# Existence of a unique “outer” normal vector

A compact set $K\subseteq\mathbb R^3$ is said to have a smooth boundary, if for all $p\in\partial K$ there is an open neighborhood $U$ of $p$ and a continuously differentiable function $\psi:U\to\mathbb R$ such that $$K\cap U=\left\{x\in U:\psi(x)\le 0\right\}\tag1$$ and $$\psi'(x)\ne0\;\;\;\text{for all }x\in U\tag2.$$ In that case, $\partial K$ is a $2$-dimensional $C^1$-submanifold of $\mathbb R^3$.

Now, if $M$ is a $k$-dimensional $C^1$-submanifold of $\mathbb R^n$, $k\le n$:

• $v\in\mathbb R^n$ is called tangent vector on $M$ at $p\in M$, if there is a $\varepsilon>0$ and a continuously differentiable function $\gamma:(-\varepsilon,\varepsilon)\to M$ such that $\gamma(0)=p$ and $\gamma'(0)=v$. The space of all such $v$ is denoted by $T_p(M)$
• $v\in\mathbb R^n$ is called normal vector on $M$ at $p$, if $v$ is orthogonal to $T_p(M)$. The space of all such $v$ is denoted by $N_p(M)$

We can show that $N_p(M)$ is $(n-k)$-dimensional for all $p\in M$. So, $N_p(\partial K)$ is $1$-dimensional for all $p\in\partial K$. We can select the unique $\nu(p)\in N_p(\partial K)$ of unit length with the following property: There is a $\varepsilon>0$ such that $p+t\nu(p)\not\in K$ for all $t\in(0,\varepsilon)$.

The map $\nu:\partial K\to\mathbb R^3$ is seen to be continuous.

Are we able to find such a unique and continuous normal field $\nu$ even for more general classes of manifolds?

For example, consider the following triangulation of a teapot (just the empty shell): Since even a single triangle is not a "compact set with smooth boundary", we cannot apply the result described above. But shouldn't we nevertheless be able to select a single "outer" unit normal vector on each point of the teapot?

## 1 Answer

Since that triangulation is not smooth, there are different normal vectors when you aproach a boundary segment from each of the two triangles of which is boundary. Hence, one cannot define a normal vector field on those segments compatible with the smooth structure of the interiors of the triangles.

• Yes, that's correct. Are we able to overcome this issue? What I've got in mind is the following: It would be sufficient for me, if $\nu$ is only well-defined on sets of positive "surface measure". The union of all edges should be a null set wrt that measure and at any other point we should be able to select a unique normal. My problem is that I'm not able to formalize this rigorously. It's not even clear me how the "surface measure" for my teapot needs to defined. I only know the construction of the surface measure for submanifolds. – 0xbadf00d Jun 30 '18 at 20:47
• If you forget about that 0-measure set, then you've got a disjoint union of open subsets of the manifold, so a normal vector field would be defined precisely by disjoint union of the vector fields on each open subset. – Javi Jun 30 '18 at 20:49
• You say "open subsets of the manifold*. Which manifold? And: Do you know where I can find the construction of the surface measure? – 0xbadf00d Jun 30 '18 at 20:52
• The manifold (surface) you are triangulating (I mean topological manifold, not smooth). By "the surface measure" I basically understand integration on surfaces, since you can use all the machinery of the Lebesgue measure (or any other measure defined on $\mathbb{R}^n$) using local charts. You can read about that in Do Carmo's Differential Geometry of Curves and Surfaces and more generally in Madsen's From Calculus to Cohomology (chapter 10, I think, it's called Integration on manifolds) – Javi Jun 30 '18 at 21:13