I can see why this is true but I'm not sure how to prove it, any help would be appreciated.

Prove that the tangent space $TS^{n}_{x}$ at a point $x$ on the $n$-sphere $S^{n}:=\{x \in \mathbb{R}^{n+1} : \lVert x\rVert=1\}$ can be identified with the space $x^{\perp}=\{v \in \mathbb{R}^{n+1} : v \cdot x=0\}$.

Where the definition of tangent space is as follows: Let $M$ be a smooth $m$-manifold. Given $x \in M$, let $c_{0}, c_{1}$ be $C^{\infty}$ curves defined on open intervals $(a_{0},b_{0}), (a_{1},b_{1})$ with $0 \in (a_{0},b_{0})$ , $0 \in (a_{1},b_{1})$ and $c_{0}(0)=c_{1}(0)=x$. Define an equivalence relation $\sim$ by $c_{0} \sim c_{1}$ $\iff$ $\frac{d}{dt}(\psi_{U}\circ c_{0}(t))\mid_{t=0} = \frac{d}{dt}(\psi_{U}\circ c_{1}(t))\mid_{t=0}$ , where $\psi_{U}:U \rightarrow \psi_{U}(U)\subseteq \mathbb{R}^{m}$ is a chart. Define the tangent space $TM_{x}$ of $M$ at $x$ as the set of all equivalence classes $[c_{0}]$.

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    $\begingroup$ a) for all differentiable curves $c$ in $S^n$ with $c(0) = x$ you have $c'(0) \in x^{\perp}$. b) for every $v \in x^{\perp}$, there is a differentiable curve $c$ in $S^n$ with $c(0) = x$ and $c'(0) = v$. $\endgroup$ Commented Oct 23, 2015 at 13:28
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    $\begingroup$ Just take $t\mapsto \frac1{\|x+tv\|}(x+tv)$ and check that it works. $\endgroup$ Commented Oct 24, 2015 at 12:26

1 Answer 1


Consider a point $ x \in S^n $ and let $ \sigma $ be a smooth curve on $ S^n $ passing through $ x $. Since $S^n\subset \mathbb{R}^{n+1}$, $\sigma$ can also be viewed as a curve in $\mathbb{R}^{n+1}$: $$ \sigma(t) = (x_1(t), x_2(t), \ldots, x_{n+1}(t)). $$

The tangent vector at $ t = 0 $ is $$ v=\sigma'(0) = \sum_{i=1}^{n+1} \dot{x}_i(0) \partial_{i}, $$

Since $\sigma(t)\cdot\sigma(t)\equiv 1$, we have $$ \sum_{i=1}^{n+1} \dot{x}_i(0) x_i(0) = 0, $$ that is, $$ v\cdot x=0.$$

We have identified $T_x \mathbb{R}^{n+1}$ with $\mathbb{R}^{n+1}$ in the equation $ v\cdot x=0$.

Therefore, $$ T_x S^n = x^{\perp}=\{v \in \mathbb{R}^{n+1} : v \cdot x=0\}. $$


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