Accumulation of algebraic subvarieties: Near one subvariety there are many others (?), 3

2011-10-11T22:01:29Z

Part 3 of this series of questions. In the meantime, I realized that there is some very simple question that was left open in http://mathoverflow.net/questions/66285/accumulation-of-algebraic-subvarieties-near-one-subvariety-there-are-many-others .

Let's work again 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: Assume that $X$ is locally a (set-theoretic) complete intersection. Does there exist a smooth projective subvariety of $\mathbb{P}^n$ contained in $\tilde{X}$ of dimension equal to the dimension of $X$?

Any ideas are welcome, also for the similar question over characteristic $p$ fields, i.e. working over a field like the completed algebraic closure of $\mathbb{F}_p((t))$, and taking a small neighborhood in the $t$-adic topology. Changing to such fields might have the advantage that bend-and-break type techniques become available, but I do not see how to make use of them.

Answer by Peter Scholze for Accumulation of algebraic subvarieties: Near one subvariety there are many others (?), 3

2011-10-30T14:39:18Z

The answer to the question for general $X$ is no: There are local obstructions coming from bad singularities.

More precisely, one can show the following: Assume that $X$ is irreducible and that there is a closed subset $A\subset X$ of codimension at least $2$, such that $X\setminus A$ is not connected. Then there are small $\epsilon$-neighborhoods $\tilde{X}$ of $X$ in $\mathbb{P}^n$ which do not contain any smooth projective subvarieties of the same dimension.

Let me sketch the argument. Let $\tilde{A}$ be a sufficiently small closed $\epsilon$-neighborhood of $A$. Choose $\tilde{X}$ such that $\tilde{X}\setminus \tilde{A}$ is still not connected. Assume that there is some smooth projective $X^\prime\subset \tilde{X}$ of the same dimension. We see that $X^\prime\setminus \tilde{A}$ is not connected. Using the sequence $$ H^{2\dim X - 1}(\tilde{A}\cap X^\prime)\rightarrow H^{2\dim X}_c(X^\prime\setminus \tilde{A})\rightarrow H^{2\dim X}(X^\prime)\ , $$ we see that $\tilde{A}\cap X^\prime$ has cohomology in degree $2\dim X -1$. If $A$ was a point, then $\tilde{A}$ would be a closed ball, and one would have a bound saying that the cohomology of $\tilde{A}\cap X^\prime$ is in degrees $\leq \dim X^\prime$. In general, one has a similar bound $\dim X^\prime + \dim A$. Now if $A$ has codimension at least $2$, one gets a contradiction.

Working slightly more carefully, one sees that the obstruction really is local in nature, and that a necessary condition is that $X$ is locally connected in codimension $1$.

On the other hand, it is clear that there are no local obstructions if $X$ is locally a set-theoretic complete intersection (as one can just deform local equations to get a local smooth deformation). Hence this argument reproves a result of Hartshorne: Set-theoretic complete intersections are locally connected in codimension $1$.

But what happens if there are no local obstructions, for example if $X$ is locally a complete intersection? I have adapted the original question accordingly.

Accumulation of algebraic subvarieties: Near one subvariety there are many others (?), 3 - MathOverflow

Accumulation of algebraic subvarieties: Near one subvariety there are many others (?), 3

Answer by Peter Scholze for Accumulation of algebraic subvarieties: Near one subvariety there are many others (?), 3