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In Terence Tao's note:

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):


$$\omega=\sum a_{i_1,\dots,i_k}({\mathbf x})\,dx^{i_1} \wedge \cdots \wedge dx^{i_k} $$

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

$$\int_S \omega =\int_D \sum a_{i_1,\dots,i_k}(S({\mathbf u})) \frac{\partial(x^{i_1},\dots,x^{i_k})}{\partial(u^{1},\dots,u^{k})}\,du^1\ldots du^k$$



is the determinant of the Jacobian.

  1. 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?

  2. 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!

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I tried to use \begin{align} to break the last equation into two lines, but it doesn't work. Any idea? Thanks! – Tim Sep 8 '11 at 6:22
I fixed it up. You have to place three backslashes at the end of each line because of how Markdown does escaping, I think. – Dylan Moreland Sep 8 '11 at 6:32
up vote 4 down vote accepted

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.

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Thanks! Yes, I am interested in knowing how the Jacobian is a coordinate-dependent expression hidden in the pullback, and some other ways of computing (which are less theoretically tidy) which are more efficient. Really appreciate your time. – Tim Sep 8 '11 at 17:50
Intuitively $\int_R dx^1 \wedge \cdots \wedge dx^n$ sounds like a definition of a volume, that I evaluate via successive "numerical" integrals $V=\int dx^1...\int dx^n$. But I need to know that I get the same answer in every coordinate frame. So I rewrite the integrand in terms of some frame $y^b$, and out pops a Jacobian determinant. I integrate again, Jacobian and all, to get some definite scalar $V'$, which I cannot simply define to be $V$. But how do we prove $V'=V$ (i.e. that Jacobians work) without a circular use of the very definition of volume which we are trying to prove consistent? – Adrian Ratnapala Jan 12 '13 at 9:21

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