Volume Forms Induced by Embedding 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?
 A: 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 $\Sigma$.
