Tensor #14. Vector and Tensor Fields텐서 #14. 벡터장과 텐서장

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 14 of the playlist, Vector and Tensor Fields. In the previous post we put one vector space at each point of spacetime, so this time I collect vector fields and tensor fields, the way to handle all those infinitely many spaces at once.

One space at each point

Every point of spacetime is named by four coordinates $x^\mu$, and each point carries a four-dimensional real vector space $\V_P$. Once there is a vector space, the dual space and countably many tensor product spaces follow, and all of them exist at every point. A structure that only names the points is like describing a building by its color, and adding the vector space at each point is what makes it something we can call architecture.

The problem is the amount. There are uncountably many points, and we cannot fix a basis space by space the way we did so far. The basis has to be fixed for all the spaces at once.

The coordinate basis

With a coordinate system we can think of functions on spacetime and of their partial derivatives. So we fix the basis of the vector space at a point P to be the partial-derivative operators taken at P.

The basis at a point Q is the same operators taken at Q. A function does not really have to exist. We are only making operators in advance to use when a function comes along someday. With this choice, a basis of four operators is placed at all points in one stroke, and it is called the coordinate basis. The lecture says it will become clear later why differential operators in particular.

The basis of the dual space

The dual space is the collection of maps that take a vector and return a real number, so we need an object that takes a differential operator and returns a real number. That is the differential $dx^\mu$, and the rule for bases is the same as before.

Then the bases of the tensor product spaces built from the vector space and the dual space at one point follow as well. For example, here is one such basis element.

Q. Is there a reason to choose $dx^\mu$ as the dual basis?
A. Apply the operator $L = a^\mu\partial_\mu$ to the coordinate function $x^\nu$ and we get $L x^\nu = a^\nu$, so $dx^\nu$ is the map that takes a component out of an operator. A map that takes a component out of a vector is exactly the definition of a dual basis, so it fits. (This explanation connects the computation of the previous post and was not covered in the lecture.)

For a realization to do its job, relations that held for the abstract basis, such as $\map{dx^\mu}{\partial_\nu} = \delta^\mu{}_\nu$, must keep holding. If only the names change, $e_\mu \to \partial_\mu$ and $e^\mu \to dx^\mu$, while the structure stays the same, it is a true realization.

Vector fields

A vector at one point P is $A^\mu \partial_\mu$, and the components $A^\mu$ are four real numbers. Now make the components functions of the coordinates, $A^\mu(x)$, and the story changes.

If the coordinates of the point P are $x_P$, then $A^\mu(x_P)\, \partial_\mu$ is a vector of $\V_P$, and putting in the coordinates of a point Q gives a vector of $\V_Q$. This single line of formula picks one vector from the space at every point. What we get by substituting values is components, and we read them together with the basis of that point. An object that fixes one vector at each point this way is called a vector field. The values come from the components being functions of the coordinates, and the basis can no longer be an abstract symbol, so it must be the differential operators.

In the toy below, pick the component functions and drag the points P and Q. The same formula gives a different vector at each point, and vectors at different points live in different spaces, so they cannot be added. With “Only point P” it is one ordinary vector, and with “Every point” it is a vector field.

Q. Is the $x$ in a vector field a component of the vector?
A. It is not a component. It is a coordinate name that tells which point we are at. Even with an upper index it is not an element of a vector space, and the words covariant and contravariant do not apply to it. We sometimes write it as plain $x$ with no index, which can actually reduce confusion. In $A^\mu(x)\partial_\mu$ the summed $\mu$ is a dummy index, and $x$ is a free variable that decides which point we are looking at.

Tensor fields

Tensors are made the same way. Say that at any point we have a tensor like this.

We do not need to mark the space point by point, because we understand that the same thing is done at every point. If we learn that the coordinates of a point Q are $y^\mu$, we put that value into the component functions and attach the basis at Q to get the tensor at Q. Once we go from one point to all points, the small marks that tell the point can be dropped and we can write just $\V$. We know that we are speaking about every point.

Most general relativity textbooks go one step further, discard the basis completely, and write only the components.

When we see this one line, we have to read two things. It carries indices and stands for a single real number, but it is really a function of the coordinates, so this is a field, and there is an unseen basis that fixes its nature as a map. There are two reasons we can compute without looking at the basis. That it is a field we know because we understand the architecture of spacetime, and the nature of the map is kept by the index gymnastics of the earlier posts. Once we have learned the gymnastics, we may skip the basis.

Q. Do the components $T^{\alpha\beta}{}_{\gamma\epsilon}$ hold all the information of the field?
A. The component functions are the part that fixes a real number at each point, and the basis is the part that fixes what map those numbers form. With only one of the two, the tensor field is not complete. Textbooks leave out the basis because the positions of the indices remember that part for us.

From now on, when we speak of a vector space we are really speaking of a vector field, whose components are real functions on spacetime instead of real numbers. The dual space is likewise a field made of the $dx^\mu$, and everything is still a map. In the next post we add one more piece of structure on top of this.