= Solution
Let $N=(0,\ldots,0,1)$ and $S=(0,\ldots,0,-1)$ be the poles of $S^n$. The open sets $U_N=S^n\setminus\{N\}$ and $U_S=S^n\setminus\{S\}$ cover the sphere. <Stereographic projection> gives coordinate charts
$$
\phi_N(r)=\frac{(r_1,\ldots,r_n)}{1-r_{n+1}},
\qquad
\phi_S(r)=\frac{(r_1,\ldots,r_n)}{1+r_{n+1}}
$$
from these sets to $\mathbb R^n$. Their inverses are
$$
\phi_N^{-1}(x)=\left(\frac{2x}{1+|x|^2},
\frac{|x|^2-1}{1+|x|^2}\right),
\qquad
\phi_S^{-1}(y)=\left(\frac{2y}{1+|y|^2},
\frac{1-|y|^2}{1+|y|^2}\right).
$$
On the overlap, the transition map is
$$
\boxed{y=\phi_S\phi_N^{-1}(x)=\frac{x}{|x|^2}},
\qquad x\neq0,
$$
which is a smooth <diffeomorphism> of $\mathbb R^n\setminus\{0\}$. These two compatible charts make $S^n$ a <smooth manifold> of dimension $n$.
Now let $G$ be an $m$-dimensional <Lie group> and choose a basis $e_1,\ldots,e_m$ of its <tangent space> $T_eG$ at the identity. Define
$$
E_i(g)=(dL_g)_e e_i,
$$
where $L_g$ is <left translation on a Lie group>. Smoothness of multiplication makes each $E_i$ a smooth <left-invariant vector field>, and invertibility of $(dL_g)_e$ makes $E_1(g),\ldots,E_m(g)$ a basis of $T_gG$ at every point. Thus the $E_i$ form a global frame and every Lie group is a <parallelizable manifold>.
The columns of a matrix in the <special unitary group> $SU(2)$ are orthonormal and its determinant is one. Consequently every element has the unique form
$$
\boxed{U=\begin{pmatrix}
z_1&-\overline z_2\\
z_2&\overline z_1
\end{pmatrix},
\qquad |z_1|^2+|z_2|^2=1}.
$$
The pair $(z_1,z_2)\in\mathbb C^2\cong\mathbb R^4$ therefore identifies $SU(2)$ diffeomorphically with $S^3$. The Lie-group construction then proves that $S^3$ is parallelizable; this is the <SU(2) as the three-sphere> identification.
Another example is $S^1$, which is the Lie group $U(1)$. Explicitly, at $(x,y)\in S^1$ the vector
$$
E(x,y)=-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}
$$
is smooth, tangent, and nowhere zero, so it is a global one-vector frame. Thus $S^1$ is another <parallelizable sphere>.
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