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:
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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.
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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?
Best Answer
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 $\Sigma$.