Elements of the tangent space $T\mathbb{S}^n$ are orthogonal to a radial point $p \in \mathbb{S}^n$

Question

I'm asking for some proper source, proof or a counter example for the following thought: Let $\mathbb{S}^n$ be the $n$ dimensional sphere. It seems quite obvious from the cases $n = 1$ (circle) and $n = 2$ (3D sphere) that given any point $p \in \mathbb{S}^n$, the elements (i.e. vector fields) of the tangent space $T_p\mathbb{S}^n$ at $p$ are orthogonal to $p$. But how would you prove this in the general case? My current hand-wavy "proof" is that by induction given any $p \in\mathbb{S}^3$ you can always take the subsphere $\mathbb{S}^{n-1}$ passing through $p$, so that $T_p\mathbb{S}^{n-1}$ is orthogonal to the lower dimensional ($n-1$) version $p'$ of $p$. Then, I suppose that you can argue that as $T_{p'}\mathbb{S}^{n-1}$ is in the hyperplane $T_{p}\mathbb{S}^n$, it follows that any other vector field spanned by the basis of $T_{p}\mathbb{S}^n$ is also orthogonal to $p$.

Edit: $T_p\mathbb{S}^n$ is defined to be the tangent space to $\mathbb{S}^n$ at a point $p$. Tangent space, to my knowledge, of an $n$-dimensional manifold is the space defined by the basis $\left\{\frac{\partial}{\partial x^1}|_p,\dots,\frac{\partial}{\partial x^n}|_p\right\}$ of the partial derivatives of the local coordinates of the manifold.

Answer 1

$\mathbb{S}^{n-1}$ is an embedded submanifold defined as the zero level set of the function $\Phi(x) = x^\top x - 1$ defined on all of $\mathbb{R}^n$. That is, $\mathbb{S}^{n-1} = \{x \in \mathbb{R}^n : \Phi(x) = 0\}$. In this case, $\Phi$ is called a defining map (see page 105 in [1]). From Proposition 5.38 in [1], we have the following

Suppose $M$ is a smooth manifold ($\mathbb{R}^n$ in our case) and $S$ is an embedded submanifold ($\mathbb{S}^{n-1}$ in our case). If $\Phi$ is any local defining map for $S$ then $\mathrm{T}_pS = \mathrm{Ker} (\mathrm{d}\Phi_p)$.

Hence it suffices to identify the kernel of $\mathrm{d}\Phi_p = 2p^\top$. That is, $\mathrm{Ker} (\mathrm{d}\Phi_p) = \{v \in \mathrm{T}_p\mathbb{R}^n : 2p^\top v = 0\} = \{v \in \mathbb{R}^n : p^\top v = 0\}$, which varifies indeed that tangent vectors in $\mathrm{T}_p\mathbb{S}^{n-1}$ are orthogonal to $p\in \mathbb{S}^{n-1}$. Here I have invoked the isomorphism $\mathrm{T}_p\mathbb{R}^n \cong \mathbb{R}^n$.

[1] Introduction to Smooth Manifolds by J. Lee.

Answer 2

Orthogonality is a metric concept, which means we must be considering some Riemannian metric on $\mathbb R^{n+1}$ such that $T_pS^n$ can be considered orthogonal to $p$. Note that we are using the canonical identification of $\mathbb R^{n+1}$ with $T_p\mathbb R^{n+1}$ to consider $p\in S^n$ as an element of $T_p\mathbb R^{n+1}$. It's implicit that we want to consider the usual inner product on $\mathbb R^{n+1}$: $$ (v,w) = \sum_i v^iw^i. $$ Then the claim is that $T_pS^n$ is orthogonal to $p$, for any $p\in S^n$. In symbols, $$ \forall p\in S^n,\ \forall v\in T_pS^n,\quad (v,p) = 0. $$ So let $p\in S^n$ and $v\in T_p S^n$ be given. If $\gamma$ is any smooth curve in $S^n$ satisfying $\gamma(0) = p$ and $\dot\gamma(0) = \sum_i \dot\gamma^i(0)\frac{\partial}{\partial x^i}|_p = \sum_i v^i\frac{\partial}{\partial x^i}|_p = v$, then differentiating the constant relation $(\gamma,\gamma) = 1$ yields $$ 0 = \frac{d}{dt}\bigg|_{t=0}(\gamma,\gamma) = 2(\dot \gamma(0),\gamma(0)) = 2(v,p), $$ as desired.