A linear isomorphism $T:\mathbb{R}^{n+1} \to \mathbb{R}^{n+1}$ induces a smooth map $\tilde{T}:\mathbb{RP}^n \to \mathbb{RP}^n$.

Question

A linear isomorphism $T:\mathbb{R}^{n+1} \to \mathbb{R}^{n+1}$ induces a smooth map $\tilde{T}:\mathbb{RP}^n \to \mathbb{RP}^n$ via $L \mapsto T(L)$. I want to understand why this is a smooth map. I am having trouble computing the coordinate representation with respect to the smooth atlas $\{(U_i,\phi_i)\}$ where $U_i$ is the collection of lines that intersect the affine hyperplane $x_i = 1$ and $\phi_i:U_i \to\mathbb{R}^n$ is defined by $[x_1,\ldots,x_{n+1}] \mapsto \left( \frac{x_1}{x_i},\ldots,\frac{x_{i-1}}{x_i}, \frac{x_{i+1}}{x_i},\ldots,\frac{x_n}{x_i}\right)$.

Answer 1

This is one instance where abstracting things further helps. Coordinates are a distraction. If $V$ is a vector space and $PV$ is the projective space over $V$, one defines an atlas $\{(U_f,\phi_f)\mid f \in V^*\setminus \{0\}\}$ by setting $$U_f =\{L \in PV \mid f[L] = \Bbb R\}$$and $\phi_f:U_f \to f^{-1}(1)$ as $\phi_f(L)=x/f(x)$ for any $x$ with $L = \Bbb R x$. For $V = \Bbb R^{n+1}$ and the projections $f = \pi_i$, we have exactly your atlas (under the homeomorphism $\pi_i^{-1}(1)\cong \Bbb R^n$ that deletes the $i$th entry). The inverse of $\phi_f$ is the span map. Any isomorphism $T:V\to V$ induces $PT:PV\to PV$ by $PT(L)= T[L]$.

Goal: show that $PT$ is smooth.

Given $f,g\in V^*\setminus \{0\}$, we compute $\phi_g\circ PT \circ \phi_f^{-1}$ as $$\phi_g\circ PT \circ \phi_f^{-1}(u) = \phi_g(PT(\Bbb R u)) = \phi_g(\Bbb R T(u)) = \frac{T(u)}{g(T(u))}.$$Since $u\mapsto T(u)/g(T(u))$ is smooth (where defined) for all $g \in V^*\setminus \{0\}$, we conclude that $PT$ is smooth.

(I'm denoting your $\widetilde{T}$ by $PT$ just to emphasize the functorial behavior of the whole thing.)