$H^{k}$ is oriented by the standard orientation of $\mathbb R^{k}$. Thus $\partial$$H^{k}$ acquires a boundary orientation. But $\partial$$H^{k}$ may be identified with $\mathbb R^{k-1}$. Show that the boundary orientation agrees with the standard orientation of $\mathbb R^{k-1}$ if and only if k is even. Symbolically, we may express this as $\partial$$H^{k}$ = $(-1)^{k}$$\mathbb R^{k-1}$.

This is Question 3.2 #6 from Guillemin and Pollack Differential Topology. If $\partial$$H^{k}$ may be identified with $\mathbb R^{k-1}$, why don't they have the same orientation? Aren't they both just represented by {$e_1$,...$e_{k-1}$}?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.