\classheader{2012-11-12}

Today we'll look at the cup product structure on the cohomology of a few spaces.

Let's start by considering $\S^1\vee\S^1\vee\S^2$ vs. $\T^2$. 

Both of these spaces have CW-complex structures with one 0-cell, two 1-cells, and one 2-cell. The difference is that the 2-cell was attached via a homotopically trivial map in $\S^1\vee\S^1\vee\S^2$, while it was attached via the the path $aba^{-1}b^{-1}$ in $\T^2$. While homotopically these are different, on the level of homology they cannot be distinguished because $H_1$ is the abelianization of the fundamental group. Surprisingly, the difference can be detected in the cup product structure.

The homology of both of these spaces is
\[H_*(\S^1\vee\S^1\vee\S^2)\cong H_*(\T^2)=\begin{cases}
\Z & \text{ in dim 0}\\
\Z^2 & \text{ in dim 1}\\
\Z & \text{ in dim 2}\\
0 & \text{ in higher dim}
\end{cases}\]
Because all of these groups are free, the Ext groups in the universal coefficient theorem are trivial, and therefore the cohomology of both spaces is isomorphic to the homology.

What about the cup product structure on $H^*$? The only interesting case is in $H^1$. We have that $H^1=\Z^2=\Hom(H_1,\Z)$ is generated by two 1-cocycles $\alpha$ and $\beta$, where $\alpha(a)=1$ and $\alpha(b)=0$, and $\beta(a)=0$ and $\beta(b)=1$.

By the anti-commutativity of the cup product in odd dimensions, for either space we have $\alpha\cupprod\alpha=-(\alpha\cupprod\alpha)$ and $\beta\cupprod\beta=-(\beta\cupprod\beta)$, and because $H^2$ has no 2-torsion, this implies $\alpha\cupprod \alpha=\beta\cupprod \beta=0$ for either space.

What about $\alpha\cupprod\beta$? 

We map $X=\S^1\vee\S^1\vee\S^2$ down to $Y=\S^1\vee\S^1$ by collapsing the 2-cell to a point:\vspace{-0.1in}
\begin{center}
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\node (c) at ($(a)!0.5!(b)$) {};
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\path[thick] (y) edge[loop,distance= 1.7cm,out=330,in=30] node[pos=0.2,auto,swap] {$b$} (y);
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\draw[->,>=latex,thick,shorten >=0.4cm,shorten <=0.4cm] (x) -- node [auto] {$f$} (y);
\node at (-1.7,3) {$X$};
\node at (-1.7,0) {$Y$};
\end{tikzpicture}
\end{center}
We have that
\[H^*(Y)=\begin{cases}
\Z & \text{ in dim 0}\\
\Z^2 & \text{ in dim 1}\\
0 & \text{ in higher dim}
\end{cases}\]
Let $\gamma,\delta\in H^1(Y)$ be generators of $H^1(Y)\cong\Z^2$. Clearly, $\gamma\cupprod\delta=0$ because $Y$ has no cohomology in dimension 2. Because we have
\[f^*(\gamma)(a)=\gamma(f_*(a))=\alpha(a),\qquad f^*(\gamma)(b)=\gamma(f_*(b))=\alpha(b)\]
and similarly with $\delta$ and $\beta$, we can conclude that
\[0=f^*(0)=f^*(\gamma\cupprod\delta)=f^*(\gamma)\cupprod f^*(\delta)=\alpha\cupprod\beta.\]

Now let's look at $\T^2$. The lines $a$ and $b$ represent homology classes which will generate $H^1(\T^2)$, and we will also draw lines $a'$ and $b'$ which will let us define the cocycles which are ``dual'' to $[a]$ and $[b]$.

\begin{center}
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\begin{scope}
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\end{scope}
\draw[thick,postaction={decorate,decoration={markings,mark=at position 0.65 with {\node (c) {};},mark=at position 0.7 with {\node (d) {};}}}] (0,0) ellipse (1.2 and 1);
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\node (f) at ($(b)!0.5!(d)$) {};
\begin{scope}[shift={(e)},x={(a)}]
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\begin{scope}[shift={(f)},x={(b)}]
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\draw[dashed,thick,red] (-1,0) arc (180:360:1 and 0.3);
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\draw [thick,blue,postaction={decorate,decoration={markings,mark=at position 0.8 with {\arrow[>=angle 90]{>}},mark= at position 0.77 with { \node [label={[black,shift={(0,-0.23)}]{\small $b$}}]{}; } }}]  (0,0.12) ellipse (1 and 0.79)  ;
\draw [thick,red,postaction={decorate,decoration={markings,mark=at position 0.8 with {\arrow[>=angle 90]{<}},mark= at position 0.775 with { \node [label={[black,shift={(0,-0.23)}]{\small $a'$}}]{}; } }}] (0,0.03) ellipse (1.11 and 0.91);
\end{tikzpicture}
\end{center}

We define cochains $\alpha,\beta$ by
\[\alpha(\sigma)=\text{(signed) \# of intersections of }a',\vspace{-0.1in}\]
\[\beta(\sigma)=\text{(signed) \# of intersections of }b'.\]
\begin{center}
\begin{tikzpicture}
\draw[thick,->,>=latex] (0,0) to (4,0) node[label=right:$\sigma$] {};
\draw[thick,->,>=latex] plot[smooth,tension=1] coordinates {(0.5,-1)  (1.5,0.5) (2.5,-0.5)  (3.5,1)} node[label={[shift={(0,0.2)}]$a'$}] {};
\node[shift={(-0.2,0.2)}] at (1,0) {{\tiny $\boldsymbol{+}$}};
\node[shift={(0.2,0.2)}] at (2,0) {{\tiny $\boldsymbol{-}$}};
\node[shift={(-0.1,0.2)}] at (3,0) {{\tiny $\boldsymbol{+}$}};
\end{tikzpicture}
\end{center}
We claim that these cochains are actually cocycles. Looking at the intersections of $a'$ with the faces of a 2-simplex,
\begin{center}
\begin{tikzpicture}
\node (a) at (0,2) [label=left:$0$] {};
\node (b) at (4,2) [label=right:$2$] {};
\node (c) at (2,0) [label=below:$1$] {};
\node (ac1) at ($(a)!0.25!(c)$) [label={[shift={(-0.27,-0.33)}]{\tiny $\boldsymbol{+}$}}] {};
\node (ac2) at ($(a)!0.50!(c)$) [label={[shift={(0,-0.6)}]{\tiny $\boldsymbol{+}$}}]{};
\node (ac3) at ($(a)!0.75!(c)$) [label={[shift={(0,-0.64)}]{\tiny $\boldsymbol{+}$}}] {};
\node (ab1) at ($(a)!0.50!(b)$) [label={[shift={(0.1,-0.56)}]{\tiny $\boldsymbol{+}$}}] {};
\node at ($(a)!0.18!(b)$) [label={[shift={(0.1,-0.56)}]{\tiny $\boldsymbol{+}$}}] {};
\node (ab2) at ($(a)!0.875!(b)$) [label={[shift={(-0.14,-0.2)}]{\tiny $\boldsymbol{-}$}}] {};
\node (bc1) at ($(b)!0.41!(c)$) [label={[shift={(0.27,-0.33)}]{\tiny $\boldsymbol{+}$}}] {};
\node (bc2) at ($(b)!0.57!(c)$) [label={[shift={(-0.23,-0.33)}]{\tiny $\boldsymbol{-}$}}] {};
\node (bc3) at ($(b)!0.75!(c)$) [label={[shift={(0,-0.64)}]{\tiny $\boldsymbol{-}$}}] {};
\node (bc4) at ($(b)!0.125!(c)$)[label={[shift={(-0.23,-0.4)}]{\tiny $\boldsymbol{-}$}}] {};
\draw[thick] (a.center) -- (b.center) -- (c.center) -- cycle;
\draw[thick,shorten >=-0.3cm,shorten <=-0.3cm,->,>=stealth] (ac1) arc (315:360:0.707);
\draw[thick,shorten >=-0.3cm,shorten <=-0.3cm,->,>=stealth] (ac3) arc (135:45:0.707);
\draw[thick,shorten >=-0.3cm,shorten <=-0.3cm,->,>=stealth] (ac2.center) to [out=30,in=265] (ab1.center); 
\draw[thick,shorten >=-0.3cm,shorten <=-0.3cm,->,>=stealth] (bc1.center) to [out=135,in=45] (2.8,1.2) to [out=235,in=135] (bc2.center); 
\draw[thick,shorten >=-0.3cm,shorten <=-0.3cm,->,>=stealth] (ab2) arc (180:235:0.353);
\end{tikzpicture}
\end{center}
and adding up all the signs on $[01]+[12]-[02]$, by the classification of 1-manifolds we know there are an even number of intersections and the sum must cancel.

Thus, $\alpha$ and $\beta$ represent classes $[\alpha],[\beta]\in H^1(\T^2)$. We have
\[[\alpha]([a])=1,\quad[\alpha]([b])=0,\quad [\beta]([a])=0,\quad [\beta]([b])=1.\]
Triangulating the torus, and drawing $a'$ and $b'$,
\begin{center}
\begin{tikzpicture}[scale=1.5]
\foreach \j in {0,...,4}
{
\foreach \i in {0,...,3}
{
\draw[thick,postaction={decorate,decoration={markings,mark=at position 0.5 with {\arrow[>=angle 90]{>}}}}] (\i,\j) -- (\i+1,\j);
}}
\foreach \j in {0,...,3}
{
\foreach \i in {0,...,4}
{
\draw[thick,postaction={decorate,decoration={markings,mark=at position 0.5 with {\arrow[>=angle 90]{>}}}}] (\i,\j) -- (\i,\j+1);
}}
\foreach \j in {0,...,3}
{
\foreach \i in {0,...,3}
{
\draw[thick,postaction={decorate,decoration={markings,mark=at position 0.5 with {\arrow[>=angle 90]{>}}}}] (\i,\j) -- (\i+1,\j+1);
}}
\draw[ultra thick,red,postaction={decorate,decoration={markings,mark=at position 0.7 with {\arrow[>=angle 90]{>}}}}] (2.75,4) node [label=above:$b'$] {} -- (2.75,0);
\draw[ultra thick,red,postaction={decorate,decoration={markings,mark=at position 0.9 with {\arrow[>=angle 90]{<}}}}] (0,0.25) -- (4,0.25) node[label=right:$a'$] {};
\end{tikzpicture}
\end{center}
Recall that, by definition, we compute $\alpha\cupprod\beta$ by having $\alpha$ eat the front face of a triangle, and $\beta$ eat the back face, i.e. $\alpha(T_{01})\beta(T_{12})$. We will compute $(\alpha\cupprod\beta)[\T^2]$ by adding up the contribution from all of the triangles in our triangulation.

However, in all but two triangles, there is no contribution because either $a'$ or $b'$ does not pass through at all, so that $\alpha$ or $\beta$ vanish. The only interesting piece is
\begin{center}
\begin{tikzpicture}[scale=3.5]
\draw[thick,postaction={decorate,decoration={markings,mark=at position 0.5 with {\arrow[>=angle 90]{>}}}}] (0,0) -- (1,0);
\draw[thick,postaction={decorate,decoration={markings,mark=at position 0.5 with {\arrow[>=angle 90]{>}}}}] (0,0) -- (0,1);
\draw[thick,postaction={decorate,decoration={markings,mark=at position 0.5 with {\arrow[>=angle 90]{>}}}}] (1,0) -- (1,1);
\draw[thick,postaction={decorate,decoration={markings,mark=at position 0.5 with {\arrow[>=angle 90]{>}}}}] (0,1) -- (1,1);
\draw[thick,postaction={decorate,decoration={markings,mark=at position 0.5 with {\arrow[>=angle 90]{>}}}}] (0,0) -- (1,1);
\node (a) at (0,0) [label=left:0] {};
\node at (0,0) [label=below:0] {};
\node (b) at (1,0) [label=below right:1] {};
\node (c) at (0,1) [label=above left:1] {};
\node (d) at (1,1) [label=above:2] {};
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\draw[ultra thick,red,postaction={decorate,decoration={markings,mark=at position 0.7 with {\arrow[>=angle 90]{>}}}}] (0.8,1) node [label=above:$b'$] {} -- (0.8,0);
\draw[ultra thick,red,postaction={decorate,decoration={markings,mark=at position 0.92 with {\arrow[>=angle 90]{<}}}}] (0,0.2) -- (1,0.2) node[label=right:$a'$] {};
\node at (0.33,0.67) {$-1$};
\node at (0.63,0.37) {$+1$};
\end{tikzpicture}
\end{center}


We see that 
\[([\alpha]\cupprod[\beta])([\T^2])=(-1)\cdot[(+1)(-1)]+(+1)\cdot[(0)(0)]=1,\] which isn't 0, so the cup product structures on $\S^1\vee\S^1\vee\S^2$ and $\T^2$ are different.


Now let's do a similar calculation for $S_g$, the closed oriented surface of genus $g$. 
\begin{center}
\begin{tikzpicture}[scale=1.5]
\begin{scope}
\clip (-0.6,-1.1) rectangle (3.5,1.1);
\draw [thick,fill=gray!40] plot [smooth,tension=0.8] coordinates { (4.5,1)  (3.5,0.5) (2.5,1) (1.5,0.5) (0.4,1) (-0.5,0) (0.4,-1) (1.5,-0.5) (2.5,-1) (3.5,-0.5) (4.5,-1)};
\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] (2.5,0) ellipse (0.4 and 0.2);
\end{scope}
\end{scope}
\node at (4.03,0) {$\cdots$};
\begin{scope}
\clip (4.5,-1.1) rectangle (6.6,1.1);
\draw [thick,fill=gray!40] plot [smooth,tension=0.8] coordinates {(3.5,1) (4.5,0.5) (5.7,1) (6.5,0) (5.7,-1) (4.5,-0.5) (3.5,-1)};
\end{scope}
\begin{scope}
\clip (5.6,0) ellipse (0.4 and 0.2);
\draw[thick,fill=white] (5.6,-0.1) ellipse (0.35 and 0.25);
\end{scope}
\begin{scope}
\clip (-0.7,0) rectangle (6.3,-1);
\draw[thick] (5.6,0) ellipse (0.4 and 0.2);
\end{scope}
\end{tikzpicture}
\end{center}
We know that 
\[H_*(S_g)=\begin{cases}
\Z & \text{ in dim 0}\\
\Z^{2g} & \text{ in dim 1}\\
\Z & \text{ in dim 2}\\
0 & \text{ in higher dim}
\end{cases}\]
Taking inspiration from what we did with the torus, we define analogous curves $a_1',\ldots,a_g'$ and $b_1',\ldots,b_g'$ on $S_g$, and then we define cocycles $\alpha_i$ and $\beta_i$ which count the signed number of intersections with $a_i'$ or $b_i'$, respectively. Thus
\begin{align*}
\alpha_i(a_i)&=1 & \beta_i(b_i)&=1\\
\alpha_i(b_j)&=0\text{ for all }j & \beta_i(a_j)&=0\text{ for all }j\\
\alpha_i(a_j)&=0\text{ for all }j\neq i & \beta_i(b_j)&=0\text{ for all }j\neq i
\end{align*}
To compute cup products, we choose a triangulation, so that the fundamental class $[S_g]$ can be represented by a sum of triangles. Then
\[([\alpha_i]\cupprod[\beta_i])([S_g])=1,\quad [\alpha_i]\cupprod[\beta_j]=0,\quad [\beta_i]\cupprod[\beta_j]=0,\quad [\alpha_i]\cupprod[\beta_j]=0\text{ for }i\neq j.\]
This is a non-degenerate, alternating (i.e. antisymmetric) pairing on $H^1$ (such a form is called symplectic). This is a general phenomenon - in even-dimensional manifolds, the cup product on middle-dimensional cohomology gives a nice pairing to the top-dimensional cohomology (which is just $\Z$ when it is orientable). If the dimension of the manifold is 0 mod 4, then the middle dimension is even, and we get a symmetric form; if the dimension of the manifold is 2 mod 4, then the middle dimension is odd, and we get an anti-symmetric form.

Let's finish with $\RP^n$. So far, we've been working with 2-dimensional manifolds, so looking for things that 1-dimensional submanifolds intersect with gave us other 1-dimensional submanifolds. In general, we need to look at 1-codimensional submanifolds.

Because $\RP^n$ is non-orientable, we can't talk about the signed number of intersections, but we can work in $\Z/2\Z$.

Viewing $\RP^n$ as $\D^n/(\pm 1\text{ on }\partial\D^n)$, there are $n$ different copies of $\RP^{n-1}$ in $\RP^{n}$, each perpendicular to one of the coordinate axes. Choose one, which we will call $a$, and define $\alpha$ by
\[\alpha(\sigma)=\#\text{ of intersections of }\sigma\text{ with }a \pmod 2\]
In particular, $\alpha(\gamma)=1\neq 0 \pmod 2$, where $\gamma$ is the coordinate axis perpendicular to the chosen copy of $\RP^{n-1}$.

\begin{center}
\begin{tikzpicture}[scale=2,every node/.style={minimum size=1cm},>=latex]
\fill[ball color=white!60] (0,0) circle (1.3);
\draw[thick,red] (0,-1.1) to (0,0);
\fill[color=blue!80,opacity=0.8] (0,0) ellipse (1.3 and 0.6);
\draw[thick,red] (0,1.1) to (0,0);
\draw[thick,dashed] (1.3,0) arc (0:180:1.3 and 0.6);
\draw[thick] (-1.3,0) arc (180:360:1.3 and 0.6);
\node[red] at (0.2,0.8) {$\gamma$};
\node[white] at (0.6,0) {$a$};
\end{tikzpicture}
\end{center}

We know that
\[H_*(\RP^n;\Z/2\Z)=\begin{cases}
\Z/2\Z & \text{ if }0\leq i\leq n,\\
0 & \text{ otherwise},
\end{cases}\]
and because $H^*(\RP^n;\Z/2\Z)=\Hom(H_*,\Z/2\Z)$, the cohomology is the same as the homology. Because $H^1(\RP^n,\Z/2\Z)=\Z/2\Z$ is a 1-dimensional vector space over $\Z/2\Z$, no matter which of the $n$ copies of $\RP^{n-1}$ we chose to define $\alpha$ with, we would get the same cohomology class in $H^1$ (the $\alpha_1,\alpha_2,\ldots$ are all different as cocycles, though).

Thus,
\[\big([\alpha_1]\cupprod[\alpha_2]\cupprod\cdots\cupprod[\alpha_n]\big)[\RP^n]=\big([\alpha]\cupprod[\alpha]\cupprod\cdots\cupprod[\alpha]\big)[\RP^n]=1,\]
and therefore
\[H^*(\RP^n;\Z/2\Z)=(\Z/2\Z)[\alpha]/[\alpha]^{n+1}\]
where $\alpha$ is a degree 1 element. In fact,
\[H^*(\RP^\infty;\Z/2\Z)=(\Z/2\Z)[\alpha].\]