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Suppose I have a holomorphic symplectic manifold, and a smooth $(1,0)$ connection on the tangent bundle which is compatible with both the complex and the symplectic structures. Say that the associated curvature has type $(1,1)$. Then there is a Rozansky-Witten form associated to the theta graph (the graph that looks like the letter theta). It is a $\overline{\partial}$ closed $(0,2)$ form. Its associated cohomology class is a Rozansky-Witten class.

Can I replace the connection with another connection with the same properties such that the Rozansky-Witten form is actually $d$ closed? Is there some way to measure any obstructions to doing this?

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  • $\begingroup$ So are you asking whether there exists a smooth (1,0) connection on TM, compatible with both the complex and symplectic structures, such that the curvature is of type (1,1) and the Rozansky-Witten form is d-closed? (Or does the replacement have to preserve other features of your original connection?) $\endgroup$
    – Clark Barwick
    Commented Jan 21, 2010 at 18:14
  • $\begingroup$ Hi, the features that you mentioned are the ones I am interested in. I think that I may unfortunately made two copies of myself on this website. Thanks, Oren $\endgroup$
    – Oren Ben-Bassat
    Commented Jan 27, 2010 at 14:53
  • $\begingroup$ In fact, I am actually willing to forget the curvature constraint, its o.k. with me if it has also the (2,0) part. $\endgroup$
    – Oren Ben-Bassat
    Commented Jan 27, 2010 at 22:16
  • $\begingroup$ seems it is closed when R^{2,0}=0 $\endgroup$
    – Oren Ben-Bassat
    Commented Jan 28, 2010 at 18:21
  • $\begingroup$ Oops, I'm lost now. Would you be so kind as to edit the question to list the conditions you seek for the connection? (I'm still not sure I'll know how to proceed, but I can try!) $\endgroup$
    – Clark Barwick
    Commented Feb 5, 2010 at 15:33

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