Tensor #30. Transformation of Forms (Part I)텐서 #30. 형식의 변환 법칙 (1부)

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 30 of the playlist, Transformation of forms - formal study (Part I). Taking the 2-dimensional 2-form example from the previous post, this time we extend it to an arbitrary dimension $n$ and degree $p$, and organize how the components of a form change under a change of basis into a formula.

Recap: the single coefficient

On a vector space with $n = 2$, a 2-form written in the tensor product basis is a $(0, 2)$ tensor with four components, and written with the wedge product it has just one coefficient. For an antisymmetric tensor the $00$ and $11$ components are 0 and the $01$ and $10$ components differ only by a sign, so the single wedge basis element $dx^0 \wedge dx^1$ spans a 1-dimensional space. Two elements of this space differ only by a constant factor, so we only need to track one coefficient.

That coefficient changes unlike an ordinary scalar function. An ordinary scalar keeps its value at the same point when the coordinates change. This coefficient changes in value by the determinant of the basis change matrix, and such a thing is called a scalar density. A determinant can have a sign, so under a reflection that flips one axis the sign of the coefficient flips. The goal of this post is to generalize this phenomenon from the single case $p = n$ to every $p$.

How to extract components

When a tensor is fed basis vectors and basis covectors, the components in that basis come out. For a $(0, p)$ tensor $T$ the components are obtained like this.

The pairing of basis covectors and basis vectors is $dx^i(\partial_j) = \delta^i{}_j$, so only one component is left. Applying this convention to a $p$-form $\omega$, the components expressed in the strict basis (the wedge basis written only in increasing order of the numbers) are read like this.

To find the components $\omega_{j'_1 \cdots j'_p}$ in the new coordinates, feed the same form $\omega$ the new basis vectors $\partial'_{j'_1}, \dots, \partial'_{j'_p}$. The form itself is one object and only the vectors we feed it change, so this is exactly the definition of the new components.

The transformation law

The difficulty is that $dx^i$ and $\partial'_j$ belong to different bases, so the delta rule cannot be used. But we already know how to compute. Feeding $p$ vectors to a wedge basis element $dx^{i_1} \wedge \cdots \wedge dx^{i_p}$ gives the determinant of the $p \times p$ matrix collecting the value of each covector acting on each vector. A new basis vector can be written as a sum of old basis vectors, $\partial'_j = \Lambda^i{}_{j'} \partial_i$, so $dx^i(\partial'_j) = \Lambda^i{}_{j'}$. Here $\Lambda^i{}_{j'} = \partial x^i / \partial x'^j$ is the covariant matrix that changes the basis. Therefore we get this.

This formula says that when we feed the new basis vectors to $\omega = \sum_I \omega_I \, dx^I$, each term is multiplied by the determinant that is the value of $dx^I$. The sum runs only over increasing sequences of numbers. In the strict basis each such sequence carries one independent component, and the terms with reordered numbers are the same component with a sign. Since all the matrix entries are entries of the transformation matrix $\Lambda$, the transformation of the components ends up written with determinants of submatrices of $\Lambda$.

Q. Why does the sum run only over increasing sequences instead of all sequences?
A. In the strict basis only the components $\omega_{i_1 \cdots i_p}$ with $i_1 < \cdots < i_p$ are independent. The components in other orders are the same value with a sign change, so adding them separately would count the same term several times.

When p = n it is the determinant

When $p$ equals the dimension $n$, the only increasing sequence is $1 < 2 < \cdots < n$, so the sum has a single term. The matrix is the whole of $\Lambda$. If the new basis vector is written as a sum of old ones as $\partial'_2 = a \, \partial_1 + b \, \partial_2 + \cdots$, then $dx^1(\partial'_2) = a$, so the $(1, 2)$ entry of the matrix is $\Lambda^1{}_{2'}$. So we get this.

Putting in $n = 2$ gives the result of the previous post. A form with $p = n$ is usually called an $n$-form, and by convention the letter $n$ stands for the dimension of the underlying vector space. This result holds in every dimension.

Writing it with minors

If $p < n$, the determinant above is not that of the whole of $\Lambda$ but of the $p \times p$ submatrix built by choosing only $p$ rows and $p$ columns. This is called a minor. A minor is a number, not a matrix, and we mark which part it is by the sequence $I$ of rows and the sequence $J'$ of columns. In these notes we write it like this.

The transformation law becomes shorter.

(This formula, which expands a product as a sum of $p \times p$ minors, has the same shape as what mathematics calls the Cauchy-Binet formula. This name is added for understanding and was not covered in the lecture.) For $p = 1$ a minor is a single entry, which gives the familiar transformation law of a covector, $\omega_{j'} = \Lambda^i{}_{j'} \omega_i$.

In the toy below, change the dimension $n$, the degree $p$, the matrix $\Lambda$, and the components $\omega_I$ of the form. Pick one component $\omega_{J'}$ of the new basis and it unfolds which terms are multiplied by which minors and added. For $p = n$ there is one term multiplied by the determinant, and for $p = n - 1$ you can also check that the same value comes out of the cofactor formula we see below.

Minors as cofactors of the inverse

The determinant lesson gave one rule. A minor of a square matrix $M$ equals the determinant of $M$ times a minor of the inverse $M^{-1}$ chosen with the complementary rows and columns, with a sign attached. The minor with this sign attached is called a cofactor. Letting $I^c, J^c$ be the sets of numbers not chosen in $I, J$, we write it like this.

($\sum I$ is the sum of all the numbers in the sequence $I$. This formula is known as Jacobi’s complementary minor theorem. The name is added for understanding and was not covered in the lecture.) At first sight it looks like it makes the job harder. But the right side contains the determinant, a value common to all minors going to the new coordinates, so it needs to be computed only once. And the closer $p$ is to $n$, the smaller the complementary part $n - p$ becomes, so the cofactor side is easier to compute.

When p = n-1 there are n components

An increasing sequence of length $n - 1$ is fixed by dropping one number $k$ from $1, \dots, n$. So the strict basis has $n$ elements, and we write the basis and the components like this.

The hat marks that the number is omitted. The count $n$ equals $\binom{n}{n-1} = \binom{n}{1}$, the number of components of a 1-form. These $n$ basis elements differ only in the number that goes into the $\hat k$ slot, and arrangements with a different order are the same form up to a sign, so they are not independent.

Now in the transformation law the minor is the $(n-1) \times (n-1)$ part with row $k$ and column $l'$ deleted. By the rule of the previous section the complementary part is a single number, so the cofactor is a single entry of the inverse matrix. The determinant of a single entry is the entry itself.

Putting this in the transformation law, the sum ends with $n$ terms, and there is no need to count over all permutations.

The sign $(-1)^{k+l}$ can be split as $(-1)^{n-k} (-1)^{n-l}$. Multiply both sides by $(-1)^{n-l}$ and define the signed component as a quantity with one upper index.

The components $A^k$ of a vector change as $A^{l'} = (\Lambda^{-1})^{l'}{}_{k} A^k$. $V$ changes by this rule with one extra factor of the determinant $\det \Lambda$. An object with this law is called a (1, 0) tensor density. The next post settles the definition.

The lecture notes that the same sign $(-1)^{n-m}$ can also be put into the basis rather than the components. If we multiply the basis $dx^1 \wedge \cdots \widehat{dx^m} \cdots \wedge dx^n$ by $(-1)^{n-m}$ and take that as the new basis, no separate sign is needed on the component side. Either way the rule is the same.

Q. Why define the sign too and make it look like a vector?
A. $\omega_{\hat k}$ itself is a minor with a sign mixed in and is hard to read. Attaching the sign $(-1)^{n-k}$ tidies the transformation law into the vector rule with one extra determinant, so what this object is can be read at a glance.

Here is an example. In $n = 2$, for a 1-form $\omega = \omega_1 dx^1 + \omega_2 dx^2$ we have $V^1 = -\omega_2, \ V^2 = \omega_1$. This pair, made by rotating the components by 90 degrees, changes like a vector multiplied by $\det \Lambda$. If $\Lambda$ is the diagonal matrix $\mathrm{diag}(a, b)$, then $\omega_{1'} = a \, \omega_1$ and $\omega_{2'} = b \, \omega_2$, and the rule above gives $V^{1'} = ab \cdot a^{-1} (-\omega_2) = -b \, \omega_2$, which agrees with the direct computation. (This example is added for understanding and was not covered in the lecture.)

General p and the next post

For general $p$ much smaller than $n$ we can still turn the minor into a cofactor of the inverse. But then the complementary part $n - p$ is large, a minor of the inverse remains, and the result does not take the clean form of a tensor density as above. The lecture ends by announcing that it will treat this general case next time together with $n - p$ forms.

In the next post we organize the transformation rules found here under the name of weight and define tensor densities formally.