Why does the halfplane model of the hyperbolic plane involve only the upper half of the plane?

Question

The Hyperbolic metric

$$ s= \pm \int \frac {\sqrt{1 +y'^2} \, dx}{y} $$

for geodesics in $ \mathbb H^2$ integrates to a full circle, but why only the upper half is considered?

The query is about either/ or/ both options. The square root allows choice of either (arbitrary) and both.

Are not both semi-circles equally valid geodesics?

Does x-axis, considered to be boundary, exclude the other half? Am assailed by such doubts about regarding disk model also.

Thanks for arguments for inadmissibility of all, or the other half.

EDIT1:

If some vagueness is tolerated, like tan or tanh function behavior there is cyclic continuity after going through $ +\infty$ as a vertical or horizontal asymptote function lines re-emerge from $ -\infty$ side (like the proverbial phoenix ! ). In a hyperbolic world... a dead end, a boundary, an edge of no return and such somehow strikes (me ) as odd, so I posted the above.

The cuspidal edges of constant negative surfaces are quite unlike the Poincare boundaries. Across the pseudospherical boundaries that Hilbert had set, we know asymptotes run across periodically repeating with geometric parametrizations. These are geodesics have stepped out of (inviolable) boundaries.

What beats me is that an application permits something that is prohibited by the Poincare model on which hyperbolic geometry as a model is based.

Answer 1

All that matters is that, from any point in the upper half plane, an arc-length parametrized geodesic does not reach the horizontal axis in finite time; using $(x,y)$ coordinates, there are two types of unit speed geodesics. Take real constant $A,$ also real constant $B > 0.$

The vertical type is $$ (A, e^t). $$ Note that $e^t$ is never infinite for finite $t,$ also $e^t$ is never zero for finite $t.$

The semicircle type is $$ (A + B \tanh t, B \operatorname{sech} t). $$ Since $ \cosh t = \frac{e^t + e^{-t}}{2},$ its reciprocal $$ \operatorname{sech} t = \frac{2}{e^t + e^{-t}}. $$ This gets very small when either $t \rightarrow +\infty$ or when $t \rightarrow -\infty,$ as the denominator becomes large. However, for $t$ finite, it does not equal $0.$

In brief, the upper half plane is its own world, if you throw in the lower half plane there is no communication between them. The horizontal axis cannot be reached in finite time.

Answer 2

We could just as well consider the lower halfplane instead of upper. Or even both, but considering both halves amounts to considering two copies of the hyperbolic plane, existing completely independently from each other.

The presence of $y$ in the denominator of the metric density makes the metric blow up as a point approaches the $x$-axis. Moreover, it blow up fast enough that the length of any smooth arc approaching the $x$-axis is infinite. There is no curve of finite length that begins in the upper half-plane and ends in the bottom half-plane; hence my assertion that considering both amounts to considering two disjoint copies of $\mathbb{H}^2$, which are unaware from each other.

Same reasoning applies to the disk and its exterior.