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I understood the definition of affine $k$-chain and that he defines $\int \limits_{\Gamma} \omega$ as $(82)$. But I can't understand the last two above examples. What does they mean?

Can anyone explain them detailed?

RFZ
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  • They're not so much examples as part of an explanation of notation. Do you know some algebraic topology? If you know simplicial or singular homology, you will know $k$-chains in that context. This is the same construct, but - as appropriate in the context - Rudin gives a different motivation. – Daniel Fischer May 01 '16 at 13:07
  • Unfortunately, i don't know the elements of algebraic topology. He says that (83) is just notation not an algebraic sum! But what about if we'll consider it as algebraic sum? There are some problems? These moments cofuses me – RFZ May 01 '16 at 13:22
  • There are only problems if one confuses different possible meanings of "$+$". If you knew some homology, we could appeal to a familiar concept in the explanation. Well, no problem, we can also explain it without, it'll just take a few words more. – Daniel Fischer May 01 '16 at 13:27
  • Sorry but I have no information aout homology :( – RFZ May 01 '16 at 13:33
  • I tried to find similiar questions in MSE but no results – RFZ May 01 '16 at 13:34
  • I'll write an answer. But it'll take a bit. – Daniel Fischer May 01 '16 at 13:37
  • @DanielFischer, Sure! I would be very thankful for your answer! – RFZ May 01 '16 at 13:47
  • Sorry, took a bit longer than anticipated. Life intervened. – Daniel Fischer May 01 '16 at 22:50

1 Answers1

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When you have a binary operation $\newcommand{\bop}{\mathop{\scriptstyle\top}}\bop \colon S\times S \to S$, that induces a corresponding operation on the space of functions $D \to S$ by applying the operation pointwise,

$$(f \bop g)(x) := f(x) \bop g(x).$$

The pointwise sum or product of real-valued functions are very familiar examples.

The same construction applied to affine $k$-simplices gives the concept of affine $k$-chains. However, one can view affine $k$-simplices as functions in different ways, and it's the non-direct way that gives rise to $k$-chains.

To avoid any ambiguity, let me use different notations for the two ways to view $k$-simplices as functions that Rudin mentions.

First, by definition an affine $k$-simplex is a function (sufficiently regular) $\sigma \colon Q^k \to \mathbb{R}^n$. We have an addition on $\mathbb{R}^n$, and that induces an addition on the space of functions $Q^k \to \mathbb{R}^n$. Let's denote this addition by $\oplus$. Then $(\sigma_1 \oplus \sigma_2) \colon u \mapsto \sigma_1(u) + \sigma_2(u)$ is an affine $k$-simplex if $\sigma_1$ and $\sigma_2$ are affine $k$-simplices. In the context of integration of differential forms, this operation is however uninteresting and rarely - if ever - considered.

The interesting concept arises when one views an affine $k$-simplex (or more generally a $k$-surface) in $E$ as a map $\Omega^k(E) \to \mathbb{R}$ via integration, where $\Omega^k(E)$ denotes the space of (continuous) $k$-forms in $E \subset \mathbb{R}^n$. Formally, since that is not exactly the same thing as the $k$-surface $\Phi$, it should be denoted differently, but doing that would be cumbersome, so it is customary to abuse notation and denote this map also by $\Phi$. For this discussion, I will however use the notation $I_\Phi$ for the map $\omega \mapsto \int_{\Phi} \omega$ to disambiguate. Thus every $k$-surface $\Phi$ in $E$ defines a map $I_{\Phi} \colon \Omega^k(E) \to \mathbb{R}$, and we have the induced addition

$$(I_{\Phi} + I_{\Psi}) \colon \omega \mapsto I_{\Phi}(\omega) + I_{\Psi}(\omega) = \int_{\Phi} \omega + \int_{\Psi} \omega.$$

It is this induced addition that pertains to affine $k$-chains. An affine $k$-chain in $E$ "is" a map $\Omega^k(E) \to \mathbb{R}$ which we can write as the sum of finitely many $I_{\sigma_i}$, where each $\sigma_i$ is an affine $k$-simplex in $E$. But out of convenience, one drops the $I$s and writes $k$-chains as $\sigma_1 + \dotsc \sigma_r$ rather than $I_{\sigma_i} + \dotsc + I_{\sigma_r}$.

In the penultimate paragraph of the section, Rudin explains that one could be tempted to interpret the notation $\sigma_1 + \sigma_2$ as $\sigma_1 \oplus \sigma_2$, but that is not how $(83)$ is to be interpreted. In the last paragraph, he gives an example illustrating that these two additions are very different. If $\sigma_1$ and $\sigma_2$ are affine $k$-simplices with $\sigma_2 = -\sigma_1$ in the sense of $(80)$, then by theorem 10.27 we have $I_{\sigma_1} + I_{\sigma_2} = 0$, and that means $\sigma_1 + \sigma_2 = 0$ in the sense of $k$-chains, but generally we don't have $\sigma_1 \oplus \sigma_2 = 0$, where the last $0$ is the constant map $0 \colon u \mapsto 0 \in \mathbb{R}^n$.

Daniel Fischer
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  • Very bright and clear answer but let me ask you some questions: 1) You wrote that in general we don't have $\sigma_1\oplus \sigma_2=0$. – RFZ May 02 '16 at 09:10
  • Lets $\sigma_1=[\mathbf{p}_0, \mathbf{p}_1, \mathbf{p}_2]$ and $\sigma_2=[\mathbf{p}_1, \mathbf{p}_0, \mathbf{p}_2]$. Moreover $\sigma_1=-\sigma_2$ in sense of $(80)$. But $\sigma_1(\mathbf{u})=\mathbf{p}_0+u_1(\mathbf{p}_1-\mathbf{p}_0)+u_2(\mathbf{p}_2-\mathbf{p}_0)$ and $\sigma_2(\mathbf{u})=\mathbf{p}_1+u_1(\mathbf{p}_0-\mathbf{p}_1)+u_2(\mathbf{p}_2-\mathbf{p}_1)$. Then $\sigma_1(0)\oplus\sigma_2(0)=\mathbf{p}_0\oplus\mathbf{p}_1\neq 0$ if $\mathbf{p}_0\oplus\mathbf{p}_1\neq 0$. Am I right? – RFZ May 02 '16 at 09:10
  • Right. And generally, $(\sigma_1 \oplus \sigma_2)(\mathbf{u}) = \mathbf{p}_0 + \mathbf{p}_1 + u_2(2\mathbf{p}_2 - \mathbf{p}_1 - \mathbf{p}_0)$, so you see you get a "degenerate" simplex. The map is independent of $u_1$, so you effectively have a $1$-simplex trivially extended to $Q^2$. If you take a $1$-simplex $\sigma_1 = [\mathbf{p}_0,\mathbf{p}_1]$, you get a constant map $\sigma_1 \oplus \sigma_2 \colon u \mapsto \mathbf{p}_0 + \mathbf{p}_1$, again effectively a simplex one dimension lower, trivially extended. That's generally the case. – Daniel Fischer May 02 '16 at 09:46
  • I little bit misunderstood your last comment. What does it show? – RFZ May 02 '16 at 14:25
  • It just expands a bit on how the map $\sigma_1 \oplus \sigma_2$ looks in general. You have established that in general $(\sigma_1 \oplus \sigma_2)(0) = \mathbf{p}_0 + \mathbf{p}_1 \neq 0$. I have looked at $(\sigma_1 \oplus \sigma_2)(\mathbf{u})$ for general $\mathbf{u} \in Q^k$ (first, with your example for $k = 2$, then briefly for $k = 1$). If you look at $\sigma_1 \oplus \sigma_2$, where $\sigma_2 = -\sigma_1$ in the sense of $(80)$, then the image of $\sigma_1 \oplus \sigma_2$ is a $(k-1)$-dimensional object. – Daniel Fischer May 02 '16 at 14:57
  • [Slight caveat: if $\sigma_1$ is already a degenerate simplex, contained in an affine subspace of dimension $d < k$, then the image of $\sigma_1 \oplus \sigma_2$, where $\sigma_2 = -\sigma_1$ in the sense of $(80)$, may still have dimension $d$. Only for nondegenerate simplices is a reduction of dimension of the image guaranteed when we add two opposite simplices.] – Daniel Fischer May 02 '16 at 14:57
  • Dear Daniel, can you help me please with the following question? http://math.stackexchange.com/questions/1770247/boundary-of-oriented-k-simplex-from-pma-rudin – RFZ May 03 '16 at 19:23
  • This moment from Rudin's book seems to me wrong. Any your answer would be appreciated – RFZ May 03 '16 at 19:52