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hallo,

I have the following question: Let $M$ be a $n-$dimensional complex manifold and $X \subset M$ be a compact $n-$dimensional totally real analytic Riemannian submanifold. Let furthermore $\alpha$ be a 2-form on $X$. Is it possible to extend $\alpha$ to a holomorphic 2-form in a neighbourhood of $X$ in $M$? I hope for answers. Thanks in advance.

bruno

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are there any references ? –  bruno Sep 5 '12 at 6:42
    
does anyone have an idea ? –  bruno Sep 5 '12 at 10:07
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Wouldn't the existence of such an extension require $\alpha$ to be real-analytic on $X$? –  Andreas Blass Sep 5 '12 at 13:34
    
@bruno: what's a totally real submanifold? –  Sándor Kovács Sep 5 '12 at 18:18
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When $\alpha$ is a function the answer is affirmative (the analytic expansion of $\alpha$ converges in a neighborhood of any point of X). For $\alpha$ a form, locally extend it using local coordinates, and then observe that by uniqueness the extension is global. –  user175348 Sep 6 '12 at 21:07

1 Answer 1

Such extension exists. This follows from the followoing three facts.

1) For every point $x\in X$ there is a neighbourhood $U(x)\subset M$ and a holomorphic map $\phi: U(x)\to \mathbb C^n$ such that $\phi (U(x)\cap X)\subset \mathbb R^n$.

2) Let $\alpha$ be a real analytic two-form defined on $\mathbb R^n$, then it can be extended to a holomorphic two-form on a neighbourhood of $\mathbb R^n$ in $\mathbb C^n$ that retracts on $\mathbb R^n$

3) If you have two extensions of $\alpha$ to a same neighbourhoods of $\mathbb R^n$ then they coincide.

I don't know a reference for 1) but it should be something standard, to get 2) and 3) you can just use Taylor decomposition of $\alpha$.

Once you know 1,2,3 the statement follows easily.

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