# Show that at regular surface is orientable if it has a smooth normal vector field

Let $S\subset\mathbb{R}^3$ be a regular surface, and let $N:S\to S^2$ be smooth. I want to show that then $S$ is orientable.

My definition of $S$ being regular is that it is a smooth surface such that if $\iota: S \to\mathbb{R}^3$ is the inclusion map, then for every local parametrization $x$ of $S$ the Jacobian of the composition $\iota\circ x$ has rank 2 at every point.

Let $p$ be a point in $S$ and let $x:U\to S$ be a local parametrization around $p$. A curve at $p$ is a smooth map $\omega:J\to S$, where $J$ is an open interval containing $0$. $T_pS$ is the set of equivalence classes of curves at $p$ under the relation $\sim$ given by $\omega\sim\tau\Leftrightarrow (x^{-1}\omega)'(0)=(x^{-1}\tau)'(0)$. This equivalence relation is independent of the chosen parametrization $x$. The map $[\omega]\mapsto(x^{-1}\omega)'(0)$ is a linear isomorphism between $T_pS$ and $\mathbb{R}^2$. Let $x_u^p$ and $x_v^p$ be the vectors in $T_pS$ that are mapped to $(1,0)$ and $(0,1)$, respectively. Then $\{x_u^p,x_v^p\}$ is a basis for $T_pS$. If $y$ is another local parametrization around $p$, we get similarly another basis $\{y_u^p,y_v^p\}$. The change of coordinate matrix from $\{x_u^p,x_v^p\}$ to $\{y_u^p,y_v^p\}$ is $D(y^{-1}x)(x^{-1}(p))$, the Jacobian of $y^{-1}x$ evaluated at $x^{-1}(p)$.

We say that $S$ is orientable it admits an atlas such that $D(y^{-1}x)$ has positive determinant whenever it is defined, for any parametrizations $x,y$.

Consider $\iota\circ\omega$ for some curve $\omega$ at $p$. We have that $(\iota\omega)'(0)=(\iota x x^{-1}\omega)'(0)=D(\iota x)(x^{-1}(p))\cdot(x^{-1}\omega)'(0)$. Since $D(\iota x)(x^{-1}(p))$ by the definition of a regular surface is of rank 2, the linear tranformation given by the matrix $D(\iota x)(x^{-1}(p))$ is a linear isomorphism onto its image. Composing with the isomorphism $\omega\mapsto(x^{-1}\omega)'(0)$, we get an isomorphism, call it $\varphi_x^p$, between $T_pS$ and a two dimensional subspace of $\mathbb{R}^3$.

Let $u_x^p=\varphi(x_u^p)$ and $v_x^p=\varphi(x_v^p)$. Then $\{u_x^p,v_x^p\}$ is a basis for $\varphi(T_pS)$. $\{u_y^p,v_y^p\}$ is another basis, and the change of coordinate matrix from $\{u_x^p,v_x^p\}$ to $\{u_y^p,v_y^p\}$ is $D(y^{-1}x)(x^{-1}(p))$.

The idea now is to make sure that $(u_x^p,v_x^p,N(p))$ is a right-hand system of vectors in $\mathbb{R}^3$, i.e. that the $(3\times 3)$-matrix $[u_x^p\;\;v_x^p\;\;N(p)]$ has positive determinant, by if necessary replacing $x$ by $x'$ given by $x'(u,v)=x(v,u)$. But how do I do this in detail? And how do I show that it indeed gives $S$ an orientation?

Thank you for any input.

• Take the atlas and use $N$ to divide it into two equivalence classes, each of which is an atlas satisfying the definition of orientability. – Neal Nov 18 '17 at 21:14

$S$ is orientable, if you can define a volume form, let $x\in S, u,v\in T_xS\subset T_x\mathbb{R}^3$, you can define $\omega_x(u,u)=\Omega(u,v,N(x))$ where $N(x)$ is the normal vector and $\Omega$ a volume form on $\mathbb{R}^3$.
• A volume form is defined by a volume $\omega_x$ at each $x\in S$ which depends smoothly of $x$, let $(f_i:U_i\subset\mathbb{R}^2\rightarrow S)$ be a smooth atlas, define $\omega_i=f_i^*\omega$, on $U_i\cap U_j$, you have $\omega_i=Jac(f_i^{-1}\circ f_j)\omega_j$, since the sign of $Jac(f_i^{-1}\circ f_j)$ is constant, you can suppose that it is $>0$. – Tsemo Aristide Nov 18 '17 at 16:28