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Motivation: I'm interested in which submanifolds of (half dimension) a complex manifold can be expressed locally in coordinates as the real part of the complex coordinates, and was wondering if the name "totally real" had any connection to this.

Definition: Let $M$ be a $2n$-dimensional real manifold with an almost complex structure $J$. A submanifold $L \subset M$ is called totally real if it is of half the dimension of $M$ and $T_q L \cap J T_q L= \{0\}$ for $q$ in $L$.

Question: I have two questions, but one is a generalization of the other.

Question 1: If $M$ is actually a complex manifold (i.e., $J$ is integrable) and $L \subset M$ is a totally real submanifold, does this imply that there exists holomorphic coordinates $(z^k= x^k + iy^k)$ on some neighborhood $U$ in $M$ such that $U \cap L = \{y^k =0\}$?

Question 2: Same as Question 1, but where $M=T^*L$ and $L$ is viewed as zero section, and it is assumed that $T^*M$ is complex manifold.

Ashley
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  • Let $M=\mathbb{C}$. If $L \subset \mathbb{C}$ is locally defined by ${y=0}$, then $L$ is the zero of an harmonic function. I think there are curves in $\mathbb{C}$ which are not locally the zero of an harmonic function, however I don't have any reference. – user10676 Aug 17 '16 at 02:01

1 Answers1

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I do not understand the 2nd question (I think, there are some assumptions missing here). Formally speaking the answer to Question 1 is negative since the existence of such a holomorphic change of variables implies that $L$ is a real-analytic submanifold (and not every totally real submanifold is real-analytic as you can see already in the case of real curves in ${\mathbb C}$). However, if you assume, additionally, that $L$ is real-analytic, then the answer is positive, see Lemma in my answer to this Mathoverflow question.

Moishe Kohan
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