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Question: Show that for $v \in T_p \mathbb{R}^{n+1} = \mathbb{R}^{n+1}$, if $v \cdot p = 0$, then $v \in T_p \mathbb{S}^{n}$, where $v \cdot p$ denotes the dot product on $\mathbb{R}^{n+1}$.

My thoughts so far:

I know that for every $p \in \mathbb{S}^{n}$, the tangent space $T_p \mathbb{S}^{n}$, is a subspace of $T_p \mathbb{R}^{n+1} = \mathbb{R}^{n+1}$. So if $v \cdot p = 0$ for $v \in T_p \mathbb{R}^{n+1} = \mathbb{R}^{n+1}$, then $v \cdot p = 0$ for $v \in T_p \mathbb{S}^{n}$ as well since every $v \in T_p \mathbb{S}^{n} \subseteq T_p \mathbb{R}^{n+1} = \mathbb{R}^{n+1}$, since this holds for every $p \in \mathbb{S}^{n}$.

I need to show that for $v \in \mathbb{R}^{n+1}$, if $v \cdot p = 0$, then $v$ is necessarily in $T_p \mathbb{S}^{n}$. By contradiction, if $v \cdot p = 0$ and $v \not\in T_p \mathbb{S}^{n}$, then $v \in T_p \mathbb{R}^{n+1} \setminus T_p \mathbb{S}^{n}$, so $v$ is the north or south pole of $\mathbb{S}^{n}$.

Then, if $v$ is the north or south pole - how do I conclude that $v \cdot p \neq 0$? Or is there a better approach?

1 Answers 1

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Hint: The sphere is given by $x_0^2+x_1^2+\cdots+x_n^2-1=0$. Taking the gradient gives the normal direction. The gradient is $\langle 2x_0,2x_1,\cdots,2x_n\rangle=2p$. The gradient points in the direction normal to the tangent plane.

This is an elementary answer, but all the more sophisticated ideas are essentially the same.

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    Why (or rather how) is $\langle 2x_0 , 2x_1 , \dots , 2x_n \rangle$ equal to $2p$, you're saying that $p = \langle x_0 , x_1 , \dots , x_n \rangle \in T_p \mathbb{S}^{n} $? So then since the gradient points in the direction normal to the tangent plane, we have that our $v = \langle 2x_0 , 2x_1 , \dots 2x_n \rangle $ is the gradient since $v \cdot p = 0$. But this $v = 2p \in T_p \mathbb{S}^{n}$, so we have that $v \in T_p \mathbb{S}^{n}$ as well. – 2017-02-22
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    Actually, if you want to be pedantic, $p$ is not in $T_p\mathbb{S}^n$, but without identifying a $p$ in $T_p\mathbb{S}^n$, then the dot product doesn't even make sense. You should really look at the line from the origin through $p$ and identify $p$ with the derivative of this line at $p$. That's why I mentioned that this answer gives the right idea, if you want an answer that is more mathematically rigorous, you need to identify what definitions you're using for various objects. – 2017-02-22