Proving that a curve is a Geodesic in the Poincaré Half-Plane

Question

Let $\mathbb{H}^2$ be the Poincaré Half-Plane, that is, $\mathbb{R}\times \mathbb{R}_+^*$ with the Riemannian metric $$\langle u,v \rangle_{(x,y)} = \frac{u \cdot v}{y^2}$$

I was asked (in a test) to prove that $\alpha \colon (-1,+\infty) \to \mathbb{H}^2$, $\alpha(t) = (1,t+1)$.

So, I started by computing $\alpha'(t)$ in coordinates (since the identity is the parametrization) and $\alpha'(t) = 0e_1 + 1e_2$, in terms of the coordinates given by the parametrization. Changing to an orthonormal reference field, $E_1 = y(p) e_1$ and $E_2 = y(p)e_2$. So $\alpha'(t) =\frac{1}{t+1}E_2$ and then $\nabla_{\alpha'(t)} \alpha'(t) = -\frac{1}{t+1} \omega_{12}(\alpha'(t))E_1$

But in a conformal manifold with ruler $g$, $\omega_{12}$ can be computed by $$\omega_{12} = -\frac{g_y}{g} dx + \frac{g_x}{g}dy$$

In this case $g_x =0$, $g_y = 1$ and then $\omega_{12} = -\frac{1}{g} dx$, but since $dx(\alpha'(t)) = 0$, $\omega_{12} (\alpha'(t)) = 0$

That is, $\nabla_{\alpha'(t)}\alpha'(t) = 0$

And then $\alpha$ is a geodesic.

I have some questions concerning this proof:

• Do we require all geodesics in a Riemannian Manifold to have constant velocity?
• Is this curve really a geodesic?
• Is my reasoning accurate?

Answer 1

No doubt vertical lines are geodesics. But as we have it, $\alpha$ is not a geodesic. If $\alpha(t) =(1 , t + 1)$, then $\alpha'(t) = (0,1)$. So: $$\| \alpha'(t)_{\alpha(t)}\| = \sqrt{\langle \alpha'(t), \alpha'(t) \rangle_{\alpha(t)}} = \sqrt{\frac{\alpha'(t) \cdot \alpha'(t)}{(t+1)^2}} = \sqrt{\frac{1}{(t+1)^2}},$$ which is not constant. As far as I know, the curve as it is, is a pre-geodesic, that is, it admits a reparametrization which makes it a geodesic (parametrization by arc-length, say).

We don't require geodesics to have constant velocity, we can prove this: $$\alpha'[\langle \alpha', \alpha' \rangle] = 2\left\langle \frac{D \alpha'}{dt}, \alpha' \right\rangle = 0.$$

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Seems to me that you got the derivative wrong. Yes, we have $\alpha'(t) = \frac{1}{1+t}E_2(t)$, then: $$\nabla_{\alpha'(t)}\alpha' = \nabla_{\alpha '(t)} \frac{1}{1+t}E_2 (t)= -\frac{1}{(1+t)^2}E_2(t) + \frac{1}{1+t} \nabla_{\alpha'(t)}E_2(t) = -\frac{1}{(1+t)^2}E_2(t) - \frac{1}{1+t} \omega_{12}(\alpha'(t))E_1(t).$$

My point is, $\alpha$ can't be a geodesic because the component in $E_2$'s direction doesn't vanish.

Answer 2

Yes, this is correct ... Almost. Geodesics must have constant speed, since the tangent vector is parallel along the geodesic and parallel translation is an isometry (preserves lengths and angles).