0
$\begingroup$

For a divisor in a complex manifold, what is known about a complementary bundle to the divisor in the manifold (either for the tangent or the cotangent bundle). Is there a description in terms of holomorphic line bundles? For a Kahler manifold it would be possible just to take the normal bundle, and split this into two line bundles over the divisor by taking the eigenspaces of the complex structure map $J$ (hope I got this right). But then what properties do these line bundles have, and how do they relate to the properties of the divisor? As a learner in algebraic geometry, is there somewhere to look to find out about this?

Secondly an apology, I am definitely not an expert in algebraic geometry so I apologise if the answer is well known (it likely is) or if my question is confused (especially the latter)! I will have to explain where it is from: I would like to look at divisors on a noncommutative complex manifold. In noncommutative differential geometry ideas of rank or codimension are rather difficult, and in general require adding extra structure. (Understandably, textbooks on classical 'commutative' manifolds are happy simply saying complex codimension one, as there is not any problem in that case.) However the rank one case (line bundles) is known in noncommutative geometry - they have several definitions as invertible bimodules, self Morita equivalences and as integer graded Hopf Galois extensions. Even the idea of 'rank 2' is rather difficult to pin down, thus my wish to try to `define' codimension one by having a complementary bundle described in terms of line bundles of known properties.

$\endgroup$
3
  • 2
    $\begingroup$ I do not understand the question. Are you trying to split the fundamental exact sequence $$\mathcal{I}/\mathcal{I}^2 \to \Omega_X|_D \to \Omega_D \to 0,$$ that is associated to a divisor $D$ inside a complex manifold $X$, with ideal sheaf $\mathcal{I}$? Or are you trying to decompose $\mathcal{I}/\mathcal{I}^2$ as a direct sum of holomorphic invertible sheaves? $\endgroup$ Commented Jun 26, 2014 at 11:37
  • $\begingroup$ Honestly, I do not know which. Whichever works nicely I guess. I had thought of the first, but anything nice to say about the second would be welcome. Modules with connection in noncommutative geometry form an abelian category, so either approach could be used in noncommutative geometry. $\endgroup$ Commented Jun 26, 2014 at 12:21
  • $\begingroup$ More thought - the direct sum by preference, if not then any related construction would be welcome. $\endgroup$ Commented Jun 26, 2014 at 12:27

0

You must log in to answer this question.

Browse other questions tagged .