\classheader{2012-10-05}

\newcommand{\dcomplex}{$\Delta$-complex}

Let $X$ be a \dcomplex. This is essentially a simplicial complex with extra structure, namely, an ordering of the vertices compatible with $\sigma$-maps.

For each $n$, 
\[\Delta_n(X)=C_n(X)=\text{free abelian group on the set of ``allowable'' maps } \Delta^n\to X.\]
For each $n$-simplex in $X$ (i.e. $\Delta^n\xrightarrow{e_\alpha}X$), there are several $\Delta$-maps
\[\xymatrix{\Delta^n \ar[r]^{e_\alpha} & X\\ \Delta^{n-1} \ar[u] \ar[ur]_{e_\beta}}\quad(\text{these are the faces of }e_\alpha)\]

Given $e_\alpha(v_0,\ldots,v_n)$, we define
\[\partial e_\alpha=\sum_{i=0}^n (-1)^ie_\alpha|_{(v_0,\ldots,\widehat{v_i},\ldots,v_n)}.\]
The key property of this construction is that $\partial\circ\partial=0$. 

Now we do some pure algebra.
\begin{definition}
A chain complex is a collection of abelian groups $C_i$ and homomorphisms $\partial_i:C_i\to C_{i-1}$ such that $\partial_{i-1}\circ \partial_i=0$. We write this as
\[\xymatrix{\cdots \ar[r] & C_{n+2} \ar[r]^{\partial_{n+2}} & C_{n+1} \ar[r]^{\partial_{n+1}} & C_n \ar[r]^{\partial_n} & C_{n-1} \ar[r]^{\partial_{n-1}} &\cdots}\]
We say that $\alpha\in C_n$ is a cycle if $\partial_n(\alpha)=0$ and $\beta\in C_n$ is a boundary if there is some $\gamma\in C_{n+1}$ such that $\partial_{n+1}(\gamma)=\beta$. In other words,
\[n\text{-cycles} = \ker(\partial_n),\quad n\text{-boundaries} = \im(\partial_{n+1}).\]
\end{definition}
All boundaries are cycles, but not all cycles are boundaries. Letting $Z_n$ be the subgroup of cycles in $C_n$, and $B_n$ the subgroup of boundaries in $C_n$, we have $B_n\subseteq Z_n$, and we define
\[H_n = Z_n/B_n.\]
Applying this to topology, we define
\[H_n(X) = \text{$n$th homology group of }(\Delta_n(X),\partial_n).\]
\begin{example}
Consider the triangle
\begin{center}
\begin{tikzpicture}[scale=2.4]
\begin{scope}[every node/.style={fill,circle,inner sep=0pt,outer sep=1pt,minimum size=0.15cm}]
\node (a) at (0,0) [label={210:$a$}] {};
\node (b) at (1,0) [label={330:$b$}] {};
\node (c) at (.5,.86) [label={90:$c$}] {};
\end{scope}
\begin{scope}[thick,outer sep=2pt,decoration={markings,mark=at position 0.57 with {\arrow[thin]{>}}}]
\draw[postaction={decorate}] (a.center) -- node[auto,swap] {$x$} (b.center);
\draw[postaction={decorate}] (b.center) -- node[auto,swap,outer sep=1pt] {$y$} (c.center);
\draw[postaction={decorate}] (c.center) -- node[auto,swap] {$z$} (a.center);
\end{scope}
\end{tikzpicture}
\end{center}
We have that
\[\Delta_0=\langle a,b,c\rangle \cong \Z^3,\quad \Delta_1 = \langle x,y,z\rangle\cong \Z^3.\]
The map $\partial_1$ sends
\[x\longmapsto b-a, \quad y\longmapsto c-b, \quad z\longmapsto c-a\]
or expressed as a matrix,
\[\partial_1=\begin{pmatrix}-1 & 0 & 1\\ 1 & -1 & 0\\ 0 & 1 & 1\end{pmatrix}.\]
Thus,
\[\ker(\partial_1)=\langle x+y-z\rangle\cong\Z.\]
Of course, we have $\ker(\partial_0)=\Z^3$,  and $\im(\partial_1)$ is the set of things which sum to zero, so $H_0\cong\Z$.
\end{example}
\begin{example}
Consider the 2-simplex
\begin{center}
\begin{tikzpicture}[scale=2.4]
\begin{scope}[every node/.style={fill,circle,inner sep=0pt,outer sep=1pt,minimum size=0.15cm}]
\node (a) at (0,0) [label={210:$a$}] {};
\node (b) at (1,0) [label={330:$b$}] {};
\node (c) at (.5,.86) [label={90:$c$}] {};
\end{scope}
\begin{scope}[thick,outer sep=2pt,decoration={markings,mark=at position 0.57 with {\arrow[thin]{>}}}]
\draw[postaction={decorate}] (a.center) -- node[auto,swap] {$x$} (b.center);
\draw[postaction={decorate}] (b.center) -- node[auto,swap,outer sep=1pt] {$y$} (c.center);
\draw[postaction={decorate}] (c.center) -- node[auto,swap] {$z$} (a.center);
\end{scope}
\node at (barycentric cs:a=0.5,b=0.5,c=0.5) {$T$};
\end{tikzpicture}
\end{center}
We have that
\[\Delta_0=\langle a,b,c\rangle \cong \Z^3,\quad \Delta_1 = \langle x,y,z\rangle\cong \Z^3,\quad \Delta_2=\langle T\rangle.\]
The map $\partial_1$ is the same as before, and $\partial_2(T)=y-z+x$.
In this case we get
\[H_0\cong\Z,\quad H_1\cong 0, \quad H_2\cong 0.\]
\end{example}
\begin{example}
Consider the torus
\begin{center}
\begin{tikzpicture}[scale=3]
\begin{scope}[every node/.style={fill,circle,inner sep=0pt,outer sep=1pt,minimum size=0.15cm}]
\node (ll) at (0,0) [label={225:$v$}] {};
\node (lr) at (1,0) [label={315:$v$}] {};
\node (ul) at (0,1) [label={135:$v$}] {};
\node (ur) at (1,1) [label={45:$v$}] {};
\end{scope}
\begin{scope}[thick,outer sep=2pt,decoration={markings,mark=at position 0.54 with {\arrow[thin]{>}}}]
\draw[postaction={decorate}]
(lr.center) -- node[auto] {$y$} (ll.center); % Bottom
\draw[postaction={decorate}]
(ur.center) -- node[auto,swap] {$y$} (ul.center); % Top
\draw[postaction={decorate}]
(ll.center) -- node[auto] {$x$} (ul.center); % Right
\draw[postaction={decorate}]
(lr.center) -- node[auto,swap] {$x$} (ur.center); % Left
\draw[postaction={decorate}]
(lr.center) -- node[auto,swap,outer sep=0pt] {$z$} (ul.center); % Diagonal
\end{scope}
\node at (barycentric cs:ll=1,ul=0.5,lr=0.5) {$A$};
\node at (barycentric cs:ur=1,ul=0.5,lr=0.5) {$B$};
\end{tikzpicture}
\end{center}
We have that
\[\Delta_0=\langle v\rangle,\qquad \Delta_1=\langle x,y,z\rangle,\qquad \Delta_2=\langle A,B\rangle\]
with $\partial_0=0$, $\partial_1=0$, and
\[\partial_2(A)=y+z-x,\quad \partial_2(B)=y-x+z\]
which implies that $\ker(\partial_2)=\langle A-B\rangle$ and $\im(\partial_2)=\langle y+z-x\rangle$. Thus,
\[H_1=\langle x,y,z\rangle/\langle y+z-x\rangle\cong \Z^2,\quad H_2=\langle A,B\rangle/\langle A-B\rangle\cong\Z.\]

\end{example}