Covariant derivative, directional derivative, and curvature tensor

Question

I'm confused about how to connect those three things together, so hopefully the question doesn't end up vague. The main problem is understanding how the curvature tensor is a commutator of covariant derivatives. As far as I understand, to take derivatives, we need to compare two vectors that are, generally speaking, elements of two different tangent spaces. The way to do that is by parallel transporting from one point to another. A related problem is that I don't understand exactly how the "connection coefficients" do that. How do they "move the vector (field)" from one point to another? And how is the path-dependence of this movement/transport encoded in the connection coefficients? Shouldn't it be? In turn, I don't understand how the covariant derivative itself encodes the path dependence of the transport process, since transporting along different paths generally yields different vectors. The directional derivative, on the other hand, seems to incorporate this path-dependence by specifying the trajectory along which we take the derivative of the vector field. I understand the covariant and directional derivative are very closely related, but I don't understand how to use one of them, or which one to use, to arrive at the curvature tensor.

Answer 1

Consider a manifold $\mathcal M$.

A parallel transport on $\mathcal M$ sends each path $\gamma:[a,b]\to \mathcal M$ to a linear map $\operatorname{Pt}_\gamma: T_a \mathcal M \to T_b\mathcal M$. With the right axioms (smoothness, locality, etc), we can show that this is given by the solutions of a first order linear equation : $$\dot \gamma^\mu\nabla_\mu X^\nu = 0 \tag1$$ The connection $\nabla_\nu$ (or covariant derivative) is related to the parallel transport in the following way : if $X^\nu(t)$ is a solution of $(1)$ then $\operatorname{Pt}_\gamma(X(a)) = X(b)$.

We can also show that, in coordinates, there exists Christoffel symbols (or connection coefficient), so that : $$\nabla_\mu X^\nu = \partial_\mu X^\nu + \Gamma_{\mu\rho}^\nu X^\rho \tag{2}$$

How do the connection coefficients "move the vector (field)" from one point to another ?

They do this by allowing us to write the parallel transport equation $(1)$, whose solution parallel transports vectors from one tangent space to another along a path.

And how is the path-dependence of this movement/transport encoded in the connection coefficients?

Path-dependence does not appear in the connection coefficients. They are purely local and have other uses beside parallel transport of vectors along curves.

The path-dependence of parallel transport appears in equation $(1)$. In full form, it reads : $$\frac{\text dX^\mu}{\text d\lambda} (\lambda) + \Gamma^\mu_{\nu\rho}(\gamma(\lambda)) \dot \gamma^\nu(\lambda) X^\rho(\lambda) = 0 \tag3$$ where $\lambda$ is the parameter along the path $\gamma$.

The directional derivative, on the other hand, seems to incorporate this path-dependence by specifying the trajectory along which we take the derivative of the vector field.

I am not sure what you mean by "directional derivative". The usual directional derivative from flat space, $X^\mu\partial_\mu Y^\nu$, is not a tensor. It can be generalized by the covariant derivative $X^\nu \nabla_\nu X^\mu$ or by the Lie derivative $ (\mathcal L_X Y)^\mu = [X,Y]^\mu = X^\nu \partial_\nu Y^\mu - X^\nu \partial_\nu X^\mu$. The latter is not related to connections or parallel transport.

Edit : Covariant derivatives

Given a function $f$ on a manifold, we can define its differential $\partial_\mu f$ (or $\text d f$ in index-free notation). As this expression is covariant, it defines a $(0,1)$ tensor (ie a $1$-form). Given a vector field, $X^\mu$, we can define the directional derivative (in the direction of $X$) as $X^\mu\partial_\mu f$ (or $X(f) = \text df (X)$ in index-free notation).

The second question is what's the difference between $\nabla_\mu V^\nu$ and $X^\mu \nabla_\mu V^\nu$ ?

This is the same difference : the first is like a "differential" (in that it has a new covariant index), while the second is like a "directional derivative" (in that this index is contracted with the direction in which we are differentiating).

Now, we wish to do the same for vectors. We want a way to differentiate a vector field $Y^\nu$, giving a $(1,1)$ tensor $\nabla_\mu Y^\nu$, which we could then contract with some $X^\mu$ to get the (covariant) directional derivatives $X^\mu \nabla_\mu Y^\nu$.

In coordinates, we can always use the same expressions from flat space, using the partial derivatives $\partial_\mu$. However, this is not covariant : $\partial_\mu Y^\nu$ is not a tensor.

It turns out that there are always ways to do this but (on a general manifold), no canonical way to do it : we need to add some new structure. A covariant derivative is an operator which does exactly that : it takes a vector field $Y^\nu$ to a $(1,1)$ tensor field $\nabla_\mu Y^\nu$ (and satisfies the Leibniz rule).

In coordinates, specifying a covariant derivative is the same as giving connection coefficients $(2)$ which transform in the right way when changing coordinates.

If the connection coefficients don't describe path dependence and are local, then what do they describe/quantify?

In short, the connection coefficient give us a way to differentiate vector fields.

Then, once we can do this, we can make sense of the sentence "the vector field $Y$ is constant along the path $\gamma$" (ie it is parallel transported). The equation transcribing this statement is $(1)$ (or $(3)$). It depends on the path $\gamma$ and on the covariant derivative we chose.

In general, there are many different connections on a manifold. However, if we have a (pseudo-)Riemaniann metric, there is a canonical way to construct a connection from it, the Levi-Civita connection.