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The problem is as follows; Consider the hyperboloid $𝑥^2 + 𝑦^2/4 − 𝑧^2/4 = 1$. $\\$ For any point on the hyperboloid, find a local chart and compute the transition map between two non-equal overlapping charts. $\\$ Now consider the function $(𝑥, 𝑦, 𝑧) → 𝑦^2$ on this surface. Write down the coordinate representation of this function with respect to two different charts of your choice.

I have found all the local charts but I struggle to see, how I could write the coordinate representation of f wrt a local chart as

$\phi^+$ : (x, y, z) → (x, y) and therefore $(\phi^+)^{-1}$ : (x, y) → (x, y, +$\sqrt{1- x^2- y^2/4}$)

But thus I cannot express $(\phi^+)^{-1} \circ f$ because I am missing a component. I named f the function $(𝑥, 𝑦, 𝑧) → 𝑦^2$.

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  • $\begingroup$ We don't know the charts you are considering, and I'm pretty sure that you haven't found all of them: there are uncountably many local charts on manifolds... $\endgroup$
    – Didier
    Commented Nov 10, 2022 at 14:54

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You did not tell in which charts it is asked to do the computations. You have said that you have found them all, which I doubt. Your $\phi^+$ presumably goes in pair with some $\phi^-$ built in the same way, but then $(\phi^+,\phi^-)$ will certainly not be an atlas. So let me first choose an atlas for $H$.

First of all, the hyperboloid is no more than a curved version of a cylinder with ellipsoidal sections: any horizontal section $H\cap\{z=c\}$ is an ellipse given by the equation $x^2 + \left(\frac{y}{2}\right)^2 = \left(\sqrt{1+\left(\frac{z}{2}\right)^2}\right)^2$. This observation leads us to notice that $$ \begin{array}{r|ccc} \varphi \colon & \Bbb S^1 \times \Bbb R & \longrightarrow & H \\ & ((a,b),z) & \longmapsto & \left(\sqrt{1+\left(\frac{z}{2}\right)^2}a, 2\sqrt{1+\left(\frac{z}{2}\right)^2}b, z\right) \end{array} $$ is a diffeomorphism. An atlas for $\Bbb S^1$ thus yields an atlas for $H$. Let us consider the stereographic atlas $\left\{\left(U_N,\psi_N\right),\left(U_S,\psi_S\right) \right\}$, where $U_N = \Bbb S^1\setminus \{N\}$, $U_S = \mathbb{S}^1\setminus \{S\}$, $N=(0,1)$ is the north pole, $S=-N$ is the south pole, and $$ \psi_N(a,b) = \frac{a}{1-b},\quad \psi_S(a,b) = \frac{a}{1+b}. $$ This now yields the two parametrizations of the circle: $$ \psi_N^{-1}(t) = \left(\frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2} \right),\quad \psi_S^{-1}(s) = \left(\frac{2s}{1+s^2},\frac{s^2-1}{1+s^2}\right). $$

Hence, we have found two local parametrizations for $H$, given by $\varphi_N(t,z) = \varphi\left(\psi_N^{-1}(t),z\right)$ and $\varphi_S(s,z)=\varphi\left(\psi_S^{-1}(s),z\right)$.

Lastly, the expression of the map $f\colon (x,y,z)\in H \mapsto y^2\in \Bbb R$ in these parametrizations are simply $f_N=f\circ \varphi_N^{-1}$ and $f_S=f\circ \varphi_S^{-1}$, that is $$ \begin{array}{r|ccc} f_N\colon & \Bbb R^2 & \longrightarrow & \Bbb R \\ & (t,z) &\longmapsto & 4\left(1+\left(\frac{z}{2}\right)^2\right) \left(\frac{1-t^2}{1+t^2}\right)^2 \end{array}, $$ and $$ \begin{array}{r|ccc} f_S\colon & \Bbb R^2 & \longrightarrow & \Bbb R \\ & (s,z) &\longmapsto & 4\left(1+\left(\frac{z}{2}\right)^2\right) \left(\frac{s^2-1}{1+s^2}\right)^2 \end{array}. $$ By chance, these two parametrizations give the same function (which isn't really a coincidence, due to the symmetries of the hyperboloid).

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