A question about a part of the proof of the existence of the exterior derivate.

Question

Let $M$ a differentiable manifold, $(U;x_1,\ldots,x_n)$ a coordinated system and $\omega$ a differential form which domain intersects $U$. In $U \cap \operatorname{Dom} \omega$, $\omega$ can be written as $\sum a_\phi \, dx_\phi$, with $\phi \in \cal{P}(\{1,\ldots,n\})$ ($a_\phi \in \cal{C}^\infty(U \cap \operatorname{Dom} \omega))$. For each $p \in U \cap \operatorname{Dom} \omega$, define $d\omega|_p=\sum da_\phi |_p \wedge \, dx_\phi |_p$. It is possible to prove that $d\omega|_p$ is well defined: changing the coordinated system does not change the result. It is possible to define, for each form $\omega$, the application $d\omega$ that sends every element $p \in U \cap \operatorname{Dom} \omega$ to $d\omega|_p$. I want to prove thar $d\omega$ is a form. It is enough to show that is a differentiable aplication. My question is: can it be shown without using charts explicitly, just reasoning which the expresion of $d\omega|_p$ is?

Answer 1

Yes, there is an expression for $d\omega$ without reference to coordinate systems.

If $\omega$ is a $k$-form, it “eats” $k$ vector fields at once and produces a function on $M$. The derivative $d\omega$ can be determined by what it does to a $(k+1)$-tuple of vectors. Remember also that vector fields on $M$ are derivations of functions on $M$.

The coordinate-free equation for $d\omega$ is \begin{multline*} d\omega(X_0,X_1,\dots,X_k) = \sum_{i=0}^k (-1)^i X_i \omega \left (X_0, \ldots, \hat X_i, \ldots,X_k \right ) \\ +\sum_{i<j}(-1)^{i+j}\omega \left (\left [X_i, X_j \right ], X_0, \ldots, \hat X_i, \ldots, \hat X_j, \ldots, X_k \right ) \end{multline*} Here $\hat X_i$ means that this vector field is skipped in this term of the sum.

Essentially we do two things: take turns feeding $\omega$ every $k$-tuple taken from $X_0,\dots,X_k$, and letting the remaining vector act on that function; and use Lie brackets to reduce the $k+1$ vector fields to $k$ vector fields, then feed that to $\omega$.

For instance, if $\theta$ is a one-form, then $$ d\theta(X,Y) = X \theta(Y) - Y\theta(X) - \theta([X,Y]) $$ Let $x^1,\dots x^n$ be a coordinate patch on $M$ Let $X = v^i \frac{\partial}{\partial x^i}$ and $Y = w^j \frac{\partial}{\partial x^j}$ be two vector fields on this patch. Then \begin{align*} [X,Y]f &= X(Yf) - Y(Xf) \\ &= v^i \frac{\partial}{\partial x^i}\left(w^j \frac{\partial f}{\partial x^j}\right) -w^j \frac{\partial}{\partial x^j}\left(v^i \frac{\partial f}{\partial x^i}\right) \\ &= v^i \left(\frac{\partial w^j}{\partial x^i}\frac{\partial f}{\partial x^j} + w^j \color{red}{\frac{\partial^2 f}{\partial x^i x^j}}\right) -w^j \left(\frac{\partial v^i}{\partial x^j}\frac{\partial f}{\partial x^i} + v^i \color{red}{\frac{\partial^2 f}{\partial x^j x^i}}\right) \\ &= v^i \frac{\partial w^j}{\partial x^i}\frac{\partial f}{\partial x^j} - w^j \frac{\partial v^i}{\partial x^j}\frac{\partial f}{\partial x^i} \\ &= \left(v^j \frac{\partial w^i}{\partial x^j}- w^j \frac{\partial v^i}{\partial x^j}\right)\frac{\partial f}{\partial x^i} \end{align*} The cancellation is due to Clairaut's theorem.

Suppose now that $\theta = f_k\,dx^k$ in these coordinates. Then \begin{align*} \theta(Y) &= (f_k\, dx^k)\left(w^j \frac{\partial f}{\partial x^j}\right) = f_k w^k \\ \implies X\theta(Y) &= v^i \frac{\partial}{\partial x^i}(f_k w^k) = v^i \frac{\partial f_k}{\partial x^i}w^k + v^i f_k \frac{\partial w^k}{\partial x^i} \\ \theta(X) &= (f_k \,dx^k)\left(v^i \frac{\partial f}{\partial x^i}\right) = f_i v^i \\ \implies Y\theta(X) &= w^j\frac{\partial}{\partial x^j}(f_i v^i) \\ &= w^j\frac{\partial f_i}{\partial x^j}v^i + w^j f_i \frac{\partial v^i} {\partial x^j} \\ \theta([X,Y]) &= f_i \left(v^j \frac{\partial w^i}{\partial x^j}- w^j \frac{\partial v^i}{\partial x^j}\right) \end{align*} After renaming dummy indices and subtracting, you get $$ d\theta([X,Y]) = (w^iv^j-v^iw^j)\frac{\partial f_i}{\partial x^j} $$

On the other hand, \begin{align*} \left(\frac{\partial f_i}{\partial x^j} dx^j \wedge dx^i\right) \left(v^k \frac{\partial}{\partial x^k},w^\ell \frac{\partial}{\partial x^\ell}\right) &= \frac{\partial f_i}{\partial x^j}\left[dx^j\left(v^k \frac{\partial}{\partial x^k}\right)dx^i\left(w^\ell \frac{\partial}{\partial x^\ell}\right) - dx^i\left(v^k \frac{\partial}{\partial x^k}\right)dx^j\left(w^\ell \frac{\partial}{\partial x^\ell}\right) \right] \\ &= \frac{\partial f_i}{\partial x^j}\left(v^jw^i - v^i w^j\right) \end{align*} These are the same thing.

So if you can show that this expression for $d\omega$ works out to your expression in local coordinates, you have shown the local expression is well-defined. It's no more complicated than the above, just with more indices!