Parallel transport along a cardioid I want to calculate an explicit example of a vector parallel transported along a cardioid to see what happens. Maybe someone could help me with that since no author of any book or pdf on the topic is capable of showing a single numerical example.
So we need a vector field on a manifold (which is the cardioid itself) $X=\frac{dx^{i}}{dt}\frac{\partial}{\partial x^{i}}$ and a curve $x^{i}=x^{i}(t)$. My problem is, I'm not sure how to make up a curve + vector field on a manifold. Let's take the parametrization of the cardioid in Cartesian coordinates as
$x(t)=a(1+2\cos t + \cos 2t)$
$y(t)=a(2\sin t + \sin 2t)$
(I think this could be written in polar coordinates which would make more sense, but I'm not sure what happens there)
So I think this should be the curve on which the vector is transported. Now I'm not sure how to make up the vector field. For the vector field I also need a function $f$, but what function? A vector function? For example could I just take $f=r(\phi, \rho)= (\rho \cos \phi, \rho \sin \phi)$ (polar coordinates) and then $X=\frac{dx^{i}}{dt}\frac{\partial}{\partial x^{i}}= \frac{dx(t)}{dt}\frac{\partial r(\phi, \rho)}{\partial \phi}+\frac{dy(t)}{dt}\frac{\partial r(\phi,\rho)}{\partial \rho}$ ? I think this looks right since the $\frac{\partial}{\partial x^{i}}$ span the tangent space. Now how exactly does the condition for parallel transport in coordinates for this looks like? 
The general formula is
$\frac{\partial X^{\mu}}{dt}+ \Gamma^{\mu}_{v\lambda} \frac{\partial x^{v}(c(t))}{dt}X^{\lambda}=0$
(I know how to calculate the Levi-Civita connection with the metric,but I'm not sure about the rest)
 A: Parallel transport on a manifold $M$, along a curve $\gamma$ in $M$, is defined only for tangent vectors to $M$ that happen to lie at a point of the image of $M$. So if $M$ is the unit sphere, and $\gamma$ is a path going around the equator, then you can take any tangent vector at the equator and translate it along $\gamma$. But you can't take a tangent vector at, say, 34 degrees north latitude and translate it along the curve. 
In your case, the manifold $M$ is the cardioid (as you said in the comments); presumably it inherits its metric from the ambient space (the plane). The path $\gamma$ is also the cardioid. 
Before I get into details, let me observe that the cardioid --- the image of the $xy$-curve that you've written down --- is not actually a manifold. At the point described by $t = \pi$, the curve has a singularity, and the image of the tangent space to $\mathbb R$ at the point $t = \pi$ is carried, by the differential of $\gamma$ at $\pi$, $dg(\pi)$, to the zero vector in $\mathbb R^2$; one cannot therefore at that point inherit the metric from $\mathbb R^2$. (If you chase the definitions carefully, the inner product of any two vectors in $d\gamma(\pi)(\mathbf{T}\mathbb R_\pi)$ turns out to be zero, so every vector ends up having length zero, which isn't allowed in a metric on what should be a 1-dimensional vector space (the tangent space to our purported 1-manifold at $\gamma(\pi)$). 
The next thing to observe is that because this (aside from the bad point) is a 1-manifold, the metric at any point $\gamma(t)$ is represented by a $1 \times 1$ matrix with a single entry, $g_{11}(t)$. That's gotten by taking the standard basis for the tangent space to $\mathbb R$ at the point $t$, namely the vector $[1]$, and pushing it forward through $d\gamma(t)$ to get a vector $A(t)$ in $\mathbb R^2$, and then computing the dot product (in $\mathbb R^2$) of that vector with itself. Let's do that. 
And to simplify, I'm gonna get rid of the constant factor $a$ throughout. 
We have:
$$
x(t)=(1+2\cos t + \cos 2t)\\
y(t)=(2\sin t + \sin 2t)
$$
so
$$
x'(t) = (-2 \sin t - 2 \sin 2t) \\
y'(t) = (2 \cos t + 2 \cos 2t).
$$
\begin{align}
A(t) &= d\gamma(t)([1]) \\
&= \begin{bmatrix}
-2 \sin t - 2 \sin 2t \\ 2 \cos t + 2 \cos 2t
\end{bmatrix}
[1] \\
&= \begin{bmatrix}
-2 \sin t - 2 \sin 2t \\ 2 \cos t + 2 \cos 2t
\end{bmatrix}.
\end{align}
So 
\begin{align}
g_{11}(t) = A(t)\cdot A(t)  &= 
&= (s \sin t + 2 \sin 2t)^2 + (2 \cos t + 2 \cos 2t)^2.
\end{align}
Now we can translate a particular tangent vector $X$ along this curve by writing out the transport equation. Let's, as an example, choose the vector 
$$
X = \begin{bmatrix}0 \\ 1\end{bmatrix}
$$
at the point $\gamma(0) = (4, 0)$. 
You're hoping to find a function $X: image(\gamma) \to \mathbb R^2$ with the properties that


*

*$X(t)$ is a vector tangent to $M = image(\gamma)$ at $\gamma(t)$. 

*$X(0)$ is the vector $X$ that I wrote down above. 

*$X(t)$ satisfies the equation you wrote down involving the Christoffel symbols. 
It may be easiest to say that $X(t) = (a(t), b(t))$, or $X(t) = (X_1(t), X_2(t))$, so that you have an explicit name for the coordinates of $X$. 
At this point, it'd be good for you to write down those Christoffel symbols. How many are there? What are their values? What are the resulting equations? 
My guess is that after reading this far, you're going to say "No, that isn't the question I was asking at all!", which is why I'm not writing out anything else just yet. 
If it really is the question you want answered, by all means do some writing, and I'll continue my answer. 
