Boundary operator of Simplex

This essay aims to give a natural approach to the simplex and singular simplex.

Simplex

Define the simplex:

$$\begin{array}{}\text{(1)}& {\mathrm{\Delta }}^n:=\{\sum _{i=0}^n{t}_i{e}_i:t_i\ge 0,\sum _{i=0}^n{t}_i=1\}\subset {\mathbb{R}}^{n+1}\end{array}$$

We will denote the ${\mathrm{\Delta }}^n$$\Delta^n$ be the simplex up to isomorphism.

Here is some example.

Each simplex has an orientation; we can denote the standard simplex as

$$\begin{array}{}\text{(2)}& [0,1,2,3,...,n]\end{array}$$

Thus, we can consider group action

$$\begin{array}{}\text{(3)}& \begin{array}{c}{\mathcal{S}}_n\times {\mathrm{\Delta }}^n\to {\mathrm{\Delta }}^n\end{array}\end{array}$$

Defined by

$$\begin{array}{}\text{(4)}& \sigma ({\mathrm{\Delta }}^n):=[\sigma (0),\sigma (1),\sigma (2),\sigma (3),...,\sigma (n)]=\mathrm{sign}(\sigma )[0,1,2,3,...,n]=\mathrm{sign}(\sigma ){\mathrm{\Delta }}^n\end{array}$$

Then, we can define ${\pi }_i$$\pi_i$. Geometrically speaking, ${\pi }_i$$\pi_i$ map ${\mathrm{\Delta }}^n$$\Delta^n$ to the face corresponding to $i$$i$.

i.e.

$$\begin{array}{}\text{(5)}& {\pi }_i([0,1,2,...,i-1,i,i+1,...,n])=[0,1,2,...,i-1,\overline{i},i+1,...n]\end{array}$$

In other words, we pull ${\mathrm{\Delta }}^n$$\Delta^n$ back to ${\mathrm{\Delta }}_i^{n-1}$$\Delta^{n-1}_i$, I will show some details at singular n-simplex.

Consider the free $\mathbb{Z}$$\mathbb Z$ module of $\{\sigma ({\mathrm{\Delta }}^n),\sigma \in S_n\}$$\{\sigma(\Delta^n),\sigma\in S_n\}$. Denote it as $S({\mathrm{\Delta }}^n)$$S(\Delta^n)$.

Then ${\pi }_i:S({\mathrm{\Delta }}^n)\to S({\mathrm{\Delta }}^{n-1})$$\pi_i:S(\Delta^n)\to S(\Delta^{n-1})$ is a module homomorphism.

$$\begin{array}{}\text{(6)}& {\pi }_{\sigma (i)}(\sigma ({\mathrm{\Delta }}^n))=\mathrm{sign}(\sigma ){\pi }_i({\mathrm{\Delta }}^n)\end{array}$$

Thus, we can define the boundary operator as follows.

$$\begin{array}{}\text{(7)}& \partial ({\mathrm{\Delta }}^n)=\sum _{i=0}^n{\pi }_i({\mathrm{\Delta }}^n)={\pi }_0[0,1,2,...,n]-{\pi }_0[1,0,2,...,n]+{\pi }_0[2,0,1,...,n]+...+(-1{)}^n{\pi }_0[n,0,1,...,n-1]\end{array}$$

It is equivalence to say

$$\begin{array}{}\text{(8)}& \partial ({\mathrm{\Delta }}^n)=[1,2,3,...,n]-[0,2,3,...,n]+[0,1,3,...,n]+...+(-1{)}^n[0,1,2,...,n-1]\end{array}$$

i.e.

$$\begin{array}{}\text{(9)}& \partial ({\mathrm{\Delta }}^n)=\sum _{i=0}^n(-1{)}^i{\partial }_i({\mathrm{\Delta }}^n)\end{array}$$

Notice that $\partial :S({\mathrm{\Delta }}^n)\to S({\mathrm{\Delta }}^{n-1})$$\part:S(\Delta^n)\to S(\Delta^{n-1})$ is sum of module homomorphism, thus it is module homomorphism as well.

For example,

An important property here is

$$\begin{array}{}\text{(10)}& {\partial }_{n-1}\circ {\partial }_n=0\end{array}$$

The reason is easy, there exist two way to get

$$\begin{array}{}\text{(11)}& [1,2,...\bar{i},...,\bar{j},...,n]\end{array}$$

You might omit $j$$j$ first when ${\partial }_n$$\part_n$ act on the simplex, then you omit $i$$i$.

The coefficient is $(-1{)}^i(-1{)}^j$$(-1)^{i}(-1)^j$.

Another way is you omit $i$$i$ first, then when you omit $j$$j$, the order of $j$$j$ becomes $j-1$$j-1$!

Thus the coefficient becomes $(-1{)}^i(-1{)}^{j-1}.=-(-1{)}^i(-1{)}^j$$(-1)^i(-1)^{j-1}.=-(-1)^i(-1)^j$.

Remark

If we go back to the definition of simplex,

$$\begin{array}{}\text{(12)}& {\mathrm{\Delta }}^n:=\{\sum _{i=0}^n{t}_i{e}_i:t_i\ge 0,\sum _{i=0}^n{t}_i=1\}\subset {\mathbb{R}}^{n+1}\end{array}$$

We could view ${\mathrm{\Delta }}^n$$\Delta^n$ as a linear map. Thus ${\partial }_i$$\part_i$ will relate to $\begin{array}{c}\frac{\partial }{\partial t_i}\end{array}$$\frac{\partial}{\partial t_i}\\$, since $\begin{array}{c}\frac{\partial }{\partial t_i}({\mathrm{\Delta }}^n)=e_i\end{array}$$\frac{\part}{\part t_i}(\Delta^n)=e_i\\$.

That is, ${\partial }_i({\mathrm{\Delta }}^n)$$\part_i(\Delta^n)$ is the kernel of $\begin{array}{c}\frac{\partial }{\partial t_i}\end{array}$$\frac{\partial}{\partial t_i}\\$. In other word ${\partial }_i({\mathrm{\Delta }}^n)$$\part_i(\Delta^n)$ is the equalizer of $0$$0$ and $\begin{array}{c}\frac{\partial }{\partial t_i}\end{array}$$\frac{\part}{\part t_i}\\$.

Observe that $t_i\frac{\partial }{\partial t_i}+{\partial }_i=I$$t_i\frac{\part}{\part t_i}+\part_i=I$ when they act on ${\mathrm{\Delta }}^n$$\Delta^n$.