Intuitive understanding of covariant derivative

Question

I've been self-studying Semi-Riemannian Geometry by Newman for a while now and have reached the section on curvature. At this point rather than rush through definitions and theorems, I want to understand the concepts properly. In accordance with SE guidelines, I'll break my doubts into multiple questions.

This question concerns intrinsic differential geometry on smooth manifolds. As a beginner to the subject, my understanding so far is:

1. A connection is a map $\nabla:\mathfrak{X}(M)\times\mathfrak{X}(M)\to\mathfrak{X}(M)$ satisfying a few properties
2. Because of those properties and a previous theorem, given a vector field $X$ and connection $\nabla$, there exists a tensor derivation $\nabla_X$ that can act on all tensors - this is the covariant derivative
3. Because of a couple of other results ($\nabla_X$ is defined pointwise), $\nabla_v$ is well-defined for any given vector $v$ on the manifold
4. Given a smooth curve $\gamma$ and $C^{\infty}(M)$ function $f$, we have the result $$\frac{d\gamma}{dt}(f)=\frac{d(f\circ\gamma)}{dt}$$ which gives me the intuition for $d\gamma/dt$ as the directional derivative operator (for a scalar field) along the curve $\gamma$
5. We can restrict some vector field $X$ to the curve to give a smooth vector field $X\circ\gamma\in\mathfrak{X}_M(\gamma)$. Then we also define the covariant derivative on $\gamma$, which is $\frac{\nabla}{dt}:\mathfrak{X}_M(\gamma)\to\mathfrak{X}_M(\gamma)$. This satisfies $$\frac{\nabla(X\circ\gamma)}{dt}(t_0)=\nabla_{(d\gamma/dt)(t_0)}(X)$$

I'm finding it difficult to get intuition about the covariant derivative. For example the intuition (for what $d\gamma/dt$ is) in point 4 can be read from the equation: we measure $f$ at $\gamma(t_0)$. Then, on making an infinitesimal change to $t$, we progress slightly "further along" the curve $\gamma$ and measure $f$ again. The change in $f$ is basically the $d\gamma/dt$ operator applied to $f$, so the operator is the directional derivative for a scalar field along $\gamma$.

Is there any similar reasoning that follows by looking at the expression for / properties of $\nabla_X$? (or $\nabla_v$ where $v\in T_p(M)$ for some $p\in M$)

Answer 1

One more thing you should add to the list is that you can define for each curve $\gamma:[a,b]\to M$, and each $a\leq t_1\leq t_2\leq b$ a linear mapping $P_{\gamma,t_1,t_2}:T_{\gamma(t_1)}M\to T_{\gamma(t_2)}M$, called the parallel-transport along $\gamma$ from $\gamma(t_1)$ to $\gamma(t_2)$. This is in fact a linear isomorphism, and it is defined by solving a certain ODE. So, now that you have this parallel-transport isomorphism, you can relate this to covariant derivatives pretty easily.

Let $v\in T_pM$ and let $X$ be a vector field on $M$. By virtue of $v$ being a tangent vector, I can find a curve $\gamma:(-\epsilon,\epsilon)\to M$ such that $\gamma(0)=p$ and $\gamma'(0)=v$. Then, you can show (essentially a 1-2 lines calculation by unwinding all the definitions...) that \begin{align} \nabla_vX&=\frac{d}{dt}\bigg|_{t=0}P_{\gamma,0,t}^{-1}(X(\gamma(t))). \end{align} What is this saying? Well, let's just unwind the formula. For each $t\in (-\epsilon,\epsilon)$ I have a vector $X(\gamma(t))\in T_{\gamma(t)}M$. Using the inverse parallel-transport map, I have $P_{\gamma,0,t}^{-1}:T_{\gamma(t)}M\to T_{\gamma(0)}M=T_pM$. Applying this on the vector $X(\gamma(t))$ gives me a resulting vector $\psi(t):=P_{\gamma,0,t}^{-1}(X(\gamma(t)))\in T_pM$; in other words, for each $t\in (-\epsilon,\epsilon)$, I have a vector $\psi(t)\in T_pM$, i.e a curve $\psi:(-\epsilon,\epsilon)\to T_pM$. This is one curve with values in a single vector space, so by basic calculus, I can calculate its derivative $\psi'(0)$. This is precisely what $\nabla_vX$ is.

So, the covariant derivative tells me how much a vector field $X$ changes (to first order... hence the first derivative) when I parallel-transport it back from various points on the curve to a single point $\gamma(0)$. So, if you were to abuse notation and ignore that things live in different vector spaces, you'd be tempted to write that for very small $t$, $X(\gamma(t))\approx X(\gamma(0))+t (\nabla_vX)$ (this is an abuse of notation since the LHS lives in $T_{\gamma(t)}M$ while the RHS lives in $T_pM$).

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I should note that this procedure is more difficult to visualize, because unless the manifold and the connection you provide are very simple, it is pretty hard to visualize first of all the tangent bundle $TM$, and second of all what the parallel-transport map along different curves looks like (note that even if two curves have same endpoints, they may induce different parallel-transport maps (a sign of non-zero Riemann-curvature)). Parallel-transport is obtained by solving an ODE, so this really is more complicated. It is only for simple cases like the Euclidean space, or spheres that we can readily visualize things.

One other special case is if you have a manifold $M$ and an embedded submanifold $S$, and a Riemannian metric on $M$. Then, the Levi-Civita connection on $S$ is the orthogonal projection of the one on $M$. In particular, if $M=\Bbb{R}^n$, then the covariant derivatives are the good old directional derivatives of vector-valued functions, so to find the covariant derivative on a submanifold $S$, you orthogonally project to the tangent spaces of $S$ (this was very quick, but see e.g. Lee's Riemannian geometry text for more information).

Answer 2

The covariant derivative is the closest substitute for the directional derivative in $\mathbb{R}^n$. Think of the directional derivative as

\begin{align*}d:\mathfrak{X}(\mathbb{R}^n)\times C^\infty(\mathbb{R}^n)&\to C^\infty(\mathbb{R}^n)\\ X,f&\mapsto df(X)=Xf, \end{align*} where the last equality is notational. This notation is justified by the convention that when $X$ is the coordinate basis $\frac{\partial}{\partial x^i}$, we recover the calculus notation $df(\frac{\partial}{\partial x^i})=\frac{df}{dx^i}$. Notice we have two inputs:

1. One input takes in a vector field $X$ to take the directional derivative in.
2. The other input and another slot which tells you to take the derivative of which object.

Furthermore, observe that the final object agrees with the object you took a derivative of. If we fix a specific point $p\in \mathbb{R}^n$, the first input only depends on the value of the vector field at $p$. In other words, we say the first input is tensorial. Explicitly, for $f,g\in C^\infty(\mathbb{R}^n)$ and $X,Y\in \mathfrak{X}(\mathbb{R}^n)$, tensorality the following equality of functions, $$(gX+hY)f=g(Xf)+h(Yf).$$ One consequence is if $X=0$, then $Xf=0$. This is wildly false for the second input; you know from calculus that a function having a zero at $p$ does not mean its derivative will vanish at $p$. Instead, as you know, being a critical point depends on the local behavior of $f$. Locality illustrated by the second input obeying the Leibniz rule, $$X(f\cdot g)=Xf\cdot g+f\cdot Xg.$$ Leibniz rule corresponding to locality can be shown by a standard bump function argument that you should attempt yourself if you haven't seen.

The next step is to allow for us to take the derivative of more objects than just functions on $\mathbb{R}^n$. For example, we may want to take directional derivative of vector fields, and the result will be another vector field. Generally, the same should hold true for any tensor field.

To generalize all this for manifolds, we simply replace in the above discussion $\mathbb{R}^n$ with $M$, and $d$ with $\nabla$ to recover the definition of a covariant derivative. The reason for bullet points $4$ and $5$ essentially come down to tensoriality of the direction input. Tensoriality allows us to restrict the input which takes in the direction vector to "any codimension we like". This is why we may take derivatives along curves with $\nabla$.

Finally, contrast this with another notion of a derivative, namely the Lie derivative. In my view, the main reason, that the Lie derivative isn't a suitable replacement for the usual derivative, is exactly because it fails to be tensorial in the direction input. There is an exercise somewhere in Lee's book demonstrating exactly this.