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Show that $i^*\omega \in \Omega^{n}(S^n)$ is an orientation form on $S^n$. Here $i:S^n \to \mathbb{R}^{n+1} - \{0\}$ and $\omega$ is a $n$-form on $\mathbb{R}^{n+1}- \{0\}$ for which $\omega_p(v_1, \dots, v_n) = \det(p,v_1, \dots, v_n).$

So to prove that this is an orientation form one needs to show that $(i^*\omega)_p \ne 0$ for every $p \in S^n$. Let $p \in S^n$ and consider $v_1, \dots,v_n \in T_pS^n$. We have that $$\begin{align*}(i^*\omega)_p(v_1, \dots, v_n) &= \omega_{i(p)}(di_p(v_1), \dots, di_p(v_n)) \\ &= \det(i(p), di_p(v_1), \dots, di_p(v_n)).\end{align*}$$

Now to show that this determinant is non-zero is where I'm getting stuck at. What options do I have here to show that this is either negative or positive?

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  • $\begingroup$ You do not need to take the $v_i$ to be arbitrary. Instead, pick them to be a basis for $T_pS^n$. Then you can consider the vectors $\{p,v_1,\dots,v_n\}$. If you have chosen the $v_i$ to be a basis for $T_pS^n$, what does this set give you? Think about this geometrically, for $S^2\subset\mathbb{R}^3$. $\endgroup$ Commented May 2, 2023 at 15:35
  • $\begingroup$ Well the $v_j$'s are linearly independent at least @QuaereVerum $\endgroup$
    – Elena
    Commented May 2, 2023 at 15:38
  • $\begingroup$ Yes. Now view them as vectors in $\mathbb{R}^{n+1}$ via $T_pS^n\subset T_p\mathbb{R}^{n+1}\cong\mathbb{R}^{n+1}$. Then they define an $n$-dimensional linear subspace. Furthermore, the vector $p$ is normal to the hypersurface $S^n$. So what can you say about the collection $\{p,v_1,\dots,v_n\}$? $\endgroup$ Commented May 2, 2023 at 16:24
  • $\begingroup$ Are you trying to get at that the set is linearly independent no matter what $p$ is? I see $p$ being the only think that would cause problems of that set not being linearly independent, but if it's always normal to $S^n$, then it cannot be expressed as linear comb of the $v_j$'s. @QuaereVerum $\endgroup$
    – Elena
    Commented May 2, 2023 at 16:35
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    $\begingroup$ Whenever there is a submanifold $N\subset M$, there is a canonical inclusion $TN\subset TM$. So when you take the differential $di_p$, you are just viewing the vectors in $T_pN$ that you're pushing forward, as vectors in the larger vector space $T_pM$. Hence, if they are linearly independent in $T_pN$, they are also linearly independent in $T_pM$. If you have further questions about this I'm happy to answer them but then you should perhaps message me directly, as we are discouraged from having entire conversations in the comments. $\endgroup$ Commented May 2, 2023 at 17:44

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