# Exterior derivative question

1. Apr 21, 2006

### Oxymoron

If I have a differential manifold, then I get the exterior derivative for free, that is, I dont have to impose any additional structure to my manifold.

However, this derivative is defined as an operator on forms. What makes the forms so special? Why doesn't a diff. manifold come with a derivative on covariant vectors? Or does it and I dont know about it?

2. Apr 21, 2006

### Hurkyl

Staff Emeritus
You get two other forms of differentiation for free:

For a vector field X and scalar field f, you have the scalar field X(f), the derivative f with respect to X.

For a pair of vector fields X and Y, you have the vector field [X, Y], the Lie derivative of Y with respect to X.

There is a dual to the exterior derivative, but it doesn't act on vector fields: it's the boundary operator on the curves/surfaces/etc of your manifold.

Last edited: Apr 21, 2006
3. Apr 21, 2006

### Oxymoron

Yes, but luckily the scalar field is also a 0-form.

For the Lie derivative we need the Lie bracket structure, I was regarding that as extra structure, but I could be wrong about that. Actually, the Lie derivative is probably as close as we are going to get to a free derivative on vector fields without adding structure to the diff. manifold.

Apparently, the operator

$$\theta^a\wedge\nabla_{X_a}$$

(where $\theta^a$ is the dual basis to the basis $X_a$ and $\nabla$ is any torsion-free connection), is identical to the exterior derivative operator, d.

This interests me greatly, so I investigated. If something involving the wedge product, the connection, and a dual basis is identical to the exterior derivative, then I want to know how.

The exterior derivative obeys four axioms:

1. It maps p-forms to (p+1)-forms.
2. $d(\omega\wedge\phi) = d\omega\wedge\phi + (-1)^p\omega\wedge d\phi$.
3. $df(X) = Xf$.
4. $d\circ d = 0$.

I did the first three without a problem, but I cant seem to show the fourth. That is, I cant seem to show that

$(\theta^a\wedge\nabla_{X_a})\circ(\theta^a\wedge\nabla_{X_a}) =0$

I know this sounds a lot like homework, but I want to continue the discussion after I understand this little problem.

I managed to get

$$= \theta^a\theta^a \wedge \theta^a\nabla_{X_a} \wedge \nabla_{X_a}\theta^a \wedge \nabla_{X_a}\nabla_{X_a}$$

but Im not sure where I can cancel and what equals zero.

Last edited: Apr 21, 2006
4. Apr 22, 2006

### Hurkyl

Staff Emeritus
The Lie bracket isn't extra structure -- you do get it for free. You also get a Lie derivative of one-forms with respect to a vector field. (I don't yet have enough intuition to tell if you get a Lie derivative of any tensor W.R.T. a vector field for free)

Well, your first problem is that you're overworking your dummy indices.

$$(\theta^a\wedge\nabla_{X_a})\circ(\theta^b\wedge\nabla_{X_b}) =0$$

That's what you want.

What exactly is the action of that operator on your forms? Is it:

$$(\theta^a \wedge \nabla_X_a)\omega = \theta^a \wedge (\nabla_X_a\omega)$$

?

Your final expression makes me uneasy -- it looks like you just formally expanded it as if $\circ$ distributes over $\wedge$, and dropped the $\circ$'s. But I'm not sure that makes sense, at least directly... for example, does $\theta^a \nabla_X_a$ make any sense, either on its own or in the context of the rest of the expression?

I think the right way to interpret it is to view $\theta^a \wedge$ as an operator that takes a one form and produces: $(\theta^a \wedge)(\omega) = \theta^a \wedge \omega$. Then, you're trying to manipulate the expression involving the composition of four operators:

$$(\theta^a \wedge) \nabla_X_a (\theta^b \wedge) \nabla_X_b$$

to which you would simply apply the usual trick: for the operators A and B, figure out how (AB)(w)=A(B(w)) and (BA)(w)=B(A(w)) relate to each other to get an algebraic relationship for AB and BA in terms of each other.

Last edited: Apr 22, 2006
5. Apr 22, 2006

### loopgrav

The Lie derivative is defined in terms of flows on a manifold. The fact that it is equal to the Lie bracket is a theorem that results from the definition. The most readable treatment I know of that doesn't get too bogged down in total rigour is Ted Frankel's "The Geometry of Physics." He also treats everything else you mention is excellent, practical detail. Another more careful treatment but one that is extremely readable is Munkres' "Analysis on Manifolds." Unfortunately, he doesn't cover Lie derivatives or differential geometry in general, but his treatment of integration of differential forms is outstanding.

6. Apr 23, 2006

### Oxymoron

I honestly dont know, and the question does not explicitly state how this operator acts on forms. I guess your guess is as good as mine.

Well, what I wrote surely $\theta^a \nabla_X_a$ does not make sense since $\nabla_X_a$ is an operator and its clearly not operating on anything.

The exterior derivative d is a map from the algebra of r-forms to r+1-forms.
The action of d on an r-form is defined by

$$d\omega = \frac{1}{r!}\left(\frac{\partial}{\partial x^{\nu}}\omega_{\mu_1\dots\mu_r\right)dx^{\nu}\wedge dx^{\mu_1}\wedge\dots\wedge dx^{\mu_r}$$

Take some r-form of the form

$$\omega = \frac{1}{r!}\omega_{\mu_1\dots\mu_r}dx^{\mu_1}\wedge\dots\wedge dx^{\mu_r}$$

Then the action of $d^2$ is simply

$$d^2\omega = \frac{1}{r!}\left(\frac{\partial^2\omega_{\mu_1\dots\mu_r}}{\partial x^{\lambda}\partial x^{\nu}}\right)dx^{\lambda}\wedge dx^{\nu} \wedge dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r}$$

which is just the action of d twice, first with nu indices and then with lambda indices. So d^2 on an r-form returns an r+2-form. But

$$\frac{\partial^2\omega_{\mu_1\dots\mu_r}}{\partial x^{\lambda}\partial x^{\nu}} = \frac{\partial^2\omega_{\mu_1\dots\mu_r}}{\partial x^{\nu}\partial x^{\lambda}}$$

because this term is symmetric with respect to $\lambda$ and $\nu$ from the properties of the partial derivative.

Furthermore,

$$dx^{\lambda}\wedge dx^{\nu} = (-1)dx^{\nu}\wedge dx^{\lambda}$$

is anti-symmetric.

This is all I know about the operator d^2. Im not sure how this proves $d\circ d = 0$ tho. Any thoughts? I figured if I can understand how d^2 =0 then perhaps I can understand how this new operator is zero.

Last edited: Apr 23, 2006
7. Apr 23, 2006

### Perturbation

The components of the exterior derivative of a p-form are

$$\left(d\tilde{\omega}\right)_{ki\cdots j}=\partial_{[k}\omega_{i\cdots j]}$$

Where the square brackets denote anti-symmetrisation. The second exterior derivative is then

$$\left[d\left(d\tilde{\omega}\right)\right]_{lki\cdots j}=\partial_{[l}\partial_{[k}\omega_{i\cdots j]]}$$

Here there is a double anti-symmetrisation. Now

$$\left[d\left(d\tilde{\omega}\right)\right]_{lki\cdots j}=-\partial_{[k}\partial_{[l}\omega_{i\cdots j]]}$$

By antisymmetry. But partials commute so

$$\left[d\left(d\tilde{\omega}\right)\right]_{lki\cdots j}=\partial_{[l}\partial_{[k}\omega_{i\cdots j]]}=-\partial_{[l}\partial_{[k}\omega_{i\cdots j]]}=0$$

For example, take the exterior derivative of a function f

$$(df)_i=\partial_if$$

Which is a one-form. Now take the second exterior derivative

$$\left[d(df)\right]_{ji}=\partial_j\partial_if-\partial_i\partial_jf=0$$

By commutativity of partials.

[I've missed the (p+1) etc. multiples off the exterior derivatives because I couldn't be arsed to keep putting them, as they're irrelevant. The factorials are accounted for in the anti-symmetrisation brackets if you're not familar with that notation.]

I'll show you what I've just said but including covariant bases as you're doing.

$$d^2\omega = \frac{1}{r!}\left(\frac{\partial^2\omega_{\mu_1\dots\mu_r}}{\partial x^{\lambda}\partial x^{\nu}}\right)dx^{\lambda}\wedge dx^{\nu} \wedge dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r}$$
$$= \frac{1}{r!}\left[\left(\frac{\partial^2\omega_{\mu_1\dots\mu_r}}{\partial x^{1}\partial x^{2}}\right)dx^{1}\wedge dx^{2}+\left(\frac{\partial^2\omega_{\mu_1\dots\mu_r}}{\partial x^{2}\partial x^{1}}\right)dx^{2}\wedge dx^{1}+\cdots \right] \wedge dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r}$$

But $dx^{2}\wedge dx^{1}=-dx^{1}\wedge dx^{2}$, so

$$\frac{1}{r!}\left[\left(\frac{\partial^2\omega_{\mu_1\dots\mu_r}}{\partial x^{1}\partial x^{2}}\right)dx^{1}\wedge dx^{2} \wedge dx^{\mu_1}+\left(\frac{\partial^2\omega_{\mu_1\dots\mu_r}}{\partial x^{2}\partial x^{1}}\right)dx^{2}\wedge dx^{1}+\cdots \right] \wedge dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r}$$
$$= \frac{1}{r!}\left[\left(\frac{\partial^2\omega_{\mu_1\dots\mu_r}}{\partial x^{1}\partial x^{2}}\right)dx^{1}\wedge dx^{2}-\left(\frac{\partial^2\omega_{\mu_1\dots\mu_r}}{\partial x^{2}\partial x^{1}}\right)dx^{1}\wedge dx^{2}+\cdots \right] \wedge dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r}$$

Partials commute so

$$\frac{1}{r!}\left[\left(\frac{\partial^2\omega_{\mu_1\dots\mu_r}}{\partial x^{1}\partial x^{2}}\right)dx^{1}\wedge dx^{2}-\left(\frac{\partial^2\omega_{\mu_1\dots\mu_r}}{\partial x^{2}\partial x^{1}}\right)dx^{1}\wedge dx^{2}+\cdots \right] \wedge dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r}$$
$$= \frac{1}{r!}\left[\left(\frac{\partial^2\omega_{\mu_1\dots\mu_r}}{\partial x^{1}\partial x^{2}}\right)dx^{1}\wedge dx^{2}-\left(\frac{\partial^2\omega_{\mu_1\dots\mu_r}}{\partial x^{1}\partial x^{2}}\right)dx^{1}\wedge dx^{2}+\cdots \right] \wedge dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r}=0$$

You had all that you needed, you just didn't put it together. Unless what you're actually asking is how one proves the operator identity $d^2=0$. This follows from the above because d is defined by its operation on differential forms, so if it holds for a general p-form then one is allowed to just right the operator identity $d^2=0$. Or it can be taken with the other properties of the exterior derivative that you gave as axioms which together uniquely determine the form of the exterior derivative operator, however I've not seen this proof, but I of know.

Last edited: Apr 23, 2006
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