# (Carroll) In $\mathbb R^d$, what is the induced volume element on a $n$-dimensional submanifold?

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.

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 $$n$$ ($$n) dimensional (embedded) submanifold with inclusion $$\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.