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Let us consider Poincaré's upper plane which is defined as $\mathbb{H} = \{ (x,y) | y>0\}$. This space has a Riemannian metric $g = \text{diag}(1/y^2, 1/y^2)$. Now let us consider a differential curve $a(t) = (t,1)$ and a vector $v \in T_{a(o)}\mathbb{H} = (0,1)$. How can $v$ be parallel transported to the curve?

I think that finding the geodesic equation is not particularly hard, but I am not sure at all what I am meant to answer in this question.

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  • $\begingroup$ UPDATE: I have resolved my issue. I still am confused with the answer that is given by @JimBelk though. $\endgroup$
    – Marion
    Jun 26, 2015 at 17:33

2 Answers 2

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I am not sure where the problem is but let me present my solution. In order to find the transport of $(0,1)$ at point $a(t)$ we need to find parallel vector field $V$ along $a$ with $V(0)=(0,1)$ and compute $V(t)$. Let:

$$V(t)=V_1(t) \partial_1(a(t)) + V_2(t) \partial_2(a(t))$$ Than: $$D_tV=\sum_{k}(V_k'+\sum_{i,j}\Gamma^k_{ij}a_i'V_j)\partial_k$$ It is trivial to compute: $$\Gamma^1_{12}=\Gamma^1_{21}=-\frac{1}{y}, \: \Gamma^2_{11}=\frac{1}{y}, \: \Gamma^2_{22}=-\frac{1}{y}$$ so equations for $V$ are:

$$V_1'(t)-\frac{1}{a_2(t)}(a_1'(t)V_2(t)+a_2'(t)V_1(t)=0$$ and $$V_2'(t)+\frac{1}{a_2(t)}(a_1'(t)V_1(t)-a_2'(t)V_2(t)=0$$

So: $V_1'(t)=V_2(t)$ and $V_2'(t)=-V_1(t)$ what gives $V_1(t)=\sin(t)$ and $V_2(t)=\cos(t)$.

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  • $\begingroup$ thanks I have already solved the problem anyway. What I am confused about is about Jim Belk's answer about the 90 degree rotated vector. I have now understood what happens to a vector moving along the specific curve but its not a 90 degree rotation. $\endgroup$
    – Marion
    Jul 3, 2015 at 9:04
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    $\begingroup$ O, so I missunderstood the question ;P $\endgroup$
    – J.E.M.S
    Jul 3, 2015 at 10:52
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For a curve $(x,y) = \bigl(x(t),y(t)\bigr)$ in $\mathbb{H}$, the parallel transport of a vector $\textbf{v}$ is determined by the following differential equation: $$ \frac{d\textbf{v}}{dt} \;=\; \frac{1}{y}\frac{dy}{dt}\,\textbf{v} \,-\, \frac{1}{y} \frac{dx}{dt}\,\textbf{v}^\perp $$ where $\textbf{v}^\perp$ is the vector obtained by rotating $\textbf{v}$ counterclockwise $90^\circ$.

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  • $\begingroup$ Hi and thanks for the answer. I have found the two geodesic equations but I am not sure why you get this equation above. I am nor sure what it represents. From my computation I get the two geodesics one for $d^2x/dt^2$ and one for $d^2y/dt^2$. How is your answer related to those geodesics? $\endgroup$
    – Marion
    Jun 23, 2015 at 17:20

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