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Here is an exercise from Guillemin and Pollack:

$H^k=\{(x_1,\dots,x_k)\in R^k: x_k \ge 0\}$ is oriented by the standard orientation of $R^k$. Thus $\partial H^k$ acquires a boundary orientation. But $\partial H^k$ may be identified with $R^{k-1}$. Show that the boundary orientation agrees with the standard orientation of $R^{k-1}$ if and only if $k$ is even.

According to the definition in the book, an orientation of a manifold is a smooth choice of orientations for all tangent spaces. The boundary orientation of $\partial H^k$ is defined by orienting $T_x(\partial H^k)$ in this way: declare the sign of a basis $(v_1,\dots, v_{k-1})$ to be the sign of the basis $(n_x, v_1,\dots,v_{k-1})$ where $n_x$ is the outward unit normal.

I do understand (by imagining pictures) that for $k=2$ the boundary orientation is the same as the standard orientation of $R^{k-1}$ and for $k=3$ it is not. But how to prove it in the general case?

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The natural basis for $T_x \mathbb{R}^k$ is $\partial_1,\dots,\partial_k$, viewed as derivations in the directions $$(1,0,\dots,0), (0,1,0,\dots,0),\dots ,(0,\dots,0,1).$$ If $x\in \partial H^k$, then $n_x= -\partial_k$ is the outward pointing normal, whereas $B=(\partial_1,\dots,\partial_{k-1})$ constitutes a basis for $T_x \partial H^k$. The orientation of $B$ obviously coincides with the orientation of $\mathbb{R}^{k-1}$. Moreover it coincides with the boundary orientation, iff $$ B' = (-\partial_k,\partial_1,\dots,\partial_{k-1}) $$ is an oriented basis of $T_x H^k$. Changing the sign of a vector and permuting two vectors flips orientation. In order to arrive at $(\partial_1,\dots,\partial_k)$ from $B'$, you have to change a sign once and permute $k-1$-times, overall the sign changes $k$-times. Thus if $k$ is even, then $B'$ has the standard orientation and if $k$ is odd it has reversed orientation.

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