Tensor #23. Operations on p-forms: The Exterior Algebra텐서 #23. p-형식의 연산과 외대수

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 23 of the playlist, Operations on p-forms. The Exterior Algebra.. Having built the spaces of p-forms in the previous post, this time we organize the operations on them and set up the exterior algebra.

Review: exterior powers

From a vector space $\V$ of dimension $n$ comes the dual space $\V^*$, whose elements are the 1-forms $\alpha = \alpha_\mu dx^\mu$. Chaining 1-forms with the wedge product gives forms of higher degree.

This 2-form can be written by collecting only the pairs with $\mu < \nu$, or by letting all values run and correcting the double counting with $1/2!$, and for degree $p$ the correction factor is $1/p!$. Which convention is used differs from book to book. Continuing this way we can build 3-forms and 4-forms, and the space where $p$-forms live is $\Lambda^p \V^*$. We have $\Lambda^1 \V^* = \V^*$, and $\Lambda^0$ is the real numbers (or real-valued functions on a manifold). The dimension is $\binom{n}{p}$, which equals $\binom{n}{n-p}$. The lecture previews that this correspondence leads to a one-to-one map, and the Hodge star at the end of this post is exactly that.

Rules of the wedge product

For 1-forms $\alpha, \beta$ we have $\alpha \wedge \beta = -\beta \wedge \alpha$ and $\alpha \wedge \alpha = 0$. Note that $\alpha \wedge \alpha = 0$ is a rule for 1-forms and cannot be carried over unchanged to forms of higher degree. Combining these two rules with associativity, in a product of 1-forms, if any two entries are the same, wherever they are, the whole product is 0, because the sign needed to bring two non-adjacent entries together is $\pm 1$ anyway. Swapping any two flips the sign, and swapping any other two flips it again. So an even number of swaps keeps the sign and an odd number flips it.

Books write this with permutations. The wedge product of 1-forms $\alpha_1, \dots, \alpha_p$ is the tensor product summed over all permutations $\sigma$, weighted by the sign.

$\mathrm{sgn}(\sigma)$ is $+1$ if the permutation is reached by an even number of swaps and $-1$ if by an odd number. The subscripts here are names of different 1-forms, not spacetime indices. The $p = 2$ definition from the last post is the first instance of this formula.

Form fields

A form field is the same idea as the tensor fields we already know, with coefficients that become real functions on spacetime. For $n = 4$ a 0-form field is a single function, and a 1-form field has the shape $a(x) dx^0 + b(x) dx^1 + c(x) dx^2 + d(x) dx^3$. A 2-form field has 6 basis elements, each with a function as coefficient.

A 3-form field has 4 basis elements, a 4-form field has one, and from 5-form fields on the basis elements would repeat, so they are 0. If we want to add a term with the order of the numbers changed, for example $dx^2 \wedge dx^1 \wedge dx^3$, we flip the sign, merge it in front of $dx^1 \wedge dx^2 \wedge dx^3$, and relabel the coefficient as a new function. In components $T_{012} = p(x)$ and relations like $T_{102} = -T_{012}$ follow.

Different degrees and the exterior algebra

Elements $\alpha \in \Lambda^2 \V^*$ and $\beta \in \Lambda^3 \V^*$ belong to different vector spaces, so $\alpha + \beta$ is not defined. At each point of spacetime the spaces from $\Lambda^0$ to $\Lambda^3$ all sit there, and an element of one space cannot be added to or compared directly with an element of another. The wedge product, on the other hand, takes a $p$-form and a $q$-form and gives a $(p+q)$-form.

In components we multiply the coefficients and chain the basis wedges, and linearity makes this well defined. If $p + q > n$ the result is 0. But to be an algebra, the product must stay inside the same vector space. Wedging two $p$-forms gives a $2p$-form, so a single $\Lambda^p$ is not closed. So we bundle the spaces of all degrees with a direct sum.

The tensor algebra has this sum going on forever, but here everything with $k > n$ is 0, so it ends at $n$. The direct sum means the elements are $(n+1)$-tuples. The first slot holds a real number, the second a 1-form, the next a 2-form, and so on. Addition is done slot by slot and the zero element is the tuple of all zeros. This gives a vector space, and the wedge product becomes the multiplication that puts the product of slot $p$ and slot $q$ into slot $p+q$, giving 0 if $p + q > n$. This algebra is the exterior algebra. (That the total dimension is $\sum_p \binom{n}{p} = 2^n$ follows from the definition of the direct sum and was not covered in the lecture.)

Q. If different degrees cannot be added, why do we treat them together in the exterior algebra?
A. The elements of the exterior algebra are tuples split into slots, and only values in the same slot are added. We do not add forms of different degrees into one form. We make a new space with several slots, and on it the wedge product becomes a multiplication that moves between slots.

The commutation law

When $\omega$ is a $p$-form and $\eta$ is a $q$-form, swapping the order of the product gives this.

The reason is counting. Moving one 1-form of $\eta$ to the front, past the $p$ factors of $\omega$, takes $p$ swaps, and we do this for all $q$ factors of $\eta$, so $pq$ swaps. The coefficients are numbers and come out unchanged. So if either $p$ or $q$ is even they commute, and if both are odd they anticommute.

Swapping $\omega$ with itself gives $\omega \wedge \omega = (-1)^{p^2} \omega \wedge \omega$, and when $p$ is odd, $\omega \wedge \omega = -\omega \wedge \omega$, so it is 0. When $p$ is even this argument yields nothing, and it need not be 0. So the statement that the wedge of a form with itself is 0 is safe only for odd degrees. In the toy below, pick the dimension $n$ and the degree $p$, and tap a dx slot to change its number. Swapping two neighboring slots flips the sign, a repeated number gives 0, and the number of basis forms $\binom{n}{p}$ appears for each degree.

Q. Doesn’t an even-degree form also become 0 when wedged with itself?
A. It need not. In $n = 4$ with $\omega = dx^0 \wedge dx^1 + dx^2 \wedge dx^3$, the two terms commute, so $\omega \wedge \omega = 2 \, dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3$, which is not 0. (This example is added for understanding and was not covered in the lecture.)

Change of basis and the determinant

Consider the linear transformation $\hat e^\alpha = \Lambda^\alpha{}_\beta \, e^\beta$ that changes the basis of 1-forms. In dimension $n$ the $n$-th exterior power is 1 dimensional, so the wedge of a basis can only become a constant multiple of the same basis element under the transformation. That constant is the determinant.

A square matrix has one reliable number, the determinant, so this result is natural. The lecture skips the proof as tricky and says only that there is a proof by induction. For a wedge of $k < n$ elements, instead of the determinant the $k \times k$ minors of $\Lambda$ appear, and the lecture calls this a generalized determinant.

The induced inner product

If $\V$ has an inner product, an inner product is induced on the dual space, and from it on every $\Lambda^p$. An inner product exists only between two elements of the same space, so it makes sense only between two $p$-forms. The inner product of the simple $p$-forms $\alpha_1 \wedge \cdots \wedge \alpha_p$ and $\beta_1 \wedge \cdots \wedge \beta_p$ is the determinant of the matrix whose entries are the inner products between 1-forms.

All entries are real numbers, and the matrix is $p \times p$. This too is a result obtained by breaking things apart and reassembling them, since everything is linear, and the lecture does not show the derivation.

The Hodge star

$\Lambda^p$ and $\Lambda^{n-p}$ have the same dimension, so a one-to-one correspondence seems possible, but there is no canonical way to choose which one. Pairing a vector space with its dual by the delta function was also only a convenience and not canonical, and it took an inner product to get a canonical correspondence. Here we borrow the inner product in the same way.

Wedging a fixed $\alpha \in \Lambda^p$ with an arbitrary $\beta \in \Lambda^{n-p}$ gives an $n$-form, an element of a 1-dimensional space, so a real multiple of the volume form $\mathrm{vol} = dx^0 \wedge \cdots \wedge dx^{n-1}$. That real number is linear in $\beta$, so it can be written as an inner product with some element $\star\alpha \in \Lambda^{n-p}$.

This $\star\alpha$ is the Hodge dual. It is a different notion from the dual space, the collection of maps from vectors to real numbers. To actually find it, we compute with a basis. Take $n = 3$ with an orthonormal basis. Set $\alpha = dx^0 \wedge dx^1$ and write its dual as $\star\alpha = a \, dx^0 + b \, dx^1 + c \, dx^2$, then insert $\beta = dx^0, dx^1, dx^2$ in turn. The wedges $\alpha \wedge dx^0$ and $\alpha \wedge dx^1$ repeat a number and vanish, so $a = b = 0$, and $\alpha \wedge dx^2 = \mathrm{vol}$ gives $c = 1$. In the same way we get the following.

Once we know the basis, linearity gives the dual of every 2-form.

This rule has the same shape as the cross product of introductory physics: $\mathbf e_1 \times \mathbf e_2 = \mathbf e_3$, $\mathbf e_1 \times \mathbf e_3 = -\mathbf e_2$, $\mathbf e_2 \times \mathbf e_3 = \mathbf e_1$, and in 3-dimensional Euclidean space we can write $\mathbf u \times \mathbf v = \star(\mathbf u \wedge \mathbf v)$. The lecture corrects itself that the Hodge star goes after the wedge product. In the same way the scalar triple product becomes $\mathbf u \cdot (\mathbf x \times \mathbf y) = \star(\mathbf u \wedge \mathbf x \wedge \mathbf y)$. The difference is that the cross product is defined only in 3 dimensions, whereas the Hodge star works between $p$ and $n - p$ in any dimension. All of this also stands on the assumption of orthonormality, that is, on an inner product. For example in $n = 7$ we have $\dim \Lambda^p = 1, 7, 21, 35, 35, 21, 7, 1$, and $p$ and $7 - p$ are paired by the Hodge star. The lecture says the heart of the computation is using the inner product in an orthonormal basis, and in higher dimensions the determinants only grow larger while the principle stays the same.

Q. Why is an inner product needed to define the Hodge star?
A. Two vector spaces of the same dimension have many one-to-one correspondences, so choosing one needs extra information. The formula $\alpha \wedge \beta = \ip{\star\alpha}{\beta} \mathrm{vol}$ has an inner product on its right-hand side, so without an inner product we cannot pick out $\star\alpha$.

In the next post we look at the exterior derivative and its uses.