I'm beginning to study $4$-dimensional Riemannian manifolds, in particular the decomposition induced by the Hodge $\star$-operator on the space of the differential two-forms. I'm considering $\mathbb{R}^4$ as an example: I understood that $\Lambda^2 \mathbb{R}^4=\Lambda_{+}\oplus \Lambda_{-}$, where $\Lambda_{\pm}$ is the eigenspace of $\star$ relative to the eigenvalue $\pm 1$.
The group $SO(4)$ acts on $\Lambda^2 \mathbb{R}^4$ in the standard way induced by the action of $SO(4)$ on $\mathbb{R}^4$, i.e. if $X\in SO(4)$ and $a\wedge b$ is a two-form, then $X(a\wedge b)=Xa\wedge Xb$. I've got three questions:
$1)$ Since $\Lambda^2 \mathbb{R}^4\cong so(4)$ (the Lie algebra of $SO(4)$), are there some relations between the action of $SO(4)$ on $\Lambda^2 \mathbb{R}^4$ and the adjoint action of $SO(4)$ on $so(4)$?
$2)$ What is the action of $SO(4)$ on $\Lambda^2 \mathbb{R}^4$ when restricted to $\Lambda_{\pm}$? I know that $so(4)\cong so(3)\oplus so(3)$, so I thought it could be useful to compute the action, but I don't know how to do it.
$3)$ I read that the restriction of the action to $\Lambda_{\pm}$ induces a surjective homomorphism $\phi:SO(4)\rightarrow SO(3)\times SO(3)$ which is two to one. How can I prove this and compute the homomorphism? I understood that there is a surjective homomorphism from $SU(2)$ to $SO(3)$ and that $SU(2)\times SU(2)$ is the universal cover of $SO(4)$ (a double cover, in particular): could it be useful?
The induced action of $SL(4,\mathbb{R})$ on $\bigwedge^2 \mathbb{R}^4$ preserves the symmetric bilinear form induced by $\bigwedge^2 \mathbb{R}^4 \times \bigwedge^2 \mathbb{R}^4 \to \bigwedge^4 \mathbb{R}^4 \cong \mathbb{R}$. This form happens to have signature $(3,3)$, so $SL(4)$ actually lands in $SO(3,3)$, and the same applies to $SO(4)$. This is enough to see that $\mathfrak{so}(4)$ lands in a copy of $\mathfrak{so}(3)\oplus \mathfrak{so}(3)$, and you can conclude (2) by dimension counting. With a little care this should yield (3) as well.
For (1), the adjoint action of $SO(n)$ on $\mathfrak{so}(n)$ is equivalent as a representation to the action of $SO(n)$ on $\bigwedge^2(\mathbb{R}^n)$. This requires writing the right map $\bigwedge^2 \mathbb{R}^n \to \mathfrak{so}(n)$: $$ v \wedge w(x)= \langle x,v \rangle w - \langle x,w \rangle v .$$