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I have an ambiguity about how to orient the boundary of a manifold.

In particular : Consider the example $M=B^2 \subset \mathbb{R^2}$ be the manifold with boundary. suppose positive orientation for M is induced by the positive orientation of $\mathbb{R^2}$ ({$(1,0),(0,1)$} is positive). According to Guillemin & Pollack text we must orient $\partial B^2=S^1$ such that : If $\mathcal{B}$ is a basis for $T_x(S^1)$, $sign \mathcal{B} = sign \{ n_x , \mathcal B \}$ as a basis for $T_x(B^2)$. where $n_x$ is the outward unit normal vector.

Now my question : let $x=e^{i\pi/4} \in S^1$. Now, $n_x=e^{i\pi/4}$ (based at $x$). I think either of the vectors $e^{i3\pi/4}$ and $e^{-i\pi/4}$ (both based at x) extend to positively oriented bases for $T_x S^1$, denote them by $\mathcal{B_1}$ and $\mathcal{B_2}$ respectively. Rotation by $\pi/4$ of {$(1,0),(0,1)$} gives us $\{ n_x , \mathcal B_1 \}$ and Rotation by $-\pi/4$ of {$(1,0),(0,1)$} gives us $\{ n_x , \mathcal B_2 \}$ . And we know that Rotation matrices have positive determinants $(det R_{\theta}=1)$ which implies that {$(1,0),(0,1)$} and $\{ n_x , \mathcal B_1 \}$ and $\{ n_x , \mathcal B_2 \}$ are all equivalently oriented.

I would be very thankful if you can clear up this ambiguity for me. Thanks!

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How do you arrive at $n_x$ by rotating $(1,0)$ first by $\pi/4$, then by $-\pi/4$? In one case you must get $-n_x$ – Thomas Apr 19 '14 at 14:27
If I rotate $\{(1,0),(0,1)\}$ by $\pi/4$ I would get $\{n_x,e^{i3\pi/4}$ \}, and if If I rotate $\{(1,0),(0,1)\}$ by $-\pi/4$ I would get $\{e^{-i\pi/4}, n_x \}$ – the8thone Apr 19 '14 at 14:31
I see ! the order is changed , $n_x$ is in the second spot. – the8thone Apr 19 '14 at 14:35
right. That's the point. – Thomas Apr 19 '14 at 14:41

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