Holomorphic functions in almost-complex geometry - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T05:25:27Z http://mathoverflow.net/feeds/question/47890 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47890/holomorphic-functions-in-almost-complex-geometry Holomorphic functions in almost-complex geometry Florin Belgun 2010-12-01T10:42:31Z 2010-12-01T13:09:42Z <p>Maximum principle implies that every holomorphic function on a compact complex manifold is constant. Is this still true if the manifold is only almost complex? </p> http://mathoverflow.net/questions/47890/holomorphic-functions-in-almost-complex-geometry/47891#47891 Answer by BS for Holomorphic functions in almost-complex geometry BS 2010-12-01T11:28:42Z 2010-12-01T13:09:42Z <p>This is still true, although as Francesco says in his comment above, it is trivially so in general : in complex dimension 2 and more, a generic almost complex structure has only constant holomorphic functions, even locally. </p> <p>Proof : if $f:(V,J)\to\mathbb{C}$ is such a function, namely $df\circ J=i\,df$, then (obviously) $d(df\circ J)=0$.</p> <p>But the second order operator $f\mapsto (d(df\circ J))^{1,1}$ from functions to $(1,1)$-forms has the "same" principal symbol at each point as in the integrable case (the "plurisubharmonic Hessian", so to speak, perhaps up to some $-2i$ factor). </p> <p>In particular you can compose it with contraction by a positive smooth $(1,1)$ form (given by any hermitian metric) to obtain a "Laplace operator", which satisfies the maximum principle. EDIT (after comment by OP): it is important to observe that the operator vanishes on constants to derive the maximum principle -- locally it writes $\sum g_{jk}(x) \partial_j\partial_k +\sum b_i(x) \partial_i$, with $g_{jk}$ symmetric positive definite. </p>