I am just starting to learn more symplectic geometry on Stein manifolds. I understand that an important class of Stein manifolds arises as the complement of a Donaldson's codimension two symplectic submanifold. I was wondering where I can find an actual proof of this important fact. Any comments or explanations will be more than welcome.
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S. K. Donaldson, Symplectic submanifolds and almost-complex geometry, J. Differential Geom. 44, 666 (1996).
For more info, and the link to Bertini's theorem, see Jonathan Evans' thesis: Symplectic topology of some Stein and rational surfaces.
answered Oct 25, 2013 at 15:29
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$\begingroup$ Thank you very much for the posting. Can you fill in with more details? In the complex projective case, the complement of a hypersurface $V$ is stein because there is a pluri-harmonic function $\log \vert\vert s\vert\vert^2$, where $s$ is a holomorphic section which vanishes exactly at $V$. In the almost complex case, how can we proceed? $\endgroup$– Yi LinOct 25, 2013 at 15:36
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$\begingroup$ I checked out Donaldson's paper and found out that Donaldson actually proved that the complement of his submanifold is Stein. An argument can be found in the proof of Lefschetz Hyper-plane theorem ( for symplectic hypersurface) given in the paper. But it is interesting to note that the notion " Stein" was never mentioned in the paper. Carlo, thank you very much for your help! $\endgroup$– Yi LinOct 26, 2013 at 2:57
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2$\begingroup$ Yi Lin: what Donaldson proves is really that the complement is a Weinstein manifold; one then has to invoke Eliashberg's theorem (as in Cieliebak-Eliashberg's book) to deform the data to be Stein. This aspect of Donaldson's theory was highlighted in work of Paul Biran. $\endgroup$ Oct 27, 2013 at 15:45