Computation 1. Let $m$ be a positive integer and $\phi:\R^m\to \C^m$ be the standard parametrization of the $m$-dimensional torus $T^m$ in $\C^m$ given by \[
\phi(x_1.\dots,x_m)=(e^{ix_1},\dots,e^{ix_m}).
\] Prove that $\phi$ is isometric (w.r.t. the standard Riemannian metric on $\R^m$ and $\C^m$).
Solution. The first thought would be to consider the map \[C:(z_1,\dots,z_m)\mapsto (\re z_1,\im z_1,\dots,\im z_m)\in \R^{2m}\] and then to consider the "local" representation of $\phi$, the $C\circ \phi$, but then it would be found that this is not of any help to compute $d\phi_p$, $p\in \R^m$.
We can only find $d\phi_p$
pointwise. Namely, let $v\in \R^m$ and let $\gamma'(0)=v$, with $\gamma(0)=p$, then
\[
d\phi_p(v)=d\phi_p(\gamma'(0))=\frac{d}{dt}{\phi\circ \gamma(t)}\bigg|_{t=0}.
\] From this we readily see that $d\phi_p$ is computable! And in fact, for $(v_1,\dots,v_m)\in \R^m$, \[
d\phi_p(v_1,\dots,v_m) =(v_1e^{ip_1},\dots,v_me^{ip_m}).
\] Therefore $\phi$ is an isometry because
\[
\inner{d\phi_p v,d\phi_p v'}_{\C^m}=\sum_{j=1}^m v'_je^{-ip_j}v_je^{ip_j}=\sum_{j=1}^mv_jv_j'=\inner{v,v'}_{\R^m}.\qed
\]
Computation 2. Let $m$ be a positive integer and
\[
\pi_m:(S^m\setminus \{(1,0,\dots,0)\},\inner{,}_{\R^{m+1}})\to \brac{\R^m,\frac{4}{(1+|x|^2)^2}\inner{,}_{\R^m}}.
\] be the stereographic projection given by
\[
\pi_m(x_0,\dots,c_m)=\frac{1}{1-x_0}(x_1,\dots,x_m).
\] Prove that $\pi_m$ is an isometry.
Solution. As before for $p\in S^{m}\setminus \{(1,0,\dots,0)\}$ and $v\in T_pS^m=(\R p)^\perp$, we have
\[
d(\pi_m)_{(p_0,\dots,p_m)} (v_0,\dots,v_m) = \frac{1}{1-p_0}(v_1,\dots,v_m) +\frac{v_0}{(1-p_0)^2}(p_1,\dots,p_m),
\] therefore
\begin{align*}
&\color{white}{=}\inner{d(\pi_m)_p (v),d(\pi_m)_p (v')}_{\pi_m(p), \brac{\R^m,\frac{4}{(1+|x|^2)^2}\inner{,}_{\R^m}}}\\
&= \frac{4}{(1+|\pi_m(p)|^2)^2} \cdot \frac{\inner{v,v'}_{\R^{m+1}}}{(1-p_0)^2}\\
&=\inner{v,v'}.\qed
\end{align*}
