# Show that $i^*\omega \in \Omega^{n}(S^n)$ is an orientation form on $S^n$.

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?

• 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 $S^2\subset\mathbb{R}^3$. Commented May 2, 2023 at 15:35
• Well the $v_j$'s are linearly independent at least @QuaereVerum Commented May 2, 2023 at 15:38
• 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 $\{p,v_1,\dots,v_n\}$? Commented May 2, 2023 at 16:24
• 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 Commented May 2, 2023 at 16:35
• 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. Commented May 2, 2023 at 17:44