$b_2$ of the blow up of a complex $3$-fold in a curve - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T12:07:43Z http://mathoverflow.net/feeds/question/75034 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75034/b-2-of-the-blow-up-of-a-complex-3-fold-in-a-curve $b_2$ of the blow up of a complex $3$-fold in a curve aglearner 2011-09-09T20:29:40Z 2011-09-10T00:55:34Z <p>Suppose that $V$ is a complex analytic manifold of dimension 3 with mild singularities, say it is an orbifold (i.e. has only quotient singularities). Let $C$ be a complex irreducible curve in $V$. Suppose that $V'$ is a blow up of $V$ along this curve that contracts only one divisor in $V'$. By this I mean that there is a holomorphic map $V'\to V$ that is an isomorphism on the preimage of $V\setminus C$ and the preimage of each point of $C$ is a curve in $V'$.</p> <p><strong>Question.</strong> How to prove that $b_2(V')-b_2(V)=1$ using preferably a purely topological reasoning?</p> <p><em>Comments.</em> I would be grateful for an idea of the proof or for a reference. Note that in the case when $C$ is smooth and does not intersect singularities of $V$, the proof is easy. I am confident that the statement is correct, and interested in a simple proof of it. If might be that the statement holds even if $V$ has more complicated singularities (though according to the answer of Remke below this is not always the case), and definitely it is not important that the dimension of $V$ is 3. I would prefer to get a proof of the statement rather to get a counterexample by relaxing the condition on singularities. </p> http://mathoverflow.net/questions/75034/b-2-of-the-blow-up-of-a-complex-3-fold-in-a-curve/75036#75036 Answer by Remke Kloosterman for $b_2$ of the blow up of a complex $3$-fold in a curve Remke Kloosterman 2011-09-09T21:02:05Z 2011-09-09T21:31:39Z <p>I am not sure whether $b_2(V')-b_2(V)=1$ always holds. Anyway, in the book of Peters and Steenbrink you can find "the Mayer-Vietoris sequence of the discriminant square". If $E$ is the exceptional divisor then you have an exact sequence of mixed Hodge structures $$H^1(V')\oplus H^1(C)\to H^1(E) \to H^2(V)\to H^2(V')\oplus H^2(C)\to H^2(E) \to H^3(V) \to \dots$$</p> <p>Now if $C$ is smooth and does not intersect the singular locus of $V$ then $E$ is a $\mathbb{P}^1$-bundle over $C$. Hence the map $H^1(C)\to H^1(E)$ is surjective. One can easily show that the map $H^2(V)\to H^2(E)$ is not the zero map; that $H^2(C)\to H^2(E)$ is injective; the image of the first map is not contained in the image of the second and that $h^2(C)=1, h^2(E)=2$ holds, so the above sequence reduces to $$0 \to H^2(V)\to H^2(V')\oplus \mathbb{C}\to \mathbb{C}^2\to 0.$$ Now, if $C$ passes through the singularities of $V$ little of the above remains true, e.g., $E$ might be a conic bundle over $C$ with reducible fibers over the points where $C$ intersects the singular locus of $V$. In this case $h^2(E)>2$ holds. The additional classes in $H^2(E)$ typically contribute to the kernel of $H^3(V')\to H^3(V)$ and reduce the dimension of the weight 2 part of $H^3$, but I expect that in some cases they might force $h^2(V')-h^2(V)>1$.</p>