X$, regardless of the choice of $Y$ there is no such deformation.
This suggests that having such an $X$ and $Y$ is very special.
On the other hand, and perhaps this is more what you would like to see, for most curves (smooth projective) of small genus this is actually doable.
The key words you should be looking for is "rational connectivity" of either moduli spaces or Hilbert schemes.
The above claims for curves follow from the fact that moduli spaces (stacks, really) for small genus are (uni)rational and hence any two generic point can be connected by a rational curve, while moduli spaces of high genus curves have non-negative Kodaira dimensions and hence they are not uniruled (so there are no rational curves through their general points).
As ulrich mentions the story starts with Harris-Mumford and Eisenbud-Harris. For current knowledge you should look at the work of Gavril Farkas. You can start here http://arxiv.org/abs/0810.0702 for example, and just look at his papers on arXiv.
Another paper to read is Verra's survey: http://arxiv.org/abs/1112.6095.
And of course, this only covers $n=1$ in your question. I believe very little is known in higher dimensions, other than that this is a very special property, even just for $X$ to be in a non-locally trivial family over $\mathbb P^1$.
answered May 8, 2012 at 8:00
-
$\begingroup$ If we relax the condition that $X$ and $Y$ should live in a family over $\mathbb{P}^1_k$, and only require that there should live in a family over some curve (possibly satisfying some additional conditions e.g. smooth), would the answer then be positive? In other words, instead of "rational connectivity", we require "algebraic connectivity". I believe in this case, any two points on the same irreducible component of the Hilbert scheme can be connected by a curve. $\endgroup$– hadimathCommented May 9, 2012 at 17:06
-
$\begingroup$ Yes, this sounds right. On the other hand, I don't know how you can decide that they belong to the same irreducible component of the Hilbert scheme other than producing such a family, so this condition may not be helpful at all. $\endgroup$ Commented May 10, 2012 at 1:01