# Recover Covariant Derivative from Parallel Transport

It is well known that one can recover the connection from the parallel transport. I struggle to understand this concept.

Since $\Gamma(\gamma)^t_s:E_{\gamma(s)}\to E_{\gamma(t)}$ is an isomorphism between the fibers, it holds that $\Gamma(\gamma)^t_s(e_s)=e_t$ for the unique parallel translate $e_t=e(\gamma(t))\in E_{\gamma(t)}$ of a section $e$ in the vector bundle along a curve $\gamma:I\to M$ . Consequently, $\nabla_{\dot{\gamma}(0)}e=\lim_{h\to 0}(\Gamma(\gamma)^0_h(e_h)-e_0)/h=0$.

However, how do we "recover" $\nabla_{X}e$ for an arbitrary section $e$ that might not be parallel along $\gamma$ where $\dot{\gamma}(0)=X$?

• @Braindead Is it true that for any section, there exists a curve to which it is parallel? I am confused since the formula linked equals zero for any section... – gofvonx May 17 '14 at 12:40
• I don't think that is true, but I don't think that's relevant to your question. Basically, the parallel transport gives you a means of comparing the value of a section over two different points in the base space by transporting one of the vectors to the other point. This allows you to take their difference over a common point. There is no reason for this to be zero all the time. – Braindead May 17 '14 at 12:47
• @Braindead But the parallel transport $\Gamma$ is defined by $\Gamma(\gamma)_s^t(e_s)=e_t$ for the solution of $\nabla_{\dot{\gamma}(0)}e=0$ and then $\Gamma(\gamma)_s^0(e_s)\equiv e_0$, so the difference in the limes is always zero, isn't it? – gofvonx May 17 '14 at 12:51
• You can apply the parallel transport to a non-parallel arbitrary section. Also, the parallel transport in the wikipedia article is not defined by that equation. See if my answer below makes sense to you. – Braindead May 17 '14 at 13:12

Although he discusses it for the tangent bundle only, John M. Lee mentions a theorem in his book Riemannian Manifolds, from which it should get clearer:

Theorem 4.11. (Parallel translation) Given a curve $\gamma:I\rightarrow M,~t_0\in I$, and an [arbitrary] vector $V_0\in T_{\gamma(t_0)}M$, there exists a unique parallel vector field $W$ along $\gamma$ s.t. $W(t_0):=W(\gamma(t_0))=V_0$

Here, I took the liberty of renaming the vector field.

Now, what does parallel vector field $W$ along $\gamma$ mean? Simple, it's $D_tW\equiv0$. What's $D_t$, I hear you wondering. Well, if $W$ is a smooth vector field on $M$, then $D_tW(t_0)=\nabla_{\dot\gamma(t_0)}W$. So, as not to wreak any more havoc in this definition jungle, we shall simply restrict to this case (since the ODE $D_tW\equiv0$ then becomes $\nabla_{\dot\gamma}W\equiv0$, which is the way it's written in your link).

What John does next is, he defines

If $\gamma:I\rightarrow M$ is a curve and $t_0,~t_1\in I$, the parallel translation defines an operator $$P_{t_0t_1}:T_{\gamma(t_0)}M\rightarrow T_{\gamma(t_1)}M$$ with $P_{t_0t_1}V_0=W(t_1)$, where $V_0\in T_{\gamma(t_0)}M$ and $W$ is the parallel vector field from the theorem.

This is, by the theorem, an isomorphism between tangent spaces and therefore gives you for each vector at each point of the curve a uniquely determined vector - at another point in the image of the curve - which 'looks like the original vector', if you will.

The formula would then, in John's book, look like this $$D_tV(t_0)=\lim_{t\rightarrow t_0}\frac{P^{-1}_{t_0t}V(t)-V(t_0)}{t-t_0}=\lim_{t\rightarrow t_0}\frac{W_t(t_0)-V(t_0)}{t-t_0},$$ where $W_t$ is the parallel vector field determined by $D_tW_t\equiv0$ and $W_t(t)=V(t)$.

Hope that cleared things up.

I am not sure of the extent of your mathematical background, so I apologize if I am writing something that's completely obvious to you.

Let $e : M \to E$ be a section of a vector bundle $E \to M$.

We want to be able to compare values of $e$ over different base points.

$e(p)$ and $e(q)$ live in different vector spaces $E_p$ and $E_q$, $e(p) - e(q)$ makes no sense. How can we compare them?

We can do it if there was some way to transport all the vectors in $E_q$ to $E_p$. To make sure all the vectors get transported in the right way, it makes sense to use an isomorphism. It also makes sense to think of this notion of transporting as being done over a path from $p$ to $q$ in $M$.

It turns out for general manifolds there is no canonical way to do this. The definition of Parallel Transport in your article reflects this: different curves in the base space give you different isomorphisms.

So pick a curve: $\gamma: [0,1] \to M$, with $\gamma(0) = p$ and $\gamma(1) = q$.

Now, $\Gamma(\gamma)_1^0 e(q) \in E_p$, so the difference $(\Gamma(\gamma)_1^0 e(q) - e(p)) \in E_p$ and it all makes sense.

There is no reason for this difference to be 0.

[Using Lee's hint]: Let $$\nabla$$ be a linear connection of $$M$$ and $$\gamma: [0,1] \to M$$ smooth curve, we want to prove:

$$D_{t}V(t_{0}) = \lim_{t \to 0} \frac{P_{t_{0},t}^{-1}V(t) - V(t_{0})}{t - t_{0}}$$

Proof: Let $$\epsilon > 0$$ with $$(U,\psi = (x^{1},...,x^{n}))$$ chart and $$\gamma([0,1] \cap (t-\epsilon,t+\epsilon)) \subset U$$. consider local basis $$\partial_{i} = \frac{\partial}{\partial x^i}$$.

Let $$(E_{i}(t))$$ fields such that $$E_{i}(0) = \partial_{i}$$ and $$D_{t}E_{i} = 0$$. Since parallel translations are isomorphisms, $$(E_{i}(t))$$ is a local basis.

Let $$V(t) = V^{i}(t) E_{i}(t)$$ for $$t \in (t-\epsilon,t+\epsilon)\cap [0,1]$$, note that $$P_{t_{0},t}^{-1} V(t) = V^{i}(t)\partial_{i}$$.

$$\frac{1}{t - t_{0}}(P_{t_{0},t}^{-1} V(t) - V(t_{0})) =$$ $$\frac{(V^{i}(t) - V^{i}(t_{0})}{t - t_{0}}\partial_{i} \to \dot{V}^{i}(t_{0})\partial_{i}$$

On the other hand

$$D_{t}V (t_{0}) = (D_{t}V^{i}(t) E_{i}(t) )(t_{0}) =$$ $$\left( \dot{V}^{i}(t_{0})E_{i}(t_{0}) + V^{i}(t_{0}) (D_{t}E_{i})(t_{0}) \right) = \dot{V}^{i}(t)\partial_{i}$$