# Standard area form on a Riemannian manifold.

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 $$S$$.

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

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 $$\omega$$.