Geodesic Uniqueness in the Hyperbolic Plane

I am researching Hyperbolic Geometry. At this component, I have actually confirmed that semicircles and also straight lines orthogonals to the actual axis are geodesics in the hyperbolic aircraft. Yet just how I evidence that this geodesics are uniques? That it does not exist others geometric areas in between 2 various factors that decreases the hyperbolic size?

I have actually seen a subject pertaining to this, yet I do not recognize absolutely nothing concerning Riemannian Geometry. If a person can show to me some referrals to this evidence (or the suggestion behind) I will certainly be really happy!

8
2022-06-07 14:33:12
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Answers: 2

There is a basic reality in Riemannian geometry which adheres to from "standard" presence and also individuality theories in the research of ODEs. Mostly, offered any kind of factor $p$ in a Riemannian manifold $M$ and also offered any kind of $v\in T_pM$, there is an one-of-a-kind geodesic specified for some open area of time around 0 beginning at $p$ towards $v$.

Thinking this, it is not also tough to show that you've located all the geodesics. You simply require to validate that for any kind of factor $p$ in the hyperbolic aircraft and also tangent vector, either the vector factors up/down (so an upright line geodesic has $v$ as tangent vector), or some semicircle via $p$ striking the $x$ - axis perpendicularly does.

Ultimately, to show there is an one-of-a-kind geodesic in between any kind of 2 factors, one says as adheres to:

Given $p$ and also $q$, if $q$ and also $p$ have the very same $x$ coordinate, an upright line travels through them. Better, no semicircle striking the $x$ - axis perpendicularly can experience both $p$ and also $q$ given that the semicircle is the chart of a function, so passes the upright line examination.

Next, offered $p$ and also $q$ with various $x$ works with, it is clear that an upright line can not connect them, so we require just locate all semicircles travelling through both vertical to the $x$ - axis. Algebraically, continue by keeping in mind the formula of such a circle is $(x-a)^2 + y^2 = r^2$ for some $a$ and also $r$. Connecting in $(x,y) = (p_1,p_2)$ and also $(x,y) = (q_1,q_2)$ specifically and also establishing both equivalent to each various other (given that they both equivalent $r^2$), one obtains $$(p_1-a)^2 + p_2^2 = (q_1-a)^2 + q_2^2.$$

By streamlining this, one sees the $a^2$ negates therefore one obtains a formula linear in $a$, so $a$ is distinctly established. (The presumption that $p_1\neq q_1$ suggests that the formula has the straight item). Recognizing $a$ conveniently indicates that $r$ is distinctly established (remembering that $r \geq 0$). This reveals that there is an one-of-a-kind semicircle vertical to the $x$ - axis experiencing $p$ and also $q$.

9
2022-06-07 14:53:23
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You can say "geometrically" as adheres to: For 2 factors $p$, $q$ on the fictional axis it is rather noticeable that the sector $[p,q]$ is the one-of-a-kind fastest contour attaching $p$ and also $q$. Currently if $p$, $q\in H$ are approximate there is a Moebius makeover $T: H\to H$ sending out $p$ and also $q$ to 2 factors $p'$, $q'$ on the fictional axis. Given that $T$ is a hyperbolic isometry it adheres to that $p$ and also $q$ have an one-of-a-kind fastest link, and also given that $T$ (currently taken into consideration as a map $\bar {\mathbb C}\to \bar {\mathbb C}$) maps circles to circles and also maintains angles in between contours it adheres to that this fastest link pushes a circle converging the actual axis orthogonally.

7
2022-06-07 14:51:47
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