I was given the following exercise:
Consider the connection on $\mathbb{R}^2$ defined by $\Gamma_{1 2}^1 =1$ and all the other $\Gamma_{jk}^i$ equal to zero wrt $(x^1, x^2) = (x,y)$. Determine the equations for parallel transport of a vector field $Y = Y^i \frac{\partial}{\partial x^i}$.
I found that the equations of motion would be given by:
$$\frac{dY^1}{dt} + \frac{dy}{dt} Y^1 = 0$$
and $$\frac{dY^2}{dt} = 0$$
I can see that the second equations implies that $Y^2$ will be constant, but I can't figure how to solve the first ODE since it contains 2 variables. The solution given to me mentions that the first equation implies that $Y^1 e^y$ is constant, but I don't see why. Furthermore, it states the following:
The components of the vector field $Y$ along a curve from $C(t_0)$ to $C(t)$ are given by :
$$ \left[ \begin {array}{c} Y^1 \left( t \right) \\ Y^2
\left( t \right) \end {array} \right]
= \left[ \begin {array}{cc} {{\rm e}^{y \left( t \right) -y \left( {
\it t_0} \right) }}&0\\ 0&1\end {array} \right] \left[ \begin {array}{c} Y^1 \left( t_0 \right) \\ Y^2
\left( t_0 \right) \end {array} \right]
$$
Which apparently follows from the first equation. Can some help clarify my confusion about this?
Best Answer
$$\frac{dY^1}{dt} + \frac{dy}{dt} Y^1 = 0$$ $$\frac{dY^1}{dt} =- \frac{dy}{dt} Y^{1}$$ $$\int \frac{dY^1}{Y^1} =- \int{dy}$$ $$\ln Y^1=-y+C$$ This gives the final result: $$Y^1=Ke^{-y}$$ Are you sure it is $\frac{dY^1}{dt} +\frac{dy}{dt} Y^{1}=0$ and not $\frac{dY^1}{dt} - \frac{dy}{dt} Y^{1}=0$ ?
Note that this DE: $$\frac{dY^1}{dt} + \frac{dy}{dt} Y^1 = 0$$ Is just: $$\frac{Y'^1}{Y^1} + y' = 0$$ That you can rewrite as: $$(\ln Y^1)'+y'=0$$ $$(\ln Y^1+y)'=0$$ Integrate both side. $$\ln Y^1+y=C$$