Recovering connection from parallel transport (doCarmo, Riemannian Geometry, p.56, Q2)
I want to prove that the Levi-Civita connection $\nabla$ is given by $$ (\nabla_X Y)(p) = \frac{d}{dt} \Big(P_{c,t_0,t}^{-1}(Y(c(t)) \Big) \Big|_{t=t_0}, $$ where $p \in M$, $c \colon I \to M$ is an integral curve of $X$ through $p$, and $P_{c,t_0,t} \colon T_{c(t_0)}M \to T_{c(t)}M$ is the parallel transport along $c$, from $t_0$ to $t$.

My approach is to use the uniqueness of the Levi-Civita connection (a theorem proved elsewhere in the textbook) and show that the RHS satisfies all of its properties, i.e.


*

*It is an affine connection,

*It is symmetric,

*It is compatible with the metric.


However, for the first part, I am stuck on proving that $$ \nabla_{fX + gY}Z = f \nabla_X Z + g \nabla_Y Z. $$ So far, I have the following
$$ f \nabla_X Z = f \Big( \frac{d}{dt} \Big( P_{c_X,t_0,t}^{-1}(Z(c_X(t)) \Big) \Big|_{t=t_0} \Big), $$
$$ g \nabla_Y Z = g \Big( \frac{d}{dt} \Big( P_{c_Y,t_0,t}^{-1}(Z(c_Y(t)) \Big) \Big|_{t=t_0}  \Big), $$
$$ \nabla_{fX + gY}Z =  \frac{d}{dt} \Big( P_{c,t_0,t}^{-1}(Z(c(t)) \Big) \Big|_{t=t_0}, $$
where $$ c_X (t_0) = c_Y (t_0) = c(t_0) = p, $$ $$ \frac{d c_X}{dt} = X(c_X(t)), $$ $$ \frac{d c_Y}{dt} = Y(c_Y(t)), $$ $$ \frac{d c}{dt} = fX(c(t)) + gY(c(t)). $$
I'm sure the solution is something simple like working in local coordinates but I'm having trouble so any direction would be appreciated.
 A: You can probably make your idea work but it won't be easy. The reason is that your formula that recovers the connection from the parallel transport is true not only for the Levi-Civita connection but also for arbitrary connections. This means that in order to identify the right hand side as the Levi-Civita connection, you will need to understand what makes the parallel transport of the Levi-Civita connection special compared with the parallel transport of a general connection. The compatibility with the metric is easy - this implies that parallel transport is an isometry. However, to understand how the symmetry affects the parallel transport is much more delicate (see here for example).
A much less painful way to solve the exercise is to use the notion of a parallel frame along $c$ (which if I remember correctly is introduced in one of the other exercises). Namely, pick some basis of $\xi_1(p), \dots, \xi_n(p)$ of $T_pM$ and extend it by parallel transport to a frame $(\xi_1, \dots, \xi_n)$ of vectors fields along $c$. Now, write the restriction of $Y$ to $c(t)$ as $Y = Y^i(t) \xi_i(c(t))$ (summation convention is in use) and note that
$$ (\nabla_X Y)(c(t)) = \frac{DY(c(t))}{dt} = \dot{Y}^i(t) \xi_i(c(t)) + Y^i(t) \frac{D\xi_i(t)}{dt} = \dot{Y}^i(t) \xi_i (c(t)) $$ 
which means that the covariant derivative relative to the frame $\xi_i$ is given simply by the regular derivative. Then,
$$ \frac{d}{dt} \left( P_{c,t_0, t}^{-1}(Y(c(t)) \right)|_{t = t_0} = \frac{d}{dt} \left( P_{c,t_0, t}^{-1}(Y^i(t) \xi_i(c(t))) \right)|_{t = t_0} \\
= \frac{d}{dt} \left( Y^i(t) \xi_i(p) \right)|_{t = t_0} = \dot{Y}^i(t_0) \xi_i(p) = (\nabla_X Y)(p).$$
