<s>[INST] Induced metric on a one-sheet hyperboloid I am trying to find the induced metric on a one-sheet hyperboloid. Suppose we use cylindrical coordinates $(r, \theta, z)$ for the ambient space in which the hyperboloid is embedded. The hyperboloid itself then consists of all points such that
$$ r^2 = R^2 + z^2 $$
where $R$ is the minimum radius, at the throat. On the hyperboloid itself, we define the following coordinates $(\rho, \varphi)$:
\begin{align}
r &= \sqrt{R^2 + \rho^2} \\
\theta &= \varphi \\
z &= \rho
\end{align}
We can now find the induced metric using the formula
$$g_{ab} = g_{\mu\nu} \partial_a X^\mu \partial_b X^\nu$$
where $a$ and $b$ describe the indices of the coordinates of the submanifold (in this case, the hyperboloid), while the $X^\mu$ and $X^\nu$ encode the embedding into the ambient space.
First, the nonzero components of the ambient metric tensor in cylindrical coordinates are
\begin{align}
g_{rr} &= 1 \\
g_{\theta \theta} &= r^2 \\
g_{zz} &= 1 \\
\end{align}
Thus we have
\begin{align}
g_{ab} &= g_{rr} \partial_a X^r \partial_b X^r + g_{\theta \theta} \partial_a X^\theta \partial_b X^\theta + g_{zz} \partial_a X^z \partial_b X^z \\
&= \partial_a X^r \partial_b X^r + r^2 \partial_a X^\theta \partial_b X^\theta + \partial_a X^z \partial_b X^z
\end{align}
Hence the nonzero components of the metric tensor on the hyperboloid are
\begin{align}
g_{\rho \rho} &= \partial_\rho X^r \partial_\rho X^r + r^2 \partial_\rho X^\theta \partial_\rho X^\theta + \partial_\rho X^z \partial_\rho X^z \\
&= \left(\partial_\rho \sqrt{R^2 + \rho^2}\right)^2 + (\partial_\rho \rho)^2 \\
&= \left(\frac{\rho}{\sqrt{R^2 + \rho^2}}\right)^2 + 1^2 \\
&= \frac{\rho^2}{R^2 + \rho^2} + 1
\end{align}
and
\begin{align}
g_{\varphi \varphi} &= \partial_\varphi X^r \partial_\varphi X^r + r^2 \partial_\varphi X^\theta \partial_\varphi X^\theta + \partial_\varphi X^z \partial_\varphi X^z \\
&= 0 + r^2 \cdot 1^2 + 0 \\
&= r^2
\end{align}
The cross-terms $g_{\rho \varphi} = g_{\varphi \rho}$ vanish. Is this derivation correct?
 [/INST] Yes, your derivation is correct.
Incidentally, you can avoid expressing the Euclidean metric in cylindrical coordinates: Parametrize the hyperboloid by
$$
\Phi(\rho, \varphi) 
  = \left(\sqrt{R^{2} + \rho^{2}} \cos\varphi, \sqrt{R^{2} + \rho^{2}} \sin\varphi, \rho\right).
$$
The partials are
\begin{align*}
\Phi_{\rho}(\rho, \varphi)
  &= \left(\frac{\rho}{\sqrt{R^{2} + \rho^{2}}} \cos\varphi, \frac{\rho}{\sqrt{R^{2} + \rho^{2}}} \sin\varphi, 1\right), \\
\Phi_{\varphi}(\rho, \varphi)
  &= \left(-\sqrt{R^{2} + \rho^{2}} \sin\varphi, \sqrt{R^{2} + \rho^{2}} \cos\varphi, 0\right),
\end{align*}
so pulling back the Euclidean metric gives the components
\begin{align*}
g_{\rho\rho} &= \Phi_{\rho} \cdot\Phi_{\rho} = \frac{\rho^{2}}{R^{2} + \rho^{2}} + 1, \\
g_{\rho\varphi} &= \Phi_{\rho} \cdot\Phi_{\varphi} = 0, \\
g_{\varphi\varphi} &= \Phi_{\varphi} \cdot\Phi_{\varphi} = \rho^{2}.
\end{align*}
</s>