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?
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1$\begingroup$ 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). $\endgroup$– Francesco PolizziCommented Dec 1, 2010 at 10:59
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$\begingroup$ A holomorphic function is one that satisfies the Cauchy-Riemann equations or, equivalently, whose derivative vanishes on the (0,1)-component of the complexified vector space. $\endgroup$– Florin BelgunCommented Dec 1, 2010 at 11:15
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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.
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$\begingroup$ Right, so the answer is the maximum principle applied to an operator with the same principal symbol as the Laplacian (The difference is a first-order term coming from the brackets [X,JX] which are non-zero 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 non-constant holomorphic function would already imply the integrability. But this is another question. $\endgroup$ Commented Dec 1, 2010 at 12:50
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$\begingroup$ 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. $\endgroup$– BS.Commented Dec 1, 2010 at 13:03
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$\begingroup$ Yes. This is the E. Hopf maximum principle from 1927 $\endgroup$ Commented Dec 1, 2010 at 14:21