Simplex Calculations for Stokes' Theorem

Oriented affine $k$-simplex $\sigma = [{\bf p}_0,{\bf p}_1,\dots,{\bf p}_k]$
A $k$-surface given by the affine function $$\sigma\left(\sum_{i=1}^k a_i {\bf e}_i \right) := {\bf p}_0 + \sum_{i=1}^k a_i ({\bf p}_i - {\bf p}_0) \tag{1},$$ where ${\bf p}_i \in \R^n$ for all $i \in \{1,\dots,k\}$.
In particular, $\sigma({\bf 0})={\bf p}_0$ and for each $i\in\{1,\dots,k\}$, $\sigma({\bf e}_i)={\bf p}_i$.
Standard simplex $Q^k := [{\bf 0}, {\bf e}_1, \dots, {\bf e}_k]$
A particular type of oriented affine $k$-simplex with the standard basis $\{{\bf e}_1, \dots, {\bf e}_k\}$ of $\R^k$. $$Q^k := \left\{ \sum_{i=1}^k a_i {\bf e}_i \Biggm| \forall i \in \{1,\dots,k\}, a_i \ge 0, \sum_{i=1}^k a_i = 1 \right\}$$

Note that an oriented affine $k$-simplex $\sigma$ has parameter domain $Q^k$.

Affine $k$-chain $\Gamma$
a finite collection of oriented affine $k$-simplexes $\sigma_1,\dots,\sigma_r$
Boundary of an oriented affine $k$-simplex $\partial \sigma$
an affine $k-1$-chain $$ \partial \sigma = \sum_{j=0}^k (-1)^j \;\underbrace{ [{\bf p}_0,\dots,{\bf p}_{j-1},{\bf p}_{j+1},\dots,{\bf p}_k]}_{{\bf p}_j \text{ removed}} \tag{2} $$

We shall make no use of the following proposition.

Proposition Let $\sigma$ be an oriented affine $k$-simplex. Then $\partial^2 \sigma = 0$.

Proof: In this $k-2$-chain, each $k-2$-simplex with ${\bf p}_i$ and ${\bf p}_j$ removed can be obtained in two ways. WLOG, assume $i < j$.

  1. ${\bf p}_i$ removed first, followed by ${\bf p}_j$. These two operations give factors $(-1)^i$ and $(-1)^{j-1}$. The “$-1$” in the exponent “$j-1$” is a result of ${\bf p}_j$’s left-shifting after removal of ${\bf p}_i$.
  2. ${\bf p}_j$ removed first, followed by ${\bf p}_i$. These two operations give factors $(-1)^j$ and $(-1)^i$. Removing ${\bf p}_j$ doesn’t affect ${\bf p}_i$’s position.

These two factors cancel each other. $$\tag*{$\square$}$$

Integral of a $0$-form over an oriented 0-simplex
Let $\sigma = \pm {\bf p}_0$ be an oriented 0-simplex. $$\int_\sigma f := \pm f({\bf p}_0)$$

This defintion seems strange to me since the integral over a singleton in $\R$ of any real valued integrable function $f$ is zero. However, I have to accept this to make a progress.

In the proof of Stokes’ Theorem, the author has taken $\sigma = Q^k$.

$$ \begin{aligned} \partial \sigma &= [{\bf e}_1,\dots,{\bf e}_k]+\sum_{i=1}^k (-1)^i \tau_i \\ &= (-1)^{r-1} \tau_0 + \sum_{i=1}^k (-1)^i \tau_i, \end{aligned} $$

where $\tau_0 = [{\bf e}_r, {\bf e}_1, \dots, {\bf e}_{r-1}, {\bf e}_{r+1},\dots, {\bf e}_k]$, and $\tau_i = [{\bf 0}, {\bf e}_1, \dots, {\bf e}_{i-1}, {\bf e}_{i+1},\dots, {\bf e}_k]$ for $i \in \{1,\dots,k\}$. Each $\tau_i$ admits $Q^k$ as its parameter domain.

Put ${\bf x} = \tau_0({\bf u})$ with ${\bf u} \in Q^{k-1}$. I need some straightforward calculations to know each coordinate of $\bf x$.

$$ \begin{aligned} {\bf x} &= \tau_0({\bf u}) \\ &= [{\bf e}_r, {\bf e}_1, \dots, {\bf e}_{r-1}, {\bf e}_{r+1}, \dots, {\bf e}_k]({\bf u}) \\ &= {\bf e}_r + u_1 ({\bf e}_1 - {\bf e}_r) + u_2 ({\bf e}_2 - {\bf e}_r) + \dots + u_{r-1} ({\bf e}_{r-1} - {\bf e}_r) \\ &+ u_r ({\bf e}_{r+1} - {\bf e}_r) + \dots + u_{k-1} ({\bf e}_k - {\bf e}_r)\\ &= \sum_{j=1}^{r-1} u_j {\bf e}_j + \left(1 - \sum_{i = 1}^{k-1} u_i \right) {\bf e}_r + \sum_{j=r+1}^k u_{j-1} {\bf e}_j \\ \therefore x_j &= \begin{cases} u_j & 1 \le j \le r-1 \\ 1 - \sum_{i = 1}^{k-1} u_i & j = i \\ u_{j-1} & r+1 \le j \le k \end{cases} \end{aligned} $$

Reference: Rudin’s PMA


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