It is known that the geodesic equations for the upper half plane equipped with the hyperbolic metric are

$$x''=\frac{2x'y'}{y},$$ $$y''=\frac{(y')^2 -(x')^2}{y}.$$

It is also well known that the geodesics are semi-circles lying on the x-axis. However, I am encountering difficulties when I attempt to verify this fact.

I am trying to follow the solution to part (b) of problem 5 on paper 1 here.

The path $\tilde \gamma(t)=(\cos(t),\sin(t))$, when re-parameterized, should be a geodesic. At the point $(\cos(t),\sin(t))$, the unit tangent vector is $$(-\sin^2(t), \sin(t)\cos(t)).$$ So we need $$x''(t)=\frac{2(-\sin^2(t))(\sin(t)\cos(t)}{\sin(t)},$$ or $$x''(t)=-2\sin^2(t)\cos(t).$$ However, $(-\sin^2(t))'=-2\cos(t)\sin(t)$, which is not quite what we want. I'm confused why this does not line up. I suppose because the velocity field does not arise from a curve.

How can this be fixed up?

  • $\begingroup$ There's no reason that the semi-circle should be centered at $0$ which is what you're assuming when using your parametrization. $\endgroup$ – Ahsan Aug 9 '15 at 9:12
  • $\begingroup$ @Ahsan I'm not sure I understand your comment. Any semicircle, regardless of center, should be a geodesic. I chose to center mine at zero for convenience. $\endgroup$ – Potato Aug 9 '15 at 9:13
  • $\begingroup$ Ignore my comment. You are right in that an appropriate semicircle, regardless of the center should satisfy the equation. Answer upcoming in a bit. $\endgroup$ – Ahsan Aug 9 '15 at 9:35
  • $\begingroup$ According to this answer below, a solution is given by $$(x(t), y(t)) = \left( \dfrac{e^{2t} - 1}{e^{2t} + 1} , \dfrac{2e^t}{e^{2t} + 1} \right), t \in \Bbb R.$$ $\endgroup$ – Watson May 14 '17 at 15:27

Recall that $$ 0=E_1(E_j,E_k)= \frac{1}{y^2} \{ \Gamma_{1j}^k + \Gamma_{1k}^j \} \Rightarrow \Gamma_{1j}^k =-\Gamma_{1k}^j $$

$$ -2y^{-3}= E_2(E_2,E_2)= \frac{1}{y^2} (2\Gamma_{22}^2) \Rightarrow \Gamma_{22}^2= - \frac{1}{y} $$

By definition,

$$ \Gamma_{12}^1 = -\frac{1}{y}$$

If $$ v= (-\sin^2 t, \sin\ t\cos\ t) :=fE_1+ g E_2$$ then $$ v (F(t))= y \frac{d}{dt} F(t) $$ and $$ \nabla_vv = \nabla_v (fE_1) + \nabla_v (gE_2) $$ $$= v(f) E_1+ v(g)E_2 + f\nabla_vE_1 + g\nabla_vE_2 $$ $$ = y(f',g') + f ( f \Gamma_{11}^i E_i + g \Gamma_{21}^i E_i) + g( f \Gamma_{12}^i E_i + g \Gamma_{22}^i E_i ) $$

$$ = y(f',g') + f^2 \Gamma_{11}^2E_2 +2f g \Gamma_{12}^1 E_1 + g^2 \Gamma_{22}^2 E_2 =0 $$

  • $\begingroup$ Thank you for your help. I am having trouble understanding why this approach works and the one I sketched above doesn't. Shouldn't they be equivalent, as the geodesic equations are derived using the procedure you indicated? $\endgroup$ – Potato Aug 9 '15 at 17:26
  • $\begingroup$ You mean that you want to solve this problem by using $x'',\ y''$ instead of solution in link ? $\endgroup$ – HK Lee Aug 9 '15 at 17:30
  • $\begingroup$ Pardon my confusion, but isn't this approach the same thing as using $x''$ and $y''$? When you take take the covariant derivative of $v$, shouldn't $x''$ and $y''$ appear? (I agree this approach is best -- because it works! -- I just don't see why it's better.) $\endgroup$ – Potato Aug 9 '15 at 17:33
  • $\begingroup$ Note that geodesic equation in the first link is about curve of unit speed. So to apply geodesic equation we must find reparametrization $c(s) = (\cos\ f(s),\sin\ f(s)),\ t=f(s),\ |c'(s)|=1$, since $(\cos\ t,\sin\ t)$ do not have a unit speed. To find $f$, we solve $(f')^2=\sin^2 f$. I think that this is not easy. $\endgroup$ – HK Lee Aug 9 '15 at 17:57
  • $\begingroup$ Indeed. My question is: isn't the formula $\nabla_{v}v$ equivalent to the geodesic equation? Your computation looks very similar to mine, except you get a different term for $x''$ and $y''$. So if the ways are equivalent, why do they get different answers? $\endgroup$ – Potato Aug 9 '15 at 18:05

You shouldn't assume that the naive parametrization $(\text{cos}(t), \text{sin}(t))$ will solve the equation. It doesn't, as it clearly doesn't satisfy even the first differential equation. However, if you assume a more general form $x(t) = \text{cos}(f(t))$, $y(t)= \text{sin}(f(t))$, which still traces out a semicircle (but with a different speed), then it should be possible to find an $f(t)$ namely a reparametrization which makes the geodesic equations satisfied. If you plug in the above ansatz, both equations tell you that you need to solve $$f''(t) = \text{cot}(f(t)) (f'(t))^2,$$ which is solved by

$$f(t) = 2\text{cot}^{-1}(e^t)$$

  • 1
    $\begingroup$ Indeed, we know it exists, but the solution I'm reading seems to claim we can find it explicitly, or that something like what I mentioned above works. I added a link that may be helpful. $\endgroup$ – Potato Aug 9 '15 at 10:04
  • $\begingroup$ Sorry, I had made a careless error in deriving the differential equation. The correct form allows us to write an explicit solution. I've fixed this in the latest version. $\endgroup$ – Ahsan Aug 9 '15 at 13:58
  • $\begingroup$ Good answer. Perhaps elaborate on your first sentence? There's an ambiguity $\endgroup$ – Neal Aug 9 '15 at 19:41

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