In Rudin, definition of differential form is
Suppose $E$ in an open set in $\Bbb R^{n}$. A differential form of order $k \geq 1$ in $E$ is a function $\omega$, symbolically represented by the sum$$\omega=\sum a_{i_{1},\dots,i_k}(x)dx_{i_{1}}\wedge\cdots\wedge dx_{i_{k}}$$
which assigns to each $k$-surface $\Phi$ in $E$ a number $\omega (\Phi) = \int_{\Phi}\omega$ according to the rule$$\int_{\Phi}\omega=\int_{D}\sum a_{i_{1},\dots,i_k}(\Phi(\mathbf u))J(\mathbf u)d\mathbf u$$
where $J$ is the Jacobian Matrix,
$$J(\mathbf u) = \frac{\partial(x_{i_{1}},…,x_{i_{k}} )}{\partial (u_{1},….,u_{k})}$$
My question is that, $d\mathbf u$ here refers to $du_{1}du_{2}…du_{k}$, which is used in Riemann integration, or refers to $du_{1}\wedge du_{2}…\wedge du_{k}$. My understanding is that $d\mathbf u$ refers to the first one, so RHS is our usual integration and we can calculate the number. But a later proof in Rudin said $du$ refers to the second one, then I don't know how to calculate it.
Thanks in advance!
Best Answer
You are correct, the symbol $d\mathbf u$ indicates the standard Riemann (or Lebesgue) integration measure over $D$. You don't know how to integrate the new object $dx_{i_1} \wedge \dots \wedge dx_{i_k}$, and Rudin wants to tell you: the equation $$ \int_{\Phi} \omega :=\int_{D} \sum a_{i_{1}, \ldots, i_{k}}(\Phi(\mathbf{u})) J(\mathbf{u}) d \mathbf{u} $$ is nothing but a definition of the symbol on the LHS (which contains $dx_{i_1} \wedge \dots \wedge dx_{i_k}$) in terms of objects you already know (everything on the RHS).
At least intuitively with this definition, $k$-forms live on $k$-surfaces, which can be parametrized by variables $\mathbf u$ in some parameter space $D \subseteq \mathbb R^k$. If you take a look at the statement of Theorem 10.24, Rudin clearly states that, besides the $k$-surface $\Phi$, he's considering the parameter domain $D$ as a $k$-surface in its own right (he calls it $\Delta$), with the trivial parametrization given by the identity: $$\Phi \xleftarrow{\ \ \mathbf x } D \xrightarrow{\mathbf u \equiv \operatorname{id}} \Delta $$ This is why he uses the symbol $du_{i_1} \wedge \cdots \wedge du_{i_k}$ for the basic $k$-form over $\Delta$: the letter after the $d$ should reflect the name of the parametrization, but since it's the identity one may just use the name of the variable in its domain. And indeed, throughout the proof, he's very careful to call the domain of integration $\Delta$ when he means the integral in the sense of the LHS, and $D$ when he means the integral in the usual sense; accordingly, the basic $k$-forms appear in the former case, $d\mathbf u$ appears in the latter.