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title: TDA_team_2

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title: TDA_team_2
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**Homologies: $H_i=ker(\partial_i)/im(\partial_{i+1})$**
$$\dim H_i=\dim~ker(\partial_i)-\dim~im(\partial_{i+1})$$ 

## Table of Contents

[TOC]

## **Example A** 
1A. Create the chain.
$$0\quad \rightarrow\quad C_1\quad \rightarrow\quad C_0\quad \rightarrow\quad 0$$

$$\partial_2 : 0 \quad \rightarrow \quad C_1$$

$$\partial_1 : C_1 \quad \rightarrow \quad C_0$$

$$\partial_0 : C_0 \quad \rightarrow \quad 0$$

For $\partial_2$ the kernel is every element which is sent to 0, thus $ker(\partial_2)= 0.$ The image of $\partial_2$ is where the transformation goes, but since it is a linear transformation on 0, it can only go to 0 thus $im(\partial_2) = 0.$

For $\partial_0$ the kernel is again every element which is sent to 0, thus $ker(\partial_0)= C_0 = Span{\{0,1,2,3\}}$. The image of $\partial_0$ is where the transformation goes, so $im(\partial_0)=0$.


2A. Describe the simplicial complexes

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

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

Since the transformation matrix for $\partial_1$ has 1 free variable and 2 pivots, the $\dim~ker(\partial_1) = 1$ and $\dim~im(\partial_1) = 2$.  Thus $ker(\partial_1)= <1 \quad 1 \quad 1>.$ 

4A. Compute homologies. 
$$\dim H_0=\dim~ker(\partial_0)-\dim~im(\partial_{1})=4-2=2$$ 

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

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

The $\dim H_0$ is the number of unconnected clusters that the system has, which is 2 in this case.

The $\dim H_1$ is the number of closed loops that they system has, which is 1 in this case.

The kernel for $\partial_1$ is composed of any closed loops in the system, which in this case is the $0-1-2$ loop.


## **Example B** 
1B. Create the chain.
$$0\quad \rightarrow\quad C_1\quad \rightarrow\quad C_0\quad \rightarrow\quad 0$$

$$\partial_2 : 0 \quad \rightarrow \quad C_1$$

$$\partial_1 : C_1 \quad \rightarrow \quad C_0$$

$$\partial_0 : C_0 \quad \rightarrow \quad 0$$

For $\partial_2$ the kernel is every element which is sent to 0, thus $ker(\partial_2)= 0.$ The image of $\partial_2$ is where the transformation goes, but since it is a linear transformation on 0, it can only go to 0 thus $im(\partial_2) = 0.$

For $\partial_0$ the kernel is again every element which is sent to 0, thus $ker(\partial_0)= C_0 = Span{\{0,1,2,3\}}$. The image of $\partial_0$ is where the transformation goes, so $im(\partial_0)=0$.


2B. Describe the simplicial complexes

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

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

Since the transformation matrix for $\partial_1$ has 1 free variable and 3 pivots, the $\dim~ker(\partial_1) = 1$ and $\dim~im(\partial_1) = 3$.  Thus $ker(\partial_1)= <1 \quad 1 \quad 1 \quad 0>.$ 

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

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

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

The $\dim H_0$ is the number of unconnected clusters that the system has, which in this case is 1.

The $\dim H_1$ is the number of closed loops that they system has, which in this case is 1.

The kernel for $\partial_1$ is composed of any closed loops in the system, which in this case is the $0-1-2$ loop.


## **Example C** 
1C. Create the chain.
$$0\quad \rightarrow\quad C_2 \quad \rightarrow \quad C_1\quad \rightarrow\quad C_0\quad \rightarrow\quad 0$$

$$\partial_3 : 0 \quad \rightarrow \quad C_2$$

$$\partial_2 : C_2 \quad \rightarrow \quad C_1$$

$$\partial_1 : C_1 \quad \rightarrow \quad C_0$$

$$\partial_0 : C_0 \quad \rightarrow \quad 0$$

For $\partial_3$ the kernel is every element which is sent to 0, thus $ker(\partial_3)= 0.$ The image of $\partial_3$ is where the transformation goes, but since it is a linear transformation on 0, it can only go to 0 thus $im(\partial_3) = 0.$

For $\partial_0$ the kernel is again every element which is sent to 0, thus $ker(\partial_0)= C_0 = Span{\{0,1,2,3\}}$. The image of $\partial_0$ is where the transformation goes, so $im(\partial_0)=0$.

2C. Describe the simplicial complexes

$$C_0:=\{0,1,2,3\},\quad C_1:=\{01,02,12, 13, 23\}, \quad C_2:=\{012\}$$

3C. Create matrices to describe boundary maps. 

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

Since the transformation matrix for $\partial_2$ has 0 free variables and 1 pivot, the $\dim~ker(\partial_2) = 0$ and $\dim~im(\partial_2) = 1$.  Thus $ker(\partial_2)= 0.$ 

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

Since the transformation matrix for $\partial_1$ has 2 free variables and 3 pivots, the $\dim~ker(\partial_1) = 2$ and $\dim~im(\partial_1) = 3$.  Thus $ker(\partial_1)= \{<1 \quad 1 \quad 1 \quad 0 \quad 0>, \quad <0 \quad 0 \quad 1 \quad 1 \quad 1>  \}.$ 

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

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

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

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

The $\dim H_0$ is the number of unconnected clusters that the system has, which in this case is 1.

The $\dim H_1$ is the number of closed loops that they system has, which in this case is 1.

The $\dim H_2$ is the number of three-dimensional closed shapes which we do not have in this case.

