Why is it important the manifold has codimension $1$ in order to prove this identity for $\operatorname{div}fV$ on $\partial M$?

Question

I've seen the following claim in some lectures notes which let me think that I might have a major misunderstanding:

The claim is that if $M$ is an embedded submanifold of $\mathbb R^d$ with boundary of codimension $1$ and $f$ and $V$ are differentiable scalar and vector fields, respectively, then $$\operatorname{div}fV=f\operatorname{div}V+\frac{\partial f}{\partial\nu}\langle V,\nu\rangle+\langle\nabla f,V_{\partial M}\tag1,$$ where $$\frac{\partial f}{\partial\nu}:=\langle\nabla f,\nu\rangle,$$ $\nu$ is the normal field and $V_{\partial M}$ is the tangential component of $V$ (i.e. the projection of $V$ onto the tangent space).

I don't understand why it is important that $M$ has codimension $1$. If $M$ is $k$-dimensional, then $\partial M$ is $(k-1)$-dimensional. If $M$ has codimension $1$, then it is $(d-1)$-dimensional and hence $\partial M$ is $(d-2)$-dimensional .... Why should this be of any use in $(1)$?

Assuming $M$ is $k$-dimensional, $(1)$ should trivially follow from $$\operatorname{div}(fV)(x)=\langle\nabla f(x),V(x)\rangle+f(x)\operatorname{div}V(x)\;\;\;\text{for all }x\in\mathbb R^k$$ and $$\langle\nabla f(x),V(x)\rangle=\langle\nabla f(x),\operatorname P_{T_x(\partial M)}V(x)\rangle+\langle V(x),\nu(x)\rangle\frac{\partial f}{\partial\nu}(x)\tag2$$ for all $x\in\partial M$, where $\operatorname P_{T_x(\partial M)}$ denotes the orthogonal projection of $\mathbb R^k$ onto the tangent space $T_x(\partial M)$ of $\partial M$ at $x\in\partial M$.

Answer 1

First, perhaps you misread the requirement: it is not itself $M$ that must have codimension 1, it is instead the boundary of $M$, denoted $\partial M$, that must have codimension 1. But the phrase "... with boundary of codimension 1... " leaves out some information which could perhaps clarify the situation: that phrase should be parsed as

... such that the boundary of $M$ has codimension in $\mathbb R^d$ equal to 1...

Equivalently, $M$ itself must have codimension in $\mathbb R^d$ equal to 0.

So, why must $\partial M$ have codimension $1$ in $\mathbb R^d$?

As soon as you write the words "$\nu$ is the normal field and $V_{\partial M}$ is the projection of $V$ onto the tangent space", one wonders: normal field of what? Tangent space of what? The only sensible answer that I can see is that $\nu$ is the normal field to $\partial M$ and $V_{\partial M}$ is the projection of $V$ onto the tangent space of $\partial M$.

And in order that $\partial M$ even possess a normal field, it must have codimension 1. So, in order for the $v$ term in equation (1) to even be defined, $\partial M$ must have codimension 1.

(It might be clearer if one rewrites equation (1) with proper quantifiers: each term should have the argument $x$ in the correct position, sort of like you did in the later equations; and the equation should hold for each $x \in \partial M$.)

The point here is that a submanifold of dimension $n \ge 2$ or higher does not have a well-defined normal field. It does have a well-defined normal bundle, which is a vector bundle of dimension $n$. For example, for a circle $C$ embedded in $\mathbb R^3$, which has codimension 2, its normal bundle has fibers of dimension $2$: at each point $x \in C$, the normal plane $N_x C$ is the 2-dimensional subspace of $T_x \mathbb R^3$ ($= \mathbb R^3$) which is normal to the 1-dimensional tangent line $T_x C$.

In general, for an codimension $m$ submanifold $B \subset \mathbb R^n$, the normal bundle is an $m$-dimensional vector bundle over $B$, whose fiber $N_x B$ is the $m$-dimensional subspace of $T_x \mathbb R^n$ (which is identified with $\approx \mathbb R^n$) that is normal to the $n-m$ dimensional subspace $T_x B$ of $T_x \mathbb R^n$. It follows that there is an orthogonal direct sum $$T_x \mathbb R^n = T_x B \oplus N_x B $$ And then, as you asked in the comments, for the case that $n=d$ and that $B = \partial M$ has codimension 1 in $\mathbb R^n$, one obtains $$T_x \mathbb R^d = T_x (\partial M) \oplus N_x (\partial M) $$