$SO(3)$ denotes the special orthogonal group, which is the (open) subset of $O(3)$ on which the determinant is one.
I have shown that every element of $SO(3)$ fixes a line in $\mathbb{R}$ pointwise and then $SO(3)$ is connected.
Also, suppose $f:G\rightarrow H$ is a smooth map of Lie groups such that $f(g_1g_2) = f(g_1)f(g_2)$ for all $g_1, g_2 \in G$ (this is called a Lie group homomorphism). Then I proved that the derivative $df_g$ has constant rank.
I am stuck at the question: Consider the map $f: S^3 \rightarrow SO(3)$ given by $r \in S^3$ maps to the element $f(r) \in SO(3)$ such that $f(r)(q) = rqr^{-1}$ for $q \in \mathbb{R}$, thought of as a purely imaginary quaternion. Show that this is a Lie group homomorphism, and compute $df_1: TS^3_1 \rightarrow TSO(3)_I$.
I am given then Hint: $TS^3_1$ is identified with the purely imaginary quaternions as it sits in $\mathbb{R}^4$ and $TSO(3)_I$ is identified with the skew-symmetric $3\times 3$ matrices as it sits in $GL_3(\mathbb{R})$.
More specifically, I am confused with these questions:
1, What is the explicit form of $r^{-1}$? I used $r^*/\|r\|^2$, where $r^* = a -bi-cj-dk$. But then it becomes cumbersome during the computation.
2, I am aware of that quaternions are not really 4-dim since it is bounded on the unit sphere. So should it be 9 x 3 matrix other than 4 x 9? Also, f is 9 dimensional, so df should be 9 columns, right?