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\author{Gabriel C. Drummond-Cole}
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\title{Deformation Theory Workshop}
\begin{document}\maketitle
\section{Kaufffman}
Here's a plan:
\begin{enumerate}
\item Introduction
\item Theme: Correlators
\item Variations
\item Deforming deformations
\end{enumerate}
There's a funny interplay here, there's a geometric lemma that will give things that you may not expect.

We've been looking at $A$ an associative algebra, and let's have it unital with unit $1$, and we'd like to study deformations of this, and it's been clear that these are governed by $HH^*(A,A)$, which has the Gerstenhaber bracket $\{,\}$.  Thinking more, one might want to pass to the chain level, where you put a $G_\infty$ structure, and this is Deligne's conjecture.  So I said a $G_\infty$ structure.  Bruno gave it the $G_\infty$ structure.  Many people prove it, by taking different $G_\infty$ structures.  This is just true for any old associative algebra.  If you have a Frobenius algebra, $A$ has an invariant pairing $\langle\,,\,\rangle$ and then more is true.  In this case, then you can still look at the Hochschild cohomology, and this has a $BV$ structure, and Morichi proved this and at the chain level it has a $BV_\infty$ structure.  This was first proven by yours truly, and there have been other proofs.  It's the same discussion as for the $G_\infty$ structure, it just means that it's homotopy $BV$.  You put the $\infty$ in, you don't have, well, you don't have a canonical choice or you do.  So here you are not making a canonical choice, you're making choices that in this case made it smaller.

So this should be called the cyclic Deligne conjecture.  You can pick your favorite presentation of $BV_\infty$ and then you get something interesting.  This should be called the $TT$ conjecture.

This is kind of the setup.  Either you can use algebraic machines to do this or you can use geometric machines.

How does this geometric machine work?  This is the theme.  The point is to use $CW$ complexes to generate operations.  The tool is correlators and graphs.  Let's talk about graphs first.  In the theme, these will just be marked ribbon graphs.  There's one distinguished vertex in each cycle, which may be of valence $2$.  A ribbon graph has a cyclic order of the incoming edges, where valence is at least three.  I will allow the graph with one edge and one vertex.  Here is a more standard picture.  A cycle is an orbit.  I want to thicken this up and make a surface.  In this simple ribbon graph, I get a surface with three boundary components.  If I mark additional points, I can mark them wherever I want, I can draw the dual graph to this on the surface.  I replace each edge with an edge that cuts it.  You can do everything in ribbon graphs, but for me you want to use the yellow graphs.  It's an exercise that you can translate these.  What is the $CW$ complex that comes from these graphs?  First of all, this has a differential, the sum of contracting edges, only allowed edges.  You do not change the topology of the graph, the number of cycles and the genus.  So from this you get a complex, so you can build an open $CW$ complex.  I can not contract all cells.  Nevertheless you have this complex, and this space is $M_{g,n}^n$, where this is the moduli space of genus $g$ curves with $n$ punctures and a vector, a tangent vector, at each puncture.

I have this moduli space so what do I do with it?  We were told briefly that there was some sort of operadic structure.  On the topological level there is a quasi operad structure, a rational operad structure.  This descends to a true operad structure on the chain level.  I don't want to use too many words but there is this structure, given by gluing graphs, and I don't want to talk about it or the lecture would be over, but it's there.

What kind of operations should I have?  I should have one operation for each graph.  That's what the correlators do for me.  For each $\Gamma$, well $Y(\Gamma)$ is a marked ribbon graph with $n$ cycles, you get a correlator $TA^{\otimes n}\to k$.  What  do they look like?  You sort of decorate the cycles, oh, you should have a cyclic trace from $A\to k$, meaning that $\int a_1\ldots a_n= \int a_n a_1\ldots a_{n-1}$, and then the formula is roughly like this:  It's the product over $n$ that are complementary regions of $\Gamma$ of $\int \prod a$, where $a$ is over an index set bounding the disks that the first product are over.  That's the theme, so I have the operations.

So I'm almost done.  I have to get rid of the almost and tell you what done means.  To treat all the cases I want I have to include some of the boundary of this in.  I need some sort of extension of what things are allowed.  Done means that this part is good enough to prove the conjectures.  I didn't explain where I got this stuff.  This is a constructive approach from geometry, and it gives you these relations.  I need one more variation, I have one more remark.  If you are \wh, you see that this is not really an operad.  This could be a cyclic operad, but I only have ins, I have no outs.  This I can do precisely if I have the Frobenius algebra business.  One great thing about having an out, you can compose.  In physics you take a propagator in between.  Now I'll take care of the almost.  I'll start with variations.  What happens if I do something to my graph that I shouldn't have done.  A typical example of what I'm going to consider is like this.  This is supposed to be a genus one surface.  I'm not allowed to take out something that goes over a bridge.  But now I allow these, and dually I allow, $\Gamma$ is a marked stable ribbon graph.  What are these?  This is going to the boundary.  These are exactly these in the Kontsevich compactification.  What's the point?  Why can't you cut these?  You have genus, you don't have disks, the complementary regions aren't disks.  Here, the edges split into disjoint unions with cyclic orders.  The picture of surfaces is much simpler.

This is Penner's description of the same situation.  I repeat this, I give correlators, and I take stable graphs now and conveniently I already have a space, and I will need to put in $e^{-\chi(S)-1}$, and I have to explain what $e$ is, and it is $\mu\Delta(1)$.  So $e$ is the Euler element, if you have a frobenius algebra, you can take the product of the coproduct of $1$.  I can still do this without the inner product.  That's more or less the end of the story.  Now the machine is done.  What I put in are graphs and what I get out are operations.  

The most important thing is the existence of that formula.  I'm not sure, well, this Euler characteristic wants to come from geometry.

I promised you some applications.  Variation two is string topology.  To turn on the machine, I need to feed it some graphs.  There's a new datum, namely you pick in and out cycles.  On the surface you partition your boundaries into in and out boundaries and allow graphs $\check{\Gamma}$ with edges only from in to out.  You can also allow out to out.  There's one more condition, that each in boundary is hit by at least one edge.  I should make this remark.  This is partly inside moduli space and partly in the boundary.  It's really a mixture.  The theorem is that on the topological level, the procedure gives you a quasiprop, where quasi means associative up to homotopy, and this induces a prop structure on the chain level, and I have a $CW$ complex here, actually a $dg$ Prop, and this acts on $(CH^*(A,A),d)$.

So what about this?  Just as a simple corollary of this, you can get these $BV_\infty$ structures.  The thing I want to stress is that it's all the same formula.

The corollary is the cyclic Deligne conjecture, and the noncyclic version.  What does the proof need for this?  First you fix genus equal to zero, one out and $n$ in, and you control, you control, you show that this part is equivalent to the framed little disk operad, and then this gives operations, the operations which in cohomology descend to the bracket.

There's another attribute you put on your graphs, exactly the $S^1$ with one point, you throw this one out.  If you take this and plug it into the machine, you get a suboperad, it's equivalent to little disks, and that uses the same formulas.  This is minimal in the sense that there is one brace operation per cell.  Can you get a topological space where each cell is exactly a brace, but then you don't have more operations, these are the minimum number of operations you need.  Is there a smooth model that is this small?  Unfortunately this is not smooth, but you can blow it up a little bit and it's smooth but not minimal in my sense.

It's not smooth but there is a blowup which is smooth.  This is again in terms of allowed graphs.  You ask this question, there is a natural geometric machine.  You pick a certain larger class of graphs, and then it gives you a smooth model.  Let's see.

The $A_\infty$ case can also be done, this is a theorem together with Rachel Schwell.  There is a graph $CW$-model whose chain acts on $CH^*(A,A)$ where $A$ is an $A_\infty$ algebra, solving the $A_\infty$ Deligne conjecture.  This was solved by [KS] with homological operation.  I have to make the cells coarser here.  The cell of a graph you need to combine cells to get the operations.  Dually what this means is that you get a new decomposition of associahedra and cyclohedra.

The $A_\infty$ Deligne says that there's a chain level action of little disks that act on Hochschild of an $A_\infty$ algebra.

All right, so I have ten minutes.  I didn't erase the formula.  The last thing I wanted to talk about was stabilization.  So let me make this, let me draw some pictures and explain what this is about.  I can look to the following sequence of graphs, and I can ask myself, any genus, one out, and any number of ins.  What operation does this picture give?  It's the cup product.  What if I apply $Y$ to this.  Here I get the bracket $\{f,g\}=f\cup_1 g- g\cup_1 f$.  This gives some sort of operation that's like a $\cup_2$.  What's the whole point.  If I take the differential, If I kill this edge I certainly see this picture.  I can see this formula by seeing this edge and not this one.  I still have the genus.  It would be $\cup_2$.  If you can identify away the genus, then you get $\cup_2$.  On the geometric level I can do whatever I want.  It's rather complicated to explain what happens, but in principle\ldots

I'm talking about any genus, one out.  So the theorem here, there is a limit of operads one out such that it ``stabilitzes'' and remains an operad.  Formally you define $o'$ which is a formal limit.  If you take the direct limit, and stabilize, that means that representatives may replace, we may forget.  In string topology, for higher genus, you need to assume $A$ is commutative, and \wh also gave operations of this type.  So here you need probably that $A$ is commutative.  And you need that $S$ is not a disk.  You forget about the genus, you erase the ridge.  So there's a way of doing this, it's not the same stabilization as in \wh but it may be related.  On the geometric side I can just define this.  What about the algebra thing, do I really want that?  It at first looks like the string brackets are all trivial.  $\cup_2$ will be the bracket.  The difference is the element $e$, the Euler characteristic, that's the difference.  Either I have this or I don't.  It's given by this factor $e^{-\chi(S)+1)$.  If $A$ is a Frobenius algebra and $e$ is invertible, then that's equivalent to the fact that $A$ is semisimple and so that would tell me that you can calculate one from the other.  If $e$ is the unit, then the formulas coincide, so $A$ is semisimple and thus $\langle e_i,e_i\rangle=1$.  In this case you have a stabilized operad, and the most important thing about the stabilized operad which contains an $E_\infty$ operad.  So not only do I get the $\cup_2$, I get all higher homotopies, and then in the stabilization business, I have the normalized thing, and now over the stabilized guy.  You have the following nice

\begin{thm}If $A$ is a normalized semisimple Frobenius algebra, then $CH^*(A,A)$ has an $E_\infty$ structure.  If I start out with a variety, I can look at the string topology, which is basically $CH^*(A,A)$, and look at $CH^*(A,A)$.  If $V$ has semisimple quantum cohomology, then there is a deformation of $A$ such that this deformation $A_{q,1}$ is $E_\infty$, or if you want it like this, the quantum string bracket vanishes, where I take $CH^*(A,Q)$ to be the quantum space.
\end{thm}
Maybe I should say a few words.  \wh.  Why do I call this the quantum loop space?  How do you get $CH^*(A,A)$.  You're sampling a loop, so that if your point, if you take one point away it goes to that or that.  These two points now coincide if they quantum coincide.  On the algebraic level this is true, and geometric level, I don't quite understand, it could be that you are looking at the quantum loop space.
\section{Park}

Thank you, and thanks to the organizers for inviting me and running this smoothly.  This is an effort to understand quantum field theory.  I want to understand QFT.  No one knows what this is, and no one has defined it.  I don't know if people can agree on it.  I will make an imaginary category of quantum field theories.  Then we will form an imaginary functor to a category of quantum algebras.  You cannot establish an equivalence against something that is not defined so it's a nonsense statement.  Physicists consider two quantum field theories.  They sometimes make a statement that two quantum field theories are physically equivalent.  If this picture holds, then our quantum algebras should be quasiisomorphic as quantum algebras.  So for example, there's a famous example, they have a string theory called topological string theory, and they have two models, the $A$ model and the $B$ model.  The famous statement is that the $A$ model on a Calabi Yau and the $B$ model on a mirror Calabi Yau are physically equivalent.  For physicists, this seems obvious, so we need this kind of statement here.  The most important thing in quantum field theory is the path integral, all of the correlation functions and so on, this method should be able to give the same kind of information.  When they say things are physically equivalent, it means that all of the things they compute agrees.  If this equivalence is true, then the quasiisomorphism should carry all of the information.  Then the burden is to define quantum algebra.  So here is the question, what is quantum algebra?  I am not claiming that $QFT$ is a category, but if it were, what is a quantum algebra.

Real physicists do experiments in the real world, I will take an experimental method against a quantum field theory.  I will do experiments there to find mathematical notions.  I will start with a known example, a particular kind of quantum algebra called a $BV$ quantum algebra, a special type.  So $k$ will be a field of characteristic zero.  $\Cc$ will be some graded $k$-module.  Then $\cdot: S^2\Cc\to \Cc$ is an associative product, then $\hbar$ is a formal parameter, and I consider the formal power series ring $k[[\hbar]]$.  Then a BV quantum algebra is a triple $(\Cc[[\hbar],\hat{K},\cdot)$.  Without $\hat{K}$ this is a graded commutative associative algebra.  And we have $\hat{K}^2=0$ and $\hat{K}=Q+\hbar K^{(1)}+\hbar^2 K^{(2)}+\cdots$  The relations between the two is that
$$\hat{K}(\hat{a},\hat{b})-\hat{K}(\hat{a})\hat{b}-\hat{a}\hat{K}(\hat{b})=\hbar(\hat{a},\hat{b})$$  Then the failure of being a derivation of the product is a derivation of the product.  Then defining the bracket in this way,  you have that $\hat{K},(,)$ is a dg Lie algebra over $k[[\hbar]]$.  So a BV quantum algebra is a $BV$ algebra over $k[[\hbar]]$ but the converse is not true.  

So Kajiura's talk here, well, first, very important example, if you have $(C,\Delta, \cdot)$, then you can associate $(\Delta, [,])$, where that is just the failure of $\Delta$ of being a derivation.  What physicists do is say $C[[\hbar]],\hbar\Delta,\cdot$, and this is a BV quantum algebra.  Then they consider $\hat{S}$ of degree zero which satisfies the quantum master equation $-\hbar \Delta e^{-\hat{S}/\hbar}=0$ or $-\hbar\Delta\hat{S}+\frac{1}{2}(\hat{S},\hat{S})=0$.  This equation in particular implies that $\hat{K}=-\hbar\Delta+(\hat{S},\,)$ squares to zero.  So in this setting we obtain a BV quantum algebra and the descendent dg Lie algebra.  Now let me also add one more remark about this picture.  Physicists say that the functional satisfying this is an action functional.  If you just consider $\hat{S}+\hat{\Gamma}$ then $\hbar\Delta e^{(-\hat{S}+\hat{\Gamma})/\hbar}=0$ which is the same as $\hat{K}\hat{\Gamma}+\frac{1}{2}(\hat{\Gamma},\hat{\Gamma})=0$.  So a word of warning, the $\Delta$ in a BV algebra, you can form the same Lie algebra but that's stupid, it doesn't control the deformations of quantum field theory.

The Lie algebra structure is a derived notion, it's a secondary object.  Let me give another example.  We consider the same setting, $(C[[\hbar]],\hat{K},\cdot)$, with the same property, and assume that the failure of $\hat{K}$ from being a derivation of $\cdot$ is divisible by $\hbar$, it's $\hbar\hat{\mu}_2$.  Then the failure of $\mu_2$ from being a derivation is divisible by $\hbar$.  The important thing is to preserve this divisibility.  Then you have a descendent $L_\infty$ algebra over $k[[\hbar]]$.  Obviously, a BV quantum algebra is a special case of that.  This is a binary quantum algebra, this will be the final case.  So what's a morphism?

$(C[[\hbar]]',\hat{K}', \cdot')\to (C[[\hbar]],\hat{K},\cdot)$ should be a morphism of complexes.  What about the product?  It's not preserved.  $\hat{K}$ does not respect anything of $\cdot$.  Well if something is an algebra map then $\hat{f}_1(\hat{a}')\hat{f}_1(\hat{b})'-\hat{f}_1(\hat{a}'\hat{b}')$ is divisible by $\hbar$.  And so on for further pieces.  So a morphism of binary quantum algebras induces an $L_\infty$ morphism on the descendent structure.  Every morphism of quantum algebras induces a descendent one, but it is not true that every $L_\infty$ morphism can be lifted.

A quasiisomorphism of a quantum algebra is just one that induces an isomorphism on homology.

The $B$ model comes in moduli space, it's the extended complex structure, there's a differential graded Lie algebra, and there are no obstructions to the deformations.  We always have a minimal $L_\infty$ structure and the Maurer Cartan equation has a solution if and only if the algebra is trivial.  There they found something called special coordinates, a very particular type of solution, and with that they were able to construct a Frobenius manifold.  Their formulation is the same as the one used by the physicists.  My point is that among all of those coordinates, theirs is the one from quantum algebra.

I hope I gave a feeling that the $L_\infty$ algebra is secondary.  Let's return to a BV quantum algebra.  If you consider the classical limit, you get $(C,Q,\cdot)$, and because the failure of being a derivation is divisible by $\hbar$, this is just a dga.  A physicist calls a representative of the classical cohomology an observable $o_\alpha$.  If you have two such, they are physically equivalent if they belong to the same equivalence class.  A quantum observable is $\hat{o}=o+\hbar{o}^{(1)}+\cdots$  so that $\hat{K}\hat{o}=0$.

It's not true that every classical observable can be extended to quantum observable.  If every classical observable goes to a quantum observable then the minimal $L_\infty$ structure on the homology of $\hat{H}$ is trivial.  This statement is slightly wrong.  This in particular means that we can find a versal solution $\hat{K}\hat{\Gamma} +\frac{1}{2}(\hat{\Gamma},\hat{\Gamma})=0$.  Then $\hat{\Gamma}=t_\alpha \hat{o}_\alpha +\frac{1}{2}t^\alpha t^\beta \hat{o}_{\alpha\beta}+\cdots$.

This is not surprising.  There is a stronger theorem than that.  The first theorem says we can associate a formal pointed graded manifold.  The second theorem says that on the tangent space of the moduli space there exists a linear pencil of flat torsion free connections.  If every classical goes to quantum, this is true.  What is a linear pencil of flat torsion free connections, I think everyone is familiar with this guy.  It's what defines a Frobenius manifold, without the invariant inner product.  On $TM$ there exists a linear pencil of flat superconnections.  This I mean there is a one form, two form, all of these things.  You can just prove, starting from this assumption.  

This statement here, from this construction, we want a more general notion of quantum algebra than $BV$ quantum algebra.

Now I want to say something about this construction related to this morphism, tell you why this should be true.  Let's start from a set of classical observables that can be extended to quantum observables.  I imagine everything can be extended, and so $o_\alpha o_\beta=m_{\alpha\beta}^\gamma o_\gamma + Q\lambda_{\alpha \beta}$.

In the quantum, we get
$\hat{K}(\hat{o}_\alpha\hat{o}_\beta)=-\hbar(\hat{o}_\alpha,\hat{o}\beta)$.  If I consider the expression
$\hat{o}_\alpha\hat{o}_\beta-m_{\alpha\beta}^\gamma\hat{o}\gamma-K\lambda_{\alpha\beta}= \hbar\hat{o}_{\alpha\beta}$.  This is divisible by $\hbar$ so I can write this in this way.  So we solved the Maurer Cartan equation.  I solved the same Maurer Cartan equation.  So I have $\hat{K}\hat{\Gamma}+\frac{1}{2}(\hat{\Gamma},\hat{\Gamma})=0$ modulo $t^3$, and you can go on like that, and but, what about $m_{\alpha\beta}$?  In the homology of $Q$, that's $m_{\alpha\beta}^\gamma$.  But then we can go to the next order, and define $\hat{c}_{\alpha\beta\gamma}:= \hat{o}_{\alpha\beta}\hat{o}_\gamma+\hat{o}_\beta\hat{o}_{\alpha\gamma}-m_{\beta\gamma}^\rho\hat{o}_{\alpha\beta}-(\hat{o}_\alpha,\lambda_{\beta\gamma})$.  I can take $\hat{K}$ of this and I get $\hbar(\hat{o}_{\alpha\beta},\hat{o}_\gamma)+\text{ cycles})$ and so you can calculate $\hat{c}_{\alpha\beta\gamma}-m_{\alpha\beta\gamma}^\rho \hat{o}_\rho - \hat{K}_{\alpha\beta\gamma}=\hbar\hat{o}_{\alpha\beta\gamma}$.  So all I did was iterate, and so I can construct $\hat{\Gamma}=t^\alpha\hat{o}_\alpha+\frac{1}{2}t^\alpha t^\beta \hat{o}_{\alpha \beta}+\frac{1}{3!}t^\alpha t^\beta t^\gamma \hat{o}_{\alpha \beta \gamma}$, and $\hat{K}\hat{\Gamma}+\frac{1}{2}(\hat{\Gamma},\hat{\Gamma})=0$ modulo $t^4$.  So then you get $\hat{\Gamma}_\alpha=\frac{\partial \hat{\Gamma}}{\partial t^\alpha}$ and $\hat{\Gamma}_{\alpha\beta}=\frac{\partial^2 \hat{\Gamma}}{\partial t^\alpha \partial t^\beta}$.  So I can write down as a full equation 
$\hbar \hat{\Gamma}_{\alpha\beta}=\hat{\Gamma}_\alpha\hat{\Gamma}_\beta -A_{\alpha\beta}^\gamma \hat{\Gamma}_\gamma -\hat{K}\Lambda_{\alpha\beta}-(\hat{\Gamma},\Lambda_{\alpha\beta})$
[Missed some]
Now rewrite the master formula in the following way, $-\hbar\delta_\alpha^\beta d+ A_\alpha^\beta (\hat{\Gamma}_\beta e^{-\hat{\Gamma}/\hbar}=\hat{K}(\theta^\alpha \Lambda_{\alpha\beta}e^{-\hat{\Gamma}/\hbar}$, where $d=\theta^\alpha \frac{\partial}{\partial t^\alpha$.  
This is what physicists take the path integral over, so $\oint \hat{\Gamma}_\beta e^{-\hat{\Gamma}/\hbar}=|\Psi_\beta\rangle$ and $(-\hbar d+ A)|\Psi\rangle =0$.  I think that unfortunately I'm at the end of this talk.  My time is up, but here, this is still an experiment.  But we more products, not just the binary one, and then you should have a general quantum algebra.  I did classical goes to quantum, but in the completely general case you will get all of these together.  The flat superconnection, you need much more product, you need product of degrees $-1$, $-2$, all of these things.  They act with $k$ in a particular way, and then you can define a natural notion of morphism, and all the things we did here are a particular example of morphism.  And if $C$ is a differential $BV$ algebra, I can make a BV quantum algebra with $Q+\hbar\Delta$ and construct the descendent structure.  They deal in [BK] the $Q\Delta-\Delta Q$ lemma, it means that classical goes to quantum, so if you put this into the machine, the solutions from quantum algebra are the special coordinates.  Perhaps this is the correct way.   This will give you more data than the path integral in that you have the homotopy path integral.  You can consider supersection of superconnection.  Classical is related to the Massey products, and so on.  I will stop my talk, thank you.

\end{document}
 his sense of minimal is what i'm trying to do for dm.  