For $t \in \Bbb R$ we give the matrix
$R(t)=\begin{pmatrix} \cos(t) & − \sin(t) &0 \\ \sin(t)& \cos(t) & 0 \\ 0 & 0 &1\end{pmatrix}$
If $p \in \Bbb R^3$ (written as a column vector), then we define $\gamma_p : \Bbb R \to \Bbb R^3 $ by $\gamma_p(t)=R(t)p$
Show that $X: S^2 \to TS^2$ defined by $X(p)=\gamma'_p(0)$ is a smooth vector field on $S^2$.
My attempt:
I have already proven that if $p \in S^2$, then $\gamma_p$ is a smooth curve in $S^2$, with $\gamma(0) = p$.
Now to prove smothness I use the Smoothness Criterion for Vector Fields by Lee
Therefore let $U_i^+:S^2 \cap \{(x^1,x^2,x^3)\in \Bbb R^3: x^i >0\}$ be a chart for $S^2$ with coordinate maps $\phi_i(x^1,x^2,x^3)=(x^1,...,\cap x^i,..,x^3)$, meaning $x^i$ is omitted.
I can just check it for an specific chart since it is analogous for the other ones:
$\phi_1(x^1,x^2,x^3)=(x^2,x^3)$ with $x^1=\sqrt{1-(x^2)^2+(x^3)^2}$ or better
$\phi_1(x,y,z)=(y,z)$ with $x=\sqrt{1-y^2-z^2}$
I just have to write to check that the components of the vector field are smooth. For this I use the curve $\gamma$
$\gamma_p(t)=R(t)p=\begin{pmatrix} \cos(t) & − \sin(t) &0 \\ \sin(t)& \cos(t) & 0 \\ 0 & 0 &1\end{pmatrix} \begin{pmatrix} x \\ y \\ z\end{pmatrix} = \begin{pmatrix} x\cos t -y\sin t \\ x\sin t +y\cos t \\ z\end{pmatrix}$
$X(p)=\gamma'_p(0)=\begin{pmatrix} -x\sin t -y\cos t \\ x\cos t -y\sin t \\ 0\end{pmatrix}|_{t=0} = \begin{pmatrix} -y \\ x\\ 0 \end{pmatrix}$ which has smooth components, therefore $X$ is smooth
It doesn't seem like I am really using the chart to conclude that $X$ is smooth in the last computation. Is this correct? Otherwise, how should it be done or what I am missing for a complete answer?