# Intuition on the covariant derivative

I'm having some trouble understanding the covariant derivative as a directional derivative for tensors. The way the covariant derivative was presented to me was by first showing that a vector field can provide a directional derivative for smooth functions on a manifold.

We then asked how we could get the directional derivative for higher rank tensors. We started by listing a number of qualities we wanted our new derivative operator to have. Since we want to have a directional derivative we need a vector to provide a direction and a tensor to derive, so the covariant derivative would be a map from a vector x tensor of type $(s,k)$ to another tensor of type $(s,k)$.

Since this was a derivative we wanted linearity, and we also wanted it to obey the Leibniz rule. We then talked how based on these rules we had some freedom in how we picked the derivative, which we later showed can be fix based on your choice of the connection coefficients. Finally, we applied the covariant derivative to a vector in some particular chart, and got the component expression on how to take the covariant derivative of a vector.

I could follow all of the algebraic manipulations up to this point, however none of this jives with my understanding of a derivative, which is something that measures how something else changes.

We've additionally talked about parallelity and the autoparallel equation $\nabla(V,V) = 0$ (where $\nabla$ is the covariant derivative and $V$ is a vector) but none of this has really helped me understand the covariant derivative as a directional derivative. Could anyone help me out? Thanks.

• Would Mathematics be a better home for this question? – Qmechanic Dec 2 '16 at 23:26
• @Qmechanic Maybe, wasn't sure. I'm learning this from the perspective of general relativity, and it seems foundational to relativity, so I thought it would be ok here. If I should move it let me know. – Dargscisyhp Dec 2 '16 at 23:31
• Two questions that I hope you don't laugh at, or take offence at. Do you understand the need for a covarient derivative in GR and do you follow the idea of vectors and derivates being equated in describing the world line . I ask this based on your comment i could follow all of the algebraic manipulations up to this point, however none of this jives with my understanding of a derivative, which is something that measures how something else changes. Does GR enter into your background, even as an example or is it purely math, as shown by your tags. – user108787 Dec 3 '16 at 0:41
• If you're still reading this @Dargscisyhp I would offer to give you a picture of $\nabla_\alpha$ as an extension of a trivial gradient-of-scalar-fields expression, via Leibniz's laws, to vector expressions. It is not too hard to see how there is a rank-3 tensor of freedom in how to choose what derivative is $\nabla$ but I'm not sure I can give you a firm physical intuition for what's going on other than "we have a lot of freedom in understanding parallel transport if we're willing to have nonvanishing torsion etc." – CR Drost Dec 3 '16 at 6:56
• Hi, @CountTo10. At this stage the course has been almost explicitly mathematical. We've defined topology, topological manifolds and differentiable manifolds, reviewed some basic linear algebra, talked about tangent spaces and vectors, tangent bundles and vector fields, and finally connections and parallel transport. The only foray we've made into physics thus far is talking about geometrized Newtonian gravity. So to answer your question, no, I don't really understand why a covariant derivative is necessary in GR. – Dargscisyhp Dec 3 '16 at 12:58

I'm not sure what exactly you're asking for, but I'll try explaining how the directional derivative operator is acting on a smooth function (at least the way I grasp this notion), which is the first thing one encounters, if one is looking at tangent vectors in the scope of topology and diff mfds.

Let $M$ be a smooth manifold. Then, we can construct a vector space $(C^\infty(M),+,\cdot)$, where $C^\infty(M)$ is the set of all infinitely many times differentiable (i.e smooth) maps $f:M\to\mathbb R$, $+$ is the pointwise addition and $\cdot$ is the pointwise s-multiplication. Now let $\gamma:\mathbb R\to M$ be a smooth curve through the point $p\in M$. W.l.o.g we can consider $p=\gamma(0)$ and we define the directional derivative operator at the point $p$ along the curve $\gamma$ as the linear map: $$X_{\gamma,p}:C^\infty(M)\tilde{\to}\mathbb R$$ $$:f\mapsto(f\circ\gamma)'(0)$$ where $f\circ\gamma:\mathbb R\to\mathbb R$ and $(f\circ\gamma)'(0)\in\mathbb R$.

We further define the set $$T_pM:\{X_{\gamma_a,p}\;\mid\;\text{all curves passing through}\; p\in M\}$$ and equip it with the closed operations $\oplus,\odot$: $$\oplus:T_pM\times T_pM\to T_pM$$ $$\odot:\mathbb R\times T_pM\to T_pM$$ which stand for vector addition and s-multiplication on $T_pM$. There is no need for further definition of the target element of these two maps for now. This set, together with these two operations, constitute the tangent vector space, i.e the space whose elements are the directional derivative operators.

As $X\in T_pM$ are operators (in the above frame), we can understand their behaviour by acting them on a smooth function $f:M\to\mathbb R$. So let us choose a chart $(U,x)$, which is an element of the smooth atlas $\mathscr A_M$ on $M$. Let us also consider $\dim(M)$-many curves $\gamma_i$, s.t $$\gamma_i:\mathrm{preim}_{\gamma_i}(U)\to U$$ with $(x^b\circ\gamma_i)(\lambda)=\delta^b_a\lambda$ for $i=1,\cdots,\dim(M)$. Obviously, $U$ is an open subset of $M$, $\lambda$ is the curve parameter, $x^b$ is the b-th component of $x$ and is smooth, because $x:U\to x(U)\subseteq\mathbb R^d$ is smooth due to the atlas being smooth. The index $i$ is just the curve index. For convenience we choose a chart, s.t $x(p)=0_{\mathbb R^d}$. As you see, we did very specific choices. We now act on smooth f with our operator: \begin{align*} X_{\gamma_i,p}(f):&=(f\circ\gamma_i)'(0)\\ &=(f\circ x^{-1}\circ x\circ \gamma_i)'(0)\\ &=\partial_b(f\circ x^{-1})(x(\gamma_i(0)))(x^b\circ\gamma_i)'(0)\\ &=\partial_b(f\circ x^{-1})(x(p))\delta^b_a\\ &=\partial_a(f\circ x^{-1})(x(p))=:\left(\dfrac{\partial}{\partial x^a}\right)_p(f) \end{align*} which means that $$\left(\dfrac{\partial}{\partial x^a}\right)_p:=X_{\gamma_i,p}$$ In order to go from the second equation to the third, we used the chain rule (funny looking that way), where $(\partial_a)_{x(p)}:C^\infty(\mathbb R^d)\to\mathbb R$. The key fact here is that in order to define the tangent vector we need a chart,i.e we need coordinates, so that the partial derivative has meaning as derivative wrt to a component of the coordinate map. On a manifold, we cannot directly have such thing. One also proves that this definition satisfies the Leibnitz rule and that $$\left(\dfrac{\partial}{\partial x^i}\right)_p\qquad,i=1,\cdots,\dim(M)$$ are a linearly independent generating system of the tangent vector space $T_pM$, a.k.a a basis. Obviously, acting this thing on vector fields requires understanding of tangent bundles $TM$, vector fields (as sections of TM),modules,rings etc.

I hope I helped and did not mess things up. Truth being said, I think it's better to stick to the definitions and try not making ill-defined schematics on our mind. It's really helpful, if you draw all these maps and see the compositions yourself.

• Thank you for your answer. You developed the concept of the tangent vector almost directly along the lines of the lectures I've been watching. Based on your definition of the tangent vector it is intuitively clear to me that vectors are indeed directional derivatives, namely because of the line $X_{\gamma_i,p}(f):=(f\circ\gamma_i)'(0)$. This line is easy to interpret as a directional derivative based on my knowledge of ordinary calculus. Unforunately, as far as I know, an analogue to this does not exist for the covariant derivative. If it did, this might be easier for me to comprehend. – Dargscisyhp Dec 3 '16 at 18:10
• the action of the covariant derivative on a smooth function is exactly the same. We have been probably watching the same lectures :) – kospall Dec 3 '16 at 19:17

# An all-to-brief review

Let's cover the basics: you take a set of points $\mathcal M$ and outfit it with a set of scalar fields $\mathcal S \subseteq (\mathcal M \to \mathbb R)$ which obey a closure axiom under the smooth functions $C^\infty(\mathbb R^n, \mathbb R),$ reinterpreted pointwise as functions $\mathcal S^n \to \mathcal S$ via a lifting operator $\mathcal L$ where $$\mathcal L_n f(s_1,\dots s_n) = p\mapsto f\big(s_1(p),\dots s_n(p)\big).$$This gives us constant fields $\mathcal L_0 1$ and pointwise $(+)$ and $(*)$ as $\mathcal L_2 (+),\mathcal L_2 (*),$ on those fields: and you say that you've seen that this induces a topology where a subset of $\mathcal M$ is called "closed" if it is the kernel of a scalar field, or "open" if its complement is closed; on this topology all of the fields $\mathcal S$ are in fact continuous maps. Then there's the "coordinate map" axiom which says that the space is $D$-dimensional: for every point $p\in\mathcal M$ there is an open set containing $p$ where all scalar fields can be expressed in terms of a set of coordinate fields $c_1, c_2, \dots c_D$ as smooth functions, $$s = \mathcal L_D \sigma(c_1, \dots c_D).$$

Now you have also come to the fundamental smart thing about vector fields, that $\mathcal V$ is the space of derivations, linear maps $\mathcal S \to \mathcal S$ obeying the generalized Leibniz law that if $V \in \mathcal V$ then $$V\big(\mathcal L_n f(s_1,\dots s_n)\big) = \sum_{k=0}^n \mathcal L_n f_{(i)}(s_1, \dots s_n) * V s_i,$$ where $f_{(i)}$ is the partial derivative of $f \in C^\infty(\mathbb R^n, \mathbb R)$ with respect to its $i$th argument. And then the covector fields $\mathcal V^\dagger$ are linear maps $\mathcal V \to \mathcal S.$ While at first this definition of "vector fields" seems like a strange definition, it retroactively makes sense because looking at that equation and the one which immediately precedes it, we see that locally every vector field has a set of $D$ scalar-field components $v_i = V c_i$ just like you'd expect, leaving $V = \sum_i v_i \partial_i.$

And now we can define valence-$[m, n]$ tensor fields as multilinear maps from $\big(\mathcal V^\dagger\big)^m\times\big(\mathcal V\big)^n \to \mathcal S,$ and with a bunch more axioms we get an abstract-index notation complete with outer products and contractions. This also gives us directly the Lie bracket $[U, V] = U \circ V - V \circ U,$ and the gradient covector for any scalar field, $\nabla_\bullet s = V \mapsto V s.$ We don't need a connection to have these more-basic aspects.

# OK, now why are vectors different?

You can interpret $\sum_i v_i \partial_i$ as the term from the transport equation, $\partial_t \rho + \vec v \cdot \nabla \rho = -\nabla \cdot \vec J + \Omega,$ describing essentially "an infinitesimal box flows downstream, the rate of change of some conserved stuff in the box is equal to the divergence of flows through the nearby boxes, plus any creation or annihilation of stuff (or flow into the system of interest from some other place). In particular this idea of "a box flows downstream" means that we're trying to analyze $\big[\rho(\vec r + d\vec r, t + dt) - \rho(\vec r, t)\big] / dt$ which of course is just the left-hand side with $d\vec r/dt = \vec v.$

Well, scalars and vectors-in-flat-space can be transported without their identity becoming ambiguous, but vectors-in-curved-space can't: imagine that you are pointing East as you walk to the North Pole, you are now pointing South (as you must, at the North Pole), and the actual longitude that you're pointing depends on how you walked there (with the straight shot, you're pointing at something 90 degrees to the East; but if you first walk sideways West you can point to increasingly Westward longitudes.

In fact we can directly anticipate that your solution will only be unique up to a $[1,2]$-valence tensor, because in practice you're going to form $w^\alpha = u^\beta \nabla_\beta v^\alpha$ and so you're taking these two vector fields $u^\beta$ and $v^\alpha$ and forming a new one $w^\alpha$, and mapping two vectors to a vector is the same as mapping two vectors and a covector to a scalar, which is a $[1,2]$-tensor field.

We can derive it quite directly; note that if you have two connections $\nabla_\alpha$ and $\hat \nabla_\alpha$ that agree on how to treat scalar fields (our universal gradient above) then the operator $\delta_\alpha = \nabla_\alpha - \hat \nabla_\alpha$ must satisfy $\delta_\alpha s = 0$ for all scalar fields $s$, hence $\delta_\alpha (s v^\beta) = s \delta_\alpha v^\beta.$ And that's a multilinear map from $\mathcal T^\beta \to \mathcal T_\alpha^\beta$ which makes it precisely a [1, 2]-tensor. In some sense this freedom means that you can use the existing connection but allow a tensor to make slight tweaks to any transporting vectors, arbitrarily. And then we've got these ideas like "torsion" which allow us hopefully to push past that.