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Let $N=(0,\dots ,0,1)$ and $S=(0,\dots ,0,-1)$ 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 chartsfrom these sets to ${\mathbb{R}}^n$. Their inverses areOn the overlap, the transition map iswhich 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 $e_1,\dots ,e_m$ of its tangent space $T_{e}G$ at the identity. Definewhere $L_g$ is left translation on a Lie group. Smoothness of multiplication makes each $E_i$ a smooth left-invariant vector field, and invertibility of $(d{L}_g{)}_e$ makes $E_1(g),\dots ,E_m(g)$ a basis of $T_{g}G$ 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 formThe 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 vectoris smooth, tangent, and nowhere zero, so it is a global one-vector frame. Thus $S^1$ is another parallelizable sphere.