I have a 3D manifold $\Omega_3$ that is $\Omega_3=S^1 \times S^2$ for which I know its distance function $d_{\Omega_3}$ (although I don't know its metric tensor):
$$d_{\Omega_3} \left( (\theta_\mathrm{i}, \theta_\mathrm{o},\phi_\mathrm{o}), (\theta_\mathrm{i}',\theta_\mathrm{o}',\phi_\mathrm{o}') \right) = \sqrt{4 \arcsin^2 \left( \sqrt{\mathrm{hav} (\theta_\mathrm{o}'-\theta_\mathrm{o})+\sin \theta_\mathrm{o}' \sin \theta_\mathrm{o} \, \mathrm{hav}(\phi_\mathrm{o}' - \phi_\mathrm{o})} \right) + (\theta_\mathrm{i}'-\theta_\mathrm{i})^2}$$
where $\mathrm{hav}(x)=\sin^2 (x/2)$ is the haversine function. I need to express the coordinates $(\theta_\mathrm{i}, \theta_\mathrm{o}, \phi)$ with another parameterization called the Rusinkiewicz parameterization $(\theta_\mathrm{h}, \theta_\mathrm{d}, \phi_\mathrm{d})$:
$$\omega_\mathrm{h}= \begin{pmatrix}\sin \theta_\mathrm{h} \cos \phi_\mathrm{h} \\ \sin \theta_\mathrm{h} \sin \phi_\mathrm{h} \\ \cos \theta_\mathrm{h}\end{pmatrix}=\frac{\omega_\mathrm{i} + \omega_\mathrm{o}}{| \omega_\mathrm{i} + \omega_\mathrm{o}|}$$
$$\omega_\mathrm{d} = \begin{pmatrix}\sin \theta_\mathrm{d} \cos \phi_\mathrm{d} \\ \sin \theta_\mathrm{d} \sin \phi_\mathrm{d} \\ \cos \theta_\mathrm{d} \end{pmatrix} = \mathrm{rot}_{\omega_\mathrm{h} \, \wedge \, \mathbf{n}, \, \theta_\mathrm{h}} (\omega_\mathrm{i})$$
with $\omega_\mathrm{i} = (-\sin \theta_\mathrm{i}, 0, \cos \theta_\mathrm{i})^\mathsf{T}$, $\omega_\mathrm{o} = (\sin \theta_\mathrm{o} \cos \phi_\mathrm{o}, \sin \theta_\mathrm{o} \sin \phi_\mathrm{o}, \cos \theta_\mathrm{o})^\mathsf{T}$, $\mathbf{n} = (0, 0, 1)^\mathsf{T}$. We can then express our new variables:
$$\theta_\mathrm{h}=\arccos(\omega_\mathrm{h}.\mathbf{n})$$ $$\theta_\mathrm{d}=\arccos(\omega_\mathrm{d}.\mathbf{n})$$ $$\phi_\mathrm{d}=\arctan(\omega_\mathrm{d}.\mathbf{b}, \omega_\mathrm{d}.\mathbf{t})$$
with $\mathbf{t} = (1, 0, 0)^\mathsf{T}$, $\mathbf{b} = (0, 1, 0)^\mathsf{T}$. I have read from Ricci and Levi-Civita (p. 128):
A manifold remains invariant to any transformation of its coordinates. The formulas and results of Absolute Differential Calculus [...] are also independent of the choice of independent variables.
Naively, I deduced that
$$d_{\Omega_3} \left( (\theta_\mathrm{i}, \theta_\mathrm{o},\phi_\mathrm{o}), (\theta_\mathrm{i}',\theta_\mathrm{o}',\phi_\mathrm{o}') \right) = d_{\Omega_3} \left( (\theta_\mathrm{h}, \theta_\mathrm{d},\phi_\mathrm{d}), (\theta_\mathrm{h}',\theta_\mathrm{d}',\phi_\mathrm{d}') \right)$$
But when I try this numerically, I see different results. Clearly I have misinterpreted the sentence. My question is:
Is it possible to deduce the new distance function after a transformation of coordinates of a manifold, and if yes, how to do it?
As for a formula for a change of a distance function, it is the following. Suppose that $:\to (,)$ is a homeomorphism, where $d=_$ is a distance function. Then the corresponding distance function $d'=_$ is given by $'(,)=((),())$. But you should not expect $d=d'$. The modern way to read the quote from from Ricci and Levi-Civita is as follows:
Suppose that $g$ is a Riemannian metric on $N$ defining the distance function $d=d_N$ and let $f: M\to N$ be a diffeomorphism of smooth connected manifolds. Let $h=f^*g$ denote the pull-back of the Riemannian metric to $M$. Then the distance function $d'$ defined by $h$ satisfies $$ d'(p,q)=d(f(p), f(q)), p, q\in M. $$