A mapping $\phi:\mathfrak{so}(3) \to \mathbb{R}^{3}$ such that $\phi(gXg^{-1}) = g \phi(X)$

Question

An element $X \in \mathfrak{so}(3)$ can be written generally written as $X = \left[ \begin{matrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{matrix} \right]$. I've been given a mapping $\phi:\mathfrak{so}(3) \to \mathbb{R}^{3}$ which is defined by its components as: $$ \phi\left( \left[ \begin{matrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{matrix} \right] \right) = \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] $$

I'm supposed to show that for any $X \in \mathfrak{so}(3)$ and any $g \in SO(3)$ the function defined above has the property $\phi(gXg^{-1}) = g \phi(X)$.

I know that you can generally write the components of any matrix in $SO(3)$ using the Rodrigues Rotation Formula, however this is a really messy formula. I've started doing the computation $gXg^{-1}$ explicitly and it get messy really fast, so I'm thinking that there might be a trick that I am missing out on.

Is there some sort of short cut I am not seeing? I feel like there could be a simpler way to do this problem. I know that any $g \in SO(3)$ can be written as $g = \exp(X)$ for some $\mathfrak{so}(3)$, and I am hoping this could be used as a shortcut somehow.

Answer 1

Consider a rotation of $\mathbb R^3$ about the axis $\vec n $ through an angle $\theta$. The matrix for this rotation is $\exp \left( \theta\phi^{-1}(\vec n) \right)$, where $\phi$ is as defined in your question.

Now do a change of basis: rotate your basis of $\mathbb R^3$ by some $g \in SO(3)$. Written with respect to the new basis, the axis of our original rotation is $g \vec n$. But by the usual change-of-basis formula, the matrix for our original rotation with respect to the new basis is $g\exp \left( \theta \phi^{-1}(\vec n) \right)g = \exp \left( \theta g\phi^{-1}(\vec n) \right)$.

Hence $g \vec n = g \phi^{-1}(\vec n) g^{-1}$.

[By the way, how do we know that $\exp \left( \theta\phi^{-1}(\vec n) \right)$ is the correct matrix for the rotation? Because $\phi^{-1}(\vec e_x), \phi^{-1}(\vec e_y), \phi^{-1}(\vec e_z)$ are the generators of the $so(3)$ Lie algebra corresponding to infinitesimal rotations about the $x, y$ and $z$ coordinate axes in the fundamental representation of $so(3)$.]

Answer 2

The matrices you have are the matrices of the linear maps $f_v(w):=v\times w$. Every element in $\mathfrak{so}(3)$ (interpreted as a linear transformation of $\mathbb{R}^3$) is expressible in this form for some $v\in\mathbb{R}^3$.

Since $g(v\times w)=(gv)\times(gw)$ for all $v,w\in\mathbb{R}^3$ and $g\in\mathrm{SO}(3)$ (for geometric reasons), we have

$$(g\circ f_v\circ g^{-1})(w)=g(v\times g^{-1}w)=(gv)\times(gg^{-1}w)=(gv)\times w=f_{gv}(w).$$

That is, conjugation by $g$ on $\mathfrak{so}(3)$ corresponds to the standard representation of $g$ on $\mathbb{R}^3$.