Basis of tangent space of a sphere Let $S$ be the two-dimensional sphere in $\mathbb R^3$. For a given $x \in S$, i.e. $\|x\|=1$, I would like to find basis vectors $b_1(x)$ and $b_2(x)$ that span the tangent space of $S$ a the point $x$. My first try was something like
$$ b_1(x) = (x_2, -x_1, 0)^T$$
and
$$ b_2(x) = (0, x_3, -x_2)^T$$
but that is troublesome, e.g. as $b_1(x=0) = 0$.
Is there a global "parametrization" of the basis vectors of $T_x  S$?
 A: Hint:
The sphere $||x||=1$ has equation $f(x,y,z)=x^2+y^2+z^2-1=0$.
The vector orthogonal to the sphere at point $P=(x_p,y_p,z_p)$ is $\mbox{grad} (f(x_p,y_p,z_p)) $ so the tangent plane at $P$ has equation $\langle \mbox{grad} (f(x_p,y_p,z_p)),(x,y,z) \rangle=0$ where $\langle \cdot,\cdot \rangle$ is the inner product.
A: Emilio's answer provides a global description of the entire tangent plane at each point.
That said, there is no global parameterization of any continuously varying basis vector of $T_x S$ on all of $S$.  This is known as the Hairy ball theorem.
On the other hand, if you remove any particular $p\in S$, then there is a global paramaterization of a pair of basis vectors $b_1$ and $b_2$ on $S\setminus\{p\}$.  The idea:  Let $\phi:S-\setminus\{p\}\rightarrow \mathbb{R}^2$ denote stereographic projection from $p$.  The map $\phi$ is a diffeomoprhism.
Then, on $\mathbb{R}^2$, we can clearly find $\tilde{b}_1$ and $\tilde{b}_2$ which span the tangent space at every point in $\mathbb{R}^2$.  Now, set $b_i = \phi^{-1}_\ast \tilde{b_i}$.
Since $\phi$ is a diffeomoprhism, $\phi^{-1}_\ast$ is a linear isomorphism at each $q\in \mathbb{R}^2$, so it maps basis vectors to basis vectors.
