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title: TDA_TEAM_3

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title: TDA_TEAM_3
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Bryten Ives, Dawn Myers, Matt Oakes

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


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

Note: these del symbols are maps, from 0 to C, C to C, etc.
Note: basis for C_1 are the edges in the picture


![](https://i.imgur.com/VohpRyE.png)


**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

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

These are the connective edges 

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

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

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

-The image of $\partial_2$ must be zero, because it is a linear transformation. 

-$\partial_0$ goes from $0$ and maps everything to zero-- the image of $\partial_0$ is zero. $ker(\partial_0)$ is $C_0$. Thus, $dim(ker(\partial_0))$=4, and dimension of the image is zero. 

-$\partial_1$ maps C_1 to C_0 by using matrices

The dimension of $H_{0}$ corresponds to the number of clusters and the dimension of $H_{1}$ corresponds to the number of loops. 

**Example B**



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

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

Reduced row echelon form 

$$ \begin{bmatrix} 1 && 0 && 1 && 0 \\
0 && 1 && 1 && 0 \\
0 && 0 && 0 && 1 \\
0 && 0 && 0 && 0\end{bmatrix} $$


$\dim~ker(\partial_1) = 1$ = number of free variables 
$\dim~im(\partial_{1})=$ 3 = number of pivots  

$\dim~ker(\partial_2) = 0$
$\dim~im(\partial_{2})= 0$

$\dim~ker(\partial_0) = 0$
$\dim~im(\partial_{0})= 4$


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

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


**Example C** extra credit 

Finish for HW

$C_0: \lbrace 0,1,2,3 \rbrace$ 

$C_1: \lbrace 01,02,23,12,13\rbrace$

$C_2: \lbrace 012 \rbrace$


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


matrix reduced row echcelon: 

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

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


dim/image and kernel calculations: 

$dim(ker(\partial_0)) = 4$
$dim(Im(\partial_0)) = 0$

$dim(ker(\partial_1)) = 2$
$dim(Im(\partial_1)) = 3$

$dim(ker(\partial_2)) = 0$
$dim(Im(\partial_2)) = 1$

$dim(ker(\partial_3)) = 0$
$dim(Im(\partial_3)) = 0$

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

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

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

Explain why the boundary map does the job: 
The boundary maps allow us to create matrices to find homologies in order to find topological features. 

