From what I can say, in most books about differential geometry and differential forms (see for example Flanders, Differential Forms with applications to the Physical Sciences), the integral of a $p$-differential form $\omega$ on a chain is defined as follows: one takes a simplex (or a cube) $\Delta$ in $\mathbb R^p$ and maps it into the manifold $M$ on which $\omega$ is defined, with a smooth mapping $\phi:\,\Delta\subseteq\mathbb R^p\to M$. Then the pullback $\phi^*\omega$ is a $p$-form on $\mathbb R^p$ which can be uniquely written as $$\phi^*\omega=A(x^1,\,x^2,\,\ldots,\,x^p)\,\mathrm dx^1\wedge\mathrm dx^2\wedge\ldots\wedge\mathrm dx^p. \label{1}\tag{1}$$ The integral is defined as $$ \int_\phi \omega=\int_\Delta A(x^1,\,x^2,\,\ldots,\,x^p)\,\mathrm dx^1\mathrm dx^2\ldots\mathrm dx^p. \label{2}\tag{2} $$ The obvious “problem” with this definition is that, at least in principle, it is dependent on a choice of coordinates. However, thanks to the change of variable formula, the invariance is ensured.

Despite the fact that this definition works fine, it seems to me that the change of variable formula should be a consequence of the wedge product rules. Instead, what happens in the definition above is that the wedge product rules just “match” with the change of variable formula, making the integral well defined.

Question: why don't we define the integral in an invariant way, such that the change of variable formula follows from the wedge product rules? For example, one could make a construction similar to the one used for the Riemann integral, breaking $\Delta$ into small rectangles, each corresponding to a $p$-vector, then letting the pullback $\phi^*\omega$ act on these $p$-vectors, then summing and taking the limit for finer and finer partitions. I know this is exactly the idea behind integration of forms, but why isn't this idea made into an explicit construction, such that the integral is already invariant from the beginning? Is there a textbook that does so? Or am I missing a step?

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    $\begingroup$ The Change of Variables Theorem is a rather deep result; you can’t just definition it away. $\endgroup$ Aug 28, 2023 at 15:30
  • $\begingroup$ You might be interested in this question: math.stackexchange.com/q/1023064/1210477 $\endgroup$
    – M W
    Aug 28, 2023 at 19:01
  • $\begingroup$ @MW thank you, is it really that difficult though? Is my sketch that naive? $\endgroup$ Aug 28, 2023 at 19:48
  • $\begingroup$ @TedShifrin is my sketch actually that naive? $\endgroup$ Aug 28, 2023 at 19:48
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    $\begingroup$ Best and funniest answer I found so far: math.stackexchange.com/a/4183216/949989 $\endgroup$
    – Kurt G.
    Aug 29, 2023 at 6:52

1 Answer 1


[Updating my previous answer which rather missed the point of your question.]

Formula \eqref{2} already could be written $$\int_\phi \omega=\int_\Delta \langle\phi^*\omega,e_1\wedge\cdots\wedge e_p\rangle\,\mathrm dx^1\mathrm dx^2\ldots\mathrm dx^p\text{,} \label{3}\tag{3}$$

which is exactly what your Riemann integral approach is describing. So your approach is not really coordinate free.

In other words, if $g\colon \Delta'\to\Delta$ is a diffeomorphism, you still need to establish that $$\int_\Delta \langle\phi^*\omega,e_1\wedge\cdots\wedge e_p\rangle\,\mathrm dx^1\mathrm dx^2\ldots\mathrm dx^p =\int_{\Delta'} \langle g^*\phi^*\omega,f_1\wedge\cdots\wedge f_p\rangle\,\mathrm dy^1\mathrm dy^2\ldots\mathrm dy^p \text{,} \label{4}\tag{4}$$ where $\{e_i\}$ and $\{f_j\}$ are bases for $\Delta$ and $\Delta'$.

Granted, the proof of \eqref{4} will eventually use rules of the wedge product at an infinitesimal level, but it is not at all trivial to establish it globally, as others have mentioned in the comments.

  • $\begingroup$ I think I understand. Could we say that the dependence on the basis in my sketch is in the choice of the partition? $\endgroup$ Aug 29, 2023 at 8:04
  • $\begingroup$ @LeonardoRossi Well, the dependence of the basis is in your choice of partition, yes, but that's not really the difficulty with your approach. You could probably argue, based on facts about wedge product and volume, that choosing a different partition (corresponding to different basis) gives the same answer. But that only shows you're invariant under a linear change in variables. (If you saw my deleted answer that's more or less the approach I was taking). $\endgroup$
    – M W
    Aug 29, 2023 at 8:16

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