Fix a set $W \subseteq \mathbb R^n$ and let $k \leq n$. Define

$S_W := \{ \Phi:U \to W: U \subseteq \mathbb R^k $ is open, $\Phi $is a smooth embedding, and$ \Phi(U)=W \}$.

Suppose that $S_W$ is not empty. For $\Phi, \Psi \in S_W$, write $\Phi \sim \Psi$ if $\Phi$ and $\Psi$ have the same orientation.

(i) Show that there are at least two equivalence classes for $\sim$.

(ii) Prove that, if $W$ is connected, then for any $\Phi, \Psi \in S_W$, either $\Phi$ and $\Psi$ have the same orientation or they have the opposite orientation; thus there are exactly two equivalence classes for $\sim$ on $S_W$.

I'm quite stuck on this. I know how to prove that $\sim$ is an equivalence relation, so I assume I would implicate this somehow into proving (i) and (ii).

We know det$(J_{\Psi^{-1}\circ\Phi(a)})>0$ for all $a \in U$ by definition of the same orientation.

Since det$^{-1}(J_{\Psi^{-1}\circ\Phi(a)})=\frac{1}{ det(J_{\Psi^{-1}\circ\Phi(a)})}$ we get det$^{-1}(J_{\Psi^{-1}\circ\Phi(a)})>0$ and the 'having the same orientation' relation is symmetric.

Would I find the determinant of two of the equivalence relations and prove that they are different, hence different equivalence relations? I examined a proof involving vectors and the change of basis formula but I'm unsure how to apply it in this case.

Any help/solutions appreciated!


(i) $(x_1,\dots,x_k)\mapsto\Phi(-x_1,\dots,x_k)$, $x\in\cdots$ has the orientation opposed to $\Phi$.

(ii) $\det(J_{\Psi^{-1}\circ\Phi(x)})$ must be $\ne 0$ for all $x\in U$. If $W$ is connected, $U$ must be connected (by homeomorphism) and this means constant sign: $>0$ or $<0$.

EDIT for (i): aplying the chain rule to $\Xi(x_1,\dots,x_k) = \Phi(-x_1,\dots,x_k)$: $$J\Xi(x_1,\dots,x_k) = J\Phi(-x_1,\dots,x_k)\pmatrix{ -1&0&0&\cdots&0\cr 0&1&0&\ddots&0\cr 0&0&1&\ddots&0\cr \vdots&\ddots&\ddots&\ddots&0\cr 0&0&\cdots&0&1\cr } $$ and $$ \left|\matrix{ -1&0&0&\cdots&0\cr 0&1&0&\ddots&0\cr 0&0&1&\ddots&0\cr \vdots&\ddots&\ddots&\ddots&0\cr 0&0&\cdots&0&1\cr }\right| = -1. $$

  • $\begingroup$ Thank you for your answer. Can I ask how (i) shows there are at least 2 equivalence classes? Does it have anything to do with the determinant? $\endgroup$ – user377174 Mar 14 '17 at 16:02
  • $\begingroup$ @smallscot, yes. Use simply the chain rule. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 14 '17 at 18:23
  • $\begingroup$ Could you show me how? I'm not sure how to apply it, sorry! $\endgroup$ – user377174 Mar 14 '17 at 21:55
  • $\begingroup$ @smallscot, I will edit my answer tomorrow. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 14 '17 at 22:28

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