Tensor #22. Invitation to p-forms텐서 #22. p-형식으로의 초대

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 22 of the playlist, Invitation to p-forms. Having seen two kinds of derivatives in the previous post, from this post on we enter the third, the exterior derivative, and the algebra underneath it. The first gate is the object the exterior derivative acts on, the p-form.

0-forms and df

A 0-form is a single function $f(x^0, x^1, x^2, x^3)$ on spacetime. Its properties are as in earlier posts, and we assume it is smooth enough that all partial derivatives exist. A 0-form has no structure worth showing, but we need it as the starting point to introduce the exterior derivative $d$. The exterior derivative of a 0-form has nothing to prove and is simply defined.

Here $dx^i$ is the coordinate basis of $\V^*$, that is, the basis of covectors. So $df$ is a covector, and given a vector $v = v^j \partial_j$ it produces a value in the same way as always.

This value is the rate at which $f$ changes in the direction of $v$. There are several notations, and $\partial_i f$ and $f_{,i}$ all mean the same thing. Putting the coordinate function $x^\mu$ in place of $f$ in the same formula gives $dx^\mu$, so the basis covectors also fit in this framework.

A 2-form is an antisymmetric (0, 2) tensor

A 0-form is a $(0, 0)$ tensor, a 1-form is a $(0, 1)$ tensor, and in the same way a p-form is a $(0, p)$ tensor. There is no new machine to build, only one more constraint on tensors we already know. First write a 2-form $\omega$ in the form of a $(0, 2)$ tensor.

Feeding it the vectors $v = v^\delta \partial_\delta$ and $w = w^\lambda \partial_\lambda$ turns the double sum into one real number. Pulling the components out by linearity and using $dx^\mu(\partial_\delta) = \delta^\mu{}_\delta$ gives the following.

The constraint added here is antisymmetry. Swapping the order of the vector arguments must flip the sign of the value.

Translating this into a condition on components gives $\omega_{\delta\lambda} v^\delta w^\lambda = -\omega_{\lambda\delta} v^\delta w^\lambda$, and since it must hold for arbitrary $v, w$ we get the following.

This is where general relativity textbooks start when they define an antisymmetric tensor as one whose sign flips when the indices are swapped. From this condition the diagonal components $\omega_{00}, \omega_{11}, \omega_{22}, \omega_{33}$ are the negatives of themselves, so they all vanish.

From 16 components to 6

With the diagonal gone, the remaining 12 components form pairs that differ only in sign. So the most general 2-form is fixed by six numbers $\omega_{01}, \omega_{02}, \omega_{03}, \omega_{12}, \omega_{13}, \omega_{23}$. For each pair the basis is bundled like this.

An ordinary $(0, 2)$ tensor is fixed by $4 \times 4 = 16$ numbers, so the single constraint of antisymmetry cut the count down to 6. In the toy below, pick an antisymmetric 2-form, a symmetric tensor, or a general tensor, and drag the vectors $v, w$. For the antisymmetric one, swapping $v$ and $w$ only flips the sign of the value, and feeding the same vector twice gives 0. (The part showing the signed area of the parallelogram is an interpretation added so you can see the antisymmetric product of two vectors, and was not covered in the lecture.)

Q. Why does an antisymmetric form give 0 when we feed it the same vector twice?
A. Because $\omega(v, v) = -\omega(v, v)$, and the only number equal to its own negative is 0. It is the same reason all the diagonal terms vanish in the components.

The wedge product

Writing out the bundles above every time is tedious, so we define a new operation between 1-forms. It is the wedge product.

With this definition the earlier expression can be written by collecting only the pairs with $\mu < \nu$.

The left sum has 6 terms. On the right, instead of restricting, we let $\mu, \nu$ run over all values, so the pairs $(\mu, \nu)$ and $(\nu, \mu)$ count the same term twice, and the $\tfrac12$ corrects for that. The shape of this definition, which looks like a plain substitution, hints at a new vector space with the wedge products $dx^\mu \wedge dx^\nu$ as its basis. Some books define the wedge product with a factor of $\tfrac12$, in which case the $\tfrac12$ above disappears, so take care when switching books.

Properties of the wedge product

The wedge product is not defined only on the basis. For two 1-forms $\alpha = a_\mu dx^\mu$ and $\beta = b_\nu dx^\nu$ we use the same definition, and the linearity of the tensor product gives the following.

This operation is linear in scalar multiplication and addition: $(a\alpha) \wedge \beta = a(\alpha \wedge \beta)$ and $(\alpha + \gamma) \wedge \beta = \alpha \wedge \beta + \gamma \wedge \beta$. Since $\alpha \wedge \beta$ is also a $(0, 2)$ tensor, it can be added to other elements of the same space. The most important property is the wedge of a form with itself.

Antisymmetry follows at once. Expanding $(\alpha + \beta) \wedge (\alpha + \beta) = 0$, the terms $\alpha \wedge \alpha$ and $\beta \wedge \beta$ vanish and this remains.

The lecture adds that in a more sophisticated treatment one takes “the wedge of a form with itself is zero” as an axiom and proves antisymmetry from it. It says it will return to this approach later.

Q. In what sense is the definition of the wedge product more than a shorthand?
A. As notation it merely shortens two $\otimes$ terms into one symbol. But all the elements the wedge product makes are antisymmetric, and these elements form a subspace of the $(0, 2)$ tensor space. There is addition and scalar multiplication, the element with the sign flipped (the inverse), and the zero tensor. That subspace gets its own name and becomes an object of study.

Exterior powers and dimension

The space of all 2-forms is called the exterior power of $\V^*$ and is written $\Lambda^2 \V^*$. Its basis is $dx^\mu \wedge dx^\nu$ with $\mu < \nu$. Here is how to count. Choose 2 distinct numbers out of 4, with no regard to order, because swapping the order changes only the sign and gives the same basis element. This is the number of combinations.

1-forms give $\binom{4}{1} = 4$, which is the dimension of the covector space itself. Doing the same argument with vectors in place of covectors gives things called 2-vectors, and in general an exterior power can be built from any vector space.

3-forms and 4-forms

A 3-form is a $(0, 3)$ tensor whose sign flips when two arguments are swapped. There are 6 ways to shuffle three arguments, so the components $T_{\alpha\beta\gamma}$ must change sign under one swap of indices. Two swaps return the sign. For example, going from $\alpha\beta\gamma$ to $\beta\gamma\alpha$ takes two swaps, so the sign is the same. The basis that bundles these 6 terms into one is $dx^\mu \wedge dx^\nu \wedge dx^\delta$. If the same number appears twice it is zero at once, so only increasing combinations remain.

There are 4 of them, matching $\binom{4}{3} = 4$. A 4-form has the single choice 0, 1, 2, 3, so it is $\binom{4}{4} = 1$ dimensional. From 5-forms on, choosing 5 numbers out of 4 makes some number repeat, so the result is always 0. In general, when $p$ exceeds the dimension $n$, $\Lambda^p$ is the zero space. A 0-form is a function and has dimension 1. Collecting the dimensions gives this.

In $n$ dimensions $\Lambda^n$ is always 1 dimensional. Also, $\binom{n}{p} = \binom{n}{n-p}$, so the space of $p$-forms and the space of $(n-p)$-forms have the same dimension. The lecture says this sets up a dual relationship between the two and previews it for the coming lectures.

Swaps and signs

Chaining more wedge products leaves the properties unchanged. Swapping two adjacent 1-forms once flips the sign of the whole, so an even number of swaps keeps the sign and an odd number flips it. For example $\gamma \wedge \omega \wedge \delta = -\gamma \wedge \delta \wedge \omega$, and reaching $\delta \wedge \gamma \wedge \omega$ takes two swaps, so the sign comes back and matches that of $\gamma \wedge \omega \wedge \delta$.

Q. Why is the result always 0 when the same number appears?
A. The wedge product is associative and antisymmetric, so after moving two copies of the same 1-form next to each other, collecting all the signs still leaves $\alpha \wedge \alpha = 0$, and the whole thing vanishes right there. It is the same statement as components with a repeated index being 0 in the antisymmetry condition. (This explanation follows from the properties above and was not covered in the lecture.)

So each degree has its own vector space, and forms of different degrees cannot be added. At the end the lecture says the motivation, why antisymmetric tensors matter, comes in the next lecture. In the next post we look at operations on p-forms and the exterior algebra.