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\newtheorem{defi}{Definition}[section]
\newtheorem{lemma}{Lemma}[section]
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\title[Symplectic Algebraic Topology]{Institute for Basic Science Center for Geometry and Physics\\ Symplectic Algebraic Topology\\ Oh Yong-Geun}
\author{Gabriel C. Drummond-Cole}

\begin{document}
\maketitle
\section{September 10, 2013}
Okay, I'm going to give a lecture in English. The grade will be based on pass or fail, but as long as you do a few homeworks, that will be fine. 

I'm going to talk about some algebraic structures that appear in the topology of symplectic manifolds. That algebraic structure is quite complex, so it requires more complicated homological algebra than classical algebraic topology. I'm not going to talk very much about symplectic manifolds. I'll talk more about homological algebra of $A_\infty$ structures.

Let me start by motivating where this kind of structure arises. The story begins with Stasheff or maybe some earlier work, but Stasheff is the one who introduced the precise concept. 

So this is Stasheff's $A_n$ structures. In the first few lectures I'm going to explain what these $A_n$ structures are about. Everything begins with a peculiar structure in the based loop space. Let's say $X$ is a topological space and $x_0$ some base point. You're given some distinguished point. Let me denote, we have $\Ll X$, the free loop space, paths from the unit interval to $X$ such that the initial and final point coincide:
\[
\Ll(X)=\{\gamma:[0,1]\to X|\gamma(0)=\gamma(1)\}
\]
Then I'll denote by $\Omega(X)$ the subset of the free loop space such that the loops start and end at the basepoint.
\[
\Omega(X)=\{\gamma\in \Ll(X)|\gamma(0)=x_0\}\]
This is the ``based loop space.'' It has a product. You can concatenate two loops.
\begin{center}
\begin{pspicture}(5,5)
\rput(2.5,0){$\bullet$}
\psline[linearc=.7](2.5,0)(5,2)(4,4)(3,2)(2.5,0)
\psline[linearc=.7](2.5,0)(0,1)(1,3)(2,2)(2.5,0)
\rput(4,2){$\beta$}
\rput(1,1.5){$\alpha$}
\end{pspicture}
\end{center}
The product $*:\Omega X\times \Omega X\to \Omega X$ works as follows:
\[
\alpha*\beta=\left\{\begin{array}{ll}\alpha(2t)&0\le t\le \frac{1}{2}\\\beta(2t-1)&\frac{1}{2}\le t\le 1\end{array}\right.\]
This is not commutative. If you do this with the other order, you will not get the same loop.

If we have a third loop $\gamma$, we could compare $(\alpha * \beta)*\gamma$, that's one way, that gives
\[
=\left\{
\begin{array}{ll}
(\alpha*\beta)(2t)&0\le t\le \frac{1}{2}\\
\gamma(2t-1)&\frac{1}{2}\le t\le 1
\end{array}
\right.
\]
and this unravels as
\[
=\left\{
\begin{array}{ll}
\alpha(4t)&0\le t\le \frac{1}{4}\\
\beta(4t-1)&\frac{1}{4}\le t\le \frac{1}{2}\\
\gamma(2t-1)&\frac{1}{2}\le t\le 1
\end{array}
\right.
\]
I divided the unit interval in half and then the left side in half again.

Now we consider $\alpha*(\beta*\gamma)$.
We get
\[
=\left\{
\begin{array}{ll}
\alpha(2t)&0\le t\le \frac{1}{2}\\
\beta(4(t-\frac{1}{2}))&\frac{1}{2}\le t\le \frac{3}{4}\\
\gamma(4(t-\frac{3}{4}))&\frac{3}{4}\le t\le 1
\end{array}
\right.
\]
in general, $(\alpha*\beta)*\gamma\ne \alpha*(\beta*\gamma)$ as an element in $\Omega X$ but thy ar homotopic and the homotopy can be given by an explicit reparameterization. That is, there exists a homotopy 
\[
\Gamma:\underbrace{[0,1]}_s\times\underbrace{[0,1]}_t\to X\]
such that $\Gamma(0,\cdot)\equiv (\alpha*\beta)*\gamma$ and $\Gamma(1,\cdot)=\alpha*(\beta*\gamma)$.

So this is a path in the loop space, and the path is given by connecting the speeds like this
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\psframe(0,0)(4,2)
\psline{->}(0,0)(0,3)
\psline{->}(0,0)(5,0)
\psline(1,2)(2,0)
\psline(2,2)(3,0)
\rput(-1,1){$s$}
\rput(-1,0){$s=1$}
\rput(-1,2){$s=0$}
\rput(2,-.5){$t$}
\end{pspicture}
We have $s$=0 at the top.

Your first homework is to find the explicit formula for $\Gamma(s,\cdot)$. The middle interval of $s$ is given by $\frac{1}{4}+\frac{s}{4}\le t\le \frac{1}{2}+\frac{s}{4}$, as a hint.

You can keep doing this with many loops and of course there are many different ways of taking the product.

This homotopy depends on $\alpha$, $\beta$, and $\gamma$ and the parameter $s$. We formalize the explicit definition of $\Gamma$, this homotopy, and define $m_3:\Omega X\times\Omega X\times \Omega X\times[0,1]\to \Omega X$ by $m_3(\alpha,\beta,\gamma,s)=\Gamma(s,\cdot)$.

Similarly we can try to construct a homotopy of homotopies. Consider four loops $\alpha$, $\beta$, $\gamma$, $\delta$, and consider their quadruple product $\alpha*\beta*\gamma*\delta$, and there are many ways of doing this, to find a parameterized based loop, you need to use some parentheses, for example
\begin{enumerate}
\item $((\alpha*\beta)*\gamma)*\delta$
\item $(\alpha*(\beta*\gamma))*\delta$
\item $\alpha*((\beta*\gamma)*\delta)$
\item $\alpha*(\beta*(\gamma*\delta))$
\item $(\alpha*\beta)*(\gamma*\delta)$
\end{enumerate}
We can see that these each share one set of parentheses with the next one and we can draw the so-called pentagon. I'll put each of these quadruple products as a vertex of the pentagon.

\begin{center}
\begin{pspicture}(-7,-2)(7,12)
\psline(0,10)(5,6)(3.5,0)(-3.5,0)(-5,6)(0,10)
\rput(0,11){$((ab)c)d$}
\rput(6,6){$(ab)(cd)$}
\rput(-6,6){$(a(bc))d$}
\rput(3.5,-1){$(a(b(cd))$}
\rput(-3.5,-1){$a((bc)d)$}
\end{pspicture}
\end{center}

On each edge we can put the explicit homotopy induced by $\Gamma$. 

So this defines a loop in $\Omega(X)$. Each point on an edge defines an element in the loop space, and this loop is actually contractible in the loop space. Again, this can be done with an explicit contraction to the barycenter of the pentagon. I'll construct that later, but my claim is that we can construct an explicit piecewise linear construction.

This pentagon is an example of the Stasheff polytope $K_4$. So for example, $K_2$ is a point, there's only one way to multiply. $K_3$ is the unit interval, which is where we defined the homotopy. $K_4$ is the pentagon there, and generally, they keep going, and in general, $K_n$ is an $(n-2)$-dimensional polytope which will be constructed inductively by some general properties.

Then, it's a fact that we have a continuous map $m_n:(\Omega X)^n\times K_n\to \Omega X$ that satisfies certain (explicit) properties sometimes called $A_n$ properties. All of this will be explained later. Such a space is called an $A_n$-space.

Then Stasheff asked the following question. What conditions on a topological space $Y$ makes it homotopy equivalent to $\Omega X$ for some topological space $X$. It has a remarkable answer. Stasheff found:

\begin{thm}
The answer to this question is exactly this $A_n$ property. A topological space $Y$ is homotopy equivalent to $\Omega X$ for some $X$ if and only if $Y$ is an $A_\infty$-space (an $A_n$-space for all $n$).
\end{thm}
This structure appeared in symplectic geometry. Well.

You can now apply a cohomological functor. Recall the loop space product $*$. Although it is not associative on the space itself, but it induces a coproduct on cohomology, $H^*(\Omega X,\rR)\to H^*(\Omega X\times \Omega X,\rR)\cong H^*(\Omega X)\otimes H^*(\Omega X)$, so that loop product induced this homomorphism, and this is an example of a so-called coproduct. I'll explain what coproducts mean explicitly next time. That is coassociative, the dual notion of associative.

In other words, $m_2$ induced a natural coproduct $m_2:H^*(\Omega X)\to H^*(\Omega X)\otimes H^*(\Omega X)$.

Maybe it would be better to work with homology, where we can get a natural product $m_2:H_*(\Omega X)\otimes H_*(\Omega X)$. Let's think about the degree. Let's look at the induced homomorphism $(m_n)_\#:(H_*(\Omega X))^{\otimes n}\to H_*(\Omega X)$. Consider $p_1,\ldots, p_n$, singular chains in $\Omega X$ of dimension or degree $d_1,\ldots, d_n$. Denote $\alpha_i=[p_i]$. Then $(m_n)_\#(\alpha_1,\ldots, \alpha_n)$, to evaluate this you apply to representatives $m_n(p_1,\ldots, p_n)$, and you have an $s$ which comes from $K_n$, and the dimension of $K_n$ is $n-2$, and so this gives us a map from something of dimension, there is a resulting chain, that dimension is $d_1+\cdots+d_n+n-2$. Maybe it's easier to see this on chains. $(m_n)_\#(p_1,\ldots, p_n)=m_n(p_1,\ldots, p_n:K_n)$, so recall that $m_n$ is a map from $(\Omega X)^n\times K_n\to \Omega X$. The image of this $m_n$ will increase the dimension by $n-2$. This defines a chain of dimension $\sum d_j + (n-2)$. So this is sort of a chain map of degree $n-2$. This $(m_n)_\#$ is a map of degree $n-2$. So you regard, well, $C_*(\Omega X)$ as a graded vector space. So $C_*(\Omega X)=\bigoplus C(\Omega X)$. This map satisfies, the $A_n$ relations induce certain relations on these. The second fact is that the $(m_n)_\#$ will satisfy certain quadratic relations called $A_\infty$ relations.

From the $A_\infty$ relations, we can derive many facts. It induces, for example, an associative product on $H_*(\Omega X)$. This is called an $A_\infty$ algebra. The conclusion is that $(H_*(\Omega X))$ with $m_n$ forms an $A_\infty$ algebra.

Maybe I can write the quadratic relations. Next time maybe I'll explain all these unexplained terminologies. Maybe I'll finish today's lecture by saying one word, which is that such an $A_\infty$ algebra also occurs in symplectic geometry for the Floer cohomology of Lagrangian submanifolds.
\section{September 12, 2013}
Missed
\section{September 24, 2013}
Maybe I should briefly remind you what we did last time. Recall that we talked about ribbon trees and also looked at rooted ribbon trees and actually the way how we defined ribbon trees is as embedded, you draw this tree but we sort of regard some extra edges as lying on the disk. We call this a ribbon tree because you construct this fictitious disk by thickening the tree. You are given the tree and you can thicken this by gluing several, so, the thickening of trees, what I mean by this, you look at the edges, instead of edges you replace them with bands, and then you are given bands like this [picture] and then what you can do is you identify the thin boundaries by taking the barycenter and then identify the half bands and do the same thing, and you will get some kind of space which is homeomorphic to the disk. By this thickening process, you give the flat metric, and since these are flat, you can identify them, and you will get some singularities, you won't have a flat metric at these barycenters. Then when you are given this metric, it canonically defines a conformal structure. The conformal structure uniquely extends over the isolated singular points. This is the way we realize this fictitious disk. That's why we call these things ribbon trees, these are the ribbons. 

Anyhow, the rooted ribbon trees, you identify one distinguished vertex $v_0$ on the boundary, and all other exterior vertices are outgoing vertices and there is an automatic orientation on each edge toward $v_0$. We denoted the set of isotopy classes of stable (no vertices of valence two) ribbon trees (or rooted ribbon trees) by $G_{n+1}$ (or $G_{n,1}$).

We gave a partial ordering, well, the ribbon tree is an embedding $(T,i)$ of the tree $T$ into the disk $D^2$ and $i^{-1}(\partial D^2)$ is the exterior vertices of $T$. In the rooted case we have $(T,v_0),i$. Denote by $[T,i]$ or $[(T,v_0),i]$ the isotopy class of $(T,i)$ or $((T,v_0),i)$. 

We denote for simplicity $\tree$ as $[T,i]$ or $[(T,v_0),i]$. We say $\tree<\tree'$ if $\tree'$ is obtained by collapsing a sequence of interior edges of $\tree$. The maximal element, we call the unique maximal element a {\em corolla}. There are no interior edges. An $n$-corolla or $n+1$-corolla or $(n,1)$-corolla has no interior edges.

Then we introduced the grafting operation, which I denote as, let $\tree_1$ be $[(T,v_0),i]$ (or let me skip this $i$, assume it) and $\tree_2=[(T',v')]$. This operation $*_i:G_{(n_1,1)}\times G_{(n_2,1)}\to G_{n_1+n_2-1,1}$ occurs by the following picture (this is a little confusing)

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\psline(0,10)(5,5)(0,0)
\psline(5,5)(10,5)(5,10)(10,5)(5,0)
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\rput(36,5){$v_0$}
\rput(16,5){$v'_0$}
\rput(34,10){$v_1$}
\rput(19,5){$v_i$}
\end{pspicture}
and you glue these together with some renumbering:

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\rput(36,5){$v_0$}
\rput(34,10){$v_1$}
\rput(23,10){$v_{i-1}$}
\rput(10,11){$v_i$}
\rput(23,0){$v_{i+n_1-1}$}
\rput(28,0){$v_{n_1+n_2-1}$}
\rput(10,1){$v_{i+n_1-1}$}
\end{pspicture}
We now go back to the axioms of the associahedron. We had two, so the third axiom is that $K_n$ is a CW-complex. The set of cells in $K_n$ is in one to one correspondence with $G_{(n,1)}$. 
Axiom four is as follows. Denote by $F(\tree)$ the face associated to $\tree\in G_{(n,1)}$. Then $F(\tree)$ is an open cell of codimension $(\#\text{ of internal edges})$. The corolla is the generic cell, and if the tree has more interior edges, then the associated cell has higher codimension. Note that this codimension is the same as the number of parentheses.

[Darko Milinkovic: What are the faces of a CW complex? Maybe I'm missing axioms one and two.] $K_n$ is an $n-2$ dimensional polyhedron. $K_n$ has a cell decomposition where each cell is the cone on its boundary.

Axiom five is as follows, maybe this is a proposition, well, given the topology of $K_n$, we have the following: The closure $\overline{F(\tree')}$ is $\bigcup_{\tree\le \tree'} F(\tree)$ and the boundary is the same except with scrict inequality:
$\partial \overline{F(\tree')}=\bigcup_{\tree< \tree'} F(\tree).$

There is one more axiom which I'll state later. But maybe I need to state axiom six to make this proposition true so I'll hold off on it.
\begin{prop}$K_n$ is homeomorphic to an $(n-2)$-cell.
\end{prop} I'll give the proof of this proposition later, but I'll give some examples. $K_2$ is a point. There's only one isotopy class.

[Some discussion of $K_1$ and $K_0$]
Let me give axiom six. There exist maps $\circ_i:K_r\times K_s\to \partial K_n\subset K_n$ where $n+1=r+s$. so that the image $\circ_i(K_r\times K_s)$ ofrms a facet of $K_n$.

The boundaries of $K_n$ is the union over $2\le s\le n-1$ and $i$ from $1$ to $s$ of the closures of $\circ_i(K_(n+1-s)\times K_s)$. These are all possible ways of putting one parenthesis in a word of length $n$ so that it's not the whole word.

That's the axioms I'll use to describe these things.

Let's go back and consider $\partial K_3$. How many boundaries? This is a zero dimensional set. This will be, $r+s=n+1=4$, and $s$ must be $2$. So there are two ways of putting one parentheses: $((ab)c)$ or $(a(bc))$. There should be two points. So let's see what $((ab)c)$ corresponds to in the graph. They should be $\circ_1(K_2\times K_2)$ and $\circ_2(K_2\times K_2)$. 

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\rput(11,5){$v_0$}
\rput(14,5){$v_1$}
\rput(26,10){$v_0$}
\rput(26,0){$v_2$}
\end{pspicture}
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\rput(21,10){$v_0$}
\rput(21,0){$v_3$}
\end{pspicture}
That's one. The other is:

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\rput(1,0){$v_3$}
\rput(26,10){$v_1$}
\rput(26,0){$v_0$}
\end{pspicture}

You can think of this as one of these turning ot the other by passing through a corolla:

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\end{pspicture}

Now I've done something axiomatic and I want to turn instead to some geometric realizations of $K_n$. One could be metric ribbon trees. Let me set some notation. For a given $\tree=(T,i)\in G_{n+1}$, we denote by $C^0(\tree)$ as the set of vertices, $C^1(\tree)$ as the set of edges, and $C^0_{int}(\tree)$ and $C^1_{int}(\tree)$ as the interior vertices and edges. Now for $\tree$ in $G_{n+1}$, we associate the open cell \[Gr(\tree)=\{\ell:C'_{int}(\tree)\to (0,\infty)\}\]
This is homeomorphic to $(0,\infty)^{\#C^1_{int}(\tree)}=(0,1)^{\#C^1_{int}(\tree)}$. Then we let $\overline{Gr(\tree)}$ to be the same thing including lengths zero and infinity: 
\[\{\ell:C'_{int}(\tree)\to [0,\infty]\}\]
This is homeomorphic to $[0,1]^{\#C^1_{int}(\tree)}$

We can make a grafting map for this. We provide a CW-structure on the union of the isotopy classes of $Gr(\tree)$. By the definition, $Gr_{(n,1)}$ is the resulting complex, which is homeomorphic to $K_n$.

Okay, for the second one, the moduli space $\overline{\mathcal{M}}_{n+1}^{main}(D^2)$. Consider $n+1$ distinct points on $S^1=\partial D^2$. Denote this by \[Conf_{n+1}(S^1)=\{(z_0,\ldots, z_n)\in (S^1)^{n+1}|z_0,\ldots, z_n\text{ are distinct and arranged counterclockwise}\}.\]
Then $PSL(2,\rR)$ acts on $S^1$. Recall that $SL(2,\rR)$ is two by two matrices of determinant one and $SL(2,\rR)$ is $PSL(2,\rR)$ modulo the relation that a matrix is equal to its negation.

Identify $D^2\backslash\{1\}$ with the upper half-plane, $z\in\cC$ such that $Im\ z\ge 0$. These are equivalent via a conformal diffeomorphism.

Then $PSL_2(\rR)$ acts on the upper half-plane by $z\mapsto \frac{az+b}{cz+d}$ if the element of $PSL(2,\rR)$ is the class of $\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$. This action acts the same by an element or its negation and preserves the upper half plane and the real line union $\{\infty\}$. For a given $(z_0,\ldots, z_n)$, we may assume $z_0=\infty$. Then we arrange $z_1<z_2<\cdots$.

So you can identify $z_1,\ldots, z_n$ around the circle. We can realize, or identify, these points, in $Conf_{n+1}(S^1)$ as a subset, if you put $z_0$ at $1$ you still have two dimensions. You can identify the point as a subset of $n$ points in $(0,1)$. I'll make this more precise next time.
\begin{defi}
We define $\mathcal{M}_{n+1}(D^2)=Conf_{n+1}(S^1)/PSL(2,\mathbb{R})$ where $PSL(2,\mathbb{R})$ acts on $(z_0,\ldots, z_n)$ by $(Az_0,\ldots, Az_n)$.
\end{defi}
We identify two points if all coordinates are moved by the same fractional linear transformation. This quotient space has a natural differential structure and the dimension is $(n+1)-3=n-2$. I'll explain the cell structure of this which tells you this is another geometric realization of the Stasheff polytope after you compactify, next time I'll talk about compactification.
\section{September 26, 2013}
Maybe just for those who are interested in looking at more details of what I'm talking about, here's a reference, the original paper of Stasheff. This is very explicit and does not use any machinery for graphs or moduli space. It's hard to motivate where these relations come from. This is the most basic reference: Homotopy Associativity of $H$-spaces, I and II, in the transactions of the AMS, 1963, volume 108. My presentation is more modern.

Axiom six is a little more involved, which I will explain using the realization of the Stasheff polytope. This is very much like the relationship between degeneracy and face maps for the simplex, so this is an analog for Stasheff polytopes.

There exist a family of maps called face operators. I already introduced these with another notation, but I want to make this more formal. This is $\partial_k(r,s)$, but I used $\circ_k$ before. I want to make every data clearly now. This is a map from $K_r\times K_s\to K_n$, and $r+s=n+1$. So $s$ is the only free parameter. $n$ is given so $r$ is determined. You should think of $K_n$ as being like a simplex. For this case, there is a map, the face map, in all dimensions. This is the analog of the inclusion of low dimensional faces into the $n$-simplex. Here $s$ runs from $2$ to $n-1$ and $k$ ranges from $1$ to $s$. This should properly be $K_s\times K_r$. Maybe the parentheses way is better:
\[
a_1\cdots a_{k-1}(a_k\cdots a_{k+r-1})a_{k+r}\cdots a_n\]
I insert $K_r$ into the $k$th spot of $K_s$. This is the spot where you attach the first tree to the second tree.

There are also degeneracy maps $s_j:K_n\to K_{n-1}$. These range from $j$ from $1$ to $n$.

You have standard relations between the face maps and the degeneracies in the simplex. I'll write down the relations here but they're much more complicated.
\begin{enumerate}
\item[i] $s_js_k=s_ks_{j+1}$ for $k\le j$ (I think this is the same as for the simplex).
\item[iia]$s_j\partial_k(r,s)=\partial_{k-1}(r-1,s)\circ (s_j\times id)$ for $j<k$ and $r>2$. Similarly:
\item[iia'] $s_j\partial_k(r,s)=\partial_k(r,s-1)\circ(1\times s_{j-k+1}$ for $s>2$ and $k\le j\le k+s$.
\item[iib] $s_j\partial_j(n-1,2)=s_{j+1}\partial_j(n-1,2)=\pi_j$ for $1<j<n$;
\item[iib'] $s_1\partial_2(2,n-1)=\pi_2=s_n\partial_1(2,n-1)$
\item[iii] $s_j\partial_k(r,s)=\partial_k(r-1,s)(s_{j-s+1}\times id)$ for $k+s\le j$
\end{enumerate}
The geometric realization makes this very clear. The ribbon graph model makes this clear.

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The degeneracy map is mapping from $Gr(n,1)$ to $Gr(n-1,1)$, just by collapsing the $k$th exterior edge followed by stabilization (deleting bivalent edges). The face map $\partial_k(r,s):\overline{Gr}_{(r,1)}\times \overline{Gr}_{(s,1)}\to \overline{Gr}_{(n,1)}$, which is nothing but the grafting map, and the image lies in the boundary. Once you know those topologies, the relations I wrote down are an immediate consequence of the compactification.

I forgot to mention one remark. By definition, the face operator gives a boundary map and you can see the boundary is 
\[\boundary K_n=\bigcup_{s=2}^{n-1}\bigcup_{k=1}^s K_r\star_k K_s\]
where $K_r*_k K_s:= \partial_k(r,s)K_r\times K_s$.

The number of faces is $\frac{n}{n-1}{2}-1$. You can count this. The similar realization of the face operator, well, that's even easier, remove the $j$th marked point. That's a little bit, maybe I'll say more about that model later.

Let's look at some special case. Well, maybe in a little bit. I'm ready to define an $A_n$-space. We say that $(Y,m,e)$ satisfies the $A_n$ relations [$m$ is a product $Y\times Y\to Y$ and $e$ a basepoint of $Y$] if there is a family of maps $M_i:K_i\times Y^i\to Y$ for $2\le i\le n$ such that (I'll interchange $Y^i$ and $K_i$ factors when it's convenient for me)
\begin{enumerate}
\item $M_2(e,y;*)=M_2(y,e;*)=y$ for $y\in Y, *\in K_2$. That is, $m=M_2$ makes $(Y,m,e)$ an $H$-space.
\item For $\rho\in K_r, \sigma\in K_s, r+s=i+1$, we have
\[
M_i(y_1,\ldots, y_i; \partial_k(r,s)(\rho,\sigma))
=M_r(y_1,\ldots, y_{k-1},M_s(y_k,\ldots, y_k+s-1;\sigma),y_{k+s},\ldots, y_i; \rho)
\]
\item $M_i(y_1,\ldots, y_{j-1}, e, y_{j+1},\ldots, y_i; \tau)=M_{i-1}(y_1,\ldots, y_{j-1},y_{j+1},\ldots, y_i;s_j(\tau))$
\end{enumerate}

We call this collection $(Y,\{M_i\})$ an $A_n$-space. For homework (from Stasheff), check that this definition is consistent with the relations between $\partial_k(r,s)$ and $s_j$ in Axiom six.

Let me explain what this relation actually means. The first $A_2$-space is nothing but an $H$-space. Why is this so? An $H$-space has $M_2$ which is a map $K_2\times Y^2\to Y$ but $K_2$ is a point, so if we set $m=M_2(*;\quad)$, this is a map $Y\times Y\to Y$, it defines a multiplication. The relation one basically says that $m(e,y)=m(y,e)=y$ so $e$ plays the role of the identity for this multiplication map. 

For general $n$, consider $i=3$. For this, we know $K_3$ is homotopic to the unit interval, and $M_3$ is a map $K_3\times Y^3\to Y$. So this is a map $[0,1]\times Y^3\to Y$. The relations say that $M_3(y_1,y_2,y_3;\partial_k(2,2)(\rho,\sigma))$ is just a point, this lies in the boundary of $K_3$ this is either $+$ or $-$, and so the second relation says that for $k=1$, this is $M_2(M_2((y_1,y_2;\sigma), y_3; \rho)$ and for $k=2$ it's $M_2(y_1,M_2(y_2,y_3,;\sigma);\rho)$. So to simplify the notation here, we denote, well, we can omit $\rho$ and $\sigma$ since $K_2$ is a point. So $M_2(x,y;*)$ as $xy$. Then this becomes, the above formula becomes 
\[=\left\{
\begin{array}{c}(y_1y_2)y_3\\y_1(y_2y_3)\end{array}
\right.\]
This is the boundary so what this formula says is that this $M_3$ which is a map from $[0,1]\times Y^3\to Y$, obviously this is a homotopy. It's a homotopy between $M_3(\{0\}\times \bullet)$ and $M_3(\{1\}\times \bullet)$. These two maps are, respectively, $(y_1y_2)y_3$ and $y_1(y_2y_3)$. So this $M_3$ provides a homotopy between these two different products. In other words, $M_3$ provides an associating homotopy between these two different triple multiplications.

Let's do one more, for $i=4$. We know $K_4$ is a pentagon and there are five ways to put a product in four letters in parentheses. So we could have an edge that is a homotopy from $a(b(cd))$ to $a((bc)d)$ and the next goes to $(a(bc))d$. This is treating $bc$ as one letter. The following one goes from that to $((ab)c)d)$, then $(ab)(cd)$, and then back to $a(b(cd))$. So there is an explicit map, restricted to the boundary we get this loop. So we have, concatenating this homotopy, we have a map $S^1\times Y^4\to Y$. That's provided by the $M_3$. Then the map $M_4$ extends this map which is already defined by $M_3$ along the boundaries and $M_4$ provides an explicit contraction of that map, extending it to $K_4\times Y^4$. Of course, $K_4$ is homeomorphic to $D^2$. This is a homotopy of a homotopy, and the higher things can be regarded as homotopies of homotopies of homotopies.

There's a trivial way of constructing these $A_n$ spaces. An $H$-space does not have to be associative. Any associative $H$-space admits an a trivial $A_n$ structure for all $n$ by setting $M_i(y_1,\ldots, y_i; \sigma)=y_1\cdots y_i$. You don't need any parentheses because the product is associative. The product comes from $m=M_2$. This satisfies, you can check, the $A_n$ relations. This map does not depend on the $K_n$ element.

Here is the motivating example, of based loop spaces. Let $X$ be a topological space with $x_0$ a basepoint. Then $Y=\Omega X=\{\gamma:[0,1]\to X|\gamma(0)=\gamma(1)=x_0\}$. Then $\Omega X$ is a (homotopy) $H$-space. Set the constant loop $e$ and then consider the standard concatenation as the product given by \[\gamma_1*\gamma_2)(t)=\left\{\begin{array}{ll}\gamma_1(2t)&0\le t\le \frac{1}{2}\\\gamma_2(2t-1)&\frac{1}{2}\le t\le 1\end{array}\right.\]

Consider $i_L,i_R:Y\to Y\times Y$ defined by $i_L(Y)=(e,\gamma)$ and $i_R(\gamma)=(\gamma,e)$. We want to check that $m\circ i_L\sim id\sim m\circ i_R$ as a map $Y\to Y$. So $m\circ i_L(\gamma)(t)$ is $m(e,\gamma)=e*\gamma$ which is
\[
\left\{
\begin{array}{ll}
*&0\le t\le \frac{1}{2}\\
\gamma(2t-1)&\frac{1}{2}\le t\le 1
\end{array}
\right.
\]
You remain constant for time $\frac{1}{2}$ and then follow $\gamma$ twice as fast. Similarly, $m\circ i_R(\gamma)(t)$ follows $\gamma$ twice as fast and then remains constant for time from $\frac{1}{2}$ to $1$:
\[\left\{
\begin{array}{ll}
\gamma(2t)&0\le t\le \frac{1}{2}\\
*&\frac{1}{2}\le t\le 1
\end{array}
\right.
\]
We have an obvious homotopy that I don't want to write from $m\circ i_L$ to the identity given by changing the $\frac{1}{2}$ to an $s$:
\[
H(s,\gamma)(t)=\left\{
\begin{array}{ll}
*&0\le t\le \frac{1-s}{2}\\
\gamma(\frac{2}{1+s}t - \frac{1-s}{1+s})&\frac{1-s}{2}\le t\le 1
\end{array}
\right.
\]
For the next couple of lectures I'll prove that $(Y,m,e)$ is an $A_\infty$ space. I'll relate this based loop space with another type of path space, $Z=\Theta(X)$ which is pairs $(r,\alpha)|r\ge 0, \alpha:[0,r]\to X$ with $\alpha(0)=x_0=\alpha(r)$. This is an associative $H$-space.  The multiplication map $\mu:Z\times Z\to Z$ is given by the following. The domain side adds and then we use concatenation. So
\[
\mu((r,\alpha),(s,\beta))=
(r+s),??\]
and I'll continue next time.
\section{October 1}
Let me remind you, let $(Y,m,e)$ be an $H$-space. We say this satisfies the $A_n$ relation if there exists a family of maps $M_i:K_i\times Y^i\to Y$ for $2\le i\le n$ satisfying the axioms (1)--(3) which I do not want to repeat. We looked at some examples. We considered $X$ a based topological space and denote $Y=\Omega X$ and $Z=\Theta(X)$. The lemma was
\begin{lemma}
$Z$ is an associative $H$-space.
\end{lemma}
Then we have a natural map. Remember that $Z=\{(r,\alpha)|r\ge 0,\alpha:[0,r]\to X\text{ with }\alpha(0)=x_0=\alpha_r\}$. The product is concatenation without reparameterization. We have a natural map $f:Y\to Z$ which is inclusion. We have a map $g:Z\to Y$ which is rescaling: $g(r,\alpha)(t)=\alpha(rt)$. These maps have the following properties. If you compose $g\circ f$ you get the identity. On the other hand, $f\circ g(r,\alpha)=f(\alpha(r\cdot))=(1,\alpha(r\cdot))$

You can easily see that this is homotopic to the identity. What is the homotopy? Indeed, the homotopy is $h:[0,1]\times Z\to Z$ and is given by $h(s,(r,\alpha))(t)=((1-s)r+s,\alpha((1-s)+sr)t)$.

Now $h(0,\cdot)=id_Z$ and $h(1,\cdot)=f\circ g$. Noting this, I'm going to derive the following theorems in general.
\begin{thm}
$(\Omega X, m,e)$ ($e$ is the constant path and $m$ the concatenation product) is an $A_\infty$ space. 
\end{thm}
This is an immediate corollary of the following theorem:
\begin{thm}
Suppose $Y$ is a topological space and let $Z$ be an associative $H$-space with multiplication $\mu:Z\times Z\to Z$. Suppose you are given maps $f:Y\to Z$ and $g:Z\to Y$ such that $f\circ g$ is homotopic to $id_Z$. Define a multiplication $m:Y\times Y$ by $m(y_1,y_2)=g(\mu(f(y_1),f(y_2)))$. Then $(Y,m,e)$ becomes an $A_\infty$-space, where $e=g(e_Z)$ and $e_Z$ is the unit in $Z$.
\end{thm}
We are given a multiplication on $Z$ and want to define one on $Y$, so we do it like this:
\[\xymatrix{
Z\times Z\ar[r]^\mu&Z\ar[d]^g\\
Y\times Y\ar[u]^{f\times f}\ar[r]_m&Y
}.\]
\begin{proof}
Denote by $\mu_k$ the $k$ multiplication map since $\mu$ is associative: $\mu_k:Z\times\cdots \times Z\to Z$. Furthermore, for simplicity, we denote $z_1\cdots z_k=\mu(z_1,\ldots, z_k)$. This will make it a little more intuitive. We'll define using $\mu_k$, $f$, $g$, and the homotopy $h$, we define an $M_i:K_i\times Y^i\to Y$ inductively. I'll describe this precisely up to $i=4$. 

The $M_2$ case, we know $K_2$ is a point, we define $M_2=m=g\mu_2(f\times f)$. In other words, $M_2(y_1,y_2)=g(f(y_1)f(y_2))$.

Second, let's see $M_3$. It's supposed to be $K_3\times Y^3\to Y$. We know $K_3$ is the unit interval $[0,1]$. What we'll do is define $M_3|_{\partial K_3\times Y^3}$ and then extend to the interior. The inductive process you define first on the boundary and then extend it to the interior. You'll recall $\partial K_3=K_2 *_1 K_2\cup K_2 *_2 K_2$ where $K_2 *_i K_2$ is the image of $\partial_i(2,2)(K_2\times K_2)$. This lies in the boundary of $K_3$. This is how the face map was defined.

We already defined $M_2$. There are two boundary points. So we define $M_3(x,y,z;0)=M_2(M_2(x,y),z)$. Similarly $M_3(x,y,z;1)=M_2(x,M_2(y,z))$. Now we choose and fix a homotopy $h$ which I assume to exist from $f\circ g$ to $Id_Z$. This will let us interpolate these two maps. So $h$ is a map from $f\circ g$ to $id_Z$, those are $h(0)$ and $h(1)$. So then we have a map $h(t)$ for time $t$. So now we interpolate the two maps. Recall how $M_2$ was defined. The first map is
\[
g(\underbrace{f(g}_{f\circ g}(f(x)f(y)))f(z))\]
and the second is
\[
g(f(x)\underbrace{f(g}_{f\circ g}(f(y)f(z)))).\]
Then we replace these with $h$. So consider first
\[
s\mapsto g(h_s(f\cdot f)\cdot f).
\]
At $s=0$ this becomes $g(f\circ g(f\cdot f)\cdot f)$ which is the first boundary map. At $s=1$ this becomes
$g(f\cdot f\cdot f)$, which you should regard as the central point. Similarly, the homotopy 
\[
g(f\cdot h_s(f\cdot f))
\]
connects $g(f\cdot f\circ g(f\cdot f))$ to $g(f\cdot f\cdot f)$. So by concatenating these homotopies you get a homotopy from one to the other.

Finally, well, denote these two homotopies as $h_1$ and $h_2$. So obviously, we define a concatenation homotopy $h$ from $g(f\circ g(f\cdot f)\cdot f)$ to $g(f\cdot f\circ g(f\cdot f))$ by setting \[h(t)=\left\{\begin{array}{ll}
h_1(2t)&0\le t\le \frac{1}{2}\\
h_2(2-2t)&\frac{1}{2}\le t\le 1
\end{array}\right.\]
Pictorially, $K_3$ is $G_{r_{3,1}}$. You should think of this as looking like
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\end{pspicture}
\end{center}
where you think of the two endpoints as being $K_2*_1 K_2$ on the left and $K_2 *_2 K_2$ on the right, with the center point.

We can then look at one more, $M_4:K_4\times Y^4\to Y$. Recall that $K_4$ is the pentagon
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\rput(7,3){$(ab)(cd)$}
\rput(4,-1){$a(b(cd))$}
\rput(2,-1){$a((bc)d))$}
\rput(3,2){$\bullet^v$}
\end{pspicture}
\end{center}
Here we first define $M_4|_{\partial K_4\times Y^4}$ using the given homotopy $h$. We denote by $m$ the midpoint of an edge and $v$ the barycenter of the $K_4$.
\begin{center}
\begin{pspicture}(10,7)
\psline(0,3)(3,5)(6,3)(4,0)(2,0)(0,3)
\rput(-1,3){$(a(bc))d$}
\rput(3,6){$((ab)c)d$}
\rput(7,3){$(ab)(cd)$}
\rput(4,-1){$a(b(cd))$}
\rput(2,-1){$a((bc)d))$}
\rput(3,2){$\bullet$}
\rput(1.3,4.2){$m$}
\rput(3,2.4){$v$}
\rput(1.5,4){$\bullet$}
\rput(4.5,4){$\bullet$}
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\rput(1,1.5){$\bullet$}
\rput(3,0){$\bullet$}
\psline(3,0)(3,2)(1,1.5)(3,2)(5,1.5)(3,2)(1.5,4)(3,2)(4.5,4)
\end{pspicture}
\end{center}
We connect $m$ to $v$ and define $M_4(\cdot, \cdot, \cdot, \cdot, m)$, which is, say, at the midpoint between $((ab)c)d$ and $(a(bc))d$, equal to $g(f\circ g(f\cdot f\cdot f)\cdot f)$. We have a similar product explicitly given for each midpoint. We get rid of the $f\circ g$ and put $g(f\cdot f\cdot f \cdot f)$ at the center point $v$, which corresponds to the corolla graph. Now using the homotopy $h$ from $f\circ g$ to $id_Z$, we define $M_4$ on the segment connecting each midpoint $m$ to $v$. So for instance,
$M_4|_{[m,v]\times Y^4}$ sends $s$ to $g(h_s(f\cdot f\cdot f)\cdot f)$. In this way we have defined $M_4$ restricted to the line segments of this subdivision [ed: the pair subdivision]. Now for any other $\lambda v$ for $\lambda\in \boundary K_4$, we interpolate linearly $M_4(\cdots ;\lambda)$ and $M_4(\cdots, v)$.

I'm using an axiom of a pentagon. $K_i$ is the cone on its boundary. On each line segment I use the homotopy $h$ to interpolate the two maps. The explicit formula is a little bit complicated but that's the idea. For higher $M_i$ we just repeat this process inductively. 

The way this was constructed, the $A_n$ relations are automatic. These relations say what happened to the restriction of $M_i$ to the boundary of $K_i$. But by construction this $M_i$ satisfies the relations for any $n$. The $A_n$ relation is nothing but the relation between, well, $M_i(y_1,\ldots, y_i;\partial_k(r,s)(\rho,\sigma))=M_r(y_1,\ldots, y_{k-1},M_s(y_k,\ldots, y_k+s-1;\sigma),y_{k+s},\ldots, y_i;\rho)$ for $r+s=i+1$. But $r$ and $s$ are already defined. That's how our inductive construction was defined.
\end{proof}
\begin{cor}
Then applying the theorem to $Y=\Omega X,Z=\Theta X$, we have proved that $\Omega X$ is an $A_\infty$ space.
\end{cor}
Not every space carries an $A_\infty$ structure, only those homotopic to a based loop space do. Now applying the cohomology functor to this space we'll get an $A_\infty$ algebra. Let's apply the cohomology functor to the $A_n$ space $(Y,\{M_i\}_{i=1}^n)$
So $M_i:K\times Y^i\to Y$. Say the coefficients for cohomology is $\rR$. Then $m_i=(M_i)^*:H^*(Y)\to H^*(K_i\times Y^i)$. Now $K_i$ is contractible so this is canonically isomorphic to $H^{*-(i-2)}(Y)$ [some discussion, there's a comment that an explanation will be forthcoming] by integration over the fiber. This integration over the fiber can be defined. If $Y$ is integrated over the fiber, well, we define a shifted complex, the cohomology ring is a graded Abelian group so denote $A^*=H^*(Y)[1]$ by defining $A^k=H^{k+1}(Y)$. This is the standard notation. Then define $m_k'$, well, I have a map $A^*\to H^{*+1}Y$, and well, I want to rewrite $m_i$ as $m_i':H^*(Y)[1]\to (H^*(Y)[1])^{\otimes i}$. The original $m_i$ has degree $i-2$ but the degree shifting gives all $m_i$ degree $1$ for all $i$. In this business, the keeping track of signs is a complicated matter, very confusing, but if you shift the degree in this way, keeping track of the degree will be much easier because the sign rule will satisfy the Koszul sign. This degree shifting is very important.

\end{document}

