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We have a complex manifold $M$ equiped with a hermitian metric, then for a complex submanifold $S \subset W$, the Wirtinger's theorem tells us that the volume form on $S$ is the restriction of a global form on $M$.

The textbook then made the remark that this is not true in the real case. Does anyone have an explicit example to explain that?

Arctic Char
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  • The length of a curve $(x(t), y(t))$ in $\mathbb{R}$ with the standard metric is given by $\int \sqrt{x(t)^2+y(t)^2} dt$. The form $\sqrt{x(t)^2+y(t)^2} dt$ is not the pullback of $1$-form on $\mathbb{R}^2$. – eggplant Oct 13 '20 at 13:57
  • @eggplant sorry I still did not see why this is not a pullback of a 1 form...? –  Oct 13 '20 at 16:07
  • You give us the $1$-form on $\Bbb R^2$. As an abstract hint, what happens to the arclength when you reverse orientation on the curve? – Ted Shifrin Oct 13 '20 at 22:21

1 Answers1

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Explicit example: Let $M = \mathbb R^2$. Let $\alpha$ be any one form on $\mathbb R^2$. Write $$\alpha = Adx + Bdy,$$ where $A, B$ are smooth functions. Consider the vector fields $X(x, y) = (B, -A)$ and let $\gamma(t)$ be any intgral curve of $X$. Then $$\gamma^*\alpha = 0$$ and thus $\alpha$ do not restrict to the volume form on $\gamma$.

Arctic Char
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