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Let $\Sigma$ be a closed surface in a Riemannian $3$-manifold $(M,g)$. I was thinking about the validity of writing $$\int_\Sigma H^2 d\mu_\Sigma,\tag{1}$$ where $H$ is the mean curvature of $\Sigma$ and $d\mu_\Sigma$ is the Riemannian volume form on $\Sigma$.

My knowledge of integration on manifolds comes from Introduction to Smooth Manifolds written by John M. Lee, and given the formalism in this book, I was wondering if I had to assume that $\Sigma$ is oriented, for (1) to be valid. I asked this question because I had heard in a post that a closed submanifold is automatically oriented, which statement I currently have no idea how to justify. Does this statement come after we first assume $M$ is oriented? Thank you.

Note: A closed manifold is, by convention, a compact manifold without boundary.

Arctic Char
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Boar
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2 Answers2

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A closed embedded hypersurface $\Sigma^n$ in $\mathbb R^{n+1}$ is orientable (see here). But for general $M$ this is not true - think of $\mathbb{RP}^2$ inside $\mathbb{RP}^3$ for example.

However, both the volume $d\mu_\Sigma$ and $H^2$ can be defined for non-orientable manifold. For $d\mu_\Sigma$, this is covered in John Lee's Riemannian manifolds. For $H^2$, note that the mean curvature vector $\vec H$ is defined for non-orientable surfaces, and if you choose a local unit normal $v$ you can define the (scalar) mean curvature $\vec H = H\vec v$. $H$ is not defined globally, but $H^2$ is since $H^2 = |\vec H|^2$.

Arctic Char
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No, volume forms cannot be defined on a non-orientable manifold. Thus, Lee on page 30 clearly states:

Another important construction provided by a metric on an oriented manifold is a canonical volume form.

To perform integration on non-orientable manifolds, one must use densities instead.

Mikhail Katz
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