I am curious if one can find a diffeomorphism between $SO(3)=\{A\in M_3(\mathbb{R}): AA^T=I_3 \text{ and det(A)=1} \}$ and $\mathbb{R}P^3$ using Grassmannian manifolds. I have seen multiple proofs regarding this using facts from Lie groups or just explicit calculations. For $m\leq n$ we denote the Grassmanian $G(m,n)=\{m-\text{dimensional subspace of }\mathbb{R}^n \}$ and it obtains a manifold structure throughout the projection map $\pi:F(m,n)\rightarrow G(m,n)$ where $F(m,n)$ are matrices of rank $m$ and also: $$ \pi([r_1 \cdots r_m]^T)=span\{r_1,\ldots,r_n \}\quad \text{for n dimensional vectors $r_i$} $$ Taking into account from linear algebra that: $$ span\{r_1,\ldots,r_m \}=span\{y_1,\ldots,y_m \}\iff \exists B\in GL_m(\mathbb{R}) \text{ such }[x_1\cdots x_m]^T=B[y_1\cdots y_m] $$ So this way we can find that the Grassmanian is the preimage of $m\times m$ submatrices of the determinant on $\mathbb{R}-\{0 \}$ and its dimension will be $m(n-m$). Genuinely the main fact we use is that:
$$ \pi^{-1}(\pi(U))=\bigcup_{B\in GL_m(\mathbb{R})} B V $$ Where U is matrix that has the unitary $I_m$ as submatrix and V also has the unitary $I_m$ as a submatrix (we may permute the columns that give us $I_m$ as a submatrix). We also know that $G(1,n+1)\cong \mathbb{R}P^{n}$. So one expects that $G(1,4)\cong \mathbb{R}P^3\cong SO(3)$. So is it possible to attain this result by proving that $G(1,4)\cong \mathbb{R}P^3$. I believe that it can be done since we use matrix descriptions for elements in the Grassmannian. Any help corrections or ideas are greatly appreciated
Think of protective space as a closed 3-ball of radius $\pi$ with antipodal points on the boundary identified. Then, each point represents a rotation about that axis with angle equal to the length of the the vector.