The projective geometry contains the affine geometry

138 Views Asked by At

I am currently reading some papers about geometric structures on manifolds and most of them state, without any details, that the affine geometry is contained in the projective geometry. So I want to know if I worked out the details correctly.

To be more precise, if $G=\mathsf{Aff}(\Bbb R^n)$ is the group of affine automorphism of $\Bbb R^n$ and if $G^\prime=\mathsf{PGL}(n+1,\Bbb R)$ is the group of projective automorphism of $\Bbb{RP}^n$ (that is the group of automorphism induced by a linear automorphism of $\Bbb R^{n+1}$), I want to find a local diffeomorphism $$\varphi:\Bbb R^n\longrightarrow \Bbb{RP}^n$$ and a group homomorphism $$\Phi:G\longrightarrow G^\prime$$ such that for every $g\in G$ the following diagram commutes

$$\require{AMScd} \begin{CD} \Bbb{R}^n @>{\varphi}>> \Bbb{RP}^n\\ @V{g}VV @VV{\Phi(g)}V\\ \Bbb{R}^n @>\varphi>> \Bbb{RP}^n \end{CD}$$

The papers say "just identify $\Bbb{R}^n$ with the hemisphere". So I think I should take $\varphi$ to be the composition of the embedding $i:\Bbb{R}^n\hookrightarrow\Bbb{R}^n\times\{1\}\subset\Bbb R^{n+1}$ and of the quotient $p:\Bbb{R}^{n+1}\to \Bbb{RP}^n$. Also if $g$ is the affine transformation with linear part $A$ and translation part $b$, I can take $\Phi(g)$ to be the transformation induced by $$M_g=\left(\begin{array}{cc}A&b\\0&1\end{array}\right)\in GL_{n+1}(\Bbb R).$$ If I do so I think it will work because I will have the commutative diagrams

$$\require{AMScd} \begin{CD} \Bbb{R}^n @>{i}>> \Bbb{R}^n\times\{1\}\\ @V{g}VV @VV{M_g}V\\ \Bbb{R}^n @>i>> \Bbb{R}^n\times\{1\} \end{CD} \quad\text{and}\qquad \require{AMScd} \begin{CD} \Bbb{R}^{n+1} @>{p}>> \Bbb{RP}^n\\ @V{M_g}VV @VV{\Phi(g)}V\\ \Bbb{R}^n @>p>> \Bbb{RP}^n \end{CD}$$ which I can glue together to make the desired diagram. The fact that $\varphi$ is a local diffeomorhism and that $\Phi$ is a morphism are ok for me.

Is this correct? Did I miss something? Thanks in advance!