TITLE: Prove a Levi-Civita connection gives $\nabla_XY(p)=\partial_t|_{t_0}[P^{-1}_{c_0,t_0,t}(Y(c(t)))]$ with $P$ parallel transport
QUESTION [18 upvotes]: I'm having trouble with the following exercise in do Carmo's Riemannian geometry.
Let $X$ and $Y$ be differentiable vector fields on a Riemannian manifold $M$. Let $p \in M$ and let $c: I \to M$ be an integral curve of $X$ through $p$, i.e. $c(t_0) = p$ and $\frac{dc}{dt} = X(c(t))$. 
Prove that the Riemannian connection of $M$ is
$(\nabla_XY \ )(p) = \frac{d}{dt} (P^{-1}_{c,t_0,t}(Y(c(t)))\ |_{t=t_0}$
where $P^{-1}_{c,t_0,t}: T_{c(t_0)}M \to T_{c(t)}M$ is the parallel transport along $c$, from $t_0$ to $t$.
I guess, I don't have enough understanding of how to handle the parallel transport (since it is only given as the unique solution to a differential equation).
Any hints would be greatly appreciated!
Thank you,
S. L.
Edit: Do Carmo first introduced the notion of an affine connection 
$\nabla: \text{Vect}(M) \times \text{Vect}(M) \to \text{Vect}(M)$,   $(X,Y) \mapsto \nabla_XY$.
With the following properties:

$\nabla_{fX + gY}Z = f\nabla_XZ + g \nabla_YZ$
$\nabla_X(Y+Z) = \nabla_XY + \nabla_XZ$
$\nabla_X(fY) = f\nabla_XY + X(f)Y$

for $X,Y,Z \in \text{Vect}(M)$ and $f,g \in C^\infty(M)$.
And then showed that there is a unique correspondence which associates to a vector field $V$ along the differentiable curve $c: I \to M$ another vector field $\frac{DV}{dt}$ along c, called covariant derivative of $V$ along $c$, with three more properties:

$\frac{D}{dt}(V+W) =  \frac{D}{dt}V +  \frac{D}{dt}W$
$ \frac{D}{dt}(fV) =  \frac{df}{dt}V +  f\frac{D}{dt}V$
If $V$ is induced by a vector field $Y \in \text{Vect}(M), then  \frac{D}{dt}V = \nabla_{dc/dt}Y$

Then he showed existence and uniqueness of the parallel transport along a curve, and went on to prove existence and uniqueness of Levi-Civita connection.
I hope this makes things clearer? Thanks for the quick answer!

REPLY [17 votes]: Let $\{e_i\}\subseteq T_{c(t_0)}M$ be a basis.  Define $E_i(t)$ as the parallel translation of $e_i$ along $c(t)$.  Prove to yourself that $\{E_i(t)\}$ forms a basis of $T_{c(t)}$ for all $t$ (hint, use the uniqueness part of solving linear ODEs).
Now, we can write $Y(c(t)) = \sum_i a_i(t) E_i(t)$.  Then, since the $E_i(t)$ are parallel, and since $P_{c,t_0,t}$ is linear, we have $P^{-1}_{c,t_0, t}(Y(c(t)) = \sum_i a_i(t) e_i$.
From here, by actually computing the limit, it's not too hard to see that $\frac{d}{dt} P^{-1}_{c,t_o,t}Y(c(t))|_{t=t_0} = \sum_i a_i'(t_0)e_i$.
Thus, the goal is to show that $\nabla_X Y(p)$ can also be written like this.
But $\nabla_X Y(p) = \nabla_X \sum_i a_i(t)E_i(t)|_{t=t_0} = \sum_i a_i(t_0) \nabla_X E_i(t)|_{t=t_0} + \sum_i a_i'(t_0) E_i(t_0)$.  (The second equality is just the Leibniz rule every connection must satisfy).  However, $\nabla_X E_i(t)$ is $\frac{D}{dt} E_i(t)$ and this is 0 since the $E_i$ are parallel.  Hence, the first term of the sum vanishes, so we get the desired result.