---
title: TDA_TEAM_2_Chris_Schmidt_Steven_White
---
**Homologies: $H_i=ker(\partial_i)/im(\partial_{i+1})$**

$$\dim H_i=\dim~ker(\partial_i)-\dim~im(\partial_{i+1})$$

**Example A.**


1. Create the chain.
$$0\quad \rightarrow\quad C_1\quad \rightarrow\quad C_0\quad \rightarrow\quad 0$$

2. Describe the simplicial complexes

$$C_0:=\{0,1,2,3\},\quad C_1:=\{01,02,12\}$$

The $\tilde{C}ech$ complex $\tilde{C}_0$ represents the vertices $0, 1, 2,$ and $3$ while $\tilde{C}_1$ represent the edges connecting the vertices. I.e. $[01]$ is the edge connecting vertex $0$ and vertex $1$.

3. Create matrices to describe boundary maps.
$$\partial_1:\quad
\begin{bmatrix}
1&1&0\\
1&0&1\\
0&1&1\\
0&0&0
\end{bmatrix}
\sim
\begin{bmatrix}
1&0&1\\
0&1&1\\
0&0&0\\
0&0&0\end{bmatrix}$$

4. Compute homologies.
$$H_1 = ker(\partial_1)/im(\partial_2)$$

$$H_1 = \big\{012\big\}/ \big\{ \emptyset\big\}$$

$$H_1=\big\{ 012\big\}$$

$$H_0 = \ker(\partial_0) / \text{ Im }(\partial_1)$$

$$H_0 =\big\{ 0,1,2,3\big\}/ \big\{0,1,2\big\}$$

$$\Rightarrow H_0 = \{3\}$$

$$\dim H_0=\dim~ker(\partial_0)-\dim~im(\partial_{1})=4-3=1$$

$$\dim H_1=\dim~ker(\partial_1)-\dim~im(\partial_{2})=1-0=1$$

5. Describe your findings in your own words. AS MUCH AS YOU CAN.

We have two clusters. Homology $0$ contains the vertex $3$ and homology $1$ is the triangle copntaining the vertices $0, 1,$ and $2$ with connecting edges $01, 02,$ and $12$ that also creates a $2D$ void, or, in plain language, a hole. Our $H_0$ tells us all the clusters that haven't been connected and are just vertices. While $H_1$ represents the set of all clusters that are connected through edges and are closed. We found that the boundary map $\partial_1$ has its kernel defined by the span$\big\{[01] + [02] + [12]\big\}$.

The idea of the homology seems related to the rank nullity theorem but not sure if that is relevant or not.

**Example B.**


1. Create the chain.
$$0\quad \rightarrow\quad C_2\quad \rightarrow\quad C_1\quad \rightarrow\quad C_0\quad \rightarrow\quad 0$$

2. Describe the simplicial complexes

$$C_0:=\{0,1,2,3\},\quad C_1:=\{01,02,12,23\}, \quad C_2:= \{012 \}$$

We add a simplicial complex $C_2$ that is the $2D$ triangle formed by vertices $0, 1$, and $2$ and a new edge $[23]$.

3. Create matrices to describe boundary maps.

$$\partial_1:
\quad \begin{bmatrix}
1&1&0&0\\
1&0&1&0\\
0&1&1&1\\
0&0&0&1
\end{bmatrix}
\sim
\begin{bmatrix}
1&0&1&0\\
0&1&1&0\\
0&0&0&1\\
0&0&0&0
\end{bmatrix}$$

$$\partial_2:
\quad \begin{bmatrix}
1\\
1\\
1\\
0
\end{bmatrix}
\sim
\begin{bmatrix}
1\\
1\\
1\\
0
\end{bmatrix}$$

4. Compute homologies.

$$\dim H_0=\dim~ker(\partial_0)-\dim~im(\partial_{1})=4-3=1$$

$$\dim H_1=\dim~ker(\partial_1)-\dim~im(\partial_{2})=1-1=0$$

$$\dim H_2=\dim~ker(\partial_2)-\dim~im(\partial_{3})=0-0=0$$

$$H_0 = \text{ ker }(\partial_0) / \text{ Im }(\partial_1) = \big\{ 0,1,2,3 \big\} / \big\{0,1,2 \big\} = \{3\}$$

$$H_1 = \text{ ker }(\partial_1) / \text{ Im }(\partial_2) = \big\{ 012 \big\} / \big\{012 \big\} = \emptyset$$

$$H_2 = \text{ ker }(\partial_2) / \text{ Im }(\partial_3) = \big\{ \emptyset \big\} / \big\{ \emptyset \big\}
= \emptyset$$

5. Describe your findings in your own words. AS MUCH AS YOU CAN.

We adjust our homologies to account for the new information that arises from adding as edge between vertices $2$ and $3$. The creates, interestingly, homology $H_2$, that is the null set and also changes homology $H_1$ to the null set, $\emptyset$, as well. We see that for $H_1$ and $H_2$ the kernel of the linear map $\partial_1$ has the same form as the image of the higher dimension linear map $\partial_2$ for $H_1$ and this holds for $H_2$ as well leading to both homologies being the null set. Is this merely coincidental?

**Example C.**


1. Create the chain.
$$0\quad \rightarrow\quad C_3\quad
\rightarrow\quad C_2\quad \rightarrow\quad C_1\quad \rightarrow\quad C_0\quad \rightarrow\quad 0$$

2. Describe the simplicial complexes

$$C_0:=\{0,1,2,3\},\quad C_1:=\{01,02,12,13,23 \}, \quad C_2:=\{012, 123\}, \quad C_3:=\{[012]\}$$
$C_0$ is the set of vertices, $C_1$ is the set of edges, $C_2$ is the set of triangles, and $C_3$ is the shaded triangle.

3. Create matrices to describe boundary maps.

$$\partial_1 \quad
\begin{bmatrix}
1&1&0&0&0\\
1&0&1&1&0\\
0&1&1&0&1\\
0&0&0&1&1
\end{bmatrix}
\sim
\begin{bmatrix}
1&1&0&0&0\\
0&1&1&1&0\\
0&0&0&1&1\\
0&0&0&0&0
\end{bmatrix}
$$

$$\partial_2 \quad
\begin{bmatrix}
1&0\\
1&0\\
1&1\\
0&1\\
0&1
\end{bmatrix}
\sim
\begin{bmatrix}
1&0\\
0&0\\
0&1\\
0&0\\
0&0\\
\end{bmatrix}
\sim \text{ row swap [02], [12]}\sim
\begin{bmatrix}
1&0\\
0&1\\
0&0\\
0&0\\
0&0\\
\end{bmatrix}
$$

$$
\partial_3 \quad
\begin{bmatrix}
1\\
0\\
\end{bmatrix}$$

4. Compute homologies.

$$\dim H_0=\dim~ker(\partial_0)-\dim~im(\partial_{1})=4-3=1$$

$$\dim H_1=\dim~ker(\partial_1)-\dim~im(\partial_{2})=2-2=0$$

$$\dim H_2=\dim~ker(\partial_2)-\dim~im(\partial_3)=0-1= -1 $$

$$H_0 = \text{ ker }(\partial_0) / \text{ im }(\partial_1) = \{3\}$$

$$H_1 = \text{ ker }(\partial_1) / \text{ im }(\partial_2) = \{02, 13, 23\}$$

$$H_2 = \text{ ker }(\partial_2) / \text{ im }(\partial_3) = \emptyset?$$

The ker$(\partial_1) = \{01, 02, 12, 13, 23\}$ and $im(\partial_2) = \{01, 12\}$.

The ker$(\partial_2) = \{\emptyset \}$ and the Im$(\partial_3) = \{012\}$ so $H_2$ should be the $\emptyset$ as well.

5. Describe your findings in your own words. AS MUCH AS YOU CAN.

In the sequence of graphs, $A, B$, and $C$, we have explored the idea of the Vietoris-Rips complex as we connected our two clusters in graph $A$ into a single cluster in graph $B$ and then to a cluster with a higher dimensional triangle in graph $C$ occuring as the radius, say epsilon, $\epsilon$, is increased.

Here in graph $C$ we have a higher dimension shaded triangle formed by the edges connecting vertices $0, 1$, and $2$ that is in $\tilde{C}$ech complex $C_3$ while the edge $[12]$ separates the two triangles that are both in $\tilde{C}$ech complex $C_2$, $\{[012], [123]\}$.

The biggest curiosity here is that dimension of $H_2$ being a negative value. What does that mean? We are fairly confident in our computation for dim$~ker(\partial_2)$ and for dim$~im(\partial_3)$ which leads us to a value of $-1$ for dim$~H_2$. A bit of internet sleuthing leads us to some complicated descriptions that seem to imply that a negative dimension of a homology is a possibility and that a reversal of direction of the edges may play a role. But, two things here, first, we have undirected graphs, and second, the information found in our cursory sleuthing is difficult to understand and may have nothing to do with our problem at hand.

It seems that dimension of $H_0$ describes how many unconnected vertices there are, $H_1$ describes the number of separate connected triangles (there are none in this example) and $H_2$ describes the number of separate connected tetrahedrons, but there are none because you've got the 123 triangle sticking off the side of it.