Integrating a differential form

214 Views Asked by At

I am currently going through Introduction to Smooth manifolds by John Lee and am a bit confused with the integration of differential forms.

So given a smooth k-form $\omega = f dx^1 \wedge ... \wedge dx^k $ on some integral domain $D \subset \mathbb{R}^k$ we can integrate $\omega$ over $D$, which is given in the book by $$\int_D f dx^1 \wedge ... \wedge dx^k = \int_D f dx^1 ... dx^k \, \, \, \, \, \, (\star)$$

What I am confused about is that if we interchange say the $dx^1$ and the $dx^2$ in both sides of the equation we get a negative side arising on the left as $dx^1 \wedge dx^2 = - dx^2 \wedge dx^1$, whilst on the right hand side there will be no sign factor as $dx^1 dx^2 =dx^2 dx^1$.

Is the reason that the above is not a contradiction a result of the fact we have fixed an orientation $(x^1, ..., x^k)$, and so with this orientation we get that $(\star)$ holds but $$\int_D f dx^2 \wedge dx^1 \wedge ... \wedge dx^k \neq \int_D f dx^2 dx^1 dx^3 ... dx^k?$$

That is, we can only integrate once we have ordered the differential form into its chosen orientation?

1

There are 1 best solutions below

3
On

Your formula should be written $$\int_D f(x)\> dx^1 \wedge ... \wedge dx^k = \int_D f(x)\>{\rm d}(x)\ , \tag{$\star$}$$ insofar as on the right hand side there is not meant a product of first order differentials of any sort, but integration with respect to standard, and in particular: orientation independent, Lebesgue measure in ${\mathbb R}^k$. The formula $(\star)$ on the other hand involves a choice of orientation: It says that the products differring by an even permutation from $dx^1 \wedge ... \wedge dx^k$ are considered positive, i.e., in line with Lebesgue measure.