If $Ω$ is any open bounded domain in $R^n$ , we then have the identity
$$\int_Ω f (x)dx_1 ∧ . . . ∧ dx_n = \int_Ω f (x) dx$$where on the left we have an integral of a differential form (with $Ω$
viewed as a positively oriented n-dimensional manifold), and on the
right we have the Riemann or Lebesgue integral of $f$ on $Ω$.
From Wikipedia (basically same as in baby Rudin):
Let
$$\omega=\sum a_{i_1,\dots,i_k}({\mathbf x})\,dx^{i_1} \wedge \cdots
\wedge dx^{i_k} $$be a differential form and $S$ a differentiable $k$-manifold over
which we wish to integrate, where $S$ has the parameterization$$S({\mathbf u})=(x^1({\mathbf u}),\dots,x^n({\mathbf u}))$$
for $u$ in the parameter domain $D$. Then (Rudin 1976) defines the
integral of the differential form over $S$ as$$\int_S \omega =\int_D \sum a_{i_1,\dots,i_k}(S({\mathbf u}))
\frac{\partial(x^{i_1},\dots,x^{i_k})}{\partial(u^{1},\dots,u^{k})}\,du^1\ldots
du^k$$where
$$\frac{\partial(x^{i_1},\dots,x^{i_k})}{\partial(u^{1},\dots,u^{k})}$$
is the determinant of the Jacobian.
-
I wonder if in the case of Wikipedia, the change of variable can be
eliminated just as in Terence Tao's, for example,\begin{align} \int_S \omega &=\int_D \sum
a_{i_1,\dots,i_k}(S({\mathbf u}))
\frac{\partial(x^{i_1},\dots,x^{i_k})}{\partial(u^{1},\dots,u^{k})}\,du^1\ldots
du^k \\ &=\int_S \sum a_{i_1,\dots,i_k}(x) \,dx^{i_1}\ldots dx^{i_k}
? \end{align}If not, when can it be?
- If the manifold $S$ is not a subset of $R^n$, the Jacobian will not make sense. Can $\int_S \omega $
still be represented by Riemann/Lebesgue integral? How is that like if yes?
Thanks and regards!
Best Answer
Most sources will take Tao's "identity" as a definition: $dx^1 \wedge \cdots \wedge dx^n$ measures volume in $\mathbf R^n$ in the a way that corresponds to our intuition and previous notions of integration in Euclidean space. It's not clear to me that he's eliminating any "change of variable".
To integrate an $n$-form $\omega$ (let's assume compactly supported so I don't have to worry about convergence) on an $n$-manifold $M$ that is neither (a) embedded in Euclidean space nor (b) parametrized by a single chart, we can do the following. If we first assume that $(U, \varphi)$ is a chart on $M$ containing the support [the points at which the fibre of $\omega$ is non-zero] of $\omega$, then we define $$ \int_M \omega = \int_{\varphi(U)} (\varphi^{-1})^*\omega $$ where $(\varphi^{-1})^*\omega$ is the pullback. The integral on the right is computed via Tao's definition.
In general, we take a finite collection of charts $(U_i, \varphi_i)$ covering the support of $\omega$ and a smooth partition of unity $\{\psi_i\}$ of $M$ subordinate to $\{U_i\}$. Then let $$ \int_M \omega = \sum_i \int_M \psi_i\omega. $$ This makes sense, because the support of $\psi_i\omega$ is contained in $U_i$. It is not too hard to show that this definition of $\int_M \omega$ is independent of the many choices we've made. For a better discussion, look in any introduction to manifolds, e.g. Lee's Introduction to Smooth Manifolds.
Note. Here the Jacobian is a coordinate-dependent expression hidden in the pullback. I may write out how this goes later if there is interest and time. Also, I would avoid using this to actually compute $\int_M \omega$. There are other ways of computing (which are less theoretically tidy) which are more efficient.