3-Sphere $S^3$ as a Lie Group

Consider 4 elements $1,i,j,k$ satisfying the following relation
$$\begin{align*} 1^2&=1,\ i^2=j^2=k^2=-1,\\ 1i&=i1=i,\ 1j=j1=j,\ 1k=k1=k,\\ ij&=-ji=k,\ jk=-kj=i,\ ki=-ik=j. \end{align*}$$
Let $\mathbb{H}$ be the algebra spanned by $1,i,j,k$ over $\mathbb{R}$
$$\mathbb{H}=\{a1+bi+cj+dk:a,b,c,d\in\mathbb{R}\}.$$
Then $\mathbb{H}\cong\mathbb{R}^4$ as a vector space over $\mathbb{R}$. Define $||q||$ of $q=a1+bi+cj+dk\in\mathbb{H}$ by
$$||q||^2:=q\bar q=a^2+b^2+c^2+d^2,$$
where $\bar q=a1-bi-cj-dk$.

Set $S^3=\{q\in\mathbb{H}: ||q||=1\}$. Then $S^3$ is a unit sphere in $\mathbb{R}^4$. $S^3$ is closed under the multiplication of $\mathbb{H}$. So, $S^3$ is a group. In fact, it is a Lie group.

The 3-Dimensional Heisenberg Groups

Set
$$G=\left\{(x,y,z):=\begin{pmatrix} 1 & y & z\\ 0 & 1 & x\\ 0 & 0 & 1 \end{pmatrix}: x,y,z\in\mathbb{R}\right\}.$$
Define a multiplication in $G$ by
$$\begin{align*} (x,y,z)\cdot (a,b,c)&=\begin{pmatrix} 1 & y & z\\ 0 & 1 & x\\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & b & c\\ 0 & 1 & a\\ 0 & 0 & 1 \end{pmatrix}\\ &=\begin{pmatrix} 1 & y+b & z+ya+c\\ 0 & 1 & x+a\\ 0 & 0 & 1 \end{pmatrix}\\ &=(x+a,y+b,z+ya+c)\in G. \end{align*}$$
The identity element is $(0,0,0)$ and that $(x,y,z)^{-1}=(-x,-y,xy-z)$. $G$ is a Lie subgroup of $\mathrm{GL}(3,\mathbb{R})$.

Let $\gamma:(-\epsilon,\epsilon)\longrightarrow G$ be a regular curve in $G$ such that $\gamma(0)=(0,0,0)=\begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 &1 \end{pmatrix}$.  Then                                        $$\begin{align*}\dot\gamma(t)&=\frac{\partial\gamma}{\partial x}\frac{dx}{dt}+\frac{\partial\gamma}{\partial y}\frac{dy}{dt}+\frac{\partial\gamma}{\partial z}\frac{dz}{dt}\\ &=\begin{pmatrix}0 & 0 & 0\\0 & 0 & 1\\0 & 0&0\end{pmatrix}\frac{dx}{dt}+\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}\frac{dy}{dt}+\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}\frac{dz}{dt}.\end{align*}$$
Let
$$\mathfrak{e}_1:=\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix},\mathfrak{e}_2:=\begin{pmatrix}0 & 1 & 0\\ 0 & 0 & 0\\0 & 0 & 0\end{pmatrix},\mathfrak{e}_3:=\begin{pmatrix}0 & 0 & 1\\0 & 0 & 0\\0 & 0 & 0\end{pmatrix}.$$
Then $\mathfrak{e}_1,\mathfrak{e}_2,\mathfrak{e}_3$ generate the Lie algebra $\mathfrak{g}$ of $G$. The Lie algebra $\mathfrak{g}$ is called Heisenberg algebra. Note the commutation relation
$$[\mathfrak{e}_1,\mathfrak{e}_3]=[\mathfrak{e}_2,\mathfrak{e}_3]=0,\ [\mathfrak{e}_1,\mathfrak{e}_2]=-\mathfrak{e}_3$$
which resembles the commutation relation $[\hat p,\hat x]=-i\hbar$ in quantum mechanics.