1
$\begingroup$

Consider the hyperbolic space as the upper half space of $\mathbb{R}^3$ $$\mathbb{H}^3=\{(x,y,z)\in\mathbb{R}^3:z>0\}$$ equipped with the metric $$ds^2=\frac{1}{z^2}(dx^2+dy^2+dz^2)$$ Prove that straight lines perpendicular to the plane $z=0$ are geodesics if they are arc-length parameterized.

Of course, such lines can be parameterized by $$\alpha(t)=(x,y,t)$$ where $(x,y)$ is an arbitrary fixed point of the plane. But, even though this is arc-length parameterized under the usual metric, it's not under the defined metric, and indeed it's not geodesic.

My problem is conceptually serious too, as I don't know how to even check it. For example, assume the parameterization given by $$\alpha(t)=(x,y,t^2+t^3)$$ Then we would have $$\alpha'(t)=(0,0,2t+3t^2)$$ Then the module is given by $$|\alpha'(t)|^2=\frac{1}{z^2}(2t+3t^2)^2$$ Now is this arc-length parameterized? I don't know, since different variables appear.

  • 0
    Is $Z$ supposed to be the same as $z$? In mathematical notation, uppercase and lowercase letters are considered different symbols. – 2017-01-24
  • 0
    Oh, it's the same thing indeed, I'll correct that thanks. – 2017-01-24
  • 1
    Hint: Write $\alpha(t) = (x,y,f(t))$, and figure out what $f$ has to be in order to make $\alpha$ unit-speed. – 2017-01-24
  • 0
    I have the same doubt with that too, though, which is that both variables $t$ and $z$ appear. I'll tell you what I've done, I arrive at $\frac{1}{z^2}f'(t)^2$, if I want that to be one, I need $f'(t)=z$ and hence $f(t)=\frac{z^2}{2}$, this looks weird though since that is not a function of $t$! – 2017-01-24
  • 1
    You have to evaluate $z$ at the point $\alpha(t)$. – 2017-01-24
  • 0
    Oh, it makes more sense now. Let's see this: Since the evaluation of $z$ in $\alpha(t)$ is $f(t)$ then I arrive at $f'(t)=f(t)$, which implies that that $f$ should be $e^t$? – 2017-01-24
  • 0
    It's not quite true that $f'(t) = f(t)$ implies $f(t)=e^t$, but that's one solution. – 2017-01-24
  • 0
    I'm aware there's a family of solutions, although I believe they all have to be exponential. It's true I didn't word it correctly though. Thanks a lot for your help, you wouldn't believe how much this has cleared my ideas up. – 2017-01-24

1 Answers 1

1

To arc-length parametrize you may integrate: $ds = dz/z$ to get $s-s_0=\pm \log(z/z_0)$ or $z= z_0 \exp( \pm (s-s_0))$.

In order to see that a vertical segment minimizes a distance (i.e. is a geodesic) consider a $C^1$ path $t\in [a,b]\mapsto \gamma(t)=(x(t),y(t),z(t))$ from $(x,y,z_0)$ to $(x,y,z_1)$ with $0

  • 0
    Thanks for the answer, I had never manipulated the differentials like that, we pretty much see it like notation, but then again it's a pretty basic course that I'm taking. This has been useful, I'll wait some more for potential answers before accepting this one. – 2017-01-24
  • 0
    You are welcome. The second part is b.t.w. a more or less standard construction of the socalled exponential map that constructs (local) geodesics emanating from a point. – 2017-01-24