# face map $d_i^n:\Delta^{n-1}\to\Delta^n$

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.

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 $$[e_0,\dotso, \hat{e_i},\dotso, e_n]:\Delta^{n-1}\to\mathbb{R}^{n+1}$$

• I think that you ought to include definitions of $d_j^n$. – Batominovski Oct 12 '18 at 8:28
• @Batominovski I did, I thought the function/notation $d_j^n$ is common. – Cornman Oct 12 '18 at 8:32
• 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: $d_i^n:\Delta^n\to \Delta^{n-1}$? – Batominovski Oct 12 '18 at 8:36
• @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. – Cornman Oct 12 '18 at 8:46
• 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. – Batominovski Oct 12 '18 at 8:48

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, $$j. Therefore, from Part (1), $$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}$$.

• 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? – Jiangnan Yu May 15 '20 at 4:53
• @JiangnanYu Thanks for pointing that out. I messed up the superscripts a little bit. I hope everything is fixed now. – Batominovski May 15 '20 at 9:46