Homomorphism between $p$-chain groups

Question

In the below paper,

Persistent Homology: An Introduction and a New Text Representation for Natural Language Processing

the author provides an visual example of the boundary of a $p$-chain as seen

Definition: The boundary of a $p$-chain is the $+_{2}$ sum of the boundaries of its simplices. The boundary of a $p$-chain is a group homomorphism from the $p$-chain group $C_{p}$ to the $(p-1)$-chain group $C_{p-1}$.

From the visuals, $p = 2$ so the homomorphism map is from $C_{2}$ to $C_{1}$.

What are the elements in $C_{2}$ then? Well, the elements are $2$-chains, by definition of a $p$-chain group.

What is the boundary of a $2$-chain then? Well, the $2$-chain is a subset of $2$-simplices in a simplicial complex. In other words, the $2$-chain is a subset of convex hulls of (2 + 1) points - i.e., subset of triangles.

For each of the $2$ - simplex, the boundary is the set of (2 - 1) - simplices face - i.e., $1$ - simplex face or edges. So a boundary of a $2$ - simplex is the blue triangle on the left. Another boundary of a $2$ - simplex is the blue triangle to the right of it. These two triangles are elements in $C_{2}$.

The definition of a group homomorphism requires that a map from a group $G$ to a group $G'$ be onto an operation preserving.

Since the homomorphism map is onto $C_{1}$ or $1$-chain group, note that a $1$-chain is a subset of $1$ - simplices. The boundary of a $1$-simplex is the set of $0$-simplices faces - i.e., point/ vertex. It appears that arriving at the blue square requires the $+_{2}$ sum of the boundaries of $1$-simplex which a $1$-chain is a subset of. I am unable to proceed. Any help is appreciated.

Answer 1

I understand it is a bit hard without seeing any algebraic topology background. The paper is (out of necessity) very terse as well.

The boundary of a triangle is three edges. In general the boundary of a $p$-simplex is the sum of all of its $(p-1)$-dimensional faces.

The first question is what do I mean by "sum". After definition 15 the author briefly mentions that addition of chains is the modulo 2 addition. There are other options in algebraic topology, but this one is very simple so we stick to it. Note that in this case a chain can be thought of as a sum of simplices, or a set of simplices - it is the same because the sum is just determined by which simplices are there.

Now that we know what a sum is, and we know how to compute the boundary of a simplex, we can also compute the boundary of a chain (that is, of a sum of simplices). If the boundary is supposed to be a group homomorphism then the only thing we can do is

$$\partial(\sigma_1+\sigma_2)=\partial(\sigma_1)+\partial(\sigma_2).$$

In your case let us denote by $\sigma_1,\sigma_2$ the two yellow triangles. Then $\sigma_1+\sigma_2$ is a chain consisting of two triangles (geometrically the square). Let us also denotes by $a,b,c,d$ the sides of that square (blue edges) and by $x$ the diagonal of the square (also blue). Then

$$\partial(\sigma_1)=a+b+x$$ $$\partial(\sigma_2)=c+d+x$$

so

$$\partial(square)=\partial(\sigma_1+\sigma_2)=\partial(\sigma_1)+\partial(\sigma_2) = a+b+x+c+d+x=a+b+c+d$$

because $2x=0$ by our convention. So the "algebraic" boundary of the square is the "geometric" boundary of the square.

Answer 2

The author is trying to illustrate how the boundary map works on an explicit example of a $2$-chain.

Our 2-chain is the square, which is made up of two 2-simplices, call them $T_1$ and $T_2$. These are the left and right yellow triangles. So our 2-chain is the formal sum $T_1+T_2$. We want to compute $\partial(T_1+T_2)$.

Let's call the edges of $T_1$ $e_1, e_2$, and $e_3$, and the edges of $T_2$ $e_3,e_4,e_5$ (so $e_3$ is the shared edge).

The fact that $\partial$ is a group homomorphism just means that $\partial(T_1+T_2) = \partial(T_1)+\partial(T_2)$. But $T_1$ and $T_2$ are simplices, so we can compute $\partial$ as the sum of their edges.

So \begin{align} \partial(T_1+T_2) = \partial(T_1)+\partial(T_2) & = e_1 + e_2 + e_3 + e_3 + e_4 + e_5 \\\\ &= e_1+e_2+2e_3+e_4+e_5 \end{align}

By $+_2$ he means the mod 2 sum, which just means that if I get an even coefficient, I set it to 0, and if I get an odd coefficient, I set it to 1. So my final computation is that $\partial(T_1+T_2) = e_1+e_2+e_4+e_5$, which is the blue square on the right.