\classheader{2012-11-14}

\subsection*{Kunneth Theorem}

Our goal is to compute $H^*(X\times Y;R)$ in terms of $H^*(X;R)$ and $H^*(Y;R)$. Recall that the product $X\times Y$ comes with natural projections
\begin{center}
\begin{tikzcd}
& X\times Y \ar{rd}{P_2} \ar{ld}[swap]{P_1} & \\
X & & Y
\end{tikzcd}
\end{center}
\begin{definition}
The $\times$ product on cohomology, $\times:H^*(X;R)\times H^*(Y;R)\to H^*(X\times Y;R)$, is defined by sending $a\times b$ to $P_1^*(a)\cupprod P_2^*(b)$.
\end{definition}

The map $\times$ is bilinear, and $\times$ induces a homomorphism of $R$-modules
\[\Psi:H^*(X;R)\tensor_R H^*(Y;R)\to H^*(X\times Y;R).\]
In fact, we can make this into a homomorphism of graded rings by \textit{defining} 
\[(a\smalltensor b)\cdot(c\smalltensor d):=(-1)^{|b||c|}(ac\smalltensor bd)\]
where $|b|$ and $|c|$ are the degrees of $b$ and $c$.

\begin{theorem}[Kunneth]
If $X$ and $Y$ are CW-complexes, and if $H^*(Y;R)$ is a free $R$-module, then $\Psi$ is an isomorphism.
\end{theorem}

\begin{example}
The cohomology ring of $\T^n$ over $\Z$ is \[H^*(\T^n,\Z)\cong\Lambda^*[x_1,\ldots,x_n]=\bigoplus_{i=0}^n\Lambda^i[x_1,\ldots,x_n]\]
where $\Lambda^i[x_1,\ldots,x_n]$ is the free abelian group on
\[\{x_{i_1}\wedge \cdots\wedge x_{i_k}\mid i_1<\cdots<i_k\},\]
so that $\rank(\Lambda^i[x_1,\ldots,x_n])=\binom{n}{i}$.
\end{example}
\begin{proof}
We induct using Kunneth:
\[H^*(\T^n,\Z)\cong H^*(\T^{n-1}\times\S^1;\Z)=H^*(\T^{n-1})\otimes H^*(\S^1).\]
For example,
\begin{align*}
H^3(\T^4)&=H^{3,0}\oplus H^{2,1}\oplus H^{1,2}\oplus H^{0,3}\\
&=[H^3(\T^3)\otimes H^0(\S^1)]\oplus [H^2(\T^3)\otimes H^1(\S^1)]\oplus[H^1(\T^3)\otimes H^2(\S^1)]\oplus[H^0(\T^3)\otimes H^3(\S^1)]\\
&=[\Z\otimes \Z] \oplus[\Lambda^2\Z^3\otimes \Z]\oplus 0\oplus 0\\
&=\Z\oplus\Lambda^2\Z^3.\qedhere
\end{align*}
\end{proof}

When $\Psi$ is an isomorphism, we have
\[H^n(X\times Y;\Z)=\bigoplus_{p+q=n}[H^p(X)\otimes H^q(Y)].\]
In Hodge theory, we see that there is a decomposition like this even when we're not looking at a product manifold.


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

\begin{definition}
A second countable, Hausdorff space $M$ is an $n$-dimensional manifold (possibly wth boundary) if for any $m\in M$, there is a neighborhood $U_m$ such that either $U_m\cong\R^n\cong B^n$ (the open ball in $\R^n$), or $U_m\cong\R_+^n=\{(x_1,\ldots,x_n)\mid x_i\geq 0\}$ (the closed upper half space). Key fact:
\[\{m\in M\text{ for which the latter holds}\}\cong\text{ an }(n-1)\text{-dimensional manifold }\partial M.\]
\end{definition}

\begin{definition}
A compact manifold $M$ is called triangulable if there is a simplicial complex $K\cong M$. Most of the manifolds which occur in real life are triangulable, and we might as well assume that all manifolds we'll use in class will be triangulable.
\end{definition}

\begin{definition}
An orientation on a simplex $[v_0\cdots v_n]$ is a choice of one of the two equivalence classes of $(n+1)$-tuples $(v_{i_0}\cdots v_{i_n})$, where two $(n+1)$-tuples are equivalent when the permutation sending one to the other is an even permutation.
\end{definition}

\begin{definition}
$M$ is orientable if one of the following equivalent statements holds.
\begin{enumerate}
\item There is a triangulation on $M$ which can be oriented compatibly.
\item All triangulations on $M$ can be oriented compatibly.
\end{enumerate}
\end{definition}

We know that under any reasonable definition, a disk ought to be orientable; giving the disk a triangulation and assigning the same orientation to each simplex, we get
\begin{center}
\begin{tikzpicture}[scale =1.7]
\begin{scope}[every node/.style={inner sep=0pt,outer sep=0pt,minimum size=5pt,fill,circle}]
\node (a) at (0,0) {};
\node (b) at (1,0){};
\node (c) at (1,1){};
\node (d) at (0,1){};
\end{scope}
\draw[thick] (a) rectangle (c);
\draw[thick] (b) -- (d);
\node[scale=1.5] at (barycentric cs:a=1.5,b=1,d=1) {$\lcirclearrow$};
\node[scale=1.5,rotate=180] at (barycentric cs:c=1.5,b=1,d=1) {$\lcirclearrow$};
\end{tikzpicture}
\end{center}
This motivates the definition of compatible orientation to be a choice of orientation on each $n$-simplex $\sigma_i$ such that if $\sigma_i\cap \sigma_j$ is a (codimension 1) face of each, then $\sigma_i$ and $\sigma_j$ induce opposite orientations on that face.

\begin{exercise}
$\text{}$

\begin{enumerate}
\item Prove that $\Sigma_g$ is orientable for all $g\geq 1$.
\item Prove that the Mobius strip and Klein bottle are not.
\item How about $\RP^n$?
\end{enumerate}
\end{exercise}

\begin{theorem}[Poincar\'e duality]
Let $M$ be a closed (i.e. compact, with no boundary), orientable manifold of dimension $n$. Then
\[H^i(M;R)\cong H_{n-i}(M;R).\]
\end{theorem}
\begin{corollary}
If $M$ is a closed manifold of odd dimension, then $\chi(M)=0$.
\end{corollary}
\begin{proof}[Proof of corollary]
Let $b_i=\dim(H_i(M;\R))$. Then we know $b_i=\dim(H^i(M;\R))$ by the universal coefficient theorem, but by Poincar\'e duality $b_i$ is also equal to $\dim(H^{n-i}(M;\R))$. Thus
\[\chi(M)=b_0-b_1+b_2-\cdots-b_{2k+1}\]
cancels out, because $b_0=b_{2k+1}$, $b_1=b_{2k}$, etc.
\end{proof}










