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Let's work over the field $\mathbb{C}$ of complex numbers, and let $X\subset \mathbb{P}^n$ be a projective variety. Let $\tilde{X}\subset \mathbb{P}^n$ be any small open neighborhood of $X$, in the complex topology.

Question: Do there exist many algebraic varieties $\subset \tilde{X}$ of dimension equal to the dimension of $X$?

Of course, $X$ itself lies in there, and one may apply some automorphism of $\mathbb{P}^n$ close to the identity to obtain some further examples. I would be interested in any construction of more subvarieties in $\tilde{X}$, or examples showing that in general these may not exist. Note that the degree of the subvarieties is allowed to be arbitrarily large, and that I do not want to consider only deformations of $X$ (which might be rigid).

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2 Answers 2

up vote 13 down vote accepted

The answer is yes (as in the case of Sasha's answer we use ramified covers)

Proof. Let $X$ be any variety in $\mathbb CP^n$. Take a section $s_m$ of $O(m)$ on $X$ such that $s_m$ is not equal to $m$-th tensor power $s^{\otimes m}$ of any section $s$ of $O(1)$ restricted on $X$. Now, let $s_m^{\frac{1}{m}}$ be the multi-section of $O(1)$ on $X$. This multi-section defines a subvarity $X_m$ in the total space of $O(1)$ on $X$, that is the cover of $X$ of degree $m$.

Finally notice that there is a family of maps from the total space of $O(1)$ on $X$ to $\mathbb CP^n$ that sends the zero section of $O(1)$ on $X$ to $X$. The image of such a map in $\mathbb CP^n$ is just the union of all lines in $\mathbb CP^n$ that join a fixed point $p$ with all points of $X$, the point $p$ itself does not belong to the image. Then the image of $X_m$ in $\mathbb CP^n$ is the desired variety. END.

We used here the fact that $O(1)$ on $\mathbb CP^n$ can be embedded in $T\mathbb CP^n$ as a subsheaf (in various ways).

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Very nice! [Although you might rename $\tilde{X}$ ;-)] –  Peter Scholze May 29 '11 at 9:48

Apart of the deformations of $X$ you can consider deformations of any subscheme supported on $X$ (that is of any subscheme $Y$ such that $Y_{red} = X$). Another choice is to consider deformations of any morphism $f:Y \to X \to {\mathbb P}^n$. Both ways will provide you with deformations of higher degree. Note that the second approach is very fruitfully used in the bend-and-break procedure.

EDIT: Let for example a $X$ be a rational curve. Then the tangent space to deformations of the embedding $X \to {\mathbb P}^n$ is $H^0(X,(T_{{\mathbb P}^n})_{|X})$ which is a globally generated vector bundle of rank $n$ and degree $(n+1)d$, where $d$ is the degree of $X$. So, the dimension of the tangent space is $n + (n+1)d$. On the other hand, if $Y$ is another rational curve and $f:Y \to X$ is an $r$-fold covering, then the same computation shows that the tangent space to deformations of $Y \to {\mathbb P}^n$ has dimension $n + (n+1)dr$ which is much bigger!

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This does potentially construct more examples, but I don't know how to prove that one such deformation problem actually admits nontrivial deformations. –  Peter Scholze May 28 '11 at 17:34
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I understand your argument, but a big tangent space does not mean a big deformation space, i.e. all deformations could be infinitesimal. Is there a way to see that you can actually get a deformation over $C[[t]]$? –  Peter Scholze May 28 '11 at 18:38
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@Sasha: In Mori's bend and break argument isn't he forced to reduce modulo some prime p and use Frobenius since if X has genus > 1 then the genus of Y will increase with r (so one cannot use Riemann-Roch to get a large deformation space)? –  ulrich May 28 '11 at 18:41
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@Peter: $T_{\mathbb{P}^n}$ is an ample vector bundle so when restricted to a $\mathbb{P}^1$ it is a sum of line bundles, each one of which is of positive degree. So $H^1$ vanishes and the deformation space is unobstructed. A similar argument works if $X$ is an elliptic curve since one can let $Y$ be an unramified covering of large degree. –  ulrich May 28 '11 at 18:49
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@ulrich. I agree, my statement that the same story has place for any $X$ was too optimistic. I have removed it. But still, sometimes it works. –  Sasha May 28 '11 at 18:54

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