The Unapologetic Mathematician

Mathematics for the interested outsider

The Exterior Derivative

The Lie derivative looks sort of familiar as a derivative, but we have another sort of derivative on the algebra of differential forms: the “exterior derivative”. But this one doesn’t really look like a derivative at first, since we’ll define it with some algebraic manipulations.

If \omega is a k-form then d\omega is a k+1-form, defined by

\displaystyle\begin{aligned}d\omega(X_0,\dots,X_k)=&\sum\limits_{i=0}^k(-1)^iX_i\left(\omega(X_0,\dots,\hat{X_i},\dots,X_k)\right)\\&+\sum\limits_{0\leq i<j\leq k}(-1)^{i+j}\omega\left([X_i,X_j],X_0,\dots,\hat{X_i},\dots,\hat{X_j},\dots,X_k\right)\end{aligned}

where a hat over a vector field means we leave it out of the list. There’s a lot going on here: first we take each vector field X_i out of the list, evaluate \omega on the k remaining vector fields, and apply X_i to the resulting function. Moving X_i to the front entails moving it past i other vector fields in the list, which gives us a factor of (-1)^i, so we include that before adding the results all up. Then, for each pair of vector fields X_i and X_j, we remove both from the list, take their bracket, and stick that at the head of the list before applying \omega. This time we apply a factor of (-1)^{i+j} before adding the results all up, and add this sum to the previous sum.

Wow, that’s really sort of odd, and there’s not much reason to believe that this has anything to do with differentiation! Well, the one hint is that we’re applying X_i to a function, which is a sort of differential operator. In fact, let’s look at what happens for a 0-form — a function f:

\displaystyle df(X)=X(f)

That is, df is the 1-form that takes a vector field X and evaluates it on the function f. And this is just like the differential of a multivariable function: a new function that takes a point and a vector at that point and gives a number out measuring the derivative of the function in that direction through that point.

As a more detailed example, what if \omega is a 1-form?

\displaystyle d\omega(X,Y)=X\left(\omega(Y)\right)-Y\left(\omega(X)\right)-\omega\left([X,Y]\right)

We’ve got two terms that look like we’re taking some sort of derivative, and one extra term that we can’t quite explain yet. But it will become clear how useful this is soon enough.

July 15, 2011 - Posted by | Differential Topology, Topology


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  2. […] extremely important property of the exterior derivative is that for all exterior forms . This is only slightly less messy to prove than the fact that is […]

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  3. […] turns out that our exterior derivative is uniquely characterized by some of its properties; it is the only derivation of the algebra of […]

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  4. […] really important thing about the exterior derivative is that it makes the algebra of differential forms into a “differential graded […]

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  5. […] turns out that there is a fantastic relationship between the interior product, the exterior derivative, and the Lie […]

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  6. […] formula in hand we can show that the Lie derivative is a chain map . That is, it commutes with the exterior derivative. And indeed, it’s easy to […]

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  7. […] . Of course we’ll really start with a -form instead of a vector field, and we already know a differential operator to use on forms. Given a -form we can send it to […]

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  8. […] is in terms of differential forms. See, if we take our vector field and consider it as a -form, the exterior derivative is already known to be (essentially) the curl. So what else can we […]

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