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Let $M$ be a manifold of dimension $n$ such that there exist two charts $(U_a,\phi_a)$ and $(U_b,\phi_b)$ such that $U_a,U_b$ are connected and $U_a\cap U_b\ne\emptyset$. Moreover the transition function $\phi_{ab}$ neither preserves the orientation nor reverses the orientation. I have to show that $M$ is not orientable.

A manifold is orientable if it admits an oriented atlas, i.e. an atlas in which all the transition functions have Jacobian with positive determinant.

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  • $\begingroup$ The exact strategy depends on your definition of "orientable", but considering the contrapositive looks like a good idea here. $\endgroup$ – Andrew D. Hwang Aug 20 '15 at 8:43
  • $\begingroup$ I added the definition above $\endgroup$ – avati91 Aug 20 '15 at 8:44
  • $\begingroup$ So...what happens when you look at the contrapositive? :) $\endgroup$ – Andrew D. Hwang Aug 20 '15 at 13:04
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    $\begingroup$ An oriented atlas for $M$ induces an oriented atlas for an arbitrary open submanifold $U_{a}$ by intersecting charts with $U_{a}$.... :) $\endgroup$ – Andrew D. Hwang Aug 20 '15 at 15:47
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    $\begingroup$ Problem solved using your last "hints" :) If you rewrite your comment as an answer, I would be glad to accept it! $\endgroup$ – avati91 Aug 21 '15 at 15:18
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Here's a fairly detailed sketch:

Let $M$ be an $n$-manifold. Suppose, contrapositively, that $M$ is orientable, and fix a maximal oriented atlas, i.e., a maximal atlas for which all the transition maps have Jacobian with positive determinant.

Lemma 1: If $(U, \phi)$ is a chart and $U$ is connected, then $(U, \phi)$ is either compatible with the orientation, or anti-compatible with the orientation.

Proof: Let $(V, \psi)$ be an arbitrary oriented chart with $U \cap V$ non-empty. The sign of the Jacobian of $\psi \circ \phi^{-1}$ in $\phi(U \cap V)$ is independent of $\psi$ because $(V, \psi)$ is selected from an oriented atlas.

Let $U^{+}$ denote the set of points of $U$ for which the Jacobian $\psi \circ \phi^{-1}$ is positive, and let $U^{-}$ denote the set of points of $U$ for which the Jacobian $\psi \circ \phi^{-1}$ is negative. The sets $U^{\pm}$ are obviously disjoint, each is open, and their union is $U$. Since $U$ is connected, one set is empty: Either $U = U^{+}$ or $U = U^{-}$.

Lemma 2: If $(U, \phi)$ is an anti-compatible chart with component functions $(\phi^{1}, \dots, \phi^{n})$, then the chart $(U, \bar{\phi})$ defined by $\bar{\phi} = (\phi^{1}, \dots, \phi^{n-1}, -\phi^{n})$ is compatible, and conversely.

Proof: The linear transformation $T(x^{1}, \dots, x^{n}) = (x^{1}, \dots, x^{n-1}, -x^{n})$ has determinant $-1$.


Let $(U_{a}, \phi_{a})$ and $(U_{b}, \phi_{b})$ be arbitrary charts with $U_{a}$ and $U_{b}$ connected, and with $U_{a} \cap U_{b}$ non-empty. If necessary, replace $\phi_{a}$ by $\bar{\phi}_{a}$ to get a compatible chart, and similarly for $\phi_{b}$. Since each chart is compatible with the oriented atlas, the transition map has positive Jacobian. It follows that the transition map $\phi_{ab}$ between the original charts has Jacobian of constant sign, i.e., either preserves orientation or reverses orientation.

Contrapositively, if there exist connected, overlapping charts $(U_{a}, \phi_{a})$ and $(U_{b}, \phi_{b})$ whose transition map $\phi_{ab}$ neither preserves nor reverses orientation, then $M$ is not orientable.

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  • $\begingroup$ Does the reverse statement holds? That is, if $M$ is non-orientable, such two charts $(U_a,\phi_a),(U_b,\phi_b)$ exists. $\endgroup$ – Fallen Apart Aug 22 '15 at 11:43
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    $\begingroup$ @Fallen: Yes, the converse is true. Intuitively, if $M$ is non-orientable, there exists a loop (closed curve) around which "the orientation reverses". Thicken the loop into a tube and cover it with two "solid cylinders" $U_{a}$ and $U_{b}$. There are details to check, and this sketch may not be the easiest strategy to make rigorous. (E.g., if no such charts exist, start with an arbitrary atlas and reverse chart orientations as needed to get an atlas of compatible charts. Again, though, there are details to check....) Perhaps post a separate question? $\endgroup$ – Andrew D. Hwang Aug 22 '15 at 12:58
  • $\begingroup$ math.stackexchange.com/questions/1406024/… You are welcome $\endgroup$ – Fallen Apart Aug 22 '15 at 15:53

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