Adjoint action of special orthogonal groups on their Lie algebras Let $SO(n)$ be the Lie group of $n\times n$ special orthogonal matrices with real coefficients, then the set of the $n\times n$ skew-symmetric matrices $\mathfrak{so}(n)$ is its Lie algebra. I know that the adjoint representation of $SO(n)$ induces an action of $SO(n)$ on $\mathfrak{so}(n)$
$$\alpha:SO(n)\times\mathfrak{so}(n)\longrightarrow\mathfrak{so}(n)$$
given by $\alpha(A,B)=ABA^{-1}$. Now let $n=4$. I know that, in this case, the Lie algebra of $SO(4)$ splits as
$$\mathfrak{so}(4)\cong\mathfrak{so}(3)\oplus\mathfrak{so}(3)$$
My question is: can I find some relations between the action of $SO(4)$ on $\mathfrak{so}(4)$ and the action of $SO(3)$ on the two different copies of $\mathfrak{so}(3)$ that appear in the direct sum? I suppose that there could be a way to assign a matrix in $SO(3)$, which acts on $\mathfrak{so}(3)$, to a matrix in $SO(4)$, which acts on $\mathfrak{so}(4)$, but I don't know how I could do that. 
 A: Well, there is a double cover map $\pi:SO(4)\rightarrow SO(3)\times SO(3)$, so every matrix in $SO(4)$ projects to a pair of matrices in $SO(3)$.  I'll write this as $\pi(A) =  (A_1,A_2)$.
Now, given $\pi$, I claim that you can choose an isomorphism $\psi:\mathfrak{so}(4)\cong \mathfrak{so}(3)\oplus \mathfrak{so}(3)$  (which I'll denote as $\psi(X) = (X_1, X_2)$) in such a way so that $\psi(AXA^{-1}) = (A_1 X_1 A_1^{-1}, A_2 X_2 A_2^{-1})$.
So, let's unravel this.  To begin with, I'm going to assume you're familiar with the double cover $\phi:SU(2)\rightarrow SO(3)$ and the double cover $\rho: SU(2)\times SU(2)\rightarrow SO(4)$.  The map $\phi$ has kernel $\{ \pm I_2\}$, while the map $\rho$ has kernel $\{\pm (I_2,I_2)\}$ (and I'm using the notation $I_n$ to denote the $n\times n$ identity matrix).
Where does $\pi$ come from?  Well, consider the composition $SU(2)\times SU(2)\xrightarrow{\rho} SO(4)\rightarrow SO(4)/\{\pm I_4\}$  Since $\rho^{-1}(I_4) = \{\pm (I_2, -I_2)$}, the kernel of the composition is $\{(\pm I_2, \pm I_n)\}$.  It follows that $SO(4)/\{\pm I_4\}\cong SO(3)\times SO(3)$.  Explicitly, the map $\pi:SO(4)\rightarrow SO(3)\times SO(3)$ is obtained as follows.  Given $A\in SO(4)$, pick $B\in \rho^{-1}(A)\subseteq SU(2)\times SU(2)$ and then apply $(\phi, \phi)$ to it.  Of course, given $A\in SO(4)$, there are two choices for $B\in \rho^{-1}(A)$, but $(\phi,\phi)$ has the same image
Now, where does $\psi$ come from?  Well, if we write $\pi_\ast$ for the differential $\mathfrak{so}(4)\rightarrow \mathfrak{so}(3)\oplus \mathfrak{so}(3)$, then we use $\psi = (\phi_\ast, \phi_\ast)\circ (\rho_\ast)^{-1}$.  Note that even though $\rho$ is not an isomorphism, $\rho_\ast$ is, so this makes sense.
