# Parametrizing $\mathbb{R}\mathrm{P}^2$ in $\mathbb{R}^5$

Let $$\mathrm{SO}(3) = \mathrm{SO}(3,\mathbb{R})$$. Suppose $$\mathrm{SO}(3) \curvearrowright \mathbb{R}^5$$ via conjugation on traceless symmetric $$3\times 3$$ real matrices: $$\begin{pmatrix} a & c & d \\ c & b & e \\ d & e & -(a+b) \end{pmatrix}.$$

The stabilizer of the vector $$M = (a,b,c,d,e) = (1,1,0,0,0)$$ can be seen to be $$\mathrm{O}(2)$$. Thus, $$\mathrm{SO}(3)/ \mathrm{O}(2) \simeq \mathbb{R}\mathrm{P}^2$$ is an embedded submanifold of $$\mathbb{R}^5$$. See for example: Are closed orbits of Lie group action embedded?.

Is there a way to explicitly parametrize $$\mathbb{R}\mathrm{P}^2$$ in $$\mathbb{R}^5$$ using the previous? That is, can I write $$x_1 = f_1(\alpha,\beta), \dots, x_5 = f_5(\alpha,\beta)$$ for some smooth functions $$f_1,\dots,f_5$$ and parameters $$\alpha,\beta$$? My thought was to write $$R \in \mathrm{SO}(3)$$ in the fundamental representation (Euler angles) and then compute $$R M R^{-1}$$ and associate the coordinates in the obvious way. However, this gives $$x_1,\dots, x_5$$ in terms of the three Euler angles, whereas we need them in terms of two parameters (since $$\mathbb{R}\mathrm{P}^2$$ has dimension $$2$$).