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\title{Dennis Sullivan Course Notes\\March 11, 2005\\Nathaniel Rounds, guest speaker}
\author{Gabriel C. Drummond-Cole}
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I'm going to talk about cohomology operations, and give the sort of basic example of them, which is Steenrod squares, and then I'll find the Wu class and the Stiefel-Witten classes, and on Monday I'll talk about Thom; his Fields medal in 1958 was for showing that a mnanifold was a boundary basically only if these things all vanish.

\section{cohomology operations}
\begin{defi}
A cohomology operation is a natural transformation of functors $\Theta:H^m(\cdot,G)\to H^m(\cdot, G')$ so that the appropriate diagram commutes:
$$
\xymatrix{
H^m(X,G)\ar[r]^\Theta & H^n(X,G')\\
H^m(Y,G)\ar[u]_{f^*}\ar[r]& H^n(Y,G')}.$$\end{defi}

There are a few obvious ones and they're actually the only obvious ones.  
\begin{example}
\begin{itemize}
\item a group homomorphism $\phi: G\to G'.$
\item $\delta:H^n\to H^{n+1}$
\item $x\mapsto x\cup x.$
\end{itemize}
\end{example}

Note that as in the last case, these need not be ring homomorphisms.

Now I want to give a more concrete example.  Those who took Dusa's class last spring will remember that we had this as the definition of cohomology.

$H^n(X,G)$ is set-isomorphic to $[X,K(G,n)]$ where $\pi_n(K(G,n))=G$ and the other homotopy groups are zero.  $G$ is a finitely generated Abelian group.  This space is unique up to homotopy type.  The only example I'm interested in is $G=\mathbb{Z}/2.$  Brackets denote homotopy classes of maps from the first to second argument.  I'll give the correspondence but will not prove it to be a bijection.

We have the Hurewicz homomorphism $h:G=\pi_n(K(G,n))\to H_n(K(G,n)),$ which is an isomorphism because there is no other homotopy, i.e., it's $n-1$-connected.  Since this an isomorphism its inverse $h^{-1}\in Hom(H_n(H(G,n),G))\cong H^n(K(G,n),G).$  Now if I call $\omega=[h^{-1}]$ then $f$ corresponds to $[f^*\omega].$

So now I claim I don't have to define this operation for all spaces and all coefficients, I just have to define it for $\omega$ on $K(G,n).$  That is, there exists a bijection between cohomology operations and $H^n(K(G,n),G').$

How do I want to prove this?  If I have a cohomology operation $\Theta$ it acts on $\omega$ to give $\Theta(\omega).$  I claim that this is a bijection, and well-defined.  Suppose $\alpha\in H^m(X,G),$ that's the same as having $f_\alpha:X\to K(G,m),$ and $\Theta_\omega\in H^n(K(G,m),G')$ so $\Theta_\omega:K(G,m)\to K(G',n).$  Composing these gives me the map I want.

The next claim is that every cohomology operation arises in this way, and that is because of naturality.

If I have a cohomology operation $H^m(\cdot, G)\to H^n(\cdot, G'),$ and $\beta\in H^m (X,G)$ then that corresponds to $f_\beta:X\to K(G,m)$ and $\Theta(\beta)=\Theta(f_\beta^*\omega)=f_\beta^*(\Theta(\omega)).$

That characterization tells us something about cohomology operations that we didn't know.  They can't lower dimension, and can only fix dimension if they are a group homomorphism because $H^m(K(G,m),G')=Hom(H_m(K(G,m)),G')=Hom(G,G').$

Now I'll define Steenrod squares:
\\
$Sq^i:H^n(X,\mathbb{Z}/2)\to H^{n+i}(X,\mathbb{Z}/2)$
\begin{enumerate}
\item naturality
\item stability: $\sigma:H^n(X)\to H^{n+1}(\Sigma X)$ 
\item ring homomorphism: $Sq^i(\alpha+\beta)=Sq^i(\alpha)+Sq^i(\beta);\ Sq^i(\alpha\cup \beta)=\sum_j Sq^j(\alpha)\cup Sq^{i-j}(\beta)$
\end{enumerate}
A few of these are easy.  $Sq^i(x)=x$ if $i=0;\ Sq^i(x)\sim\delta x$ if $i=1.$
There is a short exact sequence $0\to\mathbb{Z}_2\to \mathbb{Z}_4\to \mathbb{Z}_2\to 0$  We can take $Hom$ and get $$0\to Hom(\cdot,\mathbb{Z}_2)\to Hom(\cdot,\mathbb{Z}_4),\cdots$$

[argument ensues.]

It is $x^2=x\cup x$ for $i=\dim\ x.$  It's $0$ for $i>\dim\ x.$

[Keep going, do whatever you want.]

The main thing that I wanted to say was to give some idea of where these things come from.  I'll draw the picture that Dennis would draw.  He's going to draw that picture.

[A circle with two hemispheres attached?]

[That's a crappy picture.]

I'll try to explain that a little bit better.  I can think of $x\otimes x\in H^n(X)\otimes H^n(X)$ so $x\otimes x\in H^{2n}(X\times X).$

Now $T:X\times X\to X\times X$ swaps factors;  So $x\otimes x$ corresponds to a homotopy class of maps from $X\times X\to K(2n,\mathbb{Z}_2).$

I claim that this is homotopic to $(x\otimes x)\circ T:X\times X\to K(2n,\mathbb{Z}_2).$  So then that homotopy is a map $I\times X\times X\to K(2n,\mathbb{Z}_2).$  This is a homotopy $f$ and I can also look at $f\circ (id\times T)$  Pieces these together and get a map from $S^1\times X\times X\to K(2n,\mathbb{Z}_2).$  This is contractible, so I can fill in the homotopy and extend to $D^2\times X\times X\to K(2n,\mathbb{Z}_2).$  I use this to prescribe a map on ``the other side'' by $(s,x_1,x_2)\cong (-s,x_2,x_1).$  This gives me $S^2\times X\times X\to K(2n,\mathbb{Z}_2).$  I continue and only run into trouble at $2n.$  I claim I can get around it and get a map $S^\infty\times X\times X\to K(2n,\mathbb{Z}_2).$

[How do you get around the obstruction?]

It's not obvious that you can.  The homotopy is only $\mathbb{Z}/2,$ so you can't miss it by a lot.  The original map is injective.

[discussion.]

Now I'll say how to use such a map to get Steenrod squares.  I defined this so that it respects antipodes so this will pass to the quotient $\mathbb{RP}^\infty.$

$$\xymatrix{
S^\infty\times X\ar@{(->}@[r]\ar[d]&S^\infty\times X\times X\ar[r]& K(2n,\mathbb{Z}_2)
\\ \mathbb{RP}^\infty\times X\ar@.>@[rru]}$$

So $H^*(\mathbb{RP}^\infty,\mathbb{Z}_2)\cong \mathbb{Z}_2[\tau],\ \tau$ of degree one.  

So $\alpha\in H^{2n}(\mathbb{RP}^\infty\times X,\mathbb{Z}/2)$ we can write $f_{\alpha\otimes \alpha} =\sum_j \tau^{2n-j}\otimes a^j;$ then $Sq^j(\alpha)=a^{n+j}.$

$\alpha\otimes\alpha$ is in the $2n$ cohomology of $X\times X.$  Then the above diagram shows how to get the diagonal map of which I take a Kunneth decomposition.

I'm going to write down one thing and then we'll all leave.

I have the pairing on cohomology of a closed manifold, $H^{n-i}\times H^i\to H^n(M)\cong\mathbb{Z}_2.$  So $Sq^i\in Hom(H^{n-i},\mathbb{Z}_2).$  Then we can say, there exists $V_i$ such that $V_i\in H^i$ with $V_i\cup x = Sq^i(x)$ for all $x\in H^{n-i}(M).$

Then $W_k=\sum_{i+j=k} Sq^i(V_j)$

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