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I am reading the proof of Stokes' theorem from PMA Rudin but one moment seems to very weird. enter image description here

Why Rudin considers the case when $\sigma=[0,\mathbf{e}_1,\dots, \mathbf{e}_k]$? After all $\sigma$ may take the form $[\mathbf{p}_0, \mathbf{p}_1,\dots, \mathbf{p}_k]$.

Can anyone explain this important moment?

I can't understand it by myself.

RFZ
  • 16,814
  • He explains it in the following lines. With theorems 10.22 and 10.25, you can pull back the general case to that special case, so it suffices to prove it for the standard simplex. – Daniel Fischer May 07 '16 at 19:01
  • @DanielFischer, Sorry but how theorems 10.22 and 10.25 can be used? In What sense? – RFZ May 07 '16 at 19:07
  • Pull back via $T$. By 10.22, we have $(d\omega)T = d(\omega_T)$. By 10.25, we have $\int{\Phi} d\omega = \int_{\sigma} (d\omega)T$ and $\int{\partial \Phi} \omega = \int_{T(\partial \sigma)} \omega = \int_{\partial \sigma} \omega_T$. – Daniel Fischer May 07 '16 at 19:11
  • I understood that moment. But I am interesting why he considers the case when simplex have form $\sigma=[0,\mathbf{e}_1,..,\mathbf{e}_k]$ but not some $\sigma=[\mathbf{p}_0,\mathbf{p}_1,..,\mathbf{p}_k]$. – RFZ May 07 '16 at 19:13
  • The main question is: Why we can represent $\Phi$ as $T\circ \sigma$ where $\sigma=[0,e_1,\dots, e_k]$. That worries me :/ – RFZ May 07 '16 at 19:17
  • Because the affine simplex $[0,e_1,\dotsc,e_k]$ is given by the identity map. The set it covers is just $Q^k$, and by definition of a differentiable $k$-simplex $\Phi$ is given by a differentiable map $T$ defined in some open neighbourhood of $Q^k$. $\Phi = T\lvert_{Q^k}$, so $\Phi = \Phi \circ \operatorname{id}{Q^k} = T\circ \operatorname{id}{Q^k} = T\circ\sigma$, since $\sigma = \operatorname{id}_{Q^k}$. – Daniel Fischer May 07 '16 at 19:22
  • Let us continue this discussion in chat. – RFZ May 07 '16 at 19:26
  • @DanielFischer, Since $\Phi$ is oriented $k$-simplex of class $C''$ in open set $V\subset \mathbb{R}^m$ $\Rightarrow$ by definition 10.30 exists some mapping $T$ from $E$ into $V\subset \mathbb{R}^m$, i.e. domain of $T$ is some open set $E$. Also by definition 10.30 $\Phi$ has domain $Q^k$. Hence $\Phi=T\mid_{Q^k}$. Note that $Q^k\subset E$ because otherwise $\Phi=T\mid_{Q^k}$ would be false. Are my reasonings true? – RFZ May 08 '16 at 12:30
  • That's right so far. – Daniel Fischer May 08 '16 at 12:32
  • Hello Daniel! Can you please help with this http://math.stackexchange.com/questions/1777806/definition-of-exact-form? – RFZ May 09 '16 at 18:00
  • I think rudin utilizes the fact that compact convex open sets are homeomorphic to each other, so that open set E can be pulled back to open set around origin. – somitra Feb 16 '17 at 18:26
  • @DanielFischer, Could you help please with this topic? http://math.stackexchange.com/questions/2178128/the-erdos-and-turan-function/2178299#2178299 – RFZ Mar 09 '17 at 08:29

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