Symmetry of Fubini Theorem versus Skew Symmetry of 2-forms

207 Views Asked by At

Fubini's theorem states that if $f$ is $X \times Y$ integrable, then

\begin{equation*} \int_{X\times Y} f(x,y) \; \mathrm{d}(x,y) = \int_Y\left(\int_X f(x,y) \; \mathrm{d}x\right)\mathrm{d}y = \int_X\left(\int_Y f(x,y) \; \mathrm{d}y\right)\mathrm{d}x \end{equation*}

However, in the language of differential forms, this seems to say

\begin{align*} \int_{X \times Y} f(x,y) \; \mathrm{d}x\wedge\mathrm{d}y &= \int_Y\left(\int_X f(x,y) \; \mathrm{d}x\right)\mathrm{d}y \\ &= \int_X\left(\int_Y f(x,y) \; \mathrm{d}y\right)\mathrm{d}x \\ &= \int_{X \times Y} f(x,y) \; \mathrm{d}y\wedge\mathrm{d}x, \end{align*}

despite the fact that

\begin{equation*} \int_{X \times Y} f(x,y) \; \mathrm{d}x\wedge\mathrm{d}y = -\int_{X \times Y} f(x,y) \; \mathrm{d}y\wedge\mathrm{d}x \end{equation*}

by skew symmetry of the wedge product. What am I missing here? How do you make Fubini's theorem agree with the language of differential 2-forms?

1

There are 1 best solutions below

1
On

It just isn't true that $$\int_X \left(\int_Yf(x,y)dy\right) dx =\int_{X \times Y}f(x,y)dy \wedge dx,$$ and you have to go back to the definitions to understand why. This is just a matter of notation and definitions, but unraveling this can very well be a quagmire to the unattentive depending on how ingrained with notation abuse they are. And it is perhaps worth it to purge some overloading of variables.


Fubini's theorem says that $$\int_Y\left(y \mapsto\int_X f(\cdot,y)\right)=\int_{X \times Y} f = \int_X \left(x \mapsto \int_Yf(x,\cdot)\right).$$ It is a theorem about integrals of functions, not differential forms.

On the other hand, the definition of $\overline{\int}_{\mathbb{R}^2} \omega$ for $\omega$ a $2$-form on $\mathbb{R}^2$ (since you use $dx, dy$, I assume you are considering the case of $\mathbb{R}^2$) is as follows: Given $\omega$, there is a unique function $f$ such that $\omega= f dx \wedge dy$. We now define $$\overline{\int}_{\mathbb{R}^2}\omega = \int_{\mathbb{R} ^2}f. $$ We will indeed use a different notation in order to dispel bias. In a similar manner, we define the integral for $1$-forms in $\mathbb{R}$. (Or $\mathbb{R}^n$ for that matter.)

We then have the following chain of equations: \begin{align*} \overline{\int}_X \left(\overline{\int}_Yf(x,y)dy\right) dx &=^{(1)} \overline{\int}_X \left(x \mapsto \overline{\int}_Yf(x,\cdot)dy\right) dx \\ &=^{(2)} \overline{\int}_X\left(x \mapsto \int_Yf(x,\cdot) \right) dx \\ &=^{(3)} \int_X \left(x \mapsto \int_Yf(x,\cdot) \right) \\ &=^{(4)} \int_{X \times Y} f \\ &=^{(5)} \overline{\int}_{X \times Y} f dx \wedge dy, \end{align*} where in $^{(1)}$ we are just purging the usual "dummy variables"-loaded notation of integration, which often confuses matter in those situations. In $^{(2)}$ and $^{(3)}$ we are just using the definition of the integral of a differential form. In $^{(4)}$ we are using Fubini, and in $^{(5)}$ we are again using the definition of the integral of a differential form.