Exponential map

SU(2)

exp map onto $SU(2)$

Let an element of $\mathbb{R}i+\mathbb{R}j+\mathbb{R}k$ in the form $\theta u$, where $u$ is a unit vector, we have

$$e^{\theta u}=\cos \theta +u\sin \theta$$

that maps $\mathbb{R}i+\mathbb{R}j+\mathbb{R}k$ onto ${\mathbb{S}}^3=SU(2)$ Therefore, ${\mathbb{R}}^3$ is isomorphic a tangent space of the sphere ${\mathbb{S}}^3$ (at identity element $1$)

Proof: We prove that for each $v\in SU(2)$ there is a $u\in \mathbb{R}i+\mathbb{R}j+\mathbb{R}k$ that is being mapped to it: $v=a+bi+cj+dk=a+\frac{bi+cj+dk}{\sqrt{b^2+c^2+d^2}}\sqrt{b^2+c^2+d^2}=a+u\sqrt{b^2+c^2+d^2}$ Since $a^2+b^2+c^2+d^2=a^2+(\sqrt{b^2+c^2+d^2}{)}^2=1$, there exists $\theta$ such that $a=\cos \theta$ and $\sqrt{b^2+c^2+d^2}=\sin \theta$ (by ${\cos }^2\theta +{\sin }^2\theta =1$) $⟹$ the map is surjective

the tangent space of $SU(2)$

NOTE

Interestingly, the product rule of derivation still applies on quaternion: (because we don’t need commutativity of multiplication to prove this rule)

Let $q:\mathbb{R}\to SU(2)$ be a continuous function that maps $t$ to a unit quaternion, with $q(0)=1$

$$q\bar{q}=|q{|}^2=1⟹q(t)\overline{q(t)}=1⟹q^{\prime }(t)\overline{q(t)}+q(t)\overline{q^{\prime }(t)}=0$$

then let $t=0$, we must have

$$q^{\prime }(0)+\overline{q^{\prime }(0)}=0$$

This means every tangent vector of $SU(2)$ must satisfy this equation. Note that every $q^{\prime }(0)$ must therefore be a pure imaginary quaternion $p\in \mathbb{R}i+\mathbb{R}j+\mathbb{R}k$. (The converse statement is also true)

NOTE

$q(t)=e^{pt}$ is a path in $SU(2)$ and it is smooth (infinite differentiable)

In all cases it is easy to find the general form of a tangent vector by differentiating the defining equation of the group (in the above case $|q{|}^2=1$)