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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?

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  • $\begingroup$ Can you include the definition of orientation used. $\endgroup$ Aug 18, 2020 at 11:31
  • $\begingroup$ A surface is oriented when an equivalence class in the set of all orienting atlases is fixed. The equivalence relation on the set of atlases is defined by saying that two atlases are equivalent if their union is an orienting atlas. An orienting atlas is an atlas in which all charts are pairwise consistent. Two charts are consistent if their domains of action have empty intersection or if in the non-empty intersection the mutual transitions between one chart to the other have positive Jacobian. $\endgroup$
    – Nameless
    Aug 18, 2020 at 19:30
  • $\begingroup$ The distinction between $\phi$ and $\varphi$ is too subtle; $A_1$ and $A_2$ looked identical to my eyes until I looked at the source. So I changed the notation to make the distinction more visually obvious. $\endgroup$
    – Lee Mosher
    Aug 20, 2020 at 12:02

2 Answers 2

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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 $x$.

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  • $\begingroup$ Why does $U_i\cap U_j'$ cover $M$? $\endgroup$
    – Nameless
    Aug 20, 2020 at 12:58
  • $\begingroup$ $M=\cup _i U_i= \cup_j U'_j =\cup_i U_i \bigcap \cup_j U'_j= \cup_{i,j}U_i \cap U_j'$ an $\endgroup$
    – user6
    Aug 20, 2020 at 13:10
  • $\begingroup$ I rewrote the notation a bit, regarding the sets $U_i \cap U'_j$ and how they cover $M$. $\endgroup$
    – Lee Mosher
    Aug 20, 2020 at 14:22
  • $\begingroup$ I also reread the induction step, and it was a bit incomplete, so I filled it out somewhat. $\endgroup$
    – Lee Mosher
    Aug 20, 2020 at 14:27
  • $\begingroup$ Ok, I understood this beautiful proof. There is another gap that I would fill: could you give me a reference to study the fact that "a connected, locally path connected space is path connected", please? $\endgroup$
    – Nameless
    Aug 20, 2020 at 14:50
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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.

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  • $\begingroup$ I think (but I'm not sure) that you are using the theory of differential forms, theory that I have not yet studied. I'm following rigorously Zorich book, and the chapter on differential forms is successive. $\endgroup$
    – Nameless
    Aug 20, 2020 at 13:04
  • $\begingroup$ Ah I see. In that case, you will prefer Prof Lee's answer. But do come back to this once you have read differential forms. I studied orientation after differential forms/ homology and I guess that reflects in my answer. $\endgroup$
    – user6
    Aug 20, 2020 at 13:09

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