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Consider the Riemmanian product metric:

$$ ds^2= f(\theta)~d\theta^2 + f(r)~dr^2$$

for $\theta, r>0$ and $f(\theta)=2\sqrt{\theta}K_1(2\sqrt{\theta})$ where $K_1$ is a modified Bessel function.

I obtained this product metric from the Fisher information metric (a tool used in information geometry to obtain a metric from a given distribution function).

I'm wondering about the geometry of this "quarter plane" is. I first suspected that it had hyperbolic geometry but now I am thinking that is has Euclidean geometry based on the form of the metric above.

Is there a way to tell whether the geometry is hyperbolic or Euclidean via inspection of the metric?

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  • $\begingroup$ It should be possible to calculate Christoffel symbols, Riemann, Ricci tensor and scalar curvature. This tool docs.sympy.org/latest/modules/diffgeom.html can be used to do that with symbolic calculation. $\endgroup$
    – Kurt G.
    Commented Aug 6, 2023 at 19:46
  • $\begingroup$ The results should be: non-zero Christoffel symbols: \begin{align} \Gamma^\theta_{\theta\theta} &= - \frac{K_0\left(2 \sqrt{\theta}\right)}{2 K_1\left(2 \sqrt{\theta}\right) \sqrt{\theta}}\,,& \Gamma^r_{rr} &= - \frac{K_0\left(2 \sqrt{r}\right)}{2 K_1\left(2 \sqrt{r}\right) \sqrt{r}}\,. \end{align} Riemann tensor is zero. $\endgroup$
    – Kurt G.
    Commented Aug 7, 2023 at 6:49

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If you let $F(t)$ be an antiderivative of $\sqrt{f(t)}$ then after a change of variables $$\theta' = F(\theta), \quad r' = F(r) $$ you get $$ds^2 = (d \theta')^2 + (dr')^2 $$ and so your metric is locally Euclidean.

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  • $\begingroup$ okay so it's locally and globally euclidean, I see $\endgroup$
    – zeta space
    Commented Aug 6, 2023 at 19:59
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    $\begingroup$ I myself would be reticent to say that this metric is "globally Euclidean", because I do not know what a "modified Bessel function" is, and so I do not know whether the metric is complete. Without knowing that, I do not know whether this Riemannian manifold is isometric to the entire Euclidean plane, which is what "globally Euclidean" means to me. $\endgroup$
    – Lee Mosher
    Commented Aug 6, 2023 at 20:01

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