Fixing orientation of connected smooth manifold in $\mathbb{R}^n$ by a single chart

Question

I am studying on Zorich, Mathematical Analysis II, 1st ed. pag. 174-175. After having properly explained how orientations (equivalence classes) are defined for smooth k-dimensional surfaces in $\mathbb {R} ^ n$ that can be described with a single map, move on to the more general case by defining the meanings of:

1. consistent charts,
2. orienting atlases,
3. equivalence classes for orienting atlases (possible orientations of the surface).

Having done this, he states without proof that a connected smooth k-dimensional surface can only have two possible orientations. From this statement he immediately deduces that in order to fix an orientation on a surface of this type it is not necessary to exhibit an entire atlas of consistent charts, but it is sufficient to exhibit a single chart.

I was trying to prove why, but I can't. I assumed, by absurdity, that I had two atlases of different orientation, made of pairwise consistent charts, containing a common chart $ \varphi_1 $:

$$A_1=\{\varphi_1,\varphi_2,...,\varphi_m,...\}$$ $$A_2=\{\varphi_1,\varphi'_2,...,\varphi'_m,...\}$$

but from here I can't get to any absurdity. Can anyone help me please?

Answer 1

I'm going to use the terminology "manifold" instead of "surface", because "surface" usually means 2-dimensional.

Let me use the notation $M$ for the manifold in question.

You have to somehow make use of the hypothesis that the manifold $M$ is connected. Since manifolds are locally path connected, you can use the theorem that a connected, locally path connected space is path connected.

Consider the common chart $\varphi_1 : U_1 \to \mathbb R^k$ in $A_1 \cap A_2$, and fix a base point $p \in U_1$.

Now I'll prove directly that any chart in $A_1$ and any chart in $A_2$ are consistent at any point of their overlap.

Consider any $x \in M$, and pick charts $\phi_I : U_I \to \mathbb R^k$ in $A_1$ and $\varphi'_J : U'_J \to \mathbb R^k$ in $A_2$, such that $x \in U_I \cap U'_J$. We have to show that $\varphi_I$ and $\varphi'_J$ are consistent at the point $x$.

Using path connectivity of the manifold $M$, choose a continuous path $\gamma : [0,1]$ such that $\gamma(0)=p$ and $\gamma(1)=x$. Since the sets $\{U_i \cap U'_j\}_{i,j}$ cover $M$, their inverse images $\{\gamma^{-1}(U_i \cap U'_j)\}_{i,j}$ cover $[0,1]$. Applying the Lebesgue Number Lemma, we can choose an integer $N \ge 1$, and decompose $[0,1]$ into subintervals $I_m = [\frac{m-1}{N},\frac{m}{N}]$, $m=1,\ldots,N$, so that $\gamma(I_m)$ is a subset of one of the intersections $U_{i(m)} \cap U'_{j(m)}$.

We know that $\varphi_{i(1)}$ and $\varphi'_{j(1)}$ are both consistent with each other at $\gamma(0)=p$, because both are consistent with $\varphi_1$. Consider the path $\gamma \mid I_1$ and let $t \in I_1 = [0,1/N]$ vary from $0$ to $1/N$. As $t$ varies, the determinant of the derivative of the overlap map of the two charts $\varphi_{i(1)}$ and $\varphi'_{j(1)}$ varies continuously, it is nonzero everywhere, and it is positive at $t=0$, hence it is positive at $t=1/N$. This proves that $\varphi_{i(1)}$ and $\varphi'_{j(1)}$ are consistent at $\gamma(1/N)$.

Now we do an induction proof: assuming by induction that $\varphi_{i(m)}$ and $\varphi'_{j(m)}$ are consistent at $\gamma(m/N)$, we prove that $\varphi_{i(m+1)}$ and $\varphi'_{j(m+1)}$ are consistent at $\gamma((m+1)/N)$. Since $\varphi_{i(m)}$ and $\varphi_{i(m+1)}$ are consistent at $\gamma(m/N)$, and since $\varphi'_{j(m)}$ and $\varphi'_{j(m+1)}$ are consistent at $\gamma(m/N)$, it follows that $\varphi_{i(m+1)}$ and $\varphi'_{j(m+1)}$ are consistent at $\gamma(m/N)$. Now the proof continues as in the previous paragraph, using continuity of the determinant of the derivative of the overlap map of the two charts $\varphi_{i(m+1)}$ and $\varphi'_{j(m+1)}$ at $\gamma(t)$, as $t \in I_{m+1}$ varies from $m/N$ to $(m+1)/N$, and the consistency of those charts at $\gamma(m/N)$, to deduce consistency at $\gamma((m+1)/N)$. This completes the induction step.

To complete the proof, we have shown that $\varphi_{i(N)}$ and $\varphi'_{j(N)}$ are consistent at $\gamma(N/N)=x$. We also know that $\varphi_I$ is consistent with $\varphi_{i(N)}$, and $\varphi'_J$ is consistent with $\varphi'_{j(N)}$ at $x$. Therefore, $\varphi_I$ and $\varphi'_J$ are consistent at $x$.

Answer 2

Let $M$ be your $k$-dimensional surface oreinted with respect to the chart $\{ \varphi_i\}_i$, $\varphi_i : \mathbb R^k\rightarrow U_i \subset_{open } M $. $\exists \ \omega\in \Omega^k(M)$ such that $\omega$ is non-vanishing at every point. This is possible since $M$ is orientable. $\varphi_i^*\omega=g_i \lambda$ where $\lambda=dx_1\wedge dx_2\wedge \dots dx_n$ and $g_i:\mathbb R^k \rightarrow \mathbb R$ is a non-vanishing smooth function. Since the charts are consistent, either all $g_i$'s are positive or all negative. Assume that all the $g_i$'s are positive.

Now you have the charts $\{ \varphi_1, \varphi_j'\}_j $ As before we get $\varphi^*_1 \omega =g_1\lambda$ and ${\varphi'}_j^*\omega=h_j \lambda$. By the same logic as above, we get either $\{g_1, h_j \}_j$ are all positive functions or all negative. But since $g_1$ is positive, we get all $h_j$'s are positive. Thus you get the same orientation.