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I was reading the book Introduction to Manifold by Loring W Tu. And I am confused with a remark Tu made in his book. I need a little bit of clarification.

In Chapter 5 (differential forms), he wrote " Because integration of function on Euclidean space depends on a choice of coordinates and is not invariant under a change of coordinate, it is not possible to integrate functions on manifold. "

Can anyone tell me what does he mean by integration of function is not invariant under change of co ordinate in Euclidean space?

Timon
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There is a comment on that in Lee's "Introduction to Smooth Manifolds": consider an $n$-dimensional cube $C$ in $\mathbb R^n$, and let $f:C\to\mathbb R$ be the constant function $f(x)=1$. Then, what should happen is $\displaystyle \int_CfdV=\rm{vol}(C)$, but the volume is not invariant under changes of coordinates.

detnvvp
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    I wonder, how the volume is defined here. – SBF Jul 11 '13 at 10:57
  • @Ilya: I would say it's the Lebesgue measure of the cube (i.e. the usual concept of volume). – Daniel Robert-Nicoud Jul 11 '13 at 11:33
  • @DanielRobert-Nicoud: shan't one change the measure together with the change of coordinates then? – SBF Jul 11 '13 at 11:34
  • @Ilya: Yes, and this is sort of what differential forms do. The point here is that you cannot integrate a function on a manifold; changing the coordinates creates an effect that you cannot see by just being a function. Here, the cube $C$ could be parameterized by either $C$ or $2C$, and you are not able to choose which parameterization is "correct", so that to integrate over it. – detnvvp Jul 11 '13 at 12:50
  • @detnvvp: ok, you mean that having a smooth structure may lead us to the danger of integrating using charts (since $\Bbb R^n$ appears to have a measure structure as well), but this is not consistent since it would depend on a chart, right? – SBF Jul 11 '13 at 12:53
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    @Ilya: Exactly, that is the case if you try to integrate a function on a manifold; the result depends on the chart. – detnvvp Jul 11 '13 at 12:55