\classheader{2012-11-16}

\subsection*{Another Corollary of Poincar\'e Duality}

Here is a useful algebraic fact: any non-degenerate skew-symmetric bilinear form on $\Z^n$ is, after changing basis, of the form
\[\begin{pmatrix}
J & & \\
& \ddots & \\
& & J
\end{pmatrix}\]
where $J=\left(\begin{smallmatrix}
0 & -1 \\ 1 & 0
\end{smallmatrix}\right)$. In particular, $n$ must be even. This fact, together with Poincar\'e duality, imply
\begin{corollary}
Let $M$ be a closed, oriented, $(4k+2)$-dimensional manifold. Then because we have a non-degenerate pairing $H^{n-i}(M)\times H^i(M)\to\Z$, $\rank(H^i(M,\Z))=\chi(M)$ must be even.
\end{corollary}

\subsection*{Continued Prologue to Poincar\'e duality}

If $M$ be a closed, orientable $n$-manifold, then $H_i(M;\Q)\cong H^{n-i}(M;\Q)$. Poincar\'e's idea about this was as follows: in a $\Delta$-complex, we compute $H_i$ using $C_i^\Delta(M)$. We can build a ``dual'' CW-complex (it won't be a $\Delta$-complex in general, we may get a polygonal complex) by putting a 0-cell in the center of every $n$-simplex, adding a 1-cell connecting two $n$-simplices if they meet on a codimension 1 subface, etc. 
\begin{center}
\begin{tikzpicture}[scale=1.6]
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] (center) at (0,0) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] (a) at (36:1) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] (b) at (108:1) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] (c) at (180:1) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] (d) at (252:1) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] (e) at (324:1) {};
\draw[thick] (a.center) -- (b.center) -- (c.center) -- (d.center) -- (e.center) -- (a.center);
\draw[thick] (a.center) -- (center.center);
\draw[thick] (b.center) -- (center.center);
\draw[thick] (c.center) -- (center.center);
\draw[thick] (d.center) -- (center.center);
\draw[thick] (e.center) -- (center.center);
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] (at) at (barycentric cs:a=1/2,b=1/2,center=1/2) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] (bt) at (barycentric cs:b=1/2,c=1/2,center=1/2) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] (ct) at (barycentric cs:c=1/2,d=1/2,center=1/2) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] (dt) at (barycentric cs:d=1/2,e=1/2,center=1/2) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] (et) at (barycentric cs:e=1/2,a=1/2,center=1/2) {};
\draw[thick,dotted] (at.center) -- (bt.center) -- (ct.center) -- (dt.center) -- (et.center) -- (at.center);
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] (x) at (barycentric cs:a=1,e=1,center=-1) {};
\draw[thick] (a.center) -- (x.center);
\draw[thick] (e.center) -- (x.center);
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] (y) at (barycentric cs:a=1,x=1,e=-1) {};
\draw[thick] (a.center) -- (y.center);
\draw[thick] (x.center) -- (y.center);
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] (xt) at (barycentric cs:a=1/2,e=1/2,x=1/2) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle] (yt) at (barycentric cs:a=1/2,x=1/2,y=1/2) {};
\draw[thick,dotted] (et.center) -- (xt.center) -- (yt.center);
\end{tikzpicture}
\end{center}

Thus, $\partial\longleftrightarrow\delta$ is an isomorphism
\[C_i^\Delta(X)\cong C_{n-i}^\CW(X').\]
Thus, $\{C_i^\Delta(X),\partial\}$ computes $H_i(X)\cong H_i(M)$, and $\{C_{n-1}^{\Delta}(X),\delta\}$ computes $H^{n-i}(X')\cong H^{n-i}(M)$.

\subsection*{Classification of Surfaces}

\begin{theorem-N}
Let $M$ be a closed 2-manifold. Then $M$ is homeomorphic to one of:
\begin{itemize}
\item $\Sigma_g$ where $g\geq 0$ (when $M$ orientable), or
\item $\Sigma_g\mathbin{\#}\RP^2$ (when $M$ non-orientable).
\end{itemize}
\end{theorem-N}

\begin{theorem-N}[2-dimensional Poincar\'e conjecture]
The following are equivalent:
\begin{enumerate}
\item $\chi(M)=2$
\item $M\cong \S^2$
\item Every loop $\gamma\subset M$ (i.e. embedding $\gamma:\S^1\to M$) separates $M$ into 2 components,
\end{enumerate}
\end{theorem-N}
Statement 2 implies statement 3 by the Jordan separation theorem, and we already know that statement 2 implies statement 1.


We will be assuming the following classical result:
\begin{theorem}[Rado, 1920's]
Every surface is triangulable.
\end{theorem}


\begin{proof}[Proof of Theorem 2]
We will show that statement 1 implies statement 2. Let $K$ be a triangulation of $M$. First, pick a maximal tree $T$ in $K^{(1)}$. Note that $T\supset K^{(0)}$.

Now build a connected graph $\Gamma\subset K$ as follows: $V(\Gamma)$ consists of the 2-simplices of $K$, and $E(\Gamma)$ consists of $K^{(1)}-T^{(0)}$.

Here is an example on a tetrahedron:
\begin{center}
\begin{tikzpicture}[scale=2.4]
\begin{scope}[every node/.style={fill,blue,circle,inner sep=0pt,outer sep=1pt,minimum size=0.15cm}]
\node (v0) at (0,0) {};
\node (v1) at (0.65,-0.4) {};
\node (v2) at (1,0.2) {};
\node (v3) at (.5,0.8)  {};
\end{scope}
\begin{scope}[every node/.style={draw,thick,red,circle,inner sep=0pt,outer sep=0pt,minimum size=0.1cm}]
\node (w0) at (barycentric cs:v0=1,v1=1,v3=1.4)  {};
\node (w2) at (barycentric cs:v0=1,v1=1,v2=2.9)  {};
\node (w4) at (barycentric cs:v0=1,v3=1,v2=3.9)  {};
\node (w6) at (barycentric cs:v1=1.6,v2=1.4,v3=3.9)  {};
\end{scope}
\node[inner sep=0pt,outer sep=0pt] (w1) at (barycentric cs:v0=1,v1=1,v3=0)  {};
\node[inner sep=0pt,outer sep=0pt] (w3) at (barycentric cs:v0=1,v2=2.1,v3=0)  {};
\node[inner sep=0pt,outer sep=0pt] (w5) at (barycentric cs:v1=0,v2=1.8,v3=1)  {};




\begin{scope}
\clip (v0.center) -- (v1.center) -- (v2.center) -- (v3.center) -- cycle;
\draw[red,,double distance = 1pt] (w4) -- (w5.center) -- (w6);
\draw[red,,double distance = 1pt] (w5.center) -- (w6);
\draw[red,,double distance = 1pt] (w2) -- (w3.center) -- (w4);
\draw[densely dashed] (v0.center) -- node[auto,swap] {} (v2.center);
\draw[red,,double distance = 1pt] (w1.center) -- (w2);
\draw[red,,double distance = 1pt] (w0) -- (w1.center);
\end{scope}
\draw[thick] (v2.center) -- node[auto,swap] {} (v3.center);

\draw[thick] (v0.center) -- node[auto,swap] {} (v1.center);

\draw[blue,very thick] (v0.center) -- node[auto,swap] {} (v3.center);
\draw[blue,very thick] (v1.center) -- node[auto,swap] {} (v2.center);
\draw[blue,very thick] (v1.center) -- node[auto,swap] {} (v3.center);

\begin{scope}[every node/.style={fill,blue,circle,inner sep=0pt,outer sep=1pt,minimum size=0.15cm}]
\node (v0) at (0,0) {};
\node (v1) at (0.65,-0.4) {};
\node (v2) at (1,0.2) {};
\node (v3) at (.5,0.8)  {};
\end{scope}
\end{tikzpicture}
\end{center}

We have
\begin{align*}
\chi(M)&=V(K)-E(K)+F(K)\\
&=V(T)-(E(T)+E(\Gamma))+V(\Gamma)\\
&=[V(T)-E(T)]+[V(\Gamma)-E(\Gamma)]\\
&=\underbrace{\chi(T)}_{\leq\,1}+\underbrace{\chi(\Gamma)}_{\leq\,1}\\
&\leq 2,
\end{align*}
where we have used the following lemma.
\begin{lemma}
Let $\Omega$ be a finite connected graph. Then $\chi(\Omega)\leq 1$ with equality if and only if $\Omega$ is a tree.
\end{lemma}
\begin{proof}[Proof of lemma]
Trivial application of induction (check for yourself).
\end{proof}
Thus, we have that $\chi(M)\leq 2$, with equality if and only if $\chi(T)=\chi(\Gamma)=1$, which is the case if and only if $\Gamma$ is a tree. But if $T\subset M$ is a tree, there is a neighborhood of $T$ homeomorphic to $\D^2$. Being very careful, we can thicken $\Gamma$ and $T$ to neighborhoods each homeomorphic to $\D^2$, and such that \[\overline{\text{nbhd}(T)}\cap\overline{\text{nbhd}(\Gamma)}\cong\S^1=\partial\D_1^2=\partial\D_2^2.\]
This implies that $M\cong \S^2$.

Now, note that statement 3 implies that $\Gamma$ is a tree: if it weren't, it would have a loop $\gamma$, and $\gamma$ would have to separate $M$; but then it would separate two vertices of $K^{(0)}$, which would contradict that $\Gamma\cap T=\varnothing$.

Thus, statement 3 implies statement 1. 
\end{proof}

\begin{proof}[Proof of Theorem 1]
We are given $M$, which has $\chi(M)\leq 2$ by the above.
\begin{center}
\begin{tikzpicture}
\node[diamond,draw,thick] (a) at (0,0) {\parbox{0.48in}{\centering Does\\$\chi(M)=$\\ $2$ ?}};
\draw[thick,->,>=angle 90] (a.north east) --++ (0.5,0.5) node (b) {} --++  (2,0) node [midway,auto] {Yes} node[label={right:$M\cong\S^2$}] (d) {};
\draw[thick,->,>=angle 90] (a.south east) --++ (0.5,-0.5) node (c) {} --++ (2,0) node [midway,auto,swap] {No}  node[label={right:\parbox{1.5in}{$\exists$ a loop $\gamma$ in $M$ that does not separate $M$}}] (d) {};;
\end{tikzpicture}
\end{center}
A neighborhood of $\gamma$ is homeomorphic to either $\S^1\times[0,1]$ or the M\"obius strip (the latter is only possible if $M$ is not orientable).

\begin{center}
\begin{tikzpicture}[scale=1.2]
\begin{scope}
\clip (-0.6,-1.1) rectangle (3.6,1.1);
\draw [thick,fill=gray!40,postaction={decorate,decoration={markings,mark=at position 0.78 with {\node (a) {};}}}] plot [smooth cycle,tension=0.8] coordinates { (3.5,0) (2.7,0.9) (1.5,0.6) (0.3,0.9) (-0.5,0) (0.3,-0.9) (1.5,-0.6) (2.7,-0.9) };
\begin{scope}
\clip (0.5,0) ellipse (0.4 and 0.2);
\draw[thick,fill=white] (0.5,-0.1) ellipse (0.35 and 0.25);
\end{scope}
\begin{scope}
\clip (-0.7,0) rectangle (6.3,-1);
\draw[thick] (0.5,0) ellipse (0.4 and 0.2);
\end{scope}
\begin{scope}
\clip (2.5,0) ellipse (0.4 and 0.2);
\draw[thick,fill=white] (2.5,-0.1) ellipse (0.35 and 0.25);
\end{scope}
\begin{scope}
\clip (-0.7,0) rectangle (6.3,-1);
\draw[thick,postaction={decorate,decoration={markings,mark=at position 0.85 with {\node (b) {};}}}] (2.5,0) ellipse (0.4 and 0.2);
\end{scope}
\end{scope}
\node (c) at ($(a)!0.5!(b)$) {};
\begin{scope}[shift={(c)},x={(a)}]
\path (c) -- (a) node[rotate around={90:(c)}] (n) {};
\end{scope}
\begin{scope}[shift={(c)},x={(a)},y={(n)}]
\draw[thick,red] (1,0) node[label={[black,shift={(1.9,0)}]$\gamma$}] {} arc (0:180:1 and 0.35);
\draw[dashed,thick,red] (-1,0) arc (180:360:1 and 0.35);
\end{scope}
%
%
\draw[thick,decorate,decoration={snake,amplitude=1pt},-angle 90] (4,0) to node[midway,auto] {cut} (5,0);
%
%
\begin{scope}[shift={(6,0)}]
%
%
\begin{scope}
\clip (-0.6,-1.1) rectangle (3.6,1.1);
\draw [thick,fill=gray!40,postaction={decorate,decoration={markings,mark=at position 0.8 with {\node (a) {};},mark=at position 0.74 with {\node (c) {};}}}] plot [smooth cycle,tension=0.8] coordinates { (3.5,0) (2.7,0.9) (1.5,0.6) (0.3,0.9) (-0.5,0) (0.3,-0.9) (1.5,-0.6) (2.7,-0.9) };
\begin{scope}
\clip (0.5,0) ellipse (0.4 and 0.2);
\draw[thick,fill=white] (0.5,-0.1) ellipse (0.35 and 0.25);
\end{scope}
\begin{scope}
\clip (-0.7,0) rectangle (6.3,-1);
\draw[thick] (0.5,0) ellipse (0.4 and 0.2);
\end{scope}
\begin{scope}
\clip (2.5,0) ellipse (0.4 and 0.2);
\draw[thick,fill=white] (2.5,-0.1) ellipse (0.35 and 0.25);
\end{scope}
\begin{scope}
\clip (-0.7,0) rectangle (6.3,-1);
\draw[thick,postaction={decorate,decoration={markings,mark=at position 0.89 with {\node (b) {};},mark=at position 0.78 with {\node (d) {};}}}] (2.5,0) ellipse (0.4 and 0.2);
\end{scope}
\end{scope}
\node (e) at ($(a)!0.5!(b)$) {};
\node (f) at ($(c)!0.5!(d)$) {};
\filldraw[fill=white,draw=white,very thick] ([shift={(0.00,-0.02)}]a.center) -- ([shift={(0.04,-0.1)}]c.center) -- ([shift={(0.01,0)}]d.center) -- (b.center) -- ([shift={(0.00,-0.02)}]a.center);
\begin{scope}[shift={(e)},x={(a)}]
\path (e) -- (a) node[rotate around={90:(e)}] (n) {};
\end{scope}
\begin{scope}[shift={(e)},x={(a)},y={(n)}]
\draw[thick,shade,top color=gray!30] (1,0) node[label={[black,shift={(1.9,-0.6)}]$\gamma_2$}] {} arc (0:180:1 and 0.35) arc (180:360:1 and 0.35);
\end{scope}
\begin{scope}[shift={(f)},x={(c)}]
\path (f) -- (c) node[rotate around={90:(f)}] (m) {};
\end{scope}
\begin{scope}[shift={(f)},x={(c)},y={(m)}]
\draw[thick,shade,top color=gray!30] (1,0) node[label={[black,shift={(1.9,0)}]$\gamma_1$}] {} arc (0:180:1 and 0.35) arc (180:360:1 and 0.35);;
\end{scope}
\node at (1.5,1) {$M_1$};
\begin{scope}[rotate around={-60:(4.2,0)}]
\node (u) at (3.9,0.3) {};
\node (w) at (3.9,-0.3) {};
\node (p) at (4.5,0.3) {};
\node (q) at (4.5,-0.3) {};
\node at (4.85,0) {$M_2$};
\draw[thick,fill=gray!40] (p.center) arc (0:180:0.3 and 0.1) -- (w.center) -- (q.center) -- cycle;
\draw[thick,shade,top color=gray!30] (4.2,-0.3) ellipse (0.3 and 0.1);
\end{scope}
\end{scope}
\end{tikzpicture}
\end{center}
Let $M'$ be $M_1$ with $\gamma_1$ and $\gamma_2$ filled in by disks. Thus, $M'$ is a new closed, connected surface, and
\[\chi(M)=\chi(M_1)+\chi(M_2)-\chi(\S_1\sqcup \S_1).\]
It is easy to check that $\chi(M')>\chi(M)$; in fact, $\chi(M')=\chi(M)+2$. 
\end{proof}


