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I have the following formulas about face maps, which I try to proof.

1) $d_j^{n+1}d_i^n=[e_0,\dotso, \hat{e_i},\dotso, \hat{e_j},\dotso, e_{n+1}]:\Delta^{n-1}\to\Delta^{n+1}$ for $j>i$.

2) $d_j^{n+1}d_i^n=[e_0,\dotso,\hat{e_j},\dotso, \hat{e_{i+1}},\dotso, e_{n+1}]:\Delta^{n-1}\to\Delta^{n+1}$ for $j\leq i$

3) $d_j^{n+1}d_i^n=d_{i+1}^{n+1}d_j^n$ for $j\leq i$

Proof:

Let $(x_0,\dotso, x_{n-1})\in\Delta^{n-1}$. Then is

$d_j^{n+1}(d_i^n(x_0,\dotso, x_{n-1}))=d_j^{n+1}(x_0,\dotso, x_{i-1}, 0,\color{red}{x_i}, x_{i+1},\dotso, x_{n})=(x_0,\dotso, x_{i-1}, 0, \color{red}{x_i},x_{i+1},\dotso, x_{j-1}, 0,\color{red}{x_j}, x_{j+1},\dotso, x_{n+1})$

The second formula can be proven similary, where in the case of $i=j$ the final result is a vector which looks like this:

$(x_0, \dotso, x_{i-1}, 0, 0, x_{i+2},\dotso, x_{n+1})$

Am I right?

The 3rd formula then just follows from the first two.

Thanks in advance.

Definition: For $0\leq i\leq n-1$ is the $i-th$ face map defined as $d_i^n:\Delta^{n-1}\to\Delta^n$ induced by $[e_0,\dotso, \hat{e_i},\dotso, e_n]:\Delta^{n-1}\to\mathbb{R}^{n+1}$

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  • $\begingroup$ I think that you ought to include definitions of $d_j^n$. $\endgroup$ – Batominovski Oct 12 '18 at 8:28
  • $\begingroup$ @Batominovski I did, I thought the function/notation $d_j^n$ is common. $\endgroup$ – Cornman Oct 12 '18 at 8:32
  • $\begingroup$ You shouldn't assume that other users are going to be familiar with your notations. In fact, I am wondering if you made a mistake. Isn't $d_i^n$ the map dropping the $i$-th coordinate? Then, why does $d_i^n$ send $\Delta^{n-1}$ to $\Delta^n$? Shouldn't it be the other way around: $d_i^n:\Delta^n\to \Delta^{n-1}$? $\endgroup$ – Batominovski Oct 12 '18 at 8:36
  • $\begingroup$ @Batominovski Thats why I am asking this question. I might have done a mistake in my lecture notes, so I tried to proof it myself. I do not think, that $d_i^n$ "drops" the i-th coordinate, but sets it to $0$. But now that I think about it. I guess you are right. So it should be $\Delta^{n+1}\to\Delta^{n-1}$ in the formulas too. $\endgroup$ – Cornman Oct 12 '18 at 8:46
  • $\begingroup$ If it adds a zero, then your notation is not common at all. Usually the hat notation ($\hat{e}_i$) means the $i$-th coordinate is removed. $\endgroup$ – Batominovski Oct 12 '18 at 8:48
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I know now what is happening, and that is why one should declare what notations mean to avoid confusion. In this answer, I attempt to fix the errors and undefined terms in your question. As usual, the $k$-dimensional simplex is the set $\Delta^k\subseteq\mathbb{R}^{n+1}$ defined by $$\Delta^k:=\Big\{\left(x_0,x_1,x_2,\ldots,x_k\right)\in\mathbb{R}^{k+1}\,\Big|\,x_0,x_1,x_2,\ldots,x_k\geq 0\text{ and }x_0+x_1+x_2+\ldots+x_k=1\Big\}\,.$$

First of all, $e_0,e_1,e_2,\ldots,e_n$ are standard basis vectors of $\mathbb{R}^{n+1}$ (and by abuse of notation, $e_0,e_1,e_2,\ldots,e_{n+1}$ are also standard basis vectors of $\mathbb{R}^{n+2}$). Then, you try to identify $\Delta^{n-1}$ with the convex combination $$\left[e_0,e_1,e_2,\ldots,e_{i-1},e_{i+1},e_{i+2},\ldots,e_n\right]=\left[e_0,e_1,\ldots,\hat{e}_i,\ldots,e_n\right]\,.$$ Then, the map $d_i^n:\Delta^{n-1}\to \Delta^n$ is defined to be $$d_i^n\left(x_0,x_1,x_2,\ldots,x_{n-1}\right):=\left(x_0,x_1,\ldots,x_{i-1},0,x_i,x_{i+1},\ldots,x_{n-1}\right)\in \Delta^n$$ for all $\left(x_0,x_1,x_2,\ldots,x_{n-1}\right)\in\Delta^{n-1}$. This means the image of $d_i^n$ is precisely $\left[e_0,e_1,\ldots,\hat{e}_i,\ldots,e_n\right]$. That is, $d_i^n$ is the embedding of $\Delta^{n-1}$ into one $(n-1)$-dimensional face of $\Delta^n$. (I don't know why you wrote something like $d_j^{n+1}\circ d_i^{n}=\left[e_0,e_1,\ldots,\hat{e}_i,\ldots,\hat{e}_j,\ldots,e_{n+1}\right]$. This makes no sense. A function does not equal a simplex.)

Hence, it is easy to check all the statements. Let $\left(x_0,x_1,x_2,\ldots,x_{n-1}\right)$ be an arbitrary point of $\Delta^{n-1}$.

Now, we shall deal with Part (1). For $j>i$, $$\begin{align}\left(d_{j}^{n+1}\circ d_i^{n}\right)\left(x_0,x_1,x_2,\ldots,x_{n-1}\right)&=d_j^{n+1}\left(x_0,x_1,\ldots,x_{i-1},0,x_i,x_{i+1},\ldots,x_n\right)\\&=\left(x_0,x_1,\ldots,x_{i-1},0,x_i,\ldots,x_{j-2},0,x_{j-1},x_j,\ldots,x_{n-1}\right)\end{align}$$ has zeros in the $i$-th and $j$-th coordinates. That is, $d_j^{n+1}\circ d_{i}^{n}:\Delta^{n-1}\to \Delta^{n+1}$ satisfies $$\text{im}\left(d_j^{n+1}\circ d_i^{n}\right)=\left[e_0,e_1,\ldots,\hat{e}_i,\ldots,\hat{e}_j,\ldots,e_{n+1}\right]\,,$$ as required.

Next, we tackle Part (2). For $j\leq i$, $$\begin{align}\left(d_{j}^{n+1}\circ d_i^{n}\right)\left(x_0,x_1,x_2,\ldots,x_{n-1}\right)&=d_j^{n+1}\left(x_0,x_1,\ldots,x_{i-1},0,x_i,x_{i+1},\ldots,x_n\right)\\&=\left(x_0,x_1,\ldots,x_{j-1},0,x_j,\ldots,x_{i-1},0,x_{i},x_{i+1},\ldots,x_{n-1}\right)\end{align}$$ has zeros in the $j$-th and $(i+1)$-st coordinates. That is, $d_j^{n+1}\circ d_{i}^{n}:\Delta^{n-1}\to \Delta^{n+1}$ satisfies $$\text{im}\left(d_j^{n+1}\circ d_i^{n}\right)=[e_0,e_1,\ldots,\hat{e}_j,\ldots,\hat{e}_{i+1},\ldots,e_{n+1}]\,.$$

Finally, we justify the equality in Part (3). For $j\leq i$, we see that, $j<i+1$. Therefore, from Part (1), $d_{i+1}^{n+1}\circ d_j^{n}:\Delta^{n-1}\to\Delta^{n+1}$ satisfies $$\text{im}\left(d_{i+1}^{n+1}\circ d_j^{n}\right)=[e_0,e_1,\ldots,\hat{e}_j,\ldots,\hat{e}_{i+1},\ldots,e_{n+1}]=\text{im}\left(d_j^{n+1}\circ d_i^{n}\right)\,.$$ Additionally, $$\begin{align}\left(d_{i+1}^{n+1}\circ d_j^{n}\right)\left(x_0,x_1,x_2,\ldots,x_{n-1}\right)&=\left(x_0,x_1,\ldots,x_{j-1},0,x_j,\ldots,x_{i-1},0,x_{i},x_{i+1},\ldots,x_{n-1}\right)\\&=\left(d_{j}^{n+1}\circ d_i^{n})(x_0,x_1,x_2,\ldots,x_{n-1}\right)\end{align}$$ also by Part (1). This proves that $d_{i+1}^{n+1}\circ d_j^{n}=d_{j}^{n+1}\circ d_i^{n}$.

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  • $\begingroup$ I'm confused with your proof that, for example, in part (1): $d_i^{n-1}$ is supposed to act on $(x_0, ... , x_{n-2})$, but you apply $d_{j}^n\circ d_i^{n-1}$ to $(x_0, ... , x_{n-1})$, isn't this wrong? $\endgroup$ – Jiangnan Yu May 15 '20 at 4:53
  • $\begingroup$ @JiangnanYu Thanks for pointing that out. I messed up the superscripts a little bit. I hope everything is fixed now. $\endgroup$ – Batominovski May 15 '20 at 9:46

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