Where is the mistake in this reasoning about differential forms?

[Rasalhague]

Lee 2003: Introduction to Smooth Manifolds ( http://books.google.co.uk/books?id=eqfgZtjQceYC&printsec=frontcover#v=onepage&q&f=false ) (search eg. for "computational"), Lemma 12.10 (b), p. 304:

In any coordinate chart

$$F^*\left ( \sum_I \omega_I \; dy^{i_1} \wedge ... \wedge dy^{i_k} \right )$$

$$= F^*\left ( \sum_I \omega_I \circ F \; d(y^{i_1} \circ F) \wedge ... \wedge d(y^{i_k} \circ F) \right )$$

where $I$ is an increasing multi-index: $(i_1,...i_k)$ with each value less than or equal to all those to the right of it. This PDF of lecture notes gives the following chain rule for exterior differentiation of differential 0-forms:

[itex]d(h \circ f) = h' \circ f \; df.[/tex]

But that implies

$$d(y^{i_\alpha} \circ F) = (y^{i_\alpha})' \circ F \; dF$$

and however we interpret the primed expression, more than one $dF$ wedged together equals 0. I assume this is not the case, as the lemma is supposed to give "a computational rule for pullbacks of differential forms". Where have I made a mistake?

 

[henry_m]

In that chain rule, f is a map from a manifold to reals, and h is a map from reals to reals. Of course we know what h' and df are.

F is a map between manifolds, $f:M\to N$. The y's are coordinate charts, maps from manifolds to subsets of Euclidean space, $y:N\to \mathbb{R}^n$. What does y' mean? And what about dF? These are objects which don't make sense. You must be very careful to remember what objects these functions map between, and what operations make sense on what sort of maps.

 

[Rasalhague]

I see. Thanks for your reply, Henry.