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\title[Fukaya]{Kenji Fukaya:\\ Lagrangian correspondence 
\\and $A$-infinity functor\\
Center for Geometry and Physics (Pohang)}
\author{Gabriel C. Drummond-Cole}
\begin{document}
\maketitle
\section{August 12}
I want to talk something about constructions of functors. Let me recall what everybody knows about homology theory. You have a space $X$ and you associate this group $H(X)$, and to a continuous map $f:X\to Y$ you get a homomorphism $f_*:H(X)\to H(Y)$.

You also have something called quantum cohomology. When $X$ is a symplectic manifold, you have the quantum cohomology ring $QH^*(X)$. It is no problem that when you have a symplectic diffeomorphism, this leads to an isomorphism of quantum cohomology. But if you have any morphism, you don't get the naive guess that there is $f^*:QH(Y)\to QH(X)$. There were several attempts to make this more functorial. In the symplectic world there is a classical proposal by Weinstein. Now complex manifolds are nice, there are morphisms, but symplectic morphisms are not well-understood. He said that if you have $X_i,\omega_i$ symplectic, that $L\hookrightarrow -X_1\times X_2$, where $-X_1$ is $(X_1,-\omega_1)$, if you have such a Lagrangian submanifold you should regard this as a morphism.  This is nice for many reasons. If you have $L_{12}\subset -X_1\times X_2$ and $L_{23}\subset -X_2\times X_3$, then you can take the fiber product, and if the fiber product is transversal (this is a kind of nasty assumption necessary to make the composition well-defined) then $L_{13}$ is an immersed Lagrangian in $-X_1\times X_3$. So you can kind of compose. But then you need to go to the category of immersions, not necessarily embeddings.

The proposal by Werheim--Woodward was to construct something like a category, from $(X,\omega)$ with Lagrangian correspondence to try to construct a functor to all $A_\infty$ categories. This is called the Weinstein category. What they could do, they considered monotone Lagrangians and monotone compositions, that's what they could handle.

I want to realize this kind of project in complete generality.

This has a natural mirror. Consider $X_1$ and $X_2$ as a complex manifold. Let $\Ee$ be an object of the derived category of coherent sheaves on the product $\dD(X_1\times X_2)$. For example, if $f$ is a holomorphic map $X_1\to X_2$ then the graph of $f$ is a complex submanifold and gives an object of this derived category.

Then we have this classically established morphism, $\dD(X_1)\to \dD(X_2)$, the functor $F_\Ee$, then you take $R(\pi_2)_!(\pi_1^*(C^*)\boxtimes \Ee)$. The tensor product and push down are derived, you have the tors and things. This is called the Fourier--Mukai category.

So first our guess is that our objects are immersed Lagrangian submanifolds. For this business it's essential to go beyond the embedded case. It will be necessary to take immersions to make composition well-defined. I want to cook up the functors. I have $L_{12}$ in $-X_1\times X_2$ and $L_1$ in $X_1$, and I want to cook up $L_2$ in $X_2$, and a natural candidate is the fiber product $L_1\times_{X_1}L_{12}$. If the $L$ are Lagrangian and the fiber product is transversal, then $L_2$ is an immersed Lagrangian submanifold of $X_2$.

This is a kind of naive version. Then we should ask how this is related to Floer homology. So far we have geometric things. How does this work with Floer theory? I want to recall immersed Floer theory, due to Akahi--Joyce. Let $X$ be a symplectic manifold. Take $V$ a real oriented vector bundle over $X_{[3]}$, the $3$-skeleton. An immersed Lagrangain submanifold is $i_L:\tilde{L}\to X$, an immersion of half dimension where the symplectic form pulls back to zero. Now $\sigma$ is a spin structure on $T\tilde{L}\oplus i_L^*(V)$ on $\tilde{L}_{[3]}$. I want to assume that $\tilde{L}$ is self-clean. What does that mean? Take the fiber product $\tilde{L}\times_X \tilde{L}=L(+)$. What is $L(+)$? It's $\tilde{L}\sqcup \bigcup L(a)$. [pictures]. Here $\tilde{L}$ sits inside $\tilde{L}\times_X\tilde{L}$ as the diagonal $(x,x)$ such that $x\in \tilde{L}$. The assumption is that $L(a)$ is $C^\infty$ and $T_{p,q}L(a)=d_pi(T\tilde{L})\cap d_q i(T\tilde{L})$. But $(p,q)$ and $(q,p)$ are different points in the fiber product. So $CF(L)=\Omega(L(+))\hat{\otimes}\Lambda_0$, where $\Lambda_0$ is the Novikov ring $\{\sum a_i T^{\lambda_i}$ for $a_i\in \rR$ and $\lambda_i\ge 0$ and increasing to $\infty$.
\begin{thm}
There are $m_k:CF(L)^{\otimes k}\to CF(L)$ satisfying the $A_\infty$-relations.
\end{thm}
\begin{defi}
$L$ is \emph{unobstructed} if there is $b\in CF(L)\otimes_{\Lambda_0} \Lambda_+$ with $\sum m_k(b,\ldots b)=0$. Here $\Lambda_+=\{a_iT^{\lambda_i}:\lambda_i>0\}$.
\end{defi}
The main theorem is
\begin{thm}
If $L_1$ and $L_{12}$ are unbostructed then $L_2$ is unobstructed.\end{thm}
Just immersed Lagrangians, you might not be able to do Floer homology. You really need unobstructed.
\begin{remark}
Let $L$ be immersed in $\cC^n$, self transversal nad unobstructed. Then the number of self-intersections is at least half the total Betti number.
\end{remark}
This is not correct in general, without the unobstructed condition.

One can say the following thing. Take Hamiltonian perturbations of $L_{12}$. This can be very complicated. The topology of $L_2$ can change. The theorem says that still it is unobstructed. Further, its Floer homology is independent of the Hamiltonian perturbation.

Now we can say the following things. The Weinstein category does not seem to be good enough for Floer theory. So I want to propose the \emph{unobstructed Weinstein category} whose objects are $(X,\omega, V)$, where $V$ is the bundle on $V$.

Ah, I should have said, if $L_1$ is $V_1$-relatively spin and $L_{12}$ is $TX_1\oplus V_1\oplus V_2$-relatively spin, then $L_2$ is $V_2$-relatively spin. I need these assumptions to state the theorem. I left that out.

So $f$ is a symplectic diffeomorphism and $L_f$ is the graph of $f$. Then $L_f$ is diffeomorphic to $X_1$ but may not be spin. But it's always $TX_1$-relatively spin. With Yong-geun and the others, we studied something like Floer homology on $X\times X$. At some point we realized that something was not spin, so we tried to use this kind of relative spin.

Okay, so the morphisms are $L_{12}$ immersed in $X_1\times X_2$ with $TX_1\oplus V_1\oplus V_2$ which is relatively spin, along with a bounding cochain $b_{12}$.
\begin{thm}
There is a functor $F$ from the unobstructed Weinstein category to the category of all filtered $A_\infty$ categories. 
\end{thm}
This is kind of the main theorem of these talks. You have this triple, you can construct an $A_\infty$ category.

I think $HH(F(X_1,\omega, V))$ has an open-closed map to $H_*(X_1)$, and this is covariant with respect to $A_\infty$ functors, and you can map to $HH*F(X_2,\omega, V)$, and I claim that this square should commute.
\[\xymatrix{
HH_*F(X_1,\omega_1,V_1)\ar[d]\ar[r]&	
HH_*F(X_2,\omega_2,V_2)\ar[d]\\
H_*(X_1)\ar[r]&H_*(X_2)
}\]
Let me consider $i_L$, the immersion $\tilde{L}\to X$, and $CF(L)$ which is $\Omega(\tilde{L}\times_X\tilde{L})\hat{\otimes} \Lambda_0$.
So $\tilde{L}\times_X\tilde{L}$, let me write it as the diagonal components $\tilde{L}$ along with $\bigcup L(a)$ for $\vec{a}$ which is $(a_0,\ldots a_k)\in \Aa^k$. Now let $\circ{M}_{\vec{a}}(L,E)$ be the set of $(D^2,\vec{z},U,\gamma)$ where $\vec{z}$ is a cyclically ordered set in the boundary of $D^2$, $U:(D^2,\boundary D^2)\to (X,L)$ is holomorphic and $\gamma:\boundary D^2\backslash |\vec{z}|\to \tilde{L}$ and $i_L\circ \gamma=U$ on $\boundary D^2\backslash |\vec{z}|$. Finally, when $z$ approaches $z_i$, you get $\gamma(z_{i+1})$ is in $L(a_i)$. [picture]. The last condition is that $\int U^*\omega=E$.

Now $M$ can be compactified, and you get a map $M_{\vec{a}}(L,E)\to L(a_0)\times\cdots L(a_k)$. So you have projections
\[\xymatrix{
&M_{\vec{a}}(L,E)\ar[dr]\ar[dl]\\
L(a_1)\times\cdots\times L(a_k)&& L(a_0)
}\]
So you pull back and push out by integration along the fibers, and you get $m_{\vec{a}}^E:\Omega(L(a_1))\times\cdots\times \Omega(L(a_k))\to \Omega(L(a_0))$.

Then $m_k=\sum_{\vec{a}} T^E m^E_{\vec{a}}$. 

[pictures.]

Now I want to go to an $A_\infty$ category from the $A_\infty$ algebra. Let me assume everything is oriented. It's better to use a finite set of self-clean Lagrangian submanifolds, you have $\Ll_1=\{L_{1,i}|i=1,\ldots, N\}$. You just take the unions $L=\cup L_{1,i}$, and assume that this one is self-clean. This just means that you have, well, each one is self-clean and they have pairwise clean intersections. To go to the $A_\infty$ category you can think of this as a single Lagrangian. So you get $CF(L)^{\otimes k}\to CF(L)$, and you see that $CF(L)$ is the sum of all $CF(L_i)$ along with $\bigoplus \Omega(\tilde{L}_i\times_{X_1} \tilde{L}_j)\hat\otimes\Lambda_0$.

So you have $CF(L_i,L_j)=\Omega(\tilde{L_i}\times_{X_1}\tilde{L}_j)\hat\otimes\Lambda_0$. So $CF(L)$ is just $\bigoplus CF(L_i)\oplus \bigoplus CF(L_i,L_j)$. Then $m_k$ for $L$ is $m_k:CF(L_0,L_1)\otimes\cdots \otimes CF(L_{k-1},L_k)\to CF(L_0,L_k)$. This gives us something called an $A_\infty$ category.

This step is not actually so difficult.

Now let me make a more precise statement. Suppose you have $X_1$ with $\omega_1$, and $V_1$. Take $\Ll_1$, a finite set of immersed Lagrangian submanifolds. Take $(-X_1\times X_2,-\omega_1+\omega_2,TX_1\oplus V_1\oplus V_2)$, and then $\Ll_{12}$ a finite set of immersed Lagrangian submanifolds. Assume for all $L_{1,i}$ and all $L_{12,j}$ that this fiber product $L_{1,i}\times_{X_1}L_{12,j}$ is transversal. This you can achieve by perturbing slightly.

Now I want to consider $\Ll_2$, a finite set of a Lagrangian submanifold which contains $L_{1,1}\times_{X_1}L_{12,j}$.

From $\Ll_1$ you have this filtered $A_\infty$ category whose objects are in $\Ll_1$. This $A_\infty$-category is inconvenient, because this object may be obstructed. There are general constructions using bounding cochains. I'll go back to geometry soon, but let me do some algebra.

Let $C$ be a filtered $A_\infty$ category. What does it mean? You have a set of objects and then $c_1$ and $c_2$ are objects and $C(c_1,c_2)$ is a $\Lambda_0$-module. You have $m_k:C(C_0,C_1)\otimes\cdots\otimes C(C_{k-1},C_k)\to C(C_0,C_k)$, which satisfies the $A_\infty$ relation. So you want to define bounding cochains here because you have $m_0$ here. You have another $A_\infty$ category, with $m_0=0$. The objects of $\bar{C}$ is a pair $(c,b)$, where $c\in C$ and $b\in C(c,c)\otimes \Lambda^+$ which satisfies the Maurer-Cartani equations. I just want to start from the $A_\infty$ category. When $m_0$ is nonzero then $m_1^2$ is nonzero so you can't define Floer homology. The morphisms are just the same, but $m_k$ is different, $\bar{m}_k(x_1,\ldots x_k)$ is $m(e^{b_0}x^1e^{b_1}\cdots x_ke^{b_k})$. where $x_k$ is in $\bar{C}((c_{k-1},b_{k-1}),(c_k,b_k))$. Here $e^b$ is $1+b+b\otimes b+\cdots$. So an object is a pair, and the composition uses $b$ insertions. Then $\bar{m}_0(1)=m(e^b)=0$. 

We call such a thing strict. This is a pure algebraic construction. What we do is the following. We have $X_1$ and $\Ll_1$ the finite set of Lagrangians. Then you have $-X_1\times X_2$, and $\Ll_{12}$ and $\Ll_2$ which contains $\{L_{1,i}\times_{X_1} L_{12,j}\}$. 

Now you have $\Ll_1$ an $A_\infty$ category whose objects are $L_{1,i}\in \Ll_1$ and $F(\Ll_1)$ leads to $\bar{F}(\Ll_1)$, an $A_\infty$ category with $m_0=0$.

A theorem that I'll prove maybe next time is that 
\begin{thm}
There is a filtered $A_\infty$ bifunctor $\bar{F}(\Ll_1)\times\bar{F}(\Ll_{12})\to \bar{F}(\Ll_2)$.
\end{thm}
Maybe I think today it's better to state the things I want to prove in the first three lectures. This was proven in the assumption that everything was embedded and monotone. 

Now I want to say about the compositions. I've given the space of morphisms. The notion of $A_\infty$ bifunctor should be defined but I'll do it later, postpone it because I've done too much algebra.

Now the next thing is composability. Let me remind you that $\Ll_{12}$ is a set of Lagrangian submanifolds of $-X_1\times X_2$. I want to choose another finite set $\Ll_{23}$ of Lagrangian submanifolds of $-X_2\times X_3$. Then something I want to assume is the following. Assume that $L_{12,i}\in \Ll_{12}$ and $L_{23,j}\in \Ll_{23}$, and that $L_{12,i}\times_{X_2} L_{23,j}$ is transversal. Then the next theorem is, well, then $\Ll_{13}$ is a finite set of Lagrangian submanifolds of $-X_1\times X_3$, and I want to say that this contains all of these such fiber products. You consider the finite set of morphisms, and that transversality is okay and then a finite set containing the compositions. 
\begin{thm}
There is a bifunctor 
$\bar{F}(\Ll_{12})\times \bar{F}(\Ll_{23})\to \bar{F}(\Ll_{13})$.
\end{thm}
For example, if $L_{12}$ and $L_{23}$ are unobstructed, then so is their fiber product. The unobstructed Weinstein category is thus a kind of \wh category.

I can probably prove this in generality, but let me write a weaker version first. Let me remind that we have $\Ll_1$, $\Ll_{12}$, and $\Ll_{2}$. Then you consider $\Ll_3$ which contains all the things from $\Ll_{2}$ as well as those from $\Ll_1$.
\begin{thm}
There is a commutative square
\[\xymatrix{
	\bar{F}(\Ll_{12})\times \bar{F}(\Ll_{23})\ar[d]\ar[rrr]
	&&&\bar{F}(\Ll_{13})\ar[d]\\
Func(\bar{F}(\Ll_1),\bar{F}(\Ll_2))\times
Func(\bar{F}(\Ll_2),\bar{F}(\Ll_3))
\ar[rrr]&&& Func(\bar{F}(\Ll_1),\bar{F}(\Ll_3)
}\]
\end{thm}
[some discussion.]
What I can prove is, if I have $L_{12},b_{12}$ and $L_{23},b_{23}$, then we get $(L_{13},b_{13})$, and what we can prove, is well $(L_{12},b_{12})$ yields $W_{(L_{12},b_{12})}:F(\Ll_1)\to F(\Ll_2)$ and then 
\[W_{(L_{23},b_{23})}\circ W_{(L_{12},b_{12})}\] is homotopic to
$W_{(L_{13},b_{13})}$.

The next thing is the following thing. Suppose that $L_{12}$ and $L'_{12}$ are homotopic. Then $\varphi_t:-X_1\times X_2\to -X_1\times X_2$ is a Hamiltonian isotopy. Then somehow, what I want to say is the following. First of all, the isotopy, $b_{12}$ gives $b_{12}'$, this is general theory of Lagrangian submanifolds. What I want to claim is that $W_{L_{12},b_{12}}\sim W_{L_{12}',b_{12}'}$ over $\Lambda=\Lambda_0[T^{-1}]$. If you look at Joyce's paper about immersed Lagrangian Floer theory, there's one thing that's difficult to understand. If you have these two Lagrangians, then $HF(L,L')$ and $HF(\varphi(L),\varphi'(L'))$ are equivalent over $\Lambda$. But for immersed things, you can do isotopy locally. You have $\tilde{L}\times [0,1]\to \rR$, and things are quite delicate. If you use this Hamiltonian [picture], it's hard to understand what to do with the Floer homology. If the Hamiltonian isotopy, well, now you don't know that $L_2$ is isotopic to $L_2'$. They might not even be diffeomorphic. We can still prove the theorem about homotopy equivalence of the $W$.  

So suppose you have another $L\subset X_2$ with $b$, then we can show that $HF(W_{(L_{12},b_{12})}(L_1,b_1),(L,b'))\otimes\Lambda$ is isomorphic to
$HF(W_{(L_{12}',b_{12}')}(L_1,b_1),(L,b))\otimes\Lambda$.
There's something delicate related to the K\"unneth theorem.

[some discussion].

There is a notion of Lagrangian cobordism. In the monotone situation, cobordant Lagrangians have the same Floer homology. It seems very likely that we can extend that story in the unobstructed category. If the cobordism admits an extension of the bounding cochain, then we expect the Floer homology is the same.

\section{August 14}
[This was a national holiday and I missed the lecture.]
\section{August 17}
Suppose you have a symplectic manifold $X_1$. You cook up this category $\bar{F}(X_1)$, that has objects $(L,\sigma,b)$, where $L$ is an \emph{immersed} Lagrangian, $\sigma$ is a relative spin (I talked about this last time) and $b$ is an element in the Floer chain complex of $L$. Then for $\Ll_{12}$ which is $L_{12},\sigma_{12},b_{12}\subset -X_1\times X_2$, we want to cook up an $A_\infty$ functor $\Ww_{\Ll_{12}}:\bar{F}(X_1)\to \bar{F}(X_2)$. I want to show that this is compatible with composition of Lagrangian correspondence.

Essentially what you want to show is, suppose $L_1$ has a Maurer--Cartan solution $b_1$ and $L_{12}$ has a Maurer--Cartan solution $b_{12}$, we want to find $b_2$ which is a Maurer--Cartan solution. There are a couple of ideas. Here's one, due to \wh.

Consider a strip in $\cC$, the strip with $Im\ z$ between $0$ and $1$. You want a holomorphic map from $u_1$, the strip, to $X_1$, from $u_2$, the space above the strip, to $X_2$, from the real line to $L_1$ and from the boundary between them to $L_{12}$. [pictures, confusing].

The way I want to cook up this functor is with a $3$-step construction. Then I have $\bar{F}(X_1)$ and $\bar{F}(X_2)$, and I take the Yoneda embedding of $\bar{F}(X_2)\hookrightarrow Func(\bar{F}^{op}(X_2),ch)$. It's very natural to construct a ``K\"unneth'' functor $\bar{F}(X_1)\to Func(\bar{F}(X_2)^{op},ch)$. Then there are two things you need to prove. Consider the image of the Yoneda functor $Rep(\bar{F}(X_2)^{op},ch)$. The second theorem says that the K\"unneth functor lifts to the representables. The Yoneda theorem says that the map to representables is an isomorphism, so you have an inverse function $\theta$ to the Yoneda embedding. Then you can use this lift of the K\"unneth functor followed by a homotopy inverse functor. In this way we can bypass the complicated thing. You need to prove many things are functorial if you do this by hand. But the K\"unneth functor has nice functorial properties, so it's easier if you do it this way.

I want to define the notion of an $A_\infty$ bifunctor. K\"unneth is delicate in $A_\infty$ categories. What is the tensor product of $A_\infty$ algebras? This is not trivial, so the K\"unneth theorem is a bit complicated. An $A_\infty$ category has a set of objects, and for two objects $c_1$ and $c_2$, we have $\Cc(c_1,c_2)$ a $\Lambda_0$-module, where $\Lambda_0=\sum a_i T^{\lambda_i}$ as before. Then $m_k:B_k(c,c')\to C(c,c')$, where $B_k(c,c')=\bigoplus \bigotimes \Cc(c_{i-1},c_i)$. 
In this setting, there is no $m_0$. There are some cases where we want that. Let me say $C$ is curved if $m_0\ne 0$.

I want to explain $A_\infty$ bifunctors.  Suppose you have $\Cc_1$, $\Cc_2$, and $\Cc_3$ which are $A_\infty$ categories. Let $F_{ob}:ob(\Cc_1)\times ob(\Cc_2)\to ob(\Cc_3)$. Then 
\[F_{k_1,k_2}:B_{k_1}\Cc_1(c_1,c'_1)\otimes B_{k_2}\Cc_2(c_2,c_3)'\to \Cc_3(F_{ob}(c_1,c_2),F_{ob}(c'_1,c'_2)).\]
Note that $B\Cc=\bigoplus B_k(c,c')$ has a coalgebra sturcture $\Delta:B\Cc\to B\Cc\otimes B\Cc$, but lands in the subset of composable pairs. Then $B\Cc$ also has $\hat{d}$ where $\hat{d}(x_1\otimes\cdots\otimes x_k)=\sum x_1\otimes\cdots m(x_i,\cdots x_j)\otimes\cdots \otimes x_k)$. Then being $A_\infty$ is the same as saying that $\hat{d}\hat{d}=0$.

So $\hat{F}:B\Cc_1\otimes B\Cc_2\to B\Cc_3$, which is kind of a cohomomorphism. The tensor product of coalgebras is a coalgebra. If you just compose with the projection to $\Cc_3$ you get $F_{k_1,k_2}$.

This $F$ is an $A_\infty$-bifunctor if and only if $\hat{F}\hat{d}=\hat{d}\hat{F}$.

From this definition you can easily prove this lemma. Consider the $A_\infty$ bifunctor $\Cc_1\times \Cc_2\to \Cc_3$, that's the same as a functor $\Cc_1\to Func(\Cc_2,\Cc_3)$. I don't want to define, but there's also a similar notion of a trifunctor.

Now we have the algebraic preliminary. What do we get in the following thing? What about our K\"unneth. We have $-X_1$ and $X_2$, two symplectic manifolds. The claim is that 
\begin{thm}There is an $A_\infty$ trifunctor $\bar{F}(-X_1)\times \bar{F}(X_2)^{op}\times \bar{F}(-X_1\times X_2)\to ch$.
\end{thm} This is what I want to call the K\"unneth functor. I just want to mention an application of this.
\begin{cor}
Using the tautological identification, you get a map $F(-X_1)\times \bar{F}(-X_1\times X_2)\to Func(\bar{F}(X_2)^{op},ch)$. This is something in the Yoneda codomain of $\bar{F}(X_2)$.
\end{cor}
I want to prove the theorem. The category version follows formally from the algebra version of this theorem. I want to prove the version where I don't have the bar, so that this is a \emph{curved} $A_\infty$ functor. So say you have $L_1\subset -X_1$, $L_{12}\subset -X_1\times X_2$, and $L_2\subset X_2$, then you get $D$ an $A_\infty$ -trimodule over $CF(L_1)$, $CF(L_{12})$, and $CF(L_2)^{op}$. We proved in FOOO that if $L\subset X_1$ and $L'\subset L'$, that $CF(L,L')$ is an $A_\infty$-left module over $CF(L)$ and an $A_\infty$ right module over $CF(L')$. But it's easier to say that this is an $A_\infty$ left module over $CF(L')^{op}$. So this is like this.

So you have $u$ a map from a strip to $X$, whose boundary values are $L$ and $L'$. The bimodule structure comes from putting marked points on the strip.

The trimodule version has one more line. You put in a vertical strip with a vertical line dividing it. $U_1$ is a map from the first half of the strip to $-X_1$ and $U_2$ from the second half to $X_2$. The left boundary goes to $L_1$, the right boundary to $L_2$, and the middle line has $(U_1(z),U_2(z))\in L_{12}$. This is called a pseudoholomorphic quilt.

This looks like some different kind of moduli space. In fact, this moduli space is something equivalent to one which you already know about. Just put a strip between $L_{12}$ and $L_1\times L_2$, and define $U(z)$ as $U_1(z'),U_2(z)$ where $v=x+\sqrt{-1}y$ and $z'=-x+\sqrt{-1}y$. That makes the map to $X_1$ antiholomorphic, which means it's holomorphic to $-X_1$. So this moduli space looks slightly different, but in this simple case, you know that the moduli space is related to K\"unneth. This just relates Floer theory between $L_{12}$ and $L_1\times L_2$. You don't do too much new analysis, this case is established already.

So \wh wanted to take the limit when you let one of the strips get narrower, but if you keep the strips fixed the analysis is just the same.

Now there is a slightly different point. So now what is $D$? I wanted to construct, starting with $L_1$, $L_{12}$, and $L_2$, I wanted a chain complex with an action of all the Floer complexes.

$D$ is $\Omega(L_1\times_{X_1}L_{12}\times_{L_2}X_2)\otimes \Lambda_0$.
This is $\Omega((L_1\times L_2)\times_{X_1\times X_2}L_{12})\otimes \Lambda_0$. [Some discussion of pictures]. When the imaginary part goes to $\pm\infty$, you see a point in the fiber product. More precisely, you have the following.

You have $\Mm_{k_1,k_{12},k_2}(E,L_1,L_{12},L_2)=\{U_1,U_2,\vec{z}_1,\vec{z}_{12}\vec{z}_2|\wh,|\vec{z}_i|=k_i,\int U_1^*(-\omega)+\int U_2\omega=E\}$.
There are several evaluation maps, to $L_1^{k_1}$, to $L_{12}^{k_{12}}$, and to $L_2^{k_2}$. So you also have $ev_{-\infty}$ and $ev_{+\infty}$, both of which go to $L_1\times_{X_1} L_{12}\times_{X_2} L_2$.

You have 
\[\Omega(L_1\times_{X_1} L_{12}\times_{X_2} L_2)\otimes \Omega(L_1)^{\otimes k_1}\otimes \Omega(L_2)^{\otimes k_{12}}\otimes \Omega(L_{12})^{\otimes k_{12}}\to \Omega(L_1\times_{X_1} L_{12}\times_{X_2} L_2).\] This is basically the trimodule structure, there are some other details but let me suppress them. Then the relation is that 
\[\Delta x_1 = \sum x_1^c\otimes {x_1^c}'
\]
and then
\begin{gather*}
\sum_{c,d,e} n(n(y,x_1^cx_2^dx_{12}^e){x_1^{c}}'{x_2^d}'{x_{12}^e}')+
\\
+\sum n(y_1,(\hat{d}x_1)x_2,x_{12})
+\sum n(y_1, x_1,\hat{d}x_2,x_{12})
\sum n(y_1,x_1,x_2,\hat{d}x_{12})
=0
\end{gather*}
Now we have the trifunctor. Well, this is a trimodule, but it's kind of saying the same thing in a different way. So we get $\bar{F}(X_1)\times \bar{F}(X_{12})\to Func(\bar{F}(X_2),ch)$. There's a slight difference between this and K\"unneth. If you don't put any marked points in, then you have the same thing. But the marked points on $L_1$ and $L_2$ are only hit by differential points on $L_1$ and $L_2$ separately, whereas in the product you can hit with arbitrary differential forms on the product. So this is like the difference between $BCF(L_1)\otimes BCF(L_2)$ and $BCF(L_1\times L_2)$.

Maybe I finish the constructions after the break.

So $L_1$ is a relatively spin immersed Lagrangian with bounding chain, and $L_{12},b_{12}$ similarly, as an object of $\bar{F}(-X_1\times X_2)$, so these give us $\Ww(\Ll_1,\Ll_{12}):\bar{F}(X_2)\to Ch$. The claim is that there exists an $\Ll_2$ in $ob(\bar{F}(X_2))$ such that $Yo(\Ll_2)\cong \Ww(\Ll_1,\Ll_{12})$. Then by the inverse of the Yoneda embedding we'll get $\bar{F}(X_2)$.

The proof is two step. We have this particular geometric object $L_2^0=L_1\times_{X_1} L_{12}$. First, we want to show that there is a bounding cochain, and the second step is that $Yo(\Ll_2)\cong \Ww(\Ll_1,\Ll_{12})$.

So I want to prove this, and the key idea is that $D=CF(L_1,L_{12},L_2)=\Omega(L_1\times_{X_1} L_{12}\times_{X_2} L_2)$. The idea is to particularly choose $L_2^0$. Now I'll make a very simple geometric claim, that this is canonically diffeomorphic, $L_1\times_{X_1} L_{12}\times_{X_2} L_2^0\cong L_2^0\times_{X_2} L_2^0$. But this is something like $CF(L_2^0)$, so that $CF(L_1,L_{12},L_2^0)$ is isomorphic as a $\Lambda_0$-module to $CF(L_2^0)$. This $CF(L_1,L_{12},L_2)$ is a \emph{right} $CF(L_2)$-module. You have bounding cochains $b_1$ and $b_{12}$ then this is an uncurved right $CF(L_2)$-module. Let me give a geometric rationale. So $CF(L,L')$ is a left $CF(L)$ module and right $CF(L')$-module. This means the following. So $D$ is left $C_1$ and right $C_2$ module, then you have $n_{k_1,k_2}:B_{k_1}(C_1)\otimes D\otimes B_{k_2}(C_2)\to D$. The relations are complicated, something like\[
\sum n(x^c n({x^c}'yz^d){z^d}')+n(\hat{d}x,y,z)+n(x,y,\hat{d}z)=0.
\]
So here what's the trouble? You just see this formula. Suppose you forget $C_1$, then $D$ is not a right module structure. Let $x$ be the empty set, just $1$. Then 
\[
\sum n(n(y,z^c){z^c}')+n(y,\hat{d}z)\]
is missing the term $n(m_0(1),y,z)$. But we have the bounding chain $b$ and can correct $n$ from the left and define $n^b$ so that $n^b(\vec{x},\vec{y},\vec{z})=n(e^bx_1e^bx_2\cdots e^b,\vec{y},\vec{z})$. This gives a right $C_2$-module structure on $D$. Geometrically speaking, if you cancel your right disk bubbles by $b$, you get an honest right module structure. You can do the same thing for a trimodule. You have $L_1$, $L_{12}$, and $L_2$. You can put $b_1$ and $b_{12}$ in the same way. Then you get a right $CF(L_2)$-module. This is still curved, but only on the $L_2$-side. I want to prove that $CF(L_2)$ has a bounding cochain. After this, you can forget two of the module structures, and have a right $CF(L_2)$-module structure.

Then I'll mention a simple lemma. I like it, because it gives a way to get a bounding cochain.
\begin{lemma} Ret $D$ be a right $C$-module, where $C$ is curved. Say you have $\mathbf{1}$ in $D$ such that $n(\mathbf{1})=0\pmod \Lambda_+$, and $x\mapsto n(1,x)$ is a $\Lambda_0$-module isomorphism, then the conclusion is that there is a unique element $b$ so that $\sum n_k(1,b,\ldots,b)=0$. This implies that $\sum m_k(b,\ldots,b)=0$.
\end{lemma}
Once the statement is given, the proof is kind of obvious. The two conditions imply that you can do this by induction and it's unique.

We can actually use this lemma to obtain $b$. We're exactly in the situation in which our lemma can be applied. We have $CF(L_1,L_{12},L_2)$, and we have $b_1$ and $b_{12}$, and this is a right $CF(L_2)$-module. You can see abstractly that $\Omega(L_1\times_{X_1}L_{12}\times_{X_2}L_2^0)\cong \Omega(L_2^0\times_{X_2} L_2^0)$ abstractly. 

We can see that $L_2^0\times_{X_2} L_2^0$ is $L_2$ (the diagonal) along with a bunch of other parts. Then we say $\mathbf{1}=[L_2]$. The first property $n(1)\equiv 0\pmod \Lambda_+$ is obvious because $d\mathbf{1}=0$. Higher derivatives come from positive energy.

Now I want to see the second property. You have the abstract isomorphism, and I want to say that the abstract isomorphism (up to positive energy) is realized by this multiplication.

We consider $L_1$ and $L_{12}$ and $L_2$ in our strip, and we let one end of $L_1$ go to $1$. We want to see the other end. I want to calculate $n(1,x)\pmod \Lambda_+$. If we forget the positive energy, we should get an isomorphism. The maps $u_1$ and $u_2$ are constant. What is $x$? It's a differential form on $L_2$ or $L_2(a)$. Let me pretend that this is self-transversal. Multiplication of differential forms on $L_2$, this marked point can move everywhere. You have a constant map, and the pullbacks and pushforwards are the obvious map. Then $\Mm$ is $L_2$ with three marked points. So you get the identity map modulo $\Lambda_+$. The other option is that you're switching components. [geometric reasoning] Again, forgetting positive energy, this is the identity map.

This is a proof that your $L_2$ is unobstructed. The rest of the proof is that this particular pair represent the K\"unneth functor. We not only get the $b$, but $\mathbf{1}$ is actually a cycle. To prove the isomorphism, you use the fact that $\sum n_k(\mathbf{1},b,\ldots, b)=0$.

Let me consider that we have $\sum n(\mathbf{1},b_2^0,\ldots, b_2^0)=0$. Now suppose we have $(L_2,b_2)\in Ob(\bar{F}(X_2))$. We have
$HF((L_1,b_1),(L_{12},b_{12}),(L_2,b_2))$. Then we want to show that this is equivalent to $HF((L_2^0,b_2^0),(L_2,b_2))$.

The proof that they are isomorphic uses the following diagrams. [pictures].
\section{August 20}
So last talk I just explained that there is a functor. You have $X_1$ and $X_1$ and I constructed a functor 
\[\bar{F}(X_1)\times\bar{F}(-X_1\times X_2)\to F(X_2)\]
and this is what we did last time. You have this ``unobstructed Weinstein category'' $W$ whose objects are $(X,\omega,V)$, a symplectic manifold along with something related to its relative spin structure, and the morphisms are $L_{12}\subset -X_1\times X_2$ with a bounding cochain $b_{12}$. We want a functor from $W$ to the category of all $A_\infty$ categories. What we did so far, well, we cooked up $\bar{F}(X,V)$. Now we have $\Ll_{12}=(L_{12},b_{12})$ inducing $W_{\Ll_{12}}\bar{F}(X_1)\to \bar{F}(X_2)$. We need to show that this structure is compatible with composition of morphisms.

First we need to define composition operator on the unobstructed Weinstein category. If you have $L_{12}\subset -X_1\times X_2$, you have $L_{23}\subset -X_2\times X_3$, then $L_{13}=L_{12}\times_{X_2}L_{23}$. This fiber product might not exist. To make this a category there is some difficulty because of this. There are two ways to go around it. You can include some perturbations; then $L_{13}$ which is not fully defined, you might get non-diffeomorphic things. One way, you hav $L_{23}\sim L_{23}'$, this is an unobstructed cobordism, you have a cobordism, you have $L$ and $L'$ in $X$, they are Lagrangian cobordant if you have $\tilde{L}$ in $X\times \cC$, and outside a compact set it looks like $L\times \rR$ and $L'\times \rR$. If you assume everything is monotone, then the Floer theory of $L$ and $L'$ are the same. We're working on the non-monotone case. I didn't write the proof yet. If you have $\tilde{L},\tilde{b}$ which restricts to $L,b$ and $L',b'$, then the Floer theory of these guys will be the same, at least with coefficients in $\Lambda_0$.

Then you can make sense of the fiber product always, because Hamiltonian isotopy is a Lagrangian cobordism. Then you get an honest category and you get an $A_\infty$ functor defined up to homotopy. So you can go from the cobordism category of the Weinstein category to the homotopy category of $A_\infty$ categories. Or you could work on a topological category where everything is defined on a dense subset. Things are well-defined only up to homotopy. 

So that's not the main thing I want to explain. I want to explain two theorems.
\begin{thm}
Say $L_{12}$ has $b_{12}$ and $L_{23}$ has $b_{23}$. Then $L_{13}=L_{12}\stackrel{\pitchfork}{\times}_{X_2}L_{23}$ also has a bounding cochain $b_{13}$
\end{thm}
\begin{thm}
$W_{L_{23},b_{23}}\circ W_{L_{12},b_{12}}\sim W_{L_{13},b_{13}}$.
\end{thm}
\begin{conj}
\[\xymatrix{
\bar{F}(-X_1\times X_2)\times\bar{F}(-X_2\times X_3)\ar[r]\ar[d]&\bar{F}(-X_1\times X_3)\ar[d]
\\
F_c(\bar{F}(X_1)\bar{F}(X_2))\times F_c(\bar{F}(X_2),\bar{F}(X_3))\ar[r]
&
F_c(\bar{F}(X_1),\bar{F},X_3)
}\]
commutes. Here $F_c$ is the functor category.
\end{conj}
The first line is a categorical version of the first theorem above. The bottom line should be pure algebra. The commutativity should be a better version of the second theorem, which is commutativity at the level of objects.

We'll use the tool of $Y$ diagrams. It's a complicated quilt. Consider the following things: [picture].

The proof of the first theorem, consider a neighborhood of this puncture. Conformally change this picture [picture].

So we consider this moduli space. How do we use it? Suppose that $U_i$ is $J_i$-holomorphic and energy is finite. As $\tau$, the $\rR$-factor, goes to $\pm \infty$, you get $U_1,U_2,U_3$ go to a point in $L_{12}\times_{X_2}L_{23}\times_{X_3}L_{31}$, contracted over $X_1$. This is $L_{13}\times_{X_1\times X_3}L_{13}$.

I proved last time that $b_1$ and $b_{12}$ existing proved that there was a bounding cochain $b_2$ and this proof will be similar. You have evaluation maps
\[
\xymatrix{&\ar[dr]^{\tau\to \infty}\ar[dl]^{\tau\to-\infty}
\Mm^E(L_{12},L_{23},L_{13})\\
L_{13}\times_{X_1\times X_3}L_{13}&&L_{13}\times_{X_1\times X_3}L_{13}
}\]
So put marked points and you get a map
$CF(L_{13})\times CF(L_{12})\times CF(L_{23})\times CF(L_{13})\to CF(L_{13})$. This is a map $n_{k_1,k_2,k_3}$, so now we take $n^b_k(y,x_1,\ldots, x_k)$ to be
\[\sum n^k_{k_1,k_2,k_3}(y,b_{12}^{k_{12}},b_{23}^{k_{23}},x).\]
[picture]

This gives us a right $CF(L_{13})$ modele structure on $CF(L_{13})$. Then there is a unique $b_{13}$ so that $\sum n_k(\mathbf{1},b_{13},\ldots, b_{13})=0$. This is from the algebraic lemma I explained last time. We can then cook up the bounding cochain.

What I explained last time is a special case of this.

Now I go back to my Y diagram. There's one slighly complicated thing to understand. Recall we had $F(X_1)$ and $F(X_2)$ and $F(X_3)$, and we had $Yo:F(X_2)\to F_c(F(X_2),Ch)$ and a map from $F(X_1)$ into it lifts becaus of the inverse to the Yoneda embedding. It's hard to cook up the inverse explicitly. So to calculate the composition with a map $F(X_2)\to F_1(F(X_3),Ch)$.

I'll explain how to go around this trouble. We use the following lemma. You have $L_{12}\subset -X_1\times X_2$ and then another $L_{21}=L_{12}\subset -X_2\times X_1$. Then $W_{\Ll_{12}}:\bar{F}(X_1)\to \bar{F}(X_2)$ and $W_{\Ll_{21}}:\bar{F}(X_2)\to \bar{F}(X_1)$. These are \emph{not} inverses. But they are homotopy adjoint to one another.

What does this mean? we have $F:C_1\to C_2$ and $G:C_2\to C_1$. We say that $G$ is adjoint to $F$ if the functor $C_1^{op}\times C_2\to Ch$ given by $c_1,c_2\mapsto CF(F(C_1),C_2)$ and the functor to $C_1(c_1,G(c_2))$ are homotopic.

So Floer homology has Poincar\'e duality, $HF(W_{\Ll_{12}}(L_1,b_1),(L_2,b_2))=HF((L_{12},b_{12}),(L_1,b_1)\times (L_2,b_2))=HF((L_1,b_1),W_{L_{21}}(L_2,b_2))$.

So $L_{13}=L_{12}\times_{X_2}L_{23}$ and we have $b_{13}$. What I want to prove is that $W_{\Ll_{13}}\sim W_{\Ll_{23}}\circ W_{\Ll_{12}}$. So we'll take $(L_1,b_1)$ and $(L_3,b_3)$ in $\bar{F}(X_1)$ and $\bar{F}(X_3)$. Then I want to compare the two functors in the functor category under the Yoneda lemma. 

So look at $H(W_{\Ll_{13}}(L_1,b_1),(L_3,b_3))$ and compare to $H(W_{\Ll_{23}}\circ W_{\Ll_{12}}(L_1,b_1),(L_3,b_3))=H(W_{\Ll_{12}}(L_1,b_1),W_{\Ll_{32}}(L_3,b_3))$
. 
So we want to show that two $A_\infty$ functors are homotopic. So look at $\bar{F}(X_1)\times \bar{F}(X_3)\to Ch$. You have in the second case $(L_2,b_2)\mapsto Hom(W_{12}(L_1)(L_2),W_{23}(L_3)(L_2))$.

Let's go back to the $Y$ diagrams to see what we are doing. [pictures]

You assume that as you approach the puncture in the $Y$, you approach $[L_{13}]=\mathbf{1}$.

[pictures]

Using the moduli space of these, what do you get? You have \[\Omega(L_1\times_{X_1}L_{13}\times_{X_3}L_3)\otimes \Omega(L_1\times_{X_1}L_{12}\times_{X_2}L_2)\to \Omega(L_2\times_{X_2}L_{23}\times_{X_3}L_3)\]
What do you get?
$CF(L_1,L_{13},L_3)\to Hom(CF(L_1,L_{12},L_2),CF(L_2,L_{23},L_3))$ and the claim is that this is a chain map. If you know Floer theory, this is obvious. To prove the property is to see its boundary. The boundary is the bubbles. The boundary on the bubbles is killed with the bounding chains. The boundary on the $Y$ point is killed because $\mathb{1}$ is a cycle with respect to this bounding chain $b_3$. What remains is three ends going to $\infty$, but these are the boundary operators you are supposed to have. So this is a chain map.

What does this chain map get you? You have something $F_3(W_{13}(L_1),L_3))\to Hom(F_2(W_{12}(L_1),L_2),F_2(W_{32}(L_3),L_2))$.
You have to check functoriality. You have $CF(L_1',L_1)$. To see this, you have $L_1$, and [pictures].

So what does this mean? This gives you a canonical map $H(W_{13}(L_1),L_3)\to H\ Hom(W_{12}(L_12),W_{32}(L_{32}))$. How do we obtain this map? Rather, how do we prove it is an isomorphism?

We take the choice of $L_2$ as $L_1\times_{X_1}L_{12}$. Then we know this is representing. Then the diagram gives us $CF(W_3(L_1),L_3)\to CF(W_{32}(L_3)(L_2))$. This gives an isomorphism in homology. Why? Forget positive energy and you get the identity map.

So something I want to prove is that this is fully functorial with respect to all of these guys. You want to prove composition is associative. Then you get more complicated graphs. Collapsing edges is hard analytically. But you can do surgery, gluing a pair of pants with the correct decorations. [Picture].












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