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\begin{document}
\centerline{Math 372 lecture for Wednesday, Week 9}
\bigskip

\centerline{\bf Simplicial complexes: examples}
\bigskip

Let~$\Delta\subset2^{[n]}$ be a ~$d$-dimensional simplicial complex.  For each~$i\in\Z$ we have
the space of~$i$-dimensional chains~$C_i:=C_i(\Delta,\Q):=\Q F_i$,
the vector space of formal sums of~$i$-dimensional faces of~$\Delta$.  We have~$C_i=0$
for~$i<-1$ and~$i>d$, and since~$F_{-1}=\left\{ \emptyset \right\}$, we
have~$C_{-1}=\Q$.  The boundary mapping~$\partial_i\colon C_i\to C_{i-1}$ is
defined as follows: if~$\sigma=\overline{\sigma_1\dots \sigma_{i+1}}\in F_i$
where the~$\sigma_k$ are the vertices of~$\sigma$
and~$\sigma_1<\dots<\sigma_{i+1}$ , then
\begin{align*}
  \partial_i(\sigma)
  &= 
  \sum_{k=1}^{i+1}(-1)^{k-1}\overline{\sigma_1\dots\widehat{\sigma_k}\dots\sigma_{i+1}}\\[8pt]
  &= 
  \overline{\sigma_2\sigma_3\sigma_4\dots
  \sigma_{i+1}}-\overline{\sigma_1\sigma_3\sigma_4\dots\sigma_{i+1}}+\overline{\sigma_1\sigma_2\sigma_4\dots\sigma_{i+1}}+\cdots.
\end{align*}
for~$-1\leq 1\leq d$ and~$\partial_i=0$, otherwise.  Recall the definitions of
the reduced homology groups and Betti numbers:
\begin{align*}
  \widetilde{H}_i&:=\widetilde{H}_i(\Delta):= \ker\partial_i/\im\partial_{i+1}\\[5pt]
  \tilde{\beta}_i&:=\tilde{\beta}_i(\Delta):=\dim\widetilde{H}_i=\mathrm{nullity}(\partial_i)-\rk(\partial_{i+1}).
\end{align*}
\bigskip

{\bf Examples.}

I.1. 
\begin{center}
  \begin{tikzpicture}
    \node[ball,label={above:$\overline{1}$}] at (0,0) {};
    \node[ball,label={above:$\overline{2}$}] at (1,0) {};
    \node[ball,label={above:$\overline{3}$}] at (2,0) {};
    \node[ball,label={above:$\overline{4}$}] at (3,0) {};
  \end{tikzpicture}
\end{center}
\[
  0\to \Q^{4}\xrightarrow[\partial_0]{\left(\begin{array}{cccc}
      1&1&1&1
\end{array} \right)}\Q\to0
\]
We have~$\rk(\partial_0)=1$, and hence by the rank-nullity
theorem,~$\mathrm{nullity}=4-1=3$.  The only non-vanishing homology group is
\begin{align*}
  \widetilde{H}_0(\Delta)&= \ker\partial_0/\im\partial_{1}=\ker\partial_0
  =\mathrm{Span}\left\{
  \overline{2}-\overline{1},\overline{3}-\overline{1},\overline{4}-\overline{1}
\right\}\\[5pt]
\tilde{\beta}_0&= 3.
\end{align*}
{\bf Question:} What are the homology groups and Betti numbers
for~$\Delta=\left\{ \overline{1},\dots,\overline{n} \right\}$ for
general~$n\geq1$?
\medskip

\hrule
\medskip

\Needspace{3\baselineskip}
I.2.
\begin{center}
  \begin{tikzpicture}
    \node[ball,label={below:$\overline{1}$}] (1) at (0,0) {};
    \node[ball,label={below:$\overline{2}$}] (2) at (1,0) {};
    \node[ball,label={below:$\overline{3}$}] at (2,0) {};
    \node[ball,label={below:$\overline{4}$}] at (3,0) {};
    \node[ball,label={above:$\overline{5}$}] (5) at (0,1) {};
    \node[ball,label={above:$\overline{6}$}] (6) at (1,1) {};
    \draw (1)--(5);
    \draw (2)--(6);
  \end{tikzpicture}
\end{center}
\[
  0\to\Q^2\xrightarrow[\partial_1]{\left(\begin{array}{rr}
    -1&0\\
    0&-1\\
    0&0\\
    0&0\\
    1&0\\
    0&1\\
\end{array} \right)} 
  \Q^{6}\xrightarrow[\partial_0]{\left(\begin{array}{cccccc}
      1&1&1&1&1&1
\end{array} \right)}\Q\to0
\]
We have
\begin{align*}
  \rk(\partial_0)=1, \mathrm{nullity}(\partial_0)=6-1=5\\
  \rk(\partial_1)=2, \mathrm{nullity}(\partial_1)=0.
\end{align*}
Therefore,~$\tilde{\beta}_0=5-2=3$ and~$\tilde{\beta}_1=0$.  The same as in
example I.  

Homology:
\begin{align*}
  \widetilde{H}_0=\ker\partial_0/\im\partial_1
  &=
  \Span\{\overline{2}-\overline{1},\overline{3}-\overline{1},\overline{4}-\overline{1},\overline{5}-\overline{1},\overline{6}-\overline{1}\}
 / \Span\left\{ \overline{5}-\overline{1},\overline{6}-\overline{2}
 \right\}\\[8pt]
&\simeq 
  \Span\{\overline{2}-\overline{1},\overline{3}-\overline{1},\overline{4}-\overline{1}\}
\end{align*}
In~$\widetilde{H}_0$, we have~$\overline{5}=\overline{1}$
and~$\overline{6}=\overline{2}$, which means,~$\overline{5}-\overline{1}=0$
and~$\overline{6}-\overline{1}=\overline{2}-\overline{1}$.

{\bf Question:} How does this example generalize?
\medskip

\hrule
\medskip

II.1

\begin{center}
  \begin{tikzpicture}
    \node[ball,label={below left:{$\overline{1}$}}] (1) at (0,0) {};
    \node[ball,label={below right:{$\overline{2}$}}] (2) at (1,0) {};
    \node[ball,label={above:{$\overline{3}$}}] (3) at (0.5,0.866) {};
    \draw (1)--(2)--(3)--(1);
  \end{tikzpicture}
\end{center}
\[
  0\to\Q^3\xrightarrow[\partial_1]{\left(\begin{array}{rrr}
      -1&-1&0\\
      1&0&-1\\
      0&1&1
\end{array} \right)}
\Q^3\xrightarrow[\partial_0]{\left(\begin{array}{ccc}
    1&1&1
\end{array} \right)}
\Q\to0
\]
We have
\begin{align*}
  \rk(\partial_0)=1, \mathrm{nullity}(\partial_0)=3-1=2\\
  \rk(\partial_1)=2, \mathrm{nullity}(\partial_1)=3-2=1.
\end{align*}
Therefore,~$\tilde{\beta}_0=2-2=0$ and~$\tilde{\beta}_1=1$.  

Homology:
\[
  \widetilde{H}_1=\ker\partial_1/\im\partial_2
  =\ker\partial_1=
  \Span\{\overline{23}-\overline{13}+\overline{12}\}.
\]
A picture of the generator for~$\widetilde{H}_1$:
\begin{center}
  \begin{tikzpicture}
    \node[ball,label={below left:{$\overline{1}$}}] (1) at (0,0) {};
    \node[ball,label={below right:{$\overline{2}$}}] (2) at (1,0) {};
    \node[ball,label={above:{$\overline{3}$}}] (3) at (0.5,0.866) {};
    \draw[->-= 0.5] (1)--(2);
    \draw[->-= 0.5] (2)--(3);
    \draw[->-= 0.5] (3)--(1);
  \end{tikzpicture}
\end{center}

II.2.
\begin{center}
  \begin{tikzpicture}
    \node[ball,label={32:{$\overline{1}$}}] (1) at (36:1) {};
    \node[ball,label={108:{$\overline{2}$}}] (2) at (108:1) {};
    \node[ball,label={176:{$\overline{3}$}}] (3) at (176:1) {};
    \node[ball,label={248:{$\overline{4}$}}] (4) at (248:1) {};
    \node[ball,label={320:{$\overline{5}$}}] (5) at (320:1) {};
    \draw (0,0) circle (1);
  \end{tikzpicture}
\end{center}
\[
  0\to\Q^5\xrightarrow[\partial_1]{\left(\begin{array}{rrrrr}
      -1&0&0&0&-1\\
      1&-1&0&0&0\\
      0&1&-1&0&0\\
      0&0&1&-1&0\\
      0&0&0&1&1\\
\end{array} \right)}
\Q^3\xrightarrow[\partial_0]{\left(\begin{array}{ccccc}
    1&1&1&1&1
\end{array} \right)}
\Q\to0
\]
We have
\begin{align*}
  \rk(\partial_0)=1, \mathrm{nullity}(\partial_0)=5-1=4\\
  \rk(\partial_1)=4, \mathrm{nullity}(\partial_1)=5-4=1.
\end{align*}
Therefore,~$\tilde{\beta}_0=4-4=0$ and~$\tilde{\beta}_1=1$.  

Homology:
\[
  \widetilde{H}_1=\ker\partial_1/\im\partial_2
  =\ker\partial_1=
  \Span\{\overline{12}+\overline{23}+\overline{34}+\overline{45}-\overline{15}\}
\]
The first homology is generated by a cycle of edges.

{\bf Question:} What happens in homology if we start with the triangle and
subdivide its edges arbitrarily?
\medskip

\hrule
\medskip

III.1.
\begin{center}
  \begin{tikzpicture}
    \node[ball,label={below left:{$\overline{1}$}}] (1) at (0,0) {};
    \node[ball,label={below right:{$\overline{2}$}}] (2) at (1,0) {};
    \node[ball,label={above:{$\overline{3}$}}] (3) at (0.5,0.866) {};
    \node[ball,label={above left:{$\overline{4}$}}] (4) at (-0.5,0.866) {};
    \node[ball,label={above right:{$\overline{5}$}}] (5) at (1.5,0.866) {};
    \draw (1)--(2)--(3)--(1);
    \draw (1)--(4)--(3);
    \draw (2)--(5)--(3);
  \end{tikzpicture}
\end{center}
Let's compute the first homology.
\[
  0\xrightarrow[\partial_2]{\hspace{5mm}}\Q^7\xrightarrow[\partial_1]{
    \left(\begin{array}{rrrrrrr}
      -1&-1&-1&0&0&0&0\\
      1&0&0&-1&-1&0&0\\
      0&1&0&1&0&-1&-1\\
      0&0&1&0&0&1&0\\
      0&0&0&0&1&0&1
\end{array} \right)}\Q^5
\]
We have~$\rk\partial_1=4$, and so~$\mathrm{nullity}\partial_1=7-4=3$.
Therefore~$\tilde{\beta}_1=3$.  The homology is generated by the cycles
surrounding the three bounded faces of the complex as drawn above:
\[
  \widetilde{H}_1=\ker\partial_1=
  \Span\{\overline{23}-\overline{13}+\overline{12},\overline{34}-\overline{14}+\overline{13},
\overline{35}-\overline{25}+\overline{23}\}
\]

III.2.
\begin{center}
  \begin{tikzpicture}
    \node[ball,label={below left:{$\overline{1}$}}] (1) at (0,0) {};
    \node[ball,label={below right:{$\overline{2}$}}] (2) at (1,0) {};
    \node[ball,label={above:{$\overline{3}$}}] (3) at (0.5,0.866) {};
    \node[ball,label={above left:{$\overline{4}$}}] (4) at (-0.5,0.866) {};
    \node[ball,label={above right:{$\overline{5}$}}] (5) at (1.5,0.866) {};
    \draw (1)--(2)--(3)--(1);
    \draw[fill=gray!40] (1.center)--(4.center)--(3.center)--cycle;
    \draw (2)--(5)--(3);
  \end{tikzpicture}
\end{center}
For first homology, note that
$\partial_2(\overline{134})=\overline{34}-\overline{14}+\overline{13}$.  We get
\[
  \Q\xrightarrow[\partial_2]{
    \left(\begin{array}{r}
      0\\1\\-1\\0\\0\\1\\0
\end{array} \right)}
  \Q^7\xrightarrow[\partial 1]{
    \left(\begin{array}{rrrrrrr}
      -1&-1&-1&0&0&0&0\\
      1&0&0&-1&-1&0&0\\
      0&1&0&1&0&-1&-1\\
      0&0&1&0&0&1&0\\
      0&0&0&0&1&0&1
\end{array} \right)}\Q^5
\]
We have~$\mathrm{nullity}(\partial_1)=7-4=3$ and~$\rk{\partial_2}=1$.
Therefore~$\tilde{\beta}_1=3-1=2$.  The homology is generated by the cycles
surrounding the two unfilled bounded faces of the complex:
\begin{align*}
  \widetilde{H}_1=\ker\partial_1\im\partial_2&= 
  \Span\{\overline{23}-\overline{13}+\overline{12},\overline{34}-\overline{14}+\overline{13},
  \overline{35}-\overline{25}+\overline{23}\}/\Span\left\{ \overline{34}-\overline{14}+\overline{13} \right\}
  \\[8pt]
  &\simeq
\Span\{\overline{23}-\overline{13}+\overline{12}, \overline{35}-\overline{25}+\overline{23}\}
\end{align*}
The cycle~$\overline{34}-\overline{14}+\overline{13}$ is the boundary of
the shaded face, and thus has become~$0$ in the homology group.

{\bf Question:} How does this example generalize?
\medskip

\hrule
\bigskip

{\bf Exercise.}  Draw the following simplicial complexes, determine their Betti
numbers, and describe bases for their homology groups.  Recall that a {\em
facet} of a simplicial complex is a face that is maximal with respect to
inclusion.  So we can describe a simplicial complex by just listing its facets.
The whole simplicial complex then consists of the facets and all of their
subsets.
\begin{enumerate}
  \item $\Delta$ with facets $\overline{123}$, $\overline{24}$,
    $\overline{34}$, $\overline{45}$, $\overline{56}$, $\overline{57}$,
    $\overline{89}$, $\overline{10}$.
  \item $\Delta$ with facets $\overline{123}$, $\overline{14}$,
    $\overline{24}$, and $\overline{34}$.
  \item $\Delta$ with facets $\overline{123}$, 
    $\overline{124}$, $\overline{134}$, $\overline{234}$, 
    $\overline{125}$, $\overline{135}$, $\overline{235}$.  (Two hollow
      tetrahedra glued along a face.) 
\bigskip

\hrule
\bigskip

Let~$\Delta$ be the simplicial complex from Exercise~3, above.  It is called the
{\em equatorial bipyramid}:
\begin{center}
    \begin{tikzpicture}[scale=1.0]
      
      \coordinate (1) at (0.2,0,0.8);
      \coordinate (2) at (0.866,0,-0.5);
      \coordinate (3) at (-0.966,0,-0.6);
      \coordinate (t) at (0,1.5,0);
      \coordinate (b) at (0,-1.4,0);
      
      \draw (t)--(1);
      \draw (t)--(2);
      \draw (t)--(3);
      \draw (b)--(1);
      \draw (b)--(2);
      \draw (b)--(3);
      \draw (1)--(2)--(3)--(1);
      \shadedraw[top color=gray!50, bottom color=gray!80,opacity=0.7] (1.center)--(2.center)--(3.center)--cycle;
      \shadedraw[top color=gray!10, bottom color=gray!60,opacity=0.4] (t.center)--(1.center)--(2.center)--cycle;
      \shadedraw[top color=gray!10, bottom color=gray!60,opacity=0.4] (1.center)--(3.center)--(t.center)--cycle;
      \shadedraw[bottom color=gray!10, top color=gray!60,opacity=0.4] (1.center)--(3.center)--(b.center)--cycle;
      \shadedraw[bottom color=gray!10, top color=gray!60,opacity=0.4] (1.center)--(2.center)--(b.center)--cycle;

      \node[label=-45:{$1$},ball] at (1) {};
      \node[label=0:{$2$},ball] at (2) {};
      \node[label={180:$3$},ball] at (3) {};
      \node[label=90:{$4$},ball] at (t) {};
      \node[label=270:{$5$},ball] at (b) {};
    \end{tikzpicture}
  \end{center}
  In each dimension~$i$ except~$2$, every~$i$-cycle is an~$i$-boundary.  For
  instance, in dimension~$0$, we have that~$\overline{5}-\overline{4}$ is a~$0$-cycle since
  its boundary is~$\overline{\emptyset}-\overline{\emptyset}=0$.  However,
  since~$\overline{14}$ and~$\overline{15}$ are part of the simplicial complex,
  we see
\[
  \partial_i(-\overline{15}+\overline{14})=-(\overline{5}-\overline{1})+(\overline{4}-\overline{1})
  =\overline{5}-\overline{4}
\]
is a boundary.  Thus, in~$\widetilde{H}_1$, we have that
$\overline{5}-\overline{4}=0$.   Similarly, 
\[
  \overline{14}-\overline{34}+\overline{35}-\overline{15}
\]
is a~$1$-cycle of~$\Delta$, and it is the boundary
of~$-\overline{134}+\overline{135}$.

However,~$\widetilde{H}_2\simeq\Q^2$.  Notice, that the solid
tetrahedra~$\overline{1234}$ and~$\overline{1235}$ are not part of~$\Delta$.  We
can still imagine there boundaries, though, and these are generators
for~$\widetilde{H}_2$.  Their boundaries will be~$2$-cycles,
since~$\partial^2=0$, however, these~$2$-cycles are not boundaries of~$\Delta$.
So they are nonzero in~$\widetilde{H}_2$.

If we added~$\overline{1234}$ to~$\Delta$, we would get a simplicial complex
with~$\widetilde{H}_2\simeq\Q$.
\end{enumerate}
\medskip

\hrule
\bigskip

Let~$\Delta$ be a simplicial complex, and consider the homology of~$\Delta$ with
coefficients in~$\Z$.  This means that~$C_i:=\Z F_i$ and the boundary
mappings~$\partial_i\colon\Z F_i\to\Z F_{i-1}$ are only~$\Z$-linear.

For a hint at the difference between~$\Z$-coefficients and~$\Q$-coefficients,
note that
\[
  \Q^2/\Span_{\Q}\left\{ (0,2) \right\}\simeq\Q,
  \quad\text{and}\quad
  \Z^2/\Span_{\Z}\left\{ (0,2) \right\}\simeq \Z\times\Z/2\Z.
\]
The space~$\Z^2/\Span_{\Z}\left\{ (0,2) \right\}$ is a~$\Z$-module, meaning it is like a vector space but
with the scalars being the ring~$\Z$ rather than a field.  It has a {\em free}
part isomorphic to~$\Z$ and a {\em torsion} part, isomorphic to~$\Z/2\Z$. In
general, if~$v_1,\dots,v_k\in\Z^n$,
\[
  M:=\Z^n/\Span_{\Z}\{v_1,\dots,v_k\}\simeq\Z^r\oplus\left( \Z/d_1\Z\oplus\cdots\oplus\Z/d_{\ell}\Z \right)
\]
for some~$r$ and some integers~$d_1,\dots,d_{\ell}$.  The {\em free part}
of~$M$ is~$\Z^r$, and we say~$M$ has {\em rank}~$r$.  The {\em torsion part}
of~$M$ is $\left( \Z/d_1\Z\oplus\cdots\oplus\Z/d_{\ell}\Z \right)$. 

To start to get an idea of the difference for simplicial complexes,
let~$\Delta$ be the triangulation of the real projective plane given below:
\begin{center}
  \begin{tikzpicture}[scale=0.75]
    \node at (0,0) {$\Delta =$};
    \begin{scope}[xshift=3.5cm]
    \def\r{2.0cm}
    \draw[fill=gray!40] ({\r*cos(0*60)},{\r*sin(0*60)})--({\r*cos(1*60)},{\r*sin(1*60)})--({\r*cos(2*60)},{\r*sin(2*60)})--({\r*cos(3*60)},{\r*sin(3*60)})--({\r*cos(4*60)},{\r*sin(4*60)})--({\r*cos(5*60)},{\r*sin(5*60)})--({\r*cos(6*60)},{\r*sin(6*60)});
    
    \draw ({\r*cos(0*60},{\r*sin(0*60)}) node[label=0*60:{$5$},ball] (1) {};
    \draw ({\r*cos(1*60},{\r*sin(1*60)}) node[label=1*60:{$6$},ball] (2) {};
    \draw ({\r*cos(2*60},{\r*sin(2*60)}) node[label=2*60:{$4$},ball] (3) {};
    \draw ({\r*cos(3*60},{\r*sin(3*60)}) node[label=3*60:{$5$},ball] (11) {};
    \draw ({\r*cos(4*60},{\r*sin(4*60)}) node[label=4*60:{$6$},ball] (22) {};
    \draw ({\r*cos(5*60},{\r*sin(5*60)}) node[label=5*60:{$4$},ball] (33) {};

    \def\s{0.8cm}
    \draw ({\s*cos(210},{\s*sin(210)}) node[label=210:{$1$},ball] (4) {};
    \draw ({\s*cos(-30},{\s*sin(-30)}) node[label=-30:{$2$},ball] (5) {};
    \draw ({\s*cos(90},{\s*sin(90)}) node[label=90:{$3$},ball] (6) {};

    \draw (22)--(4);
    \draw (22)--(5)--(33);
    \draw (5)--(1)--(6)--(2);
    \draw (6)--(3)--(4)--(11);
    \draw (6)--(4)--(5)--(6);
  \end{scope}
\end{tikzpicture}
\end{center}
Where edges appear twice, it is understood that they should be glued together.
This space might look contractible (although something complicated is clearly
happening on the boundary), and in fact, over~$\Q$, all of its homology groups
are trivial.  Working over the integers, though, something interesting happens
in dimension~$2$.  Consider orienting all ten triangles in the counterclockwise
direction and adding them to get a~$2$-chain~$\alpha$.  We have
\begin{align*}
  \partial_2(\alpha)
  &= 
  \overline{45}+\overline{56}-\overline{46}+\overline{45}+\overline{56}-\overline{46}\\[5pt]
  &=2\left( \overline{45}+\overline{56}-\overline{46} \right).
\end{align*}
Letting~$\beta:=\overline{45}+\overline{56}-\overline{46}$, we
have that~$2\beta\in\im(\partial_2)$.
Therefore,~$2\beta=0\in\widetilde{H}_2:=\ker\partial_1/\im\partial_2$.  If we
were working over~$\Q$, this would also imply that~$\beta$ is~$0$
in~$\widetilde{H}_2$ (since~$\partial(\alpha/2)=\beta$).  But over~$\Z$, it turns
  out that~$\beta$ is nonzero in homology.  In fact, we have
  \[
    \widetilde{H}_2(\Delta,\Z)\simeq\Z/2\Z.
  \]
\end{document}
