# Expression for Levi-Civita Connection

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!

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What definition of the covariant derivative are you using? The connection in some sense is formally equivalent to the covariant derivative so without telling us your starting point you may have something to prove or this might be a tautology. – Ryan Budney Nov 7 '10 at 1:24
I'll edit it in the text above... – Sam Nov 7 '10 at 1:36

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.

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Ah, excellent... Now, I realise what I was missing. Choosing a "parallel basis" does the trick! This makes it possible to work with the parallel transport, which I just couldn't get a handle on. Actually, this "trick" was used in proving a proposition before... Thank you, too, Jason. – Sam Nov 7 '10 at 3:59
Can you explain on how to get $\nabla_X \sum_i a_i(t)E_i(t)|_{t=t_0} = \sum_i a_i(t_0) \nabla_X E_i(t)|_{t=t_0} + \sum_i a_i'(t_0) E_i(t_0)$? In fact, from the definition of connection, there should be two vector fields on $M$, but $E_i$ is a vector field along the curve. – user136592 May 15 at 20:00
@user136592: If you keep reading Do Carmo's book right after where Sam stopped copying it, Do Carmo explains that $\nabla_X Y$ at a point only depends on $X$ at the point and $Y$ along an integral curve of $X$. So, if you want, extend $E_i$ to all of $M$ however you would like do - you will get the same answer. – Jason DeVito May 17 at 1:15

This is a cultural comment, rather than an answer; but it is a bit long for the comment box, which is why I am writing it here.

My comment is the following: giving a connection, and giving the associated parallel transport, are essentially the same thing, and so it should be possible to establish the formula you are trying to prove by reasonably conceptual, high-level, thinking, rather than mucking about too much with complicated differential equations.

Let me elaborate a little: how would you define the derivative of a vector field along a curve? Well, the idea is that you want to form the usual Newton quotient $$\dfrac{V\bigl(c(t+\epsilon)\bigr) - V\bigl(c(t)\bigr)}{\epsilon}$$ and then let $\epsilon$ go to zero, to compute the derivative of $V$ at the point $c(t)$.

The only problem is that the subtraction in this formula doesn't make sense, because the tangent vectors being subtracted are based at different points: one is based at $c(t+\epsilon)$ and the other at $c(t)$.

Enter parallel transport: suppose we have an agreed upon way to shift, or transport, tangent vectors along a curve. Then we can apply this so as to transport $V\bigl(c(t+\epsilon)\bigr)$ from $c(t+\epsilon)$ to $c(t)$, and then form the above Newton quotient, and proceed to compute a derivative.

An agreed upon method for shifting vectors along curves is formally referred to as a choice of parallel transport, because if you have such a method, and if you define a vector field along a curve $c(t)$ by choosing some $V_0$ at $c(t_0)$ and then using the agreed upon parallel transport to define $V\bigl(c(t))$ by tranporting $V_0$ along $c$ to $c(t)$, you obtain a vector field along $c(t)$ whose derivative at each point will be zero (by construction!). Thus this vector field does not change along $c(t)$, and so consists of parallel vectors, hence the name parallel tranport. (Of course, the notions of change and parallel are not intrinsict to the vector field; they depend upon our particular choice of parallel transport.)

Of course, our choice of parallel transport should satisfy some axioms (it should be linear; it should be smooth (so that the vector fields along curves that it gives rise to are smooth); and so on). If you look in the right text, you will find these axioms written down.

As we've seen, a choice of parallel transport on our manifold gives a way of differentiating vector fields along curves. But since a tangent vector is just an infinitesimal curve, one sees that we in fact have a way of differentiating a vector field in the direction of a given vector field: extend that vector field to a curve, and then differentiate along that curve.

Thus a choice of parallel transport determines an affine connection, i.e. a way of differentiating vector fields along tangent vectors. Conversely, this is enough information to determine the parallel transport: we parallel transport a tangent vector along a curve by making sure that its derivative (using the given affine connection) at every point of the curve, in the tangent direction of the curve, vanishes.

So we have gone full circle, from parallel transport, to affine connection, back to parallel transport. The text you are reading is spelling out the second half of this circle, but is perhaps a little scanty on details regarding the first half; in fact, the exercise you are trying to solve is exactly about the first half of the circle, and its goal is to verify that you truly are going around in a circle: i.e. that you are ending up where you started from.

There really is something to verify here (i.e. the exercise is not trivial if you have not done it before, and are learning these ideas for the first time), but I hope that the above discussion may help shed some light on its meaning, and also make it seem less daunting (and perhaps more conceptual and less computational) than it might otherwise appear.

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Thanks for the comment! Indeed, in parantheses after the exercise there is the following note: "(this shows how the connection can be reobtained from the concept of parallelism)" – Sam Nov 7 '10 at 2:56
Again, thank you very much. This comment was very illuminating. The very same idea of having to transport vectors to be able to take derivatives is behind Lie derivatives, I guess. – Sam Nov 7 '10 at 4:06