Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.