If I have a Riemannian manifold $(M^m,g)$, and on it I have an embedded submanifold $i:N^{m-1} \hookrightarrow M$, we can define the surface area of $N$ to be $\text{Area}(N)=\int_N \sigma$ where $\sigma = i_n\Omega$ where $n$ is the normal vector to $N$, and $\Omega$ is the standard volume $m$-form on $M$.

My question is, is there an $(m-1)$-form $\omega$ on $M$ such that $$\text{Area}(N) = \int_N \omega = \int_M i^*\omega$$

I ask since, on a Symplectic manifold $(M^{2n},\omega)$ We have the notion of Symplectic area: $$\text{area}_\omega S := \int_S \omega $$ For any 2-dimensional submanifold $S$.

  • 1
    $\begingroup$ Isn't the $i^*\omega$ still an $(m-1)$-form and integrating it over an $m$-dimensional $M$ would be... $\text{'Area'}(N) = \int_M i^*\omega\equiv 0$? $\endgroup$
    – rych
    Oct 3 at 9:58

1 Answer 1


This could not be the case already when $M=\mathbb R^2$ and $N=\mathbb R$ is a line through the origin. The area (i.e., length) form of $N$ could not be the pullback of any 1-form $\omega$ on $M$, as can be easily seen by looking at the direction of the kernel of $\omega$.


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