I am following Carroll's book on general relativity [1]. In Eq. D.35, he states that the components of the induced volume element $\hat \epsilon_{\mu_1...\mu_{n-1}}$ on a $(d-1)$-dimensional hypersurface $\Sigma$ of a $d$-dimensional manifold $M$ are given by $$ \hat \epsilon_{\mu_1...\mu_{n-1}} = n^\lambda \epsilon_{\lambda \mu_1....\mu_{n-1}}, $$ $\epsilon$ is the volume element of $M$ and $n^\lambda$ are the components of a vector that is normal to $\Sigma$.

I am wondering about a $n$ dimensional submanifold $\mathcal M$ of the $d$-dimensional manifold $\mathbb R^d$. Were I to guess, I would say that at every point on $\mathcal M$, you could create an orthonormal (right-handed) basis $n_{(1)}^{\lambda_1}$,...,$n_{(d-n)}^{\lambda_{d-n}}$ and that the (an?) induced volume element on $\mathcal M$ should be $$ \hat \epsilon_{\mu_1...\mu_n} = n_{(1)}^{\lambda_1}...n_{(d-n)}^{\lambda_{d-n}} \epsilon_{\lambda_1...\lambda_{d-n}\mu_1...\mu_n}. $$ Is this correct?

In principle there does not seem to be a unique way to choose the unit vectors, but on the other hand, I do think that this outer product of vectors ($n_{(1)}^{\lambda_1}...n_{(d-n)}^{\lambda_{d-n}}$) is invariant under rotations in the space orthonormal to $\mathcal M$.

[1] Carroll, S. (2014). Spacetime and Geometry: Pearson New International Edition. Pearson Education Limited.


1 Answer 1


Although what Carroll says is correct, I somewhat disagree with this approach so let's look at it from another way.

Let $M$ be an $m$ dimensional manifold and $N\subseteq M$ an $n$ ($n<m$) dimensional (embedded) submanifold with inclusion $\phi:N\rightarrow M$. Suppose that $M$ is orientable and oriented and $N$ to be orientable. Note that $N$ does not inherit an orientation from $M$ in general (usually neither does a hypersurface (codimension $1$ submanifold) unless it is a boundary). Suppose furthermore that a pseudo-Riemannian metric $g$ in $M$ pulls back to $h=\phi^\ast g$ and $h$ is itself pseudo-Riemannian (so $N$ has no null points).

Recall that to define the volume form $\mu(g)$ associated to $g$ in $M$ we also need an orientation on $M$ since we need to set in each chart $(U,x)$ $$ \mu(g)=\left\{\begin{matrix} \sqrt{\mathfrak g}dx^1\wedge\dots\wedge dx^m & x\text{ is positive} \\ -\sqrt{\mathfrak{g}}dx^1\wedge\dots\wedge dx^m & x\text{ is negative}\end{matrix}\right. $$Here $\mathfrak{g}=|\det{g_{\mu\nu}}|$

Then in the submanifold $N$ if we fix an orientation we can likewise set $\mu(h)=\pm\sqrt{\mathfrak h}dy^1\wedge\dots\wedge dy^{n}$ with the $+$ sign chosen for a positive coordinate system $y$ and the negative sign for a negative coordinate system $y$.

Note that $\mu(h)$ is an $n$-form on $N$ and not on $M$, so eg. we cannot write in index notation in a coordinate system of $M$. Not by default. But because of the metric tensor, we have a reasonably unique extension of $\mu(h)$ into $M$ (by that I mean that $\mu(h)$ is extended into an $M$-tensor but is still only defined at points of $N$).

Namely, by the assumptions on the metric and the causal type of $N$ for any $p\in N$ we have $$ T_pM=(T_pN)^\perp\oplus T_pN, $$i.e. each tangent spaces decomposes into a direct sum. Let $\Pi:TM|_N\rightarrow TN$ denote the corresponding projection operator which projects tangentially.

Then define the extension $\tilde{\mu}(h)$ such that for any $p\in N$ and $n$-tuple $u_1,\dots,u_n\in T_pM$ we have $$ \tilde{\mu}(p)(u_1,\dots,u_n)=\mu(p)(\Pi u_1,\dots,\Pi u_n). $$

We'd then like to relate $\tilde \mu(h)$ to $\mu(g)$. An equivalent definition of the Riemannian volume form $\mu(g)$ is that for any point $p\in M$ and any $m$-tuple $e_1,\dots,e_m\in T_pM$ such that this ordered list forms a positively oriented orthonormal basis of $T_pM$, we have $$ \mu(g)(e_1,\dots,e_m)=1. $$This actually determines $\mu(g)$ uniquely.

If $N$ has been given an orientation, then the orthogonal complement bundle $(TN)^\perp\le TM|_N$ is also oriented as follows. Let $(n_1,\dots,n_{m-n},e_1,\dots,e_n)$ be a positively oriented orthonormal basis for $T_pM$ ($p\in N$) such that $e_1,\dots,e_n$ is a positively oriented orthonormal basis for $T_pN$, then $n_1,\dots,n_{m-n}$ is declared to be a positively oriented orthonormal basis of $(T_pN)^\perp$.

By the defining property of the volume we have $$ 1=\mu(g)(n_1,\dots,n_{m-n},e_1,\dots,e_{n})=(n_{m-n}\rfloor\dots n_1\rfloor\mu(g))(e_1,\dots,e_n). $$ This shows that $n_{m-n}\rfloor\dots n_{1}\rfloor \mu(g)$ is an $n$-form along $N$ (i.e. an $n$-form in $M$ defined only on points of $N$) which takes the value $1$ on any positively oriented orthonormal basis of $T_pN$, therefore the pullback of $n_{m-n}\rfloor\dots n_1\rfloor \mu(g)$ is $\mu(h)$.

However we then have $$ n_{m-n}\rfloor\dots n_{1}\rfloor\mu(g)=\tilde{\mu}(h), $$ since both the LHS and the RHS vanishes when contracted with any normal-directed vector (i.e. those which belong to $(T_pN)^\perp$).

It also follows from this derivation that if $n_1,\dots,n_{m-n}$ and $m_1,\dots,m_{m-n}$ are both positively oriented orthonormal bases of $(T_pN)^\perp$, then $n_{m-n}\rfloor\dots n_1\rfloor \mu(g)=m_{m-n}\rfloor\dots m_1\rfloor \mu(g)$.

Now, in OP's notation if $\mu(g)$ has components $\varepsilon_{\mu_1...\mu_m}$ and $\tilde{\mu}(h)$ has components $\varepsilon_{\mu_1...\mu_n}$, then the interior product is expressed as $$ \varepsilon_{\mu_1...\mu_n}=n^{\lambda_1}_1\dots n^{\lambda_{m-n}}_{m-n}\varepsilon_{\lambda_1...\lambda_{m-n}\mu_1...\mu_n}, $$which is of course the formula given in OP.


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