# Holomorphic functions in almost-complex geometry

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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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 Cauchy-Riemann equations or, equivalently, whose derivative vanishes on the (0,1)-component of the complexified vector space. – Florin Belgun Dec 1 '10 at 11:15

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.