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.
1 Answer
S. K. Donaldson, Symplectic submanifolds and almostcomplex 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.

$\begingroup$ 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 pluriharmonic 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

$\begingroup$ 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 Hyperplane 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

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 CieliebakEliashberg'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