I am trying to find a transitive left action of $SO(4)$ on the set $\mathcal{P}$ of oriented 2-planes in $\mathbb{R}^4$.
Each element in $P$ in $\mathcal{P}$ is a 2 dimensional subspace of $\mathbb{R}^4$ can be represented as a equivalence class of pairs of orthonormal vectors $(v_1,v_2)$ where $(v_1,v_2)\sim (w_1,w_2)$ iff they are related by an element of $SO(4)$.
I think the left action of $A\in SO(4)$ on $[(v,w)]\in \mathcal{P}$ should be defined by $$A\cdot [(v,w)]=[(Av,Aw)].$$ And I think I should show that for any $[(v_1,v_2)],[(w_1,w_2)]\in \mathcal{P}$, we should have $A\in \mathcal{P}$ such that $A\cdot [(v_1,v_2)]=[(w_1,w_2)]$ and naturally( I believe ), $(Av_2,Av_2)=(w_1,w_2)$. Can we build such matrix in terms of $v_1,v_2,w_1,w_2$?
Any explanation would be very thankful.
I am very new to Geometry so it is possible that I made a mistake even in defining the action. So please let me know if I am wrong.
The definition of the action is correct. This is a transitive action, because given $[(v_1,v_2)]$ and $[(w_1,w_2)]$ in $\mathcal{P}$, you may complete $(v_1,v_2)$ and $(w_1,w_2)$ to positive orthonormal bases of $\Bbb R^4$: $$(v_1,v_2,v_3,v_4)\quad\mbox{and}\quad(w_1,w_2,w_3,w_4).$$This way, by construction we have that the linear map $A$ satisfying $Av_i = w_i$ for $1 \leq i \leq 4$ is in ${\rm SO}(4)$, and we have $A\cdot [(v_1,v_2)] = [(w_1,w_2)]$. The map $A$ is not unique -- indeed, you have the freedom of applying counter-clockwise rotations in the $2$-plane normal to $[(v_1,v_2)]$. So, if you want an explicit map $A$, you need to make a choice of basis completion for both $(v_1,v_2)$ and $(w_1,w_2)$.