# Prove there exists a smooth unit normal at the boundary of the following manifold

Let $M$ be a compact subset of $\mathbb{R}^3$ with smooth boundary $S=\partial M$. Consider M with the standard orientation $\mu=\mu_{0}$ from $\mathbb{R}^3$ and $S$ with the boundary orientation $\partial \mu$ induced from M.

Prove that there exists a smooth unit normal field $n(x)=(n_{1}(x),n_{2}(x),n_{3}(x))$ ($x \in S$) on $S$ such that for any $x \in S$, if $\{v_{1},v_{2}\}$ is a basis for $T_{x}S$ with $\partial \mu(x)=[v_{1},v_{2}]$, then $[v_{1},v_{2},n(x)]=\mu_{0}.$

Where $[v_{1},v_{2}]$ denotes the equivalence class of $\{v_{1},v_{2}\}$ under the equivalence relation $\{v_{1},v_{2}\}\sim \{w_{1},w_{2}\}$ iff $\{v_{1},v_{2}\}$ and $\{w_{1},w_{2}\}$ have the same orientation.

I'm really unsure on how to do this, so any help would be greatly appreciated.

• Hint: If I give you a vector $w_1 \in \mathbb{R}^2$, do you know how to construct a unit vector $w_2 \in \mathbb{R}^2$ orthogonal to $w_1$ so that $w_1,w_2$ are positively oriented? Next, if I give you two linearly independent vectors $v_1,v_2 \in \mathbb{R}^3$, do you know how to construct a unit vector $v_3$ orthogonal to $v_1,v_2$ so that $v_1,v_2,v_3$ are positively oriented? – Lee Mosher May 20 '16 at 12:21
• What do you mean by positively oriented vectors? Do you mean that they have the same orientation as the standard orientation on $\mathbb{R}^2$? – jon27 May 21 '16 at 2:48
• To answer your question, would you extend $w_{1}$ to a basis $\{w_{1},w_{2}\}$ for $\mathbb{R}^2$ and then define $w'_{2}$ using the Gram-Schmidt process? – jon27 May 21 '16 at 2:51
• Yes to both comments. – Lee Mosher May 21 '16 at 16:19
• Right so as $T_xS$ is a 2-dimensional sub manifold of $\mathbb{R}^2$, we may extend $\{v_{1},v_{2}\}$ to a basis $\{v_{1},v_{2},v_{3}\}$ for $\mathbb{R}^3$. Using the Gram-Schmidt process, we may define $v'_{3}$ so that it is orthogonal to both $v_{1},v_{2}$ and is of unit length. Moreover $\{v_{1},v_{2},v'_{3}\}$ will have the same span as $\{v_{1},v_{2},v_{3}\}$. However I don't understand why this has the same orientation as $\mu_{0}$ and how is it a function of $x$? – jon27 May 22 '16 at 11:37

Generally speaking, given $n-1$ linearly independent vectors $v_1,...,v_{n-1}$ in $\mathbb{R}^n$, there is a unique unit vector $v_n$ which is orthogonal to each of $v_1,...,v_{n-1}$ and such that the basis $v_1,...,v_{n-1},v_n$ is positively oriented. This vector $v_n$ can be written as a function of the coordinates of $v_1,...,v_{n-1}$. Furthermore, each coordinate of $v_n$ is a smooth function of the coordinates of $v_1,....,v_{n-1}$.
For example, in dimension 2, if $v_1=\langle a,b \rangle$ then $v_2 = \langle -b,a \rangle \, / \, \sqrt{a^2+b^2}$. You can check the positive orientation condition by making a matrix with first row $v_1$ and second row $v_2$ and then computing the determinant to verify that it is positive.
In dimension 3, one can do the same with the cross product, taking $v_3 = (v_1 \times v_2) \, / \, \| v_1 \times v_2 \|$.
So then, near each point of $\partial M$ choose a coordinate patch for $\partial M$ which is positively oriented (with respect to the boundary orientation). As long as the coordinate functions of your coordinate patch are at least $C_1$, you obtain a positively oriented field of bases $v_1(p),...,v_{n-1}(p) \in T_p \partial M$ for $p$ in that patch, and the coordinate functions of each vector $v_i(p)$ are smooth. Thus the coordinte functions for $v_n(p)$ are smooth.
Now although $v_n(p)$ has been constructed patch-by-patch, where two patches overlap the values of $v_n(p)$ will be unique, because of the uniqueness of the construction of the unit orthogonal vector.