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This is a sequel to the question Accumulation of algebraic subvarieties: Near one subvariety there are many others (?) .

Let $Y$ be some projective variety, over $\mathbb{C}$. Let $X\subset Y$ be some closed subvariety, and let $\tilde{X}$ be some small open neighborhood of $X$, in the complex topology.

Naive Question: Are there many varieties $\subset \tilde{X}$ of dimension equal to the dimension of $X$?

In the previous question, this was answered positively if $Y=\mathbb{P}^n$, using crucially the fact that the tangent sheaf of $\mathbb{P}^n$ is ample, which allows to construct deformations of infinitesimal thickenings of $X$.

On other hand, the answer in the general case is obviously no, because one might take as $Y$ the blow-up of $\mathbb{P}^2$ at a point, and for $X$ the exceptional divisor. In that case, any $X^\prime\subset \tilde{X}$ has to project to the point in $\mathbb{P}^2$, for otherwise it meets any line in $\mathbb{P}^2$, which however may be outside of the image of $\tilde{X}$. But then $X^\prime$ is set-theoretically equal to $X$.

But in this case, the obstruction already exists if we take $\tilde{X}$ to be a small neighborhood in the Zariski topology.

Question: Assume that within any Zariski neighborhood of $X$, there are many projective subvarieties of the same dimension. Then is the same true for $\tilde{X}$?

Finally, note that the condition that within Zariski neighborhoods, there are many subvarieties of the same dimension is e.g. guaranteed by the following numerical criterion.

Lemma: Assume that $X$ meets every subvariety of complementary dimension. Then within any Zariski neighborhood of $X$, there are many projective subvarieties of the same dimension.

For the proof, just note that the complement of any Zariski neighborhood of $X$ has codimension smaller than $\mathrm{dim} X$, and hence by taking an intersection of generic hyperplane sections, you get subvarieties which do not meet the boundary. Also note that the condition of the Lemma is trivially verified if $Y=\mathbb{P}^n$.

However, note that one can not in general expect to have complete intersections inside of $\tilde{X}$, already if $Y=\mathbb{P}^n$ and $X$ is smooth. For the proof, note that if we take $\tilde{X}$ small enough, it will have the same (singular, say) cohomology as $X$; in particular, for any $X^\prime\subset \tilde{X}$, we get a map $H^i(X,\mathbb{Q})\rightarrow H^i(X^\prime,\mathbb{Q})$. It is compatible with cup-product, and injective in top-degree (compare with cohomology of $\mathbb{P}^n$), thus injective. But $H^i(X^\prime,\mathbb{Q})$ is zero below the middle dimension if $X^\prime$ is a complete intersection.

One might also ask the following weaker question (say, $Y$ is smooth in order to have a nice cup-product):

Question: Assume that the intersection pairing of $X$ with any subvariety of complementary dimension is positive. Are there many subvarieties in $\tilde{X}$ of the same dimension?

Counterexamples to these questions are most welcome, on the other hand I would be particularly interested in any positive result or method to construct such subvarieties, so feel free to make additional assumptions.

EDIT: Both questions above are shown to have negative answers by the nice example of Dmitri. Therefore, let me ask the following, pretty vague, question:

Question: Is it possible to formulate any positive result in this direction, possibly assuming that 1) the tangent or normal sheaf is in some sense positive or ample, 2) some positivity condition as already required above + 3) whatever you need.

share|improve this question
    
Just a nitpick. I guess in the first (not naive) question the condition is that each Zariski neighbourhood contains many projective subvarieties, (because surelly there will be plenty of closed)? –  Dmitri May 29 '11 at 12:43
    
Of course... Thanks! –  Peter Scholze May 29 '11 at 12:59

1 Answer 1

One little counterexample (to both versions of the question). On a quintic $3$-fold in $\mathbb CP^4$ there are $2875$ lines. You can take any of these lines. An analytic neighbourhood of such a line is the one of the small resolution of the cone $x_1^2+x_2^2+x_3^2+x_4^2=0$. So there can be no curves in the neighbourhood of such a line, since the cone is affine.

This example is not surprising, since the normal bundle to such a line is $O(-1)\oplus O(-1)$, one needs at least a bit of positivity...

share|improve this answer
    
Hmm... OK, I see. Thanks for the example! –  Peter Scholze May 29 '11 at 15:01
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John, I don't believe that there can be a counterexmple when the normal bundle to $Y$ is ample. I would first prove that for any $Y$ and an ample bundle $E$ on it, $E$ has lost of multi-sections (this should be true, should not it?). Of course generically no analytical neighbourhood of $Y$ in $X$ is isomorphic to a neighbourhood of $Y$ in the total space of its normal bundle, but still... Anyway, you are the specialist in this area (judging by your last preprint) :)). So I hope you would be able to say something in the positive direction (anyway my answer should not have been accepted). –  Dmitri May 29 '11 at 23:24
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In the paper "Positivity and excess intersection" by Fulton and Lazarsfeld they construct an ample bundle on $\mathbb{P}^2$ with no multi-sections except for the zero section. This does not give a counterexample to the original question since the condition on Zariski neighbourhoods is not satisfied, but at least says that even with some positivity assumptions the extra condition is definitely needed. –  ulrich May 30 '11 at 5:38
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Just curious: Would the etale topology be enough to see the obstruction in your example? –  Peter Scholze May 30 '11 at 11:34
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@Dmitri: By the way, there are no examples like that of Fulton and Lazarsfeld with $Y$ a curve and $E$ of rank $2$; this was proved by Barlet, Peternell and Schneider in "On two conjectures of Hartshorne's". So a guess might be that for curves in threefolds the ampleness of the normal bundle suffices for some multiple of the curve to move as a cycle. –  ulrich May 30 '11 at 13:40

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