I'm trying to get my head around the fact that $\phi$ is orientation preserving, due to $d\phi$, i.e. $d\phi$ sends outward vectors on $\partial \mathcal{M}$ to outward vectors on $\mathbb{H}^n$.

Here $\mathcal{M}$ is a smooth orientable $n$-dimensional manifold, and $\mathbb{H}^n$ is simply $\mathbb{R}^n$ with the restriction that $x^n \geq 0$.

So $\partial \mathcal{M}$ has the induced orientation by outward vectorfields, and clearly an outward vector on $\mathbb{H}^n$ is one where $x^n$ is negative.

The idea that $d\phi$ must send oriented basis to oriented basis and thus $\phi$ is orientation preserving, somehow simply escapes me.

EDIT, excerpt of the statement: Since $d\phi$ takes outward-pointing vectors on $\partial \mathcal{M}$ to outward-pointing vectors on $\mathbb{H}^n$, it follows $\phi|_{U\cap \partial\mathcal{M}}$ is an orientation-preserving diffeomorphism onto $\phi(U)\cap\partial \mathbb{H}^n$.

$\phi$ is positively oriented by assumption.

  • $\begingroup$ What's your definition of "outward vector on $\partial \mathcal{M}$"? $\endgroup$ – Vectornaut Jun 12 '15 at 10:13
  • $\begingroup$ Under your conditions, it seems like the claim "any boundary chart $\phi \colon \mathcal{M} \to \mathbb{H}^n$ must be orientation-preserving" is obviously false. For example, let $\mathcal{M}$ be $\mathbb{H}^3$ with the standard orientation, and let $\phi \colon \mathcal{M} \to \mathbb{H}^3$ be the map $\phi(u^1, u^2, u^3) = (-u^1, u^2, u^3)$. Observe that $\phi$ is an orientation-reversing boundary chart on $\mathcal{M}$. What have I done wrong? $\endgroup$ – Vectornaut Jun 12 '15 at 10:20
  • $\begingroup$ I agree, but assuming $\mathcal{M}$ is oriented, s.t. the boundary has the induced orientation, this should somehow make sense, but I'm falling a little short in the argument of why. $\endgroup$ – mort Jun 12 '15 at 10:23
  • $\begingroup$ I'm confused. If you agree that the statement you're trying to prove has an obvious counterexample, why are you trying to prove it? When you say "this should somehow make sense," what does "this" refer to? $\endgroup$ – Vectornaut Jun 12 '15 at 10:26
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    $\begingroup$ Listen I understand, but the more evident that there's a counterexample, the more sure I am my question has been stated with too few restrictions. I'm using intro to smooth manifolds by J.M. Lee, where this explicitly is used. $\endgroup$ – mort Jun 12 '15 at 10:49

The statement you've excerpted comes from the proof of Stokes's theorem in Lee's Introduction to Smooth Manifolds. In that part of the proof, Lee is "assuming without loss of generality that $\phi$ is an oriented chart" (see the beginning of the paragraph where the statement appears). Without that assumption, as you've noticed, the statement is false.

When you read a statement that seems obviously wrong, it's generally a good idea to go back and look for assumptions you might have missed, or definitions that might be different than what you expected.

  • $\begingroup$ wauw thanks sooo much for sticking with me on this :) So by the assumption of positive orientation of $\phi$, we restrict the map to a positive frame. From there it's evident that outwardpointing vectors correspond, s.t. $\phi$ is orientation preserving. Does this suffice? I just want to say you're one hell of a trooper, and sorry I've taken this much of your time :( $\endgroup$ – mort Jun 13 '15 at 11:20
  • $\begingroup$ Guess I'm a broken record, but help me close the gap please. Assuming $\phi$ is oriented, then we know the coordinate frame $\partial / \partial x^i$ is oriented. This to me is the coordinate domain, so how that this ensure $\phi$ doesn't reverse orientation on the codomain (or did I get this wrong)? Simply put how am I sure, from the assumptions, that $d\phi ( \partial / \partial x^i )$ has a positive determinant? $\endgroup$ – mort Jun 14 '15 at 9:58
  • $\begingroup$ Say I have the frame $\partial / \partial x^i$ with my oriented map $\phi$, then $(d\phi(\partial / \partial x^i))$ is a frame for $\mathbb{H}^n$. I'm blind as to why this frame is positively oriented. Or am I silly here? $\endgroup$ – mort Jun 14 '15 at 10:06
  • $\begingroup$ @mortjt: The manifold $\mathbb{R}^n$ comes with a standard orientation, which $\mathbb{H}^n$ inherits, so the coordinate domain of a chart always has an orientation; no extra assumptions are necessary to make this happen. A chart on an oriented manifold is called an oriented chart if it's orientation-preserving. $\endgroup$ – Vectornaut Jun 14 '15 at 17:37
  • $\begingroup$ @mortjt: I think you might be misunderstanding what Lee is trying to do in that part of the proof. Lee is showing that if $\phi \colon \mathcal{M} \to \mathbb{H}^n$ is an oriented (that is, orientation-preserving) chart for $\mathcal{M}$, then the restriction $\phi \colon \partial \mathcal{M} \to \partial \mathbb{H}^n$ is an oriented chart for $\partial \mathcal{M}$. $\endgroup$ – Vectornaut Jun 14 '15 at 17:43

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