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The definitions which we need for the proof of the theorem. We define the standard simplex $Q^k$ to be the set of all $u$ $\in$ $R^k$ of the form $u$ = $\sum_{i=1}^k$ $\alpha_i$ $e_i$.

Assume now that $p_0$, $p_1$,...$p_k$ are points of $R^n$.

The oriented affine $k$-simplex $\sigma$ $=$ [$p_0$, $p_1$,...$p_k$] is defined to be the $k$-surface in $R^n$ with parameter domain $Q^k$ which is given by the affine mapping $\sigma$($\sum_{i=1}^k$ $\alpha_i$ $e_i$) $=$ $\sigma(u)$ $=p_0$ + $\sum_{i=1}^k$ $\alpha_i$($p_i$ - $p_0$). Note that $\sigma$ is characterized by $\sigma(0)$ = $p_0$, $\sigma(e_i)$ = $p_i$ (for $1$ $\leq$ $i$ $\leq$ $k$).

For $k$ $\geq$ $1$, the boundary of the oriented affine $k$-simplex

$\sigma$ $=$ [$p_0$, $p_1$,...$p_k$] is defined to be the affine ($k-1$)-chain

$\partial$$\sigma$ = $\sum_{j=0}^k$ $(-1)^j$[$p_0$,...,$p_{j-1}$,$p_{j+1}$,..,$p_k$]. For $1$ $\leq$ $j$ $\leq$ $k$, observe that the simplex $\sigma_j$ = [$p_0$,...,$p_{j-1}$,$p_{j+1}$,..,$p_k$] has $Q^{k-1}$ as its parameter domain and that is defined by

$\sigma_j(u)$ = $p_0$ + $Bu$ ( $u$ $\in$ $Q^{k-1}$)

The class $\mathscr C'$ means the class of continuously differentiable functions and etc.

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$x_j$ = ${ \begin{cases} {u_j (1 \leq j \lt r),} \\ {1 - (u_1 + ... + u_{k-1}) (j=r),} \\ {u_{j-1} (r \lt j \leq k). } \end{cases} } $

( this is $(98)$).

$x_j$ = ${ \begin{cases} {u_j (1 \leq j \lt i),} \\ { 0 (j=i),} \\ {u_{j-1} (i \lt j \leq k). } \end{cases} } $ .

( this is $(99)$).

I dont'understand how do we get the $(98)$ and $(99)$.

Any help would be appreciated.

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  • $\begingroup$ Maybe you should write what is $Q^{k-1}$ and how $[e_1,e_2,\dots,e_k](\mathbf{u})$ ist defined as a map. $\endgroup$
    – hal4math
    Commented Mar 10, 2022 at 10:40
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    $\begingroup$ @hal4math Check it. I did it. $\endgroup$
    – JohnNash
    Commented Mar 10, 2022 at 11:26

1 Answer 1

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You have $x = \tau_0(u) = e_r + \sum_{i=1}^{k-1} u_i(p_i - e_r)$. So if $1 \leq j < r$, then you get

$$x_j = \left(e_r + \sum_{i=1}^{k-1} u_i p_i -\sum_{i=1}u_i e_r\right)_j = u_j,$$

since for $j < r$ we have $p_j = e_j$. However, for $j > r$ we have $p_{j-1} = e_{j}$, so that in that case

$$x_j = \left(e_r + \sum_{i=1}^{k-1} u_i p_i -\sum_{i=1}u_i e_r\right)_j = u_{j-1}.$$

For the case $j=r$, it is

$$ \begin{aligned} x_j &= \left(e_r + \sum_{i=1}^{k-1} u_i p_i -\sum_{i=1}^{k-1} u_i e_r\right)_j \\ &= \left(e_r + \sum_{i=1}^{k-1} u_i p_i -\sum_{i=1}^{k-1} u_i e_r\right)_j \\ &= \left(e_r -\sum_{i=1}^{k-1} u_i e_r\right)_j \\ &= \left((1 -\sum_{i=1}^{k-1} u_i) e_r\right)_j \\ &= 1 -\sum_{i=1}^{k-1} u_i. \end{aligned} $$

Hope that helps!

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  • $\begingroup$ sorry I don't understand how did you get $u_j$ in the fist line. please explain it more explicitly. $\endgroup$
    – JohnNash
    Commented Mar 11, 2022 at 11:34
  • $\begingroup$ You have $\sum_{i=1}^{k-1} u_i p_i = u_1p_1 + \dots u_{r-1}p_{r-1} + \dots u_{r}p_{r} +\dots u_{k-1}p_{k-1}$. So plugging in the $p_i$s you get $u_1e_1 + \dots u_{r-1}e_{r-1} + \dots + u_{r}e_{r+1} +\dots + u_{k-1}e_{k}$. There you see that the $j$-th component is $u_j$. Notice that I should not have summed up to $k$ but only $k-1$. $\endgroup$
    – hal4math
    Commented Mar 11, 2022 at 11:47
  • $\begingroup$ the j-th component is $u_j$$e_j$ . and why are the other other components equal of zero? @hal4math $\endgroup$
    – JohnNash
    Commented Mar 11, 2022 at 11:49
  • $\begingroup$ They are not zero, but they are not in the j-th component? Also the $j$-th component is $u_j$ and not $u_j e_j$. That is still a vector. $\endgroup$
    – hal4math
    Commented Mar 11, 2022 at 11:51
  • $\begingroup$ okay I get it. but could you tell why is it still a vector? $\endgroup$
    – JohnNash
    Commented Mar 11, 2022 at 12:04

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