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?

What do you mean by "holomorphic function" on an almost complex manifold? In general, you do not have complex coordinates $z_i$, unless the almost complex structure is integrable (i.e, the manifold is complex). – Francesco Polizzi Dec 1 '10 at 10:59

A holomorphic function is one that satisfies the CauchyRiemann equations or, equivalently, whose derivative vanishes on the (0,1)component of the complexified vector space. – Florin Belgun Dec 1 '10 at 11:15
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.
Proof : if $f:(V,J)\to\mathbb{C}$ is such a function, namely $df\circ J=i\\,df$, then (obviously) $d(df\circ J)=0$.
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).
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.

Right, so the answer is the maximum principle applied to an operator with the same principal symbol as the Laplacian (The difference is a firstorder term coming from the brackets [X,JX] which are nonzero in general). About the local question, I agree that if there are n independent functions around a point (dim M=2n), then the J is integrable around that point. Probably, if the J satifies further restrictions (e.g, nearly Kähler in dimension 6), then the existence of ONE nonconstant holomorphic function would already imply the integrability. But this is another question. – Florin Belgun Dec 1 '10 at 12:50

You are right that it is important that the operator vanishes on the constants (i.e. has "no order 0 part") to conclude a maximum principle from the principal symbol. – BS. Dec 1 '10 at 13:03
