Part 3 of this series of questions. In the meantime, I realized that there is some very simple question that was left open in 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.

  • $\begingroup$ Some singularities are non-smoothable, so at least if you want the variety $\tilde{X}$ to be in a family containing $X$, I think that the answer is that what you ask does not happen in general. $\endgroup$
    – M P
    Commented Oct 12, 2011 at 17:40
  • $\begingroup$ Of course. For this reason, the question is formulated in the weakest possible sense -- instead of $X$ you might deform any morphism $Y\rightarrow \mathbb{P}^n$ with set-theoretical image equal to $X$, for example. It might also be that one can answer this question via some kind of approximation method, i.e. without using deformation theory... Or, the answer might be no, even for this weak question. $\endgroup$ Commented Oct 12, 2011 at 18:43

1 Answer 1


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.

  • $\begingroup$ Perhaps I'm confused, but do you mean X is connected rather than irreducible? You can never disconnect an irreducible subvariety. Sorry for possibly being pedantic 3 years later! $\endgroup$
    – jacob
    Commented Jul 3, 2014 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.