Tensor #18. The Covariant Derivative, Continued텐서 #18. 공변미분 이어서

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 18 of the playlist, The covariant derivative continued. In the previous post we obtained the component formula of the covariant derivative from parallel transport, so this time I look at the same object in two more abstract ways. The lecture says this part is optional, but it gives the template for the other derivatives that come later.

Recap of the previous post

The covariant derivative of a vector field $X^\alpha$ is the partial derivative plus a connection term.

For the components of a covector the connection term gets the opposite sign. For a general tensor, each upper index gets this rule and each lower index gets the opposite one, while the partial-derivative term appears only once for the whole tensor. The connection coefficients $\Gamma^\alpha{}_{\beta\mu}$ are functions on spacetime, and they fix the standard of parallel transport that compares the vector spaces at different points. Without that standard we cannot speak of a rate of change between two spaces. The lecture summarizes the logical flow like this. The stress-energy tensor goes through the Einstein equation to fix the metric tensor $g_{\mu\nu}$, and once there is a metric, a natural connection follows from it.

The tensor the covariant derivative makes

$\nabla_\mu X^\alpha$ is the component of a (1,1) tensor with one upper and one lower index. The same thing is also written with a semicolon, $X^\alpha{}_{;\mu}$. With the basis written out it reads as follows.

So the covariant derivative takes a (1,0) tensor and returns a (1,1) tensor. In the formula above the partial-derivative term and the connection term are not tensors separately, but their sum is. The reason the covariant derivative exists is to make this sum transform like a tensor when the coordinates change.

Q. Neither term is a tensor, so how can the sum be one?
A. Under a change of coordinates, the amount by which the partial-derivative term departs from the tensor law is matched by an exactly opposite departure of the connection coefficients, and they cancel in the sum. (This explanation states the standard result. The lecture only says the sum is a tensor, and this was not covered in the lecture.)

A machine with three slots

Misner, Thorne, and Wheeler’s Gravitation (page 254) defines the covariant derivative abstractly from the start. At a point, the covariant derivative is a machine with three slots. A tensor is also a machine with slots, but this machine is not a tensor.

• The first slot takes a covector $\theta$ at the point $P$.
• The second slot takes a vector field $X$. It is a field, not a vector at one point.
• The third slot takes a vector $u$ at the point $P$.

With all three slots filled, one real number comes out. How this number is made is almost forced. A covector eats a vector and gives a real number, so we must first make a vector from the vector field $X$ and the vector $u$. Calling that vector $\nabla_u X$, we have the following.

Q. Is the subscript $\mu$ in $\nabla_\mu X^\alpha$ the same as the subscript $u$ in $\nabla_u X$?
A. They are different symbols. $\mu$ is a spacetime index and $u$ is a whole vector. The two are linked by $u = u^\mu \partial_\mu$. The lecture also says the two symbols are often confused in books, and many books tell them apart with bold type.

Reading it in components

In a coordinate basis, writing $\nabla_u X$ in components is the same as letting the (1,1) tensor of the previous post eat the vector $u$.

The index $\gamma$ is contracted with $u$ and disappears, leaving only the basis $\partial_\alpha$, so the result is a vector. The slots of this machine may also be filled only in part.

• Filling all three gives the real number $\nabla(\theta, X, u)$.
• Leaving only the first slot empty gives the vector $\nabla_u X$, the covariant derivative of $X$ in the direction $u$.
• Putting in only the vector field $X$ and leaving the rest empty gives the (1,1) tensor $\nabla X$. We got it in the previous post, and it is also called the gradient of $X$.

The toy below does this computation at one point in polar coordinates. Pick a vector field and a direction $u$, and the partial-derivative term, the connection term, and the sum appear separately. The constant vector field $\partial_x$ of the plane has polar components that change, yet the sum is 0. Switch the connection to $\Gamma = 0$ and this disappears.

Why this machine is not a tensor

For a tensor, a function multiplying a slot must come out to the front unchanged. This holds for the first and third slots. The trouble is the second slot. If we put in the vector field $fX$ multiplied by a function $f$, the product rule produces one extra term.

$u f$ is the value of the vector $u$ acting on the function $f$, that is, the directional derivative of $f$. Because of the leading $(u f) X$, the machine is not linear in the second slot. The difference line at the bottom of the toy above is exactly this term. There is one more reason. The slots of a tensor take vectors at a single point, whereas the second slot of this machine takes a vector field. For these two reasons the covariant derivative machine is not a tensor.

Extracting the connection coefficients

Feed the machine basis vectors one by one and the connection coefficients come out as numbers.

The $e_\beta$ in the middle slot is a basis vector field, with one basis vector placed at every point. So these numbers are functions on spacetime, and they are not tensors either. Reading this formula as a vector shows the meaning of the connection coefficients most clearly.

It tells us, as a vector, how much the basis vector $e_\beta$ changes when we move a little in the direction $\gamma$ from the point $P$. The amount by which a vector changes is itself a vector. In a coordinate basis, $e \to \partial$.

Q. If the connection coefficients are not a tensor, what is different?
A. Under a change of coordinates their values change without following the tensor law. In the plane, the connection coefficients of Cartesian coordinates are all 0, but in polar coordinates there are nonzero values such as $\Gamma^r{}_{\theta\theta} = -r$. (This example uses the values in the toy above and was not covered in the lecture.)

Defining it by axioms

The last viewpoint is from Chandrasekhar’s The Mathematical Theory of Black Holes. It starts abstractly, but not as far out as Misner, Thorne, and Wheeler. Here a connection is an object that takes a vector field $X$ and returns “a map sending a vector field $Y$ to the vector field $\nabla_X Y$”. It must obey four rules.

1. It is function-linear in $X$. $\nabla_{fX + gZ} Y = f\, \nabla_X Y + g\, \nabla_Z Y$
2. It preserves addition in $Y$. $\nabla_X (Y + W) = \nabla_X Y + \nabla_X W$
3. Acting on a function, it gives the value of the vector field acting as a differential operator. $\nabla_X f = X f$
4. It obeys the product rule. $\nabla_X (f Y) = (X f)\, Y + f\, \nabla_X Y$

Something that satisfies these is called a derivation, and the lecture points out that there are three ways in mathematics to solve these rules: the covariant derivative, the Lie derivative, and the exterior derivative. Now take dual bases $e_i, e^j$ and expand $Y = Y^j e_j$ with the product rule.

In the second term, $\nabla_X e_j$ is a vector, so it has to be expanded in the basis, and since it is linear in $X$, the components of $X = X^k e_k$ come out in front. Writing the leftover coefficients as $\omega^l{}_{jk}$ we get the following.

In a coordinate basis, rewrite $\omega$ as $\Gamma$ and the formula of the previous post comes out exactly. The axioms only fix the structure, and the coefficients $\Gamma$ must be supplied separately. That is the job of choosing the connection of spacetime.

Three kinds of derivatives

The lecture steps back here to compare. There are three derivatives that satisfy the rules above, and none of them comes for free. Each needs something more to be assumed.

• The covariant derivative needs the connection coefficients added. Without a notion of parallelism we cannot define a derivative.
• The Lie derivative needs one preferred vector field.
• The exterior derivative applies only to antisymmetric objects.

In the next post we step away from derivatives for a moment and review algebraic structures.