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Let $\Omega \subset \mathbb{R}^d$ be open and with $C^1$ boundary $\Gamma$. For any given point $x_0 \in \Gamma$ we know there's a neighborhood where $\Gamma$ is the graph of some $C^1$ function $\gamma : \mathbb{R}^{d - 1} \longrightarrow \mathbb{R}^d, x' \longmapsto \gamma ( x') = x_d$. We can use it to straighten the boundary with the local diffeomorphism

$$ T ( x', x_d - \gamma ( x')) = ( x', x_d - \gamma ( x')), $$

and its differential $D T$ has a nice $( d - 1) \times ( d - 1)$ identity matrix as first block and a bottom row $\nabla T_d = ( - \nabla \gamma, 1)$ which is proportional to the vector $\vec{n}$ normal to $\Gamma$ at each point, say $c ( x) \vec{n} ( x) = \nabla T_d ( x)$, where $c ( x) = - \| \nabla T_d ( x) \|$.

For my calculations in concrete examples with parametrized domains, etc., I want $\nabla T_d$ to actually be the outward pointing normal: I need this $c ( x)$ to be $- 1$. If I try to impose the condition after constructing $T$, then I have to integrate expressions which I'm just not capable of. I can try to throw it at some symbolic integration software, but there has to be some other way, right? In almost every book on PDEs it's stated that this $T$ may be normalized so as to have the property I mention. But how?

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I think the implicit function theorem could give you the answer. – Jose27 Nov 18 '12 at 0:10
With the implicit function theorem I can infer existence of $\gamma$ representing locally $\Gamma$ as the nullset of some $F ( x', x_N) = 0$ to which the theorem applies. Then using the chain rule I see that $$ \frac{\partial F}{\partial x_i} = - \frac{\partial F}{\partial x_d} \frac{\partial \gamma}{\partial x_i}, $$ and this allows me to write the normal vector using this function $F$, but I don't see how I can use this to explictly normalize my transformation $T$ with actual coordinates. – AnCo Nov 18 '12 at 0:29
Yeah, I was being too optimistic. I don't think this is an easy problem: For curves the property you need is given by using the arc-length parametrization which I don't think has a higher dimensional analog. Could you give a reference where this statement is used? – Jose27 Nov 18 '12 at 1:24
It turned out it was easier than we both thought… As to the reference, I admit I can't give any without looking it up, but I've seen it stated often enough and thanks to Lukas Geyer we now know it's definitely true :) – AnCo Nov 18 '12 at 16:51
up vote 2 down vote accepted

If $\phi(x')$ denotes the $d$-th component of the normal vector at $(x',\gamma(x'))$, then first of all it is immediate from the graph structure that $\phi(x') \ne 0$. Let $S(x',y_d) = (x',\phi(x')y_d)$, and write $\tilde{T} = S \circ T$. Then $\tilde{T}$ is a $\mathcal{C}^1$ diffeomorphism which straightens the boundary, and it is normalized, as can be checked easily with the chain rule: $$ DS (x',y_d) = \left[ \begin{array}{c|c} \mathrm{Id} & 0 \\ \hline \nabla \phi(x') y_d & \phi(x') \end{array} \right] $$ In particular $$ DS (x',0) = \left[ \begin{array}{c|c} \mathrm{Id} & 0 \\ \hline 0 & \phi (x') \end{array} \right] $$ So on the boundary $\Gamma$ you get $$ D\tilde{T}(x',\gamma(x')) = DT(x',\gamma(x')) DS(x',0) =\left[ \begin{array}{c|c} \mathrm{Id} & 0 \\ \hline -\phi(x')\nabla \gamma(x') & \phi (x') \end{array} \right] $$ I.e., the last row is a multiple of the outer normal, and since the $d$-th entry is the same, it is equal to the outer normal.

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Perfect, thanks! It's painfully easy now I see it (as usual…). – AnCo Nov 18 '12 at 16:47
I just realized that there is a little regularity problem with this proof if you only have $\mathcal{C}^1$ boundary, since then you only know that $\phi$ is continuous, not necessarily differentiable. You might have to assume $\mathcal{C}^2$ boundary for it to work. – Lukas Geyer Nov 19 '12 at 18:59
But wouldn't the Gauss map be as differentiable as the manifold? Globally only if it's orientable of course, but locally at least? – AnCo Nov 20 '12 at 0:07
The Gauss map in coordinates involves first partial derivatives, so if the surface is $\mathcal{C}^k$, its Gauss map should be $\mathcal{C}^{k-1}$. – Lukas Geyer Nov 20 '12 at 0:49
Ok. I'll just stick with $C^2$. Thanks for the help! – AnCo Nov 21 '12 at 22:15

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