Accumulation of algebraic subvarieties: Near one subvariety there are many others (?) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T05:38:44Z http://mathoverflow.net/feeds/question/66285 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66285/accumulation-of-algebraic-subvarieties-near-one-subvariety-there-are-many-others Accumulation of algebraic subvarieties: Near one subvariety there are many others (?) Peter Scholze 2011-05-28T15:31:39Z 2011-05-29T09:51:27Z <p>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.</p> <blockquote> <p>Question: Do there exist many algebraic varieties $\subset \tilde{X}$ of dimension equal to the dimension of $X$?</p> </blockquote> <p>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).</p> http://mathoverflow.net/questions/66285/accumulation-of-algebraic-subvarieties-near-one-subvariety-there-are-many-others/66300#66300 Answer by Sasha for Accumulation of algebraic subvarieties: Near one subvariety there are many others (?) Sasha 2011-05-28T17:31:33Z 2011-05-28T18:52:21Z <p>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.</p> <p>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! </p> http://mathoverflow.net/questions/66285/accumulation-of-algebraic-subvarieties-near-one-subvariety-there-are-many-others/66347#66347 Answer by Dmitri for Accumulation of algebraic subvarieties: Near one subvariety there are many others (?) Dmitri 2011-05-29T09:16:04Z 2011-05-29T09:51:27Z <p>The answer is yes (as in the case of Sasha's answer we use ramified covers)</p> <p><strong>Proof.</strong> 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 $m$.</p> <p>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. <strong>END.</strong></p> <p>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). </p>