Tensor #29. Transformations of Tensors and p-forms텐서 #29. 텐서와 p-형식의 변환

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 29 of the playlist, Transformations of tensors and p-forms (part review). The previous post covered the formal definition of the wedge product and minors, so this time we look at how the components of tensors and p-forms change under a change of coordinates. The lecture explains it with p-forms, but the analysis for p-vectors is the same. As the title says review, the first part is review, and what we newly gain is that components in the strict basis change by a different rule.

The rule depends on the basis you write in

A p-form is a $(0, p)$ tensor, so we already know its transformation rule. The tensor products of dual basis vectors $dx^{\mu} \otimes dx^{\nu} \otimes \cdots$ form a basis, this basis changes with the inverse matrix, and the components on it change with the covariant matrix. As long as we use all $n^p$ basis elements, it is no different from a general tensor. But a p-form is totally antisymmetric, so its components are tied to each other, and in fact it lives in a subspace of dimension $\binom{n}{p}$. As the basis of that space we use the strict basis $dx^{\mu} \wedge dx^{\nu} \wedge \cdots$ ($\mu < \nu < \cdots$). Components written in this basis do not follow the rule above. The goal of this post is to write the same tensor in the two bases and see how the two rules fit together.

A change of coordinates produces two matrices

At each point of spacetime there is a tangent space $V$ and its dual. The coordinate basis is $\partial_i$ and the dual basis is $dx^i$. If the new (primed) coordinates are given as smooth functions of the old ones and the reverse also holds, partial derivatives give two matrices.

The one that moves the basis vectors is called the covariant matrix, and here I take $\Lambda$ to be that.

Components change in the opposite direction from the basis. For a $(0, 2)$ tensor $T = T_{ij} \, dx^i \otimes dx^j$, the lower-index components follow the basis and change with $\Lambda$, and upper-index components change with $\Lambda^{-1}$.

Putting basis and components together, $\Lambda$ and $\Lambda^{-1}$ multiply to the identity, so the tensor itself is unchanged. Physics books take this rule as the definition of a tensor, and call something that does not obey this formula not a tensor. One such matrix is needed in each tangent space, one per point, so $\Lambda$ itself is a function of the coordinates.

Q. Is it fixed which of the two matrices is called the covariant matrix?
A. It is a convention. The matrix that moves the basis vectors is called the covariant matrix, and one picks $\partial x/\partial x'$ or $\partial x'/\partial x$ depending on which direction one wants to change. The lecture finds the direction from old coordinates to new ones natural and takes $\partial x/\partial x'$ as the covariant matrix.

An example: polar coordinates in two dimensions

In a two-dimensional spacetime, take old coordinates $(x^0, x^1)$ and new coordinates $(r, \theta)$. The lecture writes $x'^0, x'^1$ for the new coordinates too, but here I write $r, \theta$ for readability.

The covariant matrix comes from differentiating $x^i$ with respect to $r, \theta$. The row is $i$ and the column is $i'$.

There is a reason to rewrite it in the old coordinates as on the right. The components of the tensor to be transformed are given as functions of the old coordinates, so $\Lambda$ must be known in the old coordinates at the same point before we can multiply. The lecture differentiates $\arctan$ and tidies up with trigonometry, and in this form all the transcendental functions disappear. The inverse is the opposite matrix, from differentiating $r, \theta$ with respect to $x^i$.

Moving a (0, 2) tensor by the rule

Pick a tensor to move.

Since it is a $(0, 2)$ tensor, we use $Q_{i'j'} = \Lambda^i{}_{i'} \Lambda^j{}_{j'} Q_{ij}$. Only two terms survive, $Q_{01}$ and $Q_{10}$.

In the same way $Q_{1'0'} = -r$, and the diagonal components vanish because the two terms cancel, as in $Q_{0'0'} = \Lambda^0{}_{0'} \Lambda^1{}_{0'} (Q_{01} + Q_{10}) = 0$. The result is this.

Q. Is a component that was 0 in the old coordinates also 0 in the new ones?
A. No. A general tensor gets nonzero components in the new coordinates. Here the diagonal components stay 0 only because $Q_{01} = -Q_{10}$ makes the two terms cancel exactly. Components are fickle numbers that can change a lot when the coordinates change.

In the strict basis there is only one component

The $Q$ above is also a wedge product. By the definition, $dx^0 \wedge dx^1 = dx^0 \otimes dx^1 - dx^1 \otimes dx^0$, so $Q = dx^0 \wedge dx^1$. In two dimensions the space of 2-forms has dimension $\binom{2}{2} = 1$, so the only strict basis element is $dx^0 \wedge dx^1$ and every 2-form is a constant multiple $a \, dx^0 \wedge dx^1$. The component of $Q$ is $a = 1$. Reading the earlier result in the same shape, in the new basis $Q = r \, dr \wedge d\theta$, that is $a' = r$.

In the tensor-product basis there are four components (only $Q_{01}, Q_{10}$ are nonzero), and the rule multiplies $\Lambda$ twice. In the strict basis there is a single component and it changes like this.

It is the same tensor, but the rule for the components depends on the basis. A component written in the strict basis does not follow the tensor rule above, so in the lecture’s words it is not a tensor but something else; it is usually called a tensor density, though the lecture says it could be another name. The naming is postponed to later posts. Three dimensions work the same way: every 3-form has the shape $c \, dx^0 \wedge dx^1 \wedge dx^2$, so the constant $c$ is multiplied by $\det \Lambda$.

In the toy below, dragging the point $P$ shows $\Lambda$ and $\det \Lambda$ at that point, and changing the four components of $T$ computes $T_{i'j'}$ by the rule. The single component of the antisymmetric part, $a = (T_{01} - T_{10})/2$, is always multiplied by $\det \Lambda$, while the symmetric part does not follow this rule.

Q. Is the multiplier $\det \Lambda$ for other coordinate changes than polar ones?
A. For a form with $p = n$ it is the same for any smooth coordinate change. The multiplier is the determinant of $\Lambda = \partial x/\partial x'$, that is, the Jacobian determinant. The lecture says that doing the same in three dimensions gives the same rule.

Why a determinant is multiplied

The lecture does not explain this separately, but it can be read from a picture. The parallelogram formed by the new basis $\partial_r, \partial_\theta$ has area $\det \Lambda = r$, and the square formed by the old basis $\partial_0, \partial_1$ has area 1. The 2-form $dx^0 \wedge dx^1$ is a device that measures the oriented area formed by two vectors. With the pairing rule of the previous post, the value on the two new basis vectors is exactly $\det \Lambda$.

$Q(\partial_r, \partial_\theta)$ is also the new-basis component $Q_{r\theta}$, so the component is multiplied by $r$ exactly as the area is. (This explanation applies the pairing rule of the previous post and was not covered in the lecture.)

This rule holds only when p equals the dimension

This is the limit the lecture stresses at the end. The factor $\det \Lambda$ seen so far holds only when $p = n$, so that there is a single strict basis element. For example, with $p = n - 1$ there are $\binom{n}{n-1} = n$ components, which follow a different rule. How the strict-basis components change for general $p$ is the subject of the next post. From the minor formula of the previous post one can guess that the $p \times p$ minors of $\Lambda$ will appear. For $p = n$, with a single element, the minor is all of $\Lambda$, which shows up as $\det \Lambda$. (This guess follows from the previous post’s formula and was not covered in the lecture.) In the next post we look at this general rule.