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Let $(M, g)$ be a Riemannian Manifold of dimension $d$, $g$ naturally gives rise to an invariant volume form $V_M \in \Omega^d(M)$.

Let $\Sigma$ be a smooth embedded submanifold of dimension $d-1$ in $M$. One can pull back the metric $g$ by the embedding map and construct am invariant volume form $V_\Sigma \in \Omega^{d-1}(\Sigma)$.

Question: Is it true that

\begin{equation} V_\Sigma = i_nV_M \end{equation}

where $n|_p$ is the unit normal with respect to $g$ at $p \in \Sigma$.

Comments:

  • I think it is well known (and easy to check) that the above equation is true for surfaces in $\mathbb R^3$, but I am not sure if it holds in general.

  • If not true, is there a general equation of this type where one can express the "induced volume element" in terms of the global volume element, without explicit reference to the metric?

zzz
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1 Answers1

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Think back to what the Riemannian volume form is. It's a form that, when fed an oriented orthonormal frame, spits out 1. (Of course every manifold here is oriented.)

What are the oriented orthonormal frames on $\Sigma \subset M$, where $\Sigma$ is a (Riemannian) submanifold? Because it's a Riemannian submanifold, if $(x_1, \dots, x_{d-1})$ is an oriented orthonormal frame of $T_p \Sigma$, it's also orthonormal in $T_p M$ - we're just missing one vector. By definition, $n$ is normal to all of these, and (modulo orientation conventions), $(n, x_1, \dots, x_{d-1})$ is an oriented orthonormal frame of $T_p M$. Therefore $$i_nV_\Sigma(x_1, \dots, x_{d-1}) = V_M(n, x_1, \dots, x_{d-1}) = 1,$$ so that $i_nV_\Sigma$ does exactly as a volume form should.

You cannot refer to the induced volume form without at least having chosen a nonvanishing normal field. The Riemannian metric gives you a Riemannian structure on the normal bundle of $\Sigma$, which is what you need to define the induced volume form on $\Sigma$.

  • Great, this clears up everything, thanks! As a followup, can you explain or give a reference on how the Riemannian metric gives a Riemannian structure on the normal bundle? – zzz Nov 08 '15 at 01:40
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    @bechira: I phrased things a little poorly there. The Riemannian metric identifies $N\Sigma$ with $(T\Sigma)^\perp$, the line bundle of vectors perpendicular to $T\Sigma$. Then, because this is a subbundle of $TM$ (restricted to $\Sigma$), you get an actual Riemannian metric on this - you know exactly what it means for a normal vector to have length 1. So you can "pick out" a normal vector to feed into your volume element, like you did above. If you have some other method of picking a normal field, you have another way of getting a volume form on $\Sigma$. –  Nov 08 '15 at 01:42
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    To be a bit more explicit: Any manifold with a volume form is orientable, and a hypersurface of an orientable $M$ (which should have been stipulated) is orientable iff there is a smooth global unit normal field. – Ted Shifrin Nov 08 '15 at 07:40