Can any one give an example of a 3-fold X which contains an embedded ample divisor $D \cong CP^2$ with normal bundle $O(3)$ in X?
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1$\begingroup$ Could you provide some motivation for this problem? $\endgroup$– J.C. OttemCommented Apr 20, 2010 at 16:34
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1$\begingroup$ If E is any vector bundle on Y, then E is the normal bundle for the embedding of Y in E as the zero section. $\endgroup$– mdelandCommented Apr 20, 2010 at 16:38
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$\begingroup$ You are right. I did not ask my question clearly. I was not looking for some thing like answer bellow. Here is precisely what I want : There is a Thm which says if X is Kahler manifold and D is an ample smooth divisor in it then $X-D$ retracts to an isotropic CW complex L (You can think of it as a Legrangian sub-manifold in many cases) for example if you consider the an smooth quadratic surface in projective space, its complement retracts on a copy of real projective space. So I have to ask following question: Can you find D and X as above such that D is ample? $\endgroup$– Mohammad Farajzadeh-TehraniCommented Apr 20, 2010 at 17:18
2 Answers
Let me give an attempt of a proof of the fact that such example does not exist.
Proof. Suppose $X$ is such a $3$-fold and let $L$ be the line bundle corresponding to the divisor $\mathbb CP^2$. First we will prove that $Pic(X)=\mathbb Z$ and then will get a contradiction.
Notice that $L$ has a lot of sections. In particular in a neighbourhood of $\mathbb CP^2$ for every two points $x,y$ there is section of $L$ that contains $x$ but does not contain $y$. And also notice that for every point of $X$ there is a section that does not contain it. Hence we have a morphism $X\to P(H^0(L)^*)$. This morphism can not contract anything since $L$ is ample. The map is an embedding on the neighbourhood of $D$ and so the image can have singularities at most in codimension $3$.
From this it should follow (I guess), that we can apply Lefshetz that says $Pic (X)=Pic(D)=\mathbb Z$. So $X$ is a Fano with $Pic=\mathbb Z$. Torsten Ekedahl explained how now one can deduce contrudiction, see his comment.
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$\begingroup$ I did a mistake, I had to say an ample divisor D. In this example zero section is not ample. $\endgroup$ Commented Apr 20, 2010 at 17:17
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$\begingroup$ Notice, that this reasoning does not work if you replace $(X^3, \mathbb CP^2)$ by $(X^2, \mathbb CP^1)$ since Hard Lefshetz does not tell us much in this case... (It will just tell us that $\pi_1(X^2)=0). $\endgroup$ Commented Apr 20, 2010 at 18:36
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2$\begingroup$ Completing Dmitri's argument: By the Lefschetz hyperplane theorem (not the hard one...) $\mathrm{Pic}(X)=\mathrm{Pic}(D)=\mathbb Z$ and by the adjunction formula $(K_X+D)_{|D}=\mathcal{O}(-3)$ and hence $K_X=\mathcal{O}(-6)$ making $X$ a Fano variety of index $6$ which is not possible by a theorm of Fujita (Thm. 3.1.14 of Enc. of Mathematics, Algebraic Geometry V). $\endgroup$ Commented Apr 20, 2010 at 18:47
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$\begingroup$ Dear Torsten, thank you very much for this comment! I correted the answer accordingly. $\endgroup$ Commented Apr 21, 2010 at 5:59
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1$\begingroup$ Notice that the Lefschetz hyperplane theorem works for any ample divisor. Hence, you do not need to know that $L$ has lots of sections in order to apply it. $\endgroup$ Commented Apr 21, 2010 at 6:41
There exists such variety which is singular with Gorenstien singularities.
Let $S$ be a del Pezzo surface of degree $d$ and let ${\mathcal L} = O_S(−K_S)$. Consider the $\mathbb P^1$ -bundle $\mathbb{P} = \mathbb{P}_S (\mathcal{O}_S \bigoplus \mathcal{L})$.
Now the variety $X$ can be constructed as a contruction of a zero divisor. The map $\mathbb{P}\to X$ given by the linear system ${\mathcal{O}}_{\mathbb{P}}(n), \quad n ≫ 0.$
It contracts the negative section. Since $−K_{\mathbb P} ∼ O_{\mathbb P}(2),$ the variety $X$ is a Fano threefold of index $2$ and degree $8d$ with canonical Gorenstein singularities. For $S = {\mathbb P}_2$ we have $−K^3_X = 72$ and $X ≃ \mathbb{P}(3, 1, 1, 1)$ is a weighted projective space.