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

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 $$x\in\partial M$$.

• How do you define the normal field if the codimension is not 1? Jul 10 '20 at 15:36
• @peek-a-boo In the same way as described in my other question: math.stackexchange.com/q/3748993/47771. Isn't it only important the codimension of $\partial M$ is $1$ (which it always is)? Jul 10 '20 at 15:47

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 $$T_x \mathbb R^d = T_x (\partial M) \oplus N_x (\partial M)$$

• Thank you for your answer. But doesn't isn't the codimension of the boundary $\partial M$ (which is a $k-1$-dimensional submanifold) of a $k$-dimensional submanifold with boundary $M$ always $(1)$? And we have the orthogonal decomposition $\mathbb R^k=T_x(\partial M)\oplus N_x(\partial M)$. Jul 10 '20 at 16:00
• I've rewritten the first paragraph of my answer a bit, in order to clarify the notion of "codimension" for purposes of your question and my answer. Does this clear things up? Jul 11 '20 at 16:15
• I also added some lines at the end regarding the orthogonal decomposition. Jul 11 '20 at 16:18
• I'm sorry, I don't know what I was thinking. For some reason, I thought about $\partial M$ as a submanifold of $\mathbb R^k$; which it is not. It is a submanifold of $\mathbb R^d$ with dimension $k-1$ and so the normal space has dimension $d-(k-1)$. Jul 11 '20 at 17:16