Tensor #24. The Exterior Derivative: A Formal Introduction텐서 #24. 외미분의 형식적 도입

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 24 of the playlist, The Exterior Derivative - formal introduction. Having set up the exterior algebra in the previous post, this time we build the operation that differentiates on it, the exterior derivative.

Review and where this post is going

The exterior algebra gathers the $p$-th exterior powers $\Lambda^p$ of a vector space $\V$ into a direct sum and takes the wedge product as its multiplication. The wedge product is $\Lambda^p \times \Lambda^q \to \Lambda^{p+q}$, it is associative, and $\omega \wedge \eta = (-1)^{pq} \, \eta \wedge \omega$. The lecture points out that we usually apply this to the dual space but the construction works for any vector space. The Hodge star maps $\Lambda^p$ to $\Lambda^{n-p}$ when there is a metric, and the star goes on the left, as in $\star\omega$ (the lecture corrects the right-hand placement in the last lecture).

Now for the calculus. We already have two derivatives, the Lie derivative and the covariant derivative. The third is the exterior derivative, the operation $d$ that sends a $p$-form to a $(p+1)$-form. The lecture says it will pile up definitions and facts in order this time and attach intuition slowly afterwards. The judgment is that this order is better than starting from intuition and building the forms. This post follows the same order.

The differential df of a function

A 0-form is a smooth function $f$ on spacetime. Its derivative $df$ has to be a 1-form, so it takes a vector $v$ and returns a real number. We have only two ingredients, a vector acting on a function, $v(f)$, and a form acting on a vector, and joining the two leaves exactly one way to go.

That is, $df$ is the 1-form that, by swallowing a vector, does what the vector would do by swallowing the function. In a coordinate basis this is the following formula.

To check it, insert $v = a^j \partial_j$. We get $df(v) = \partial_i f \, a^j \, dx^i(\partial_j) = \partial_i f \, a^j \delta^i{}_j = a^j \partial_j f = v(f)$. Applied to a coordinate function $x^i$ itself this gives $d(x^i) = dx^i$, so the basis 1-forms $dx^i$ we have been using are differentials of coordinate functions.

Three requirements on the exterior derivative

The operation $d$ is a linear map from $\Lambda^p$ to $\Lambda^{p+1}$, and we require three things.

• (1) On 0-forms it is the formula $df = \partial_i f \, dx^i$ above.
• (2) The anti-Leibniz rule: if $\omega$ is a $p$-form and $\eta$ is a $q$-form, then $d(\omega \wedge \eta) = d\omega \wedge \eta + (-1)^p \, \omega \wedge d\eta$.
• (3) $d(d\omega) = 0$, written $d^2 = 0$ for short.

Here is how to read the sign in (2). An ordinary product rule would give $d\omega \wedge \eta + \omega \wedge d\eta$, but between $\omega$ and $d\eta$ the 1-form slot made by $d$ has to pass the $p$ slots of $\omega$. Each crossing flips the sign once, which gives $(-1)^p$. Only the degree $p$ of the front form fixes the sign, and the degree $q$ of $\eta$ does not matter. Antisymmetry twists the product rule this way. Rule (3) is also a consequence of antisymmetry: differentiating twice always gives 0. The lecture notes that this touches the topological fact that the boundary of a boundary is empty (the boundary of a disc is a circle, and a circle has no boundary), and moves on, promising to return once the definition is understood.

If we can show that a $d$ satisfying these three requirements (a) is at most one and (b) really exists, then the exterior derivative is not something we chose but something the requirements force. The rest of this lecture is those two proofs.

Exact forms and closed forms

First some vocabulary. If $\omega = d\sigma$ for some $\sigma$, then $\omega$ is called exact, and if $d\omega = 0$ it is called closed. The word has nothing to do with closed sets in topology. An exact form is always closed, because $\omega = d\sigma$ gives $d\omega = d(d\sigma) = 0$. The converse fails, since not every closed form is exact. The lecture says it should have mentioned these words in the last lecture, and that they become important when we reach differential equations.

Q. Are there forms that are closed but not exact?
A. Yes. On the plane with the origin removed, the 1-form $\dfrac{-y \, dx + x \, dy}{x^2 + y^2}$ gives 0 under $d$, but it is not the differential of any function on the whole region. The angle function does not return to its value after one loop around the origin. (This example is added for understanding and was not covered in the lecture.)

Uniqueness (1): the exterior derivative of a basis form is 0

We write the basis of $p$-forms with ascending numbers $I = (i_1 < i_2 < \cdots < i_p)$, as in $dx^I = dx^{i_1} \wedge \cdots \wedge dx^{i_p}$. An out-of-order product such as $dx^3 \wedge dx^2 \wedge dx^4$ is $\pm$ the properly ordered basis form, so we do not need it separately. This notation, which writes the whole list $I$ as a single letter, is called a multi-index, and many books use it.

The claim is $d(dx^I) = 0$. With rule (3) it looks obvious, but we do not yet know that $dx^I$ is the derivative of some form. We show it by mathematical induction on $p$.

For $p = 1$, $dx^i = d(x^i)$, so $d(dx^i) = d(d(x^i)) = 0$. Suppose it holds up to $p - 1$, and wedge the 0-form $x^{i_1}$ in front of $dx^{i_2} \wedge \cdots \wedge dx^{i_p}$ and use rule (2).

Since it is a 0-form the sign is $(-1)^0 = 1$. The second term is 0 by the induction hypothesis, so $dx^I$ is the exterior derivative of some form. By rule (3) its $d$ is 0, and $d(dx^I) = 0$.

Q. May we wedge the function $x^{i_1}$ with a form?
A. Yes, because a function is a 0-form. Wedging a 0-form with a $p$-form gives a $p$-form, and its value is just the function times the form, $f \wedge \omega = f \, \omega$. The lecture says this wedge looked strange to the lecturer at first.

Uniqueness (2): $d\omega$ of a general p-form is fixed

A general $p$-form is $\omega = \sum_I \omega_I \, dx^I$, with coefficients $\omega_I$ that are functions on spacetime. Since $d$ is linear we can go term by term and apply rule (2) to one term. The function $\omega_I$ is a 0-form, so the sign is 1.

The second term was 0 in the previous section. What remains, $d\omega_I$, is the derivative of a 0-form, computed by rule (1).

There is no gap in this formula. Once there is a coordinate system the coefficient functions and partial derivatives are fixed, and all we used were rules (1), (2) and (3). So if a $d$ satisfying the requirements exists it is unique, and the formula above is what it looks like. In words: differentiate the coefficient of each term by every coordinate and attach that $dx^j$ at the front. If $dx^I$ already contains the same number the wedge is 0, and otherwise a sign appears from moving it into ascending order.

Check this in the toy below by changing the coefficients. Pick the dimension and the degree, and tap a cell to change its coefficient function. You can see which derivative attaches to which basis element for each term, and where the slots collide and give 0. At the bottom you can watch the terms of $d(d\omega)$ collected by basis element and cancelling each other.

(Here is one example. On the plane, if $\omega = y \, dx + x^2 \, dy$ then $d\omega = dy \wedge dx + 2x \, dx \wedge dy = (2x - 1) \, dx \wedge dy$. This example is added for understanding and was not covered in the lecture.)

Existence: does rule (2) hold?

Taking the formula above as the definition, we must check that all three rules hold. Rule (1) holds because on 0-forms the formula is the definition. For rule (2), linearity lets us look at one term at a time. Let $\omega = g \, dx^I$ be a $p$-form and $\eta = h \, dx^J$ a $q$-form, so $\omega \wedge \eta = gh \, dx^I \wedge dx^J$, and by the formula we get the following.

The derivative of a product has split in two. The first term is $(\partial_k g \, dx^k \wedge dx^I) \wedge (h \, dx^J) = d\omega \wedge \eta$. In the second term we must move $dx^k$ past the $p$ 1-forms of $dx^I$ to reach the shape of $d\eta$, and that costs $(-1)^p$.

So $d(\omega \wedge \eta) = d\omega \wedge \eta + (-1)^p \omega \wedge d\eta$. The sign we wrote in rule (2) comes straight out of this computation.

Existence: does rule (3) hold?

Finally $d^2 = 0$. Applying it twice to one term $\omega = g \, dx^I$ gives the following.

The first $d$ put $dx^k$ at the front, and the second put $dx^l$ at the front. The terms with $k = l$ vanish because $dx^k \wedge dx^k = 0$. For $k \neq l$ the $(k, l)$ term and the $(l, k)$ term form a pair. The wedge products have opposite signs, $dx^l \wedge dx^k = -dx^k \wedge dx^l$, and for a smooth function the order of partial derivatives does not matter, $\partial_l \partial_k g = \partial_k \partial_l g$. The two terms cancel exactly, so the whole sum is 0. One side is the partial derivative, whose value stays the same when the order changes, and the other is the wedge product, whose sign flips when the order changes, which is why differentiating twice gives 0.

Q. What does it mean that $d^2 = 0$ comes from antisymmetry?
A. In the sum above, what cancels is the symmetry of partial derivatives $\partial_l \partial_k = \partial_k \partial_l$ colliding with the antisymmetry of the wedge $dx^l \wedge dx^k = -dx^k \wedge dx^l$. If the wedge product were symmetric the two terms would add up and would not give 0.

Summary

The lecture ends by retracing the whole path. On 0-forms we fixed $df(v) = v(f)$. On top of that we required the anti-Leibniz rule and $d^2 = 0$. If a $d$ satisfying these exists, its value on a general form is fixed by a single formula, so it is unique. Taking that formula as the definition, we computed that rules (2) and (3) really hold, which also shows existence. The uniqueness proof found the object as well, so what remained was checking that the object satisfies the rules. In the next post we compute some examples of the exterior derivative and apply it to electromagnetism.