Where does the invariant expression for the exterior derivative come from?

Question

So I've just spent about four TeXed pages (plus about a dozen TeXed pages of discarded work) proving the identity

\begin{align*} d \omega(\zeta_1, \ldots, \zeta_{k+1}) &= \sum_{i=1}^{k+1} (-1)^{i-1} \zeta_i \cdot \omega(\zeta_1, \ldots, \hat{\zeta_i}, \ldots, \zeta_{k+1}) \\ &\qquad + \sum_{1 \leq i < j \leq k + 1} (-1)^{i+j} \omega([\zeta_i, \zeta_j], \zeta_1, \ldots, \hat{\zeta_i}, \ldots, \hat{\zeta_j}, \ldots, \zeta_n). \end{align*}

Obviously this is not how this identity was discovered. But I don't see any explanation written down anywhere -- just things along the lines of "it can be laboriously checked that the right hand side is in fact linear in $\zeta_1 \wedge \cdots \wedge \zeta_{k+1}$, and then the identity can be verified on an appropriate basis."

Questions:

1. Is there an easy way to see that the right-hand side is a linear function of $\zeta_1 \wedge \cdots \wedge \zeta_{k+1}$?
2. Is there a reasonable explanation for how one could come up with this identity?

Regarding the first question: In particular, I notice that, on a basis, the first sum gives the desired results and the second sum vanishes; it's simply the case that neither sum is actually a linear function of $\zeta_1 \wedge \cdots \wedge \zeta_{k+1}$, but combining the two gives a linear map. Is there some way to regard the second sum as the result of some "correction" process applied to the first that makes it linear? (Compare for instance situations where you average over a group of symmetries to get an invariant map, for instance.)

Answer 1

For your first question, consider that you look at the RHS as a map taking $k+1$ vector fields in argument and returning a real number. This map is clearly a linear form defined on $\Gamma(TM)^{\otimes k+1}$. What we want is to show that : 1) this map is antisymmetric: this follows quite immediately from the fact that $\omega$ itself is antisymmetric and that the bracket is ; 2) that this map is tensorial: this is really the crucial point; we want that the value of this map over a point $x$ of $M$ only depends on the values of the vector fields at $x$ (but of course, it does not depend only on the value of $\omega$ at $x$). In order to show that, the tensoriality lemma says that it is equivalent to check that it linear over the functions; that is: if you take a smooth $f$ and multiply one of your vector fields by $f$, then the result should be multiplied by $f$. And this works !

This is not so magical: the idea is that brackets appear everywhere when you want to obtain natural tensors on your manifold. Maybe the simplest example of this is the torsion operator of a connection. Take $\nabla$ a connection on your tangent space. If you have two vector fields, you can compare $\nabla_X Y$ and $\nabla_Y X$ but $T(X,Y) = \nabla_X Y - \nabla_Y X$ is not tensorial in $X$ and $Y$. Indeed, $T(fX,Y) = f(\nabla_X Y - \nabla_Y X) - (Y.f)X$. The last term involves the derivative of $f$ and you want to get rid of it.

Now observe that $[fX,Y] = f[X,Y] - (Y.f)X$ so the same term appears in the non-tensoriality of the bracket. This leads to the good definition of the torsion: $T(X,Y) = \nabla_X Y - \nabla_Y X - [X,Y]$ and this is tensorial in $X$ and $Y$.

If you don't know what a connection is, this may seem unclear but the phenomenon is really the same that in your case. In some sense, the bracket is the "universal default of non-tensoriality". I think that the notion of tensoriality is really the good one to understand these formulas and this should be more emphasized in a first lesson in differential geometry.

For your second question, once you know that the RHS is tensorial in the vector fields, you can take vector fields with nice properties, that is with brackets vanishing. This is always possible locally (just take vector fields coming from a system of coordinates); then the computation is more simple because there is a simple expression for $d\omega$ if one expresses $\omega$ in a basis coming from coordinates.

That is: if $\omega = \sum_I \alpha_I dx^I$, then $d\omega = \sum_{j,I} (\partial_j.\alpha_I) dx^j \wedge dx^I$ (here $I$ is a multi-index, $j$ an index). This gives the identity you are looking as an easy computation in multilinear algebra (not in geometry: when you know that things are tensorial, everything is algebra).

Curiously, there is a situation when the RHS is used differently. When you are on a Lie group, then you can take a basis of left-invariant vector fields, and look at a left-invariant differential form. Then the first term in the RHS will be zero and only the second term is important. And in fact, you obtain in this way the usual differential that you get in Lie algebra cohomology.

A last remark: there is in fact no need to show the antisymmetry of the RHS. You can look at is just as a multilinear map and once you know that this is equal to LHS, this proves the antisymmetry.

Answer 2

I have also struggled with this quite a bit. I do not have a complete answer, but here is at least a start.

Let $\omega$ be a $1$-form on a manifold $M$. It eats vector fields and spits out functions. I need to try to figure out a way to cook up a $2$-form out of this.

Remember that the invariant way to define vectors are as derivations of the algebra of smooth functions. So a vector field $X$ will eat a function and spit out a function (which we think of as the directional derivatives of $f$ in the directions given by $X$). In particular we want $df(X) = X(f)$.

An attempt at $d\omega(X,Y)$ is $X(\omega(Y)) - Y(\omega(X))$. This looks like pretty much the only way I will be able to get a function. This is alternating at least, and is multilinear in $X$ and $Y$ (check it). The only other thing I want out of the exterior derivative is that $d\circ d=0$, so let's check that out using our formula above:

$d(df)(X,Y) = X(df(Y)) - Y(df(X)) = X(Y(f)) - Y(X(f)) = [X,Y](f) = df([X,Y])$, not $0$ as we had wished.

So it makes sense to redefine $d(\omega(X,Y)) = X(\omega(Y)) - Y(\omega(X)) - \omega([X,Y])$

I believe (but do not have the fortitude to show) that similar reasoning will bootstrap you to the definition of the higher level exterior derivatives.

Answer 3

Here is a neat proof that I like.

Let $\nabla$ be any torsion free connection (you can always find one say by fixing a Riemannian metric and taking $\nabla$ to be its Levi-Civita connection). Remember that torsion free means that for any vector fields $X,Y$ we have $$ \nabla_X Y -\nabla_Y X = [X,Y]. $$ Let $\omega$ be a $1$-form for simplicity then $$ \nabla_X (\omega(Y)) = (\nabla_X\omega)(Y) + \omega(\nabla_X Y). $$ Swapping the role of $X,Y$ in the above and taking the difference, together with the fact that $\nabla$ is torsion free, we get $$ (\nabla_X\omega)(Y) - (\nabla_Y \omega)(X) = \nabla_X (\omega(Y)) - \nabla_Y (\omega(X)) - \omega([X,Y]). $$ The LHS is by definition precisely $d\omega(X,Y)$ (this is one way of defining the exterior derivative, namely, as the skew-symmetrisation of $\nabla \omega$) and the RHS is just $X (\omega(Y)) - Y (\omega(X)) - \omega([X,Y])$. When $\omega$ is a $(k+1)$-form, essentially the same proof works since $\nabla_X (\omega(Y_1,...,Y_k))=(\nabla_X \omega)( Y_1,...,Y_k)+\omega(\nabla_X Y_1,..., Y_k)+\cdots + \omega(Y_1,...,\nabla_X Y_k)$.

Answer 4

A wedge product of a k-form $\omega$ and a l-form $\eta$ is defined as $\omega \wedge \eta (X_1,\ldots,X_{k+l}) = \frac{1}{k!l!} \sum_{\sigma} sgn(\sigma) \omega(X_{\sigma(1)},\ldots,X_{\sigma(k)}) \eta(X_{\sigma(k+1)},\ldots,X_{\sigma(k+l)})$. The exterior derivative is a bit like $d\wedge \omega$, where $d(X_{\sigma(1)}) \omega(X_{\sigma(2)},\ldots,X_{\sigma(k+1)})$ is understood as $X_{\sigma(1)} \omega(X_{\sigma(2)},\ldots,X_{\sigma(k+1)})$. But this expression is not tensorial in the sense that if you multiply a function $f$ to one of the vectors, say $X_p$, you get an additional term $\sum_{i\neq p} (-1)^{i+1} (X_i f) \omega(X_1,\ldots,\hat{X_i},\ldots,X_{k+1})$. You can view the second sum as a "correction" to make $d\omega$ a tensor.