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\title[Quantum Monday 2016]{CGP Quantum Monday Seminar, 2016}
\author{Gabriel C. Drummond-Cole}
\begin{document}
\maketitle
\section{April 11: Ben Knudsen: Higher enveloping algebras and configuration spaces of manifolds}
Thanks for coming, it's a pleasure to be here. The motivating object for me is the configuration spaces in the title. For me $M$ is going to be an $n$-manifold and I'm going to write $B_k(M)$ for the space of unordered configurations of $k$ points in the manifold. This is $k$-tuples of points in $M$ such that no two of them are the same, quotiented by the action of $\Sigma_k$. These spaces are interesting and complicated. 

For an example of how these spaces are interesting and complicated, there's a theorem of Longoni and Salvatore that says that $B_2(L_{7,1})$ is not homotopy equivalent to $B_2(L_{7,2})$ even though the two spaces are homotopy equivalent (but not homeomorphic).

I'll lump these configuration spaces together, denoting the disjoint union 
\[
B(M)=\amalg_{k\ge 0}B_k(M).
\]
Now $B(M_1\sqcup M_2)=B(M_1)\times B(M_2)$. If I have five points in the disjoint union, then I have four in one side and one in the other or something similar.

My goal is to find something about the cohomology of configuration spaces with field coefficients where the field has characteristic zero. The strategy will be to understand the structure locally on the manifold and then globalize that information.

Let me say a little something about the sense in which these are local objects on the manifold.

I'll write $U(M)$ for the partially ordered set of opens in the manifold and I'll think of rational singular chains on $B$ (I'll omit the coefficients) as a functor to chain complexes. So $A$ takes an open set to singular chains on that open set.

Within this set $U(M)$ I'll single out a subset $D(M)$, whose objects are the open sets which are homeomorphic to a nonempty disjoint union of Euclidean spaces. The morphisms of this poset are inclusions which are $\pi_0$-surjections.

A little bit of notation, for an object in the poset I'll write $U_I$ for the disjoint union $\amalg_I U_i$, where each look like Euclidean space. This is a poset, but it's more than a poset, it has a ``partially defined'' symmetric monoidal structure given by disjoint union. If I have $U_I$ and $U_J$, then I can combine those into a tuple $U_{I\sqcup J}$ if $U_i\cap U_j$ is empty for all $i,j$. All I'm saying is that I can take the disjoint union of open sets if they're disjoint.

From this point of view, the functor $A$ is symmetric monoidal. Why is that? I'm supposed to look at $A(U_I\sqcup U_J)$ and that's by definition $C_*(B(U_I\sqcup U_J))$ which is $C_*(B(U_I)\times B(U_J))$ which I can identify with $C_*(B(U_I))\otimes C_*(B(U_J))$ which is $A(U_I)\otimes A(U_J)$.

Now I said I'd say something about locality. The key observation about this functor $A$ is that if I look at the collection $B(U)$ where $U$ is an object of $D(M)$, this collection is a basis for the topology of $B(M)$. Why is this good? It means that if I take the homotopy colimit over $D(M)$ of the functor $A$, that naturally maps to $C_*(B(M))$, and because this is homotopy colimit, this map is a quasi-isomorphism. This is useless or useful depending on how well we understand that local data.

Let me give a little bit of jargon. I'll call $A$ a \emph{locally constant $D(M)$-algebra} and $C_*(B(M))$ is its \emph{factorization homology}. Locally constant means that if I have an inclusion $U_I\subset U_J$, and if these have the same number of components (this is then a union of disjoint inclusions of Euclidean space into Euclidean space), then this is a homotopy equivalence.

A $D(M)$-algebra, this means that $\amalg$ is sent to $\otimes$. By factorization homology I just mean I'm taking the homotopy colimit of this thing.

So far all we've done is to package the question in a way that boils down to local data. Now I want to talk about what the local structure is. I submit to you that locally on the manifold, I'm looking at a neighborhood $U$ that is Euclidean, and locally I want to say that $A(U)$ has something like the structure of a Hopf algebra. What does a Hopf algebra have? It has a multiplication (given by stacking points next to each other), meaning that given an inclusion of two other neighborhoods $V\sqcup W$ into $U$, I get a structure map $A(V\sqcup W)\to A(U)$, and $A(V\sqcup W)$ is $A(V)\otimes A(W)$ and up to quasi-isomorphism this is $A(U)\otimes A(U)$. What does this look like in terms of a picture? [picture]. 

This is not commutative (it's commutative at the level of homology). What else does a Hopf algebra have? It has a comultiplication, for that I want to split points apart. I wnat to go $A(U)\to A(U)^{\otimes 2}$ and what I want to do heuristically is take the set of points $\{x_1,\ldots, x_k\}$ and split into all the ways I can split into $i$ and $j$ points and take open sets containing the chosen $i$ and the chosen $j$. You can convince yourself that those are compatible. This, we should note, is cocommutative. 

What about an antipode? I flip $\rR^n$ about some hyperplane. You'll note that this is not uniquely defined, there's an orthogonal group worth of antipodes. So this is like a Hopf algebra, but not a Hopf algebra.

Why do I want to say this is like a Hopf algebra? A theorem of Milnor--Moore says that (I won't say this precisely) almost all cocommutative Hopf algebras are universal enveloping algebras of Lie algebras. Whenever you see one in the wild you should look for the Lie algebra that goes with it.

I'm about to tell you a theorem but let me first remind you of something about Lie homology. If $\gGg$ is a Lie algebra, then 
\[
H^{\Ll}(\gGg)\coloneqq \Tor^{U(\gGg)}(\fF,\fF),
\]
but this is computed by something nice, $H(CE(\gGg))$, the ``Chevalley--Eilenberg'' complex.
\[
CE(\gGg)=(\Sym(\gGg[1]),d_\gGg+d_{[,]}),
\]
with the second differential determined on $2$-tensors by the bracket and extended to be a coderivation.

This is a coalgebra under shuffle product.

The Chevalley--Eilenberg complex is part of an adjunction between Lie algebras and conilpotent cocommutative coalgebras.

So now I would like to tell you a theorem.
\begin{thm}(Knudsen) 
There is a homotopical adjunction (at this point I should come clean about my foundations, I'm thinking about infinity categories, by which I mean quasi-categories; that shouldn't be so important but feel free to ask for clarification) between $\Alg_\Ll(\ch)$ and $\Alg^{\loc}_{D(\rR^n)}(\ch)$, and there's a forgetful functor called $[n-1]$ and there's an adjoint $U_n$, I'll call it ``higher enveloping algebra.'' This can be computed by $U_n(\gGg)\cong CE(\Omega_c()\otimes \gGg)$.
\end{thm}
\begin{remark}
\[
\CE(\Omega_c(\rR^n)\otimes\gGg)\cong \Sym(H(\gGg)[1-n])
\]
which I'll think of as a Poincar\'e--Birkhoff--Witt theorem.
\end{remark}
I'll say something eventually about the proof but first I want to talk about how to use it.

So I'll say that $C_*(B(\quad))$ is free as a $D(\rR^n)$-algebra, so locally on $M$, my configuration spaces look like higher enveloping algebras of a free Lie algebra.

Once you have that local equivalence you can take a homotopy colimit and on one side you get the configuration spaces of $M$ and on the other side you get the homotopy colimit of these Chevalley--Eilenberg complexes which are easier to handle.

It turns out after being careful about various things, you get a bigraded isomorphism
\begin{thm}(Knudsen)
\[
H_*(B(M))\cong H^\Ll(\gGg_M)
\]
where the left side is bigraded by degree and number of points where
\[
\gGg_M=\left\{
\begin{array}{ll}
H_*(M)\text{ with no brackets}&n\text{ odd}
\\
H^{-*}_c(M,F^\omega)\otimes v \oplus H^{-*}_c(M,\fF)\otimes [v,v]&n\text{ even}
\end{array}
\right.
\]
with bracket in the latter case $[\alpha\otimes v,\beta\otimes v]\mapsto \pm \alpha\cup \beta\otimes [v,v]$ where $|v|=n-1$ and $\wt(v)=1$.
\end{thm}
You can get some immediate corollaries from this result
\begin{cor}
$H_*(B_k(M)))$ depends only on 
\begin{itemize}
\item $H_*(M)$ if $n$ is odd
\item the twisted compactly supported cup product if $n$ is even.
\end{itemize}
\end{cor}
If $M$ is compact oriented it depends only on the cohomology ring. This was known in some cases, there's a theorem of B\"odigheimer--Cohen--Taylor in odd dimensions, of B\"odigheimer--Cohen for punctured surfaces, and F\'elix--Thomas for complete oriented nilpotent manifolds.

The next corollary is not as immediate but not too far
\begin{cor}
$H_i(B_k(M))$ is independent of $k$ for $k$ large. 
\end{cor}
You can say precisely when it stabilizes. This is a theorem of Tom Church and Oscar Randal-Williams. You look at the Chevalley--Eilenberg complex and see when it stabilizes.

Other things you can do with this result is compute a whole lot of things. Let me do an example. I'll compare two different manifolds. One is the once punctured torus and the other is the twice punctured plane. These two manifolds are homotopy equivalent. However, we'll show that the second configuration space of $M_1$ and the second configuration space of $M_2$ are not equivalent by computing $H_2$. 

These are even dimensional so we're in the second case. We need the compactly supported cohomology ring of each. So
\[
H^{-*}_c(M_1)\cong \tilde{H}(T^2)\cong \fF\langle a,b,c\rangle, ab=c, |a|=|b|=-1.
\]
On the other hand,
\[
H^{-*}_c(M_2)\cong \tilde{H}(S^2\vee S^1\vee S^1)\cong \fF\langle a,b,c\rangle
\]
with all products zero. There are a bunch of shifts, I'll be abusive and still call things $a$, $b$, and $c$ after shifting them. We're interested in weight two and degree two. 

\begin{tabular}{l|ccc}
weight $2$&$c^2$&$\tilde{c}, ac, bc$& $\tilde{a},\tilde{b},ab$
\\
weight $1$&$c$&$a,b$
\\
weight $0$&$1$
\\
\hline
& degree $0$&degree $1$&degree $2$ 
\end{tabular}
The differential sends $ab$ to $\tilde{c}$ if $j=1$ and $0$ if $j=2$. Then $\dim H_2(B_2(M_j))$ is either $2$ or $3$ depending on $j$.

You might say this is a silly example, you can do better than that. Ongoing joint work with Drummond-Cole is to compute every Betti number of every configuration space of every surface. So for example you can show that the dimension of $H^{101}(B_{100}(\Sigma_2))=65,024$, that's a computation that you can do.

Enough applications, let me tell you something about the proofs.

If there aren't any more questions, let me take a brief interlude into the \emph{Ran space}.

If $X$ is a topological space this is a definition, if it's something else it's a heuristic. As a set, this is $S\subset X$ such that $0<|S|<\infty$. I'll topologize as the colimit over $X^I$ where I take the colimit over the opposite category of finite sets and surjections. You use diagonals parameterized by the surjections. In other words, I'm gluing all configuration spaces of $X$ together along various subdiagonals. So for example, I have $X^2$, and I have the inclusion of $X$ via the diagonal, and I have the inclusion of the ordered configurations of two points of $X$, and $X^2$ is glued from the open stratum of the configuration space and the closed stratum of $X$, the diagonal.

So Beilinson--Drinfeld were motivated by considerations out of conformal field theory, they considered an algebraic structure on a $D$-module $F$ and because the Ran space is this colimit, a $D$-module, sheaf, cosheaf, anything, has a component at level $k$, there's $F_k$, compatible sheaves or cosheaves or $D$-modules, one for each $k$.

What a $D$-module is plays no role in the talk. This is just historically where this came from. This algebraic structure admits two equivalent descriptions, in terms of the dichotomy between open strata and the closed diagonal. In terms of the open strata, this is what is called a \emph{factorization algebra}, and the condition is that if I take $F$ and restrict it from level $k$ to the open configuration space $j_k^* F_k$, then I get $j_k^*F_1^{\boxtimes k}$. 

The description in terms of closed diagonals is a little different, it's a sheaf or whatever equipped with maps, for example
\[
j_{2*}j_2^*(F_1\boxtimes F_1)\to \Delta_*F_1
\]
and this is called a \emph{chiral algebra}, or rather this is part of the data that specifies a chiral algebra. Once given actual definitions, these are equivalent pictures. So this side I'll think of as gluing data.

The book was published in 2004, and sometime later these ideas were reformulated by John Francis and Dennis Gaitsgory, and the factorization side was interpreted as a cocommutative coalgebra with an extra property and the chiral side as a Lie algebra with an extra property, and the correspondence is Koszul duality. Under Koszul duality, the Lie structure becomes the differential, which is literally the glue that holds the strata together.

What is the connection to topology and configuration spaces and the poset $D(M)$? It's basically this theorem.
\begin{thm}(Lurie)
The map of posets $D(M)\to U(\Ran(M))$ given by taking $U_I$, the collection over $I$ of disjoint Euclidean neighborhoods, to points $S$ in the Ran space such that $S\subset \amalg_I U_i$ and $S\cap U_i\ne 0$ for all $i\in I$, and this induces a fully faithful functor from locally constant functors from $D(M)$ to chain complexes to the target infinity category of cosheaves on the Ran space with values in chain complexes.

The essential image of this functor is the \emph{constructible cosheaves} (the restriction to each \wh is locally constant on that configuration space).
\end{thm}
We wanted locally constant functors with a symmetric monoidal property, so we want something on the other side that is like that, and the symmetric monoidal property (two disjoint opens go to the tensor product) look like the factorization algebra condition (disjoint means off the diagonal and off the diagonal we go to the tensor product). If we're looking for a factorization algebra, we look for a cocommutative coalgebra under Francis and Gaitsgory's characterization.

So the question is then, how is a symmetric monoidal functor like a cocommutative coalgebra. 

To answer this, let me introduce (right) \emph{Day convolution}. Let $\Cc$ and $\Dd$ be symmetric monoidal categories, then I define a tensor product on functors from $\Cc$ to $\Dd$, so $F$ and $G$ are such functors, then $F\otimes G$ is defined by, I can get a functor $F\times G$ to $D\times D$, and then tensor together pointwise to $D$. On the other hand I can tensor in $C$. Then I take the right Kan extension in this diagram:
\[
\xymatrix{
	\Cc\times \Cc\ar[d]_{\otimes_\Cc}\ar[r]^{F\times G}&\Dd\times\Dd\ar[r]^{\otimes_\Dd}&\Dd\\
	\Cc\ar@{.>}[urr]_{F\otimes G}
}
\]
and the formula is
\[
(F\otimes G)(c)=\lim_{c\to c_1\otimes c_2}F(c_1)\otimes G(c_2)
\]
What is a commutative coalgebra here? [missed a little]. The punchline is that in right Day convolution, oplax functors from $\Cc$ to $\Dd$ are equivalent to cocommutative coalgebras under right Day convolution. Inside that is the subcategory of strong symmetric monoidal functors. This is what we want to see, motivated by Beilinson--Drinfeld. One reason people don't talk about right Day convolution is because it's not really well-behaved, it's usually not symmetric monoidal because monoidal structures don't distribute over limits, so the limits should be very special (which is a property of $\Cc$).

I'm almost done. Let me just leave you with kind of a final board. We were looking for a forgetful functor. We needed this functor to make the theorem go. Maybe I'll give you a formula for the tensor product. We set $\Cc$ to be $D(M)$ and $\Dd$ to be chain complexes, and then the formula is \[F\otimes^{\amalg} G(U_I)\cong
\bigoplus_{I=I_1\sqcup I_2} F(U_{I_1})\otimes G(U_{I_2})\]
and this is the disjoint tensor product reminiscient of something from Beilinson and Drinfeld. We also have an overlapping tensor product. I have
\[\xymatrix{
\Alg^{\loc}_{D(\rR^n)}(\ch)\ar[d]^\subset\\
 \Coalg_{\com}(\Fun(D(\rR^n),\ch)_\amalg)\ar[d]\\
  \Alg_\Ll(\Fun(\Dd(\rR^n),\ch)_\amalg)\ar[d]
  \\
  \Alg_\Ll(\Fun(\Dd(\rR^n),\ch)_\cup)\\
  \Alg_\Ll(\ch)\ar[u]
  }\]
and you want to let the composition here (where you can lift along the last map) is the forgetful functor.
\section{May 16: Alexander Voronov: Quantum master equation and deformation theory}

It's a pleasure to be here, it seems such an exciting place to do mathematics and in addition it's in the paradise of this neighborhood, you have the ocean, good weather, all you need to do mathematics.

My talk is meant to be some kind of motivation for Martin Markl's talk on Wednesday and also to some extent, I wanted to do some things so that people who do homotopy perturbation theory will see something in common.

You guys all know the quantum master equation (I'll review it later) but let me start with the classical master equation, also known as the Maurer--Cartan equation, the CME (classical master equation) or MCE (Maurer--Cartan equation).

The equation is
\[
d\omega + \frac{1}{2}[\omega,\omega]=0
\]
and this arose first in differential geometry, where $G$ is a Lie group with a closed Lie subgroup $H$, and I look at a principal $H$-bundle which is $G$ over $G/H$ and for $\omega$ you take the Maurer--Cartan form $\omega=g^{-1}dg$, which defines a ``Cartan connection'' on this principal $H$ bundle $G\to G/H$. Then it turns out that it satisfies this equation, which is the curvature equation, which is $d\omega+\frac{1}{2}[\omega,\omega]=0$. This is in $\Omega^1(G,\gGg)$. The Maurer--Cartan equation means the curvature is zero.

This equation turns out to be the cornerstone of deformation theory. In algebraic deformation theory, originating in the works of Deligne, Schlessinger--Stasheff, Goldman--Millson, and Kontsevich--Soibelman, the starting point is the following meta-theorem; I don't know if you know what a meta-theorem is, but you will see, it's a theorem which is true even in the cases you haven't thought about yet. It says 

\begin{quote}
The formal deformation theory of a mathematical object is described by a suitable differential graded Lie algebra $(\gGg,[\,,\,],d)$ via the MCE in $\gGg$.
\end{quote}

This is an equation for $S\in \gGg^1$ (the differential is degree $1$) such that $dS+\frac{1}{2}[S,S]=0$. Here I'm talking about the situation where I fix a ground field of characteristic zero.

The formal deformation theory is described by the solutions to this equation. You could also write $S\in \lambda \gGg^1[[\lambda]]$ such that $dS+\frac{1}{2}[S,S]=0$ and then $\lambda$ is the deformation parameter and $S=S(\lambda)$ is a deformation of something, which is the deformation of the thing that the Lie algebra is supposed to describe.

For instance, let me indicate how this theorem is proven. The proof should be a meta-proof. It should go by examples. We teach students not to prove things by examples in calculus, but this is how you prove metatheorems. Say $M$ is a complex manifold and you want to deform the complex structure on the manifold. The deformations of the complex structure on $M$, you define the following differential graded Lie algebra, which is $\Omega^{0,\cdot}(M,T_M)$, the  smooth $(0,q)$-forms on $M$ valued in the holomorphic tangent bundle to $M$. There's a grading (let me skip it, it's shifted by $1$), and this has differential $\bar\partial$ acting on smooth sections, which squares to zero, and then also, the bracket is given by taking the bracket of holomorphic vector fields and wedging the $(0,q)$-forms. It comes from the commutator of holomorphic vector fields. This describes deformations of the complex structure on $M$ in the following way. 

You expect $\bar\partial'=\bar\partial +\mu(\lambda)$, where $\mu(\lambda)$ is $\lambda\mu_1+\lambda^2\mu_2+\cdots$ in $\lambda\Omega^{0,1}(M,T_M)[[\lambda]]$ which is $\lambda C^\infty(M,\Hom(\bar{T}_M,T_M))[[\lambda]]$.  The integrability of the new complex structure, Newlander--Nirenberg structure, is equivalent to $\bar\partial \mu +\frac{1}{2}[\mu,\mu]=0$. Thereby deformations of complex structure are described by setting up this Maurer--Cartan equation.

Now instead suppose $A$ is an associative algebra, and suppose you want to deform the product on this vctor space. Then the differential graded algebra will be $\gGg=C^\cdot(A,A)=\bigoplus \Hom(A^{\otimes n}, A)$, and this has a differential (the Hochschild differential $d$) and the bracket as the Gerstenhaber bracket.

In this example, which is not a metatheorem but just algebra, one takes $m\in \lambda C^2(A,A)[[\lambda]]$, and then the original multiplication $m_0:A\otimes A\to A$ is also in $C^2(A,A)$, and the bracket and differential are defined in a way so that $d=[m_0,\quad]$, and then when you consider a deformation $m'=m_0+m$, a new product defined by a deformation,  and this equation $[m',m']=0$ is given my $dm + \frac{1}{2}[m,m]=0$, and this is again the Maurer--Cartan equation for $m$, which describes a formal deformation for the multiplication.

For instance, if you look at the first term, if you write $m=\lambda m_1+\lambda^2 m_2+\cdots$, then this equation, if you collect terms of degree $\lambda^1$, you get $dm_1=0$, which means that $m_1\in H^2(A,A)$, the \emph{infinitesimal} or \emph{first order} deformation, and this relates to the fact that first order deformations of an algebra structure are described by $H^2(A,A)$. In the complex case, it was the first cohomology group of the tangent bundle, this is the Kodaira--Spencer theorem.

Then for $\lambda^2$ you get
\[
dm_2+\frac{1}{2}[m_1,m_1]=0,
\]
and you read this as $[m_1,m_1]$ defines an element in $H^3(A,A)$, and this is the \emph{primary obstruction} because the condition to extending a first order deformation to one of second order is that this vanishes.

You can continue and analyze this classical deformation theory perturbatively. We're dealing with formal deformations because $\lambda$ was formal.

Gabriel is right that in principle one should think about leaving the commutative world for the Frobenius world in the spirit of John Terilla's paper, ``Quantizing deformation theory.'' I want to touch ``quantum deformation theory.'' You can think of this as continuing Terilla's paper, I could call this ``quantizing deformation theory II.'' I don't know the precise relation to Terilla.

I will concentrate on certain things generalizing notions from classical deformation theory. I want to talk about a geometric interpretation of classical deformation theory. If you start with a differential graded Lie algebra $\gGg$, then, well, I should make a remark to those people, for the metatheorem. If you find another, a way that a metatheorem works, if you find something that contradicts the metatheorem, then your approach is wrong. It is like a religion.

I believe that it has some meaning. If you start with a differential graded Lie algebra, I want to do the following. Start with the Lie algebra, set up the equation, and study the properties of the equation and its solutions. Now suppose I have a differential graded Lie algebra, I can think of it as identifying a coderivation $D$ on the graded commutative coalgebra, the symmetric algerba of the shifted vector space $\gGg[1]$, a coderivation which is a codifferential, this $D$ involves both $d$ and the bracket, $D=d+[\,,\,]$, seen as a map $S^{\le 2}\gGg[1]\to \gGg[1]$. Because this coalgebra is in a suitable sense cofree, a coderivation on it is determined by a map to the cogenerators. The compatibility with the bracket is nicely written down in $D^2=0$, and the degree is $1$. I want to say as well that $D$ preserves the coaugmentation which takes $\fieldk$ to $S^\cdot g[1]$, placing it in the zero component. $D$ preserves this means that $D(1)=0$.

You can think of the dual picture, $S(\gGg[1])$ as linearly dual to $S(\gGg^*[-1])$, at least when $\gGg$ is finite dimensional, just to think of this as easier to understand. Then this is the algebra of polynomial functions on $\gGg[1]$, which I can call $\Spec^*(S(\gGg[1]))$. Then if you have a graded commutative coalgebra, I'm thinking of a mysterious space which has this coalgebra as the coalgebra of distributions or generalized functions on it.

Then in these terms, the linear dual $D^*:S(\gGg^*[-1])\to S(\gGg^*[-1])$ is a derivation compatible with the dual of the coaugmentation, $S\gGg^*[-1]\to \fieldk$, and this augmentation defines a basepoint. When you have a morphism from your polynomial algebra to your ground field, that's the same as a $\fieldk$-point of our space. This is killing all positive powers to zero. Then the point defined by this map is the point $0$.

So what we have is a pointed differential graded formal manifold.

Now we get back to deformation theory. Really I haven't changed much, my differential graded Lie algebra was $\gGg$. My bracket and differential were encoded in $D$, but now I've switched the viewpoint. I'm thinking about this not as a vector space but as a manifold.

Why is this useful? Because if I look at the deformation theory associated to $(\gGg[1], D, 0)$, this is given by the Maurer--Cartan equation $\MC_{\gGg}=\{S\in \lambda \gGg^1[[\lambda]]|d_+\frac{1}{2}[S,S]=0\}$. Better, I can think of a functor on complete local commutative rings which assigns to $(R,m)$,  the Maurer--Cartan elements $\MC_{\gGg}(R)$ which are Maurer--Cartan elements in $\gGg^1\otimes m$. For example, take $R=\fieldk[[\lambda]]$ and $m=(\lambda)$. Then the theorem is that this is representable.
\begin{thm} (Quillen) 
This functor is represented by $(\gGg[1], D,0)$, i.e., there is a natural map
\[
\MC_{\gGg}(R)\xrightarrow{\sim} \Map((\Spec\ R,0,m),\gGg[1],D,0).
\]
Geometrically it means that the set of Maurer--Cartan solutions with values in $R$ is represented by pointed maps $\Spec\ R\to \gGg[1]$.
\end{thm}
The interpretation is that the functor of solutions is described by, describes such maps, so y're talking about curves in the space. Our diferential gradde manifold makes sense of the derived moduli space of this deformation problem. An $R$ deformation means you have such a map. 

The meaning is, this is also the derived moduli space, this is actually a dg manifold so it's derived, for the deformation problem defined by $\gGg$.
 
Going backwards from a codifferential $D$, a square zero coderivation on the coalgebra $S(\gGg[1])$, of degree $1$, and $D(1)=0$, this is an $L_\infty$ structure on the vector space $\gGg$ (by definition). There's also a natural isomorphism, well I should change some of my definitions, if we can think of this scheme as a generalization. This is something like $\Mor_{L_\infty}((m/m^2)[-1],\gGg)$.

But I should put this in quotes, this is wrong, this is not my $R$, without quotes it's $\Mor_{dg\coalg}(R^*,S(\gGg[1]))$.

Let me continue. Just to emphasize Jae-Suk's point that this theorem is trivial, let me give the idea of the proof. A morphism from $(R^*,0)$, a pointed dg coalgebra, to $(S(\gGg[1]),D)$, is the same as, by the cofreeness, a linear map $R^*\to (\gGg[1])^0=\gGg^1$, respecting $D$, or an element $S$ in $R\widehat{\otimes}\gGg^1$, the process of going back and forth between $R^*\to S\gGg[1]$ and $R^*\to \gGg[1]$ is exponentiation, and in this formulation respecting the basepoint puts it in $m\widehat{\otimes}\gGg^1$ and then the respecting the differential says that $D(e^S)=0$, which is the same as $dS+\frac{1}{2}[S,S]=0$. So this is saying that the Maurer--Cartan equation is satisfied.

This basically proves the theorem.

Now let me move on to the quantum, to quantum deformation theory, this is a sequel to the papers of Terilla and Terilla with Gabriel and Jae-Suk, Kontsevich--Katzarkov--Pantev, Costello.

Anyway, this starts with the \emph{quantum} master equation $dS+\hbar\Delta S + \frac{1}{2}\{S,S\}=0$ on elements in $S\in V[[\hbar]]^0$, and this can be defined in a differential graded $BV$ algebra, a differential graded commutative algebra, and has a second order differential of degree $-1$, called $\Delta$, that satisfies $[d,\Delta]=0$. Being a differential means that $\Delta^2=0$, it should also be a second order differential operator, so that $[[[\Delta,L_a],L_b],L_c]=0$. 

You can set this up assuming that $|\hbar|=2$, I haven't explained the antibracket. You can consider different versions of this with $\Delta$ and $\hbar$ of various degrees, with $\hbar$ of degree $0$ and $\Delta$ of degree $1$, or other variations.

The following thing, the antibracket, $\{S_1,S_2\}$ for $S_1$ and $S_2$ in $V$, is defined as 
\[(-1)^{|S_1|}\left(\Delta(S_1S_2)-\Delta(S_1)S_2-(-1)^{|S_1|}S_1\Delta(S_2)\right)\]
and this is a \emph{shifted} Lie bracket. This equation arises in quantum field theory, in homotopy probability theory, et cetera.

This equation, when $\hbar$ approaches zero, becomes $dS+\frac{1}{2}\{S,S\}=0$. Maybe here I should have taken $S\in \lambda V[[\hbar,\lambda]]^0$ or $m\otimes V[[\hbar]]^0$ where $m$ is a maximal ideal in a complete local ring. So now I recover the classical master equation and the quantum master equation is a quantization of classical deformation theory.

Now one can consider in this vein a functor on local rings $R\mapsto QME_V(R)=\{S\in m\otimes V[[\hbar]]|dS+\hbar\Delta S+\frac{1}{2}\{S,S\}=0\}$. Now we're taking solutions to the quantum master equation.

We want a metatheorem, something like:
\begin{quote} The quantum deformation theory of a quantum math object is described by a suitable differential graded BV algebra.
\end{quote}
The proof is extremely simple, it's vacuously true. I don't know any quantum deformation problem so it's vacuously true. On the other hand, I do know at least one quantum deformation problem, and that's the following. Say you have a differential graded Lie algebra, $\gGg$, to it we can associate the coalgebra we talked about before, but let me shift it by $S(\gGg[-1])$, and this becomes a differential graded BV algebra. This is actually a differential graded $BV$ algebra. The differential is the induced differential from $\gGg$, with internal $d$, the product is exterior product, and the BV operator is the Chevalley--Eilenberg differential, and if you set up the classical Maurer--Cartan equation, if $\gGg$ is not graded but just a Lie algebra, you can still do this, but the internal differential is zero, and you still have the Chevalley--Eilenberg operator, and then the Maurer--Cartan equantion is $\{S,S\}=0$ for $S\in S^2(\gGg[-1])$, which is known as the classical Yang--Baxter equation, the solution is an $r$-matrix. The quantum version, if $\gGg$ is differential graded Lie, then the quantum master equation is, $dS+\hbar\Delta S+\frac{1}{2}\{S,S\}=0$, even the classical master equation, $dS+\frac{1}{2}\{S,S\}=0$, this is, $S$ is an $r$-matrix in the dg setting, now deformed (we studied this with my student Bashkirov) in the $L_\infty$ setting), this is some kind of quantum $r$-matrix. We can consider other geometric settings, multivector fields on a manifold, say, and in the presence of, even though it has a bracket and a differential, well, it doesn't have a differential, but the BV operator exists if you have a volume form, but then I don't know what those quantum deformation theories describe.

Now moving on, I want this metatheorem but we don't have enough examples to talk about. We can set up this functor of solutions to the quantum master equation, the geometric interpretation, Jae-Suk told me a few minutes ago that ``quantum deformation'' was coined by him and---

\begin{prop}
The functor $QME_V$ is naturally isomorphic to the functor $\Map_{dg\Man}/\Spec \fieldk[[\hbar]]((\Spec R[[\hbar]],0,m),(V,[[\hbar]],\widehat{\Delta},0))$, where $\widehat{\Delta}=d\hbar\Delta:V[[\hbar]]\to V[[\hbar]]$.
\end{prop}
With this $\widehat{\Delta}$, then $V[[\hbar]]$ is a differential graded Lie algebra, because both $d$ and $\Delta$ are derivations of the bracket and $\widehat{\Delta}$ squares to zero.

So except that now we have extended our constants, we're in the same situation. In another words, these are morphisms of dg coalgebras from $R^*$ to the symmetric algebra on an appropriately shifted $V$ (over $\fieldk[[\hbar]]$). 

This is then not so interesting but what is interesting is the representability question, which goes as follows.

The functor $QME_V(R)$ can be represented, the category on the right hand side, $V$ was a differential graded BV algebra. The replacement of $V$ was an $L_\infty$ algebra. Here we have a differential graded BV algebra. Here we are using $\Mor_{BV_\infty}(R^*,V)$, or perhaps, $\Maps_{BV_\infty-\Man}(\Spec V, \Spec R^*)$. The problem is that then we should make sense out of those notions. So $R^*$, if you take $R$ just a ring, a commutative ring, then $R^*$ will need extra structure to be called a $BV_\infty$ algebra. So nothing is defined.

When I started the geometric interpretation, what do I mean by a BV manifold? When I start with a dg $BV$ algebra, I want to interpret this as functions on a dg BV manifold called ``$\Spec V$,'' so I look for a scheme with this as an algebra of functions, this is a fixed second order differential operator which squares to zero, on functions on this space. This might be a right geometric interpretation here, it should be some interpretation, a representability theorem in this context, representing this functor by $V$ or corepresenting by $\Spec\ V$. 

The right hand side doesn't make sense since $R^*$ is not an algebra and $\Spec R^*$ is not defined.

This is another argument for studying (commutative) $BV_\infty$ algebras and morphisms.

This study was done by Cieliebak--Latschev and Cieliebak--Fukaya--Latschev, and also Bashkirov--V. and also Markl--V.

So C.--L. considered $BV_\infty$ morphisms from $(S(U),\Delta)\to (W,\Delta')$. So the difference is that $S(U)$ has underlying algebra a free graded commutative algebra. They didn't have a category. Now Cieliebak--Fukaya--Latschev considered maps $(S(U),\Delta)\to (S(U'),\Delta')$, where by definition an $IBL_\infty$ structure on $U$ is a $BV_\infty$ operator on $S(U)$. Then in this case, with Markl, we looked at $MV$-morphism, $V\to V'$, between $MV$-algebras, which included all sorts of things which were here before, including $IBL_\infty$ algebras and all $BV_\infty$ algebras.

In particular, one can talk about (in the $MV$ category) morphisms of the following kind:
\[
(R^*,0)\to (V,\Delta).
\]
Then if you start with, if $R$ is a complete local ring, then $R^*$ is a coalgebra (naturally) and then it has an $MV$-algebra structure given by the fact that $R^*$ is an algebra and a coalgebra (everything is commutative), where the algebra is trivial, where multiplication, elements in $m^*$ multiply to zero. The remaining part, the field part, has a nonzero multiplication. This is an MV algebra, basically, and there is an augmentation and coaugmentation, so the augmentation for the algebra structure is a counit and the coaugmentation is a unit for the algebra structure. There should also be a $\Delta$, a linear operator of appropriate degree that kills $1$, so on.

With this data, this is roughly speaking an MV algebra, still not a complete definition, and one can take, given $(R^*,0)$, you can define this MV algebra structure on $R^*$ and given a $BV$ algebra or (commutative) $BV_\infty$ algebra, you also get an MV algebra on $V[[\hbar]]$, with the given algebra structure and you can define a coalgebra, a trivial comultiplication on this, and then the $MV$-morphisms are pretty general things, these are roughly speaking, $\phi:(V,\Delta) \to (V',\Delta')$, is practically anything that commutes with the differentials, $\Delta'\phi=\phi\Delta$ and $\Phi(1)=1$. 

The theorem that we proved with Martin, in rough version since we haven't given complete definitions (I forgot to mention that we named $MV$ algebras after Markl and Voronov), is
\begin{thm}(Markl--V.) Quantum master equation solutions for $V$ a dg BV algebra (there is a version for commutative $BV_\infty$ algebras.)
\begin{gather*}
QME_{V((\hbar))}(R)\cong\\
\Mor_{MV}/\fieldk((\hbar))(R^*((\hbar)),V((\hbar)))
\\
\cong \Mor_{MV\Man}/D^\circ(\Spec(V)\times \Spec(R)\times D^\circ,\Spec (R^*)\times D^\circ)
\end{gather*}
\end{thm}
Here an $MV$ manifold $\Spec V$ is a space whose algebra of functions is $V$. Since it has other structure, the distributions on $V$ should also have multiplication. Its a formal space on which distributions have multiplication. On the other hand, you can consider an $MV$ algebra $V$, you can associate an $MV$-manifold $\Spec V$ and also an $MV$-manifold $\Spec^* V$, the dual $MV$ manifold, and this guy is, functions on it, distributions on it are the coalgebra $V$.

In this sense, continuing the theorem, this is also
\[
\Mor_{\text{dual }MV\Man}/D^\circ(D^\circ\times \Spec R\to \Spec^* V((\hbar)))
\]
and this is like the old picture [pictures].

This theorem is related to the quantum master equation being related to $\Delta(e^{S/\hbar})=0$. This is why we needed to localize with respect to $\hbar$. The proof comes from the observation about $dS+\hbar\Delta S+\frac{1}{2}\{S,S\}=0$. 

This is now the analogue of Quillen's theorem in this context. I'm talking about the same category but viewing, I'm embedding my category of complete local rings into this category of MV spaces or MV algebras.
\section{May 18: Martin Markl: The $MV$ formalism for $IBL_\infty$ and $BV_\infty$ algebras}
This is joint with Sasha but based on others' work, Cieliebak--Fukaya--Latschev, Gabriel, others.

This was all based on trying to understand $IBL_\infty$ algebras. The sources are so heavily technical and appalling that I wanted to make it easier. I visited Sasha Voronov and he was working on similar things from a different perspective. We were able to simplify the calculations and predict some other things also.

Let me as motivation to indicate what $IBL_\infty$ algebras are, this is a structure that arises in string field theory, and basically what I want, I wish to any isomorphism class of Riemann surface with two sets of holes, input holes and output holes, of genus $g$, I wish to associate to it a linear operator $\ell^g_{m,n}:S^m(U)\to S^n(U)$, for some Hilbert space $U$. Definitely I wish these things to satisfy a condition that I will indicate roughly. To formulate, basically what I wish to say is that if I compose the surfaces in all possible ways, the sum will be zero. I'll be more precise. Given this sequence of linear maps (I should have said these are degree $1$) I associate a generating function $\Delta$ on $S(U)[[h]]$, this is the completed tensor product $S(U)\widehat{\otimes}\fieldk[[h]]$, and instead of giving a precise definition, I'll give you an idea, if I take $\Delta$ I'll apply it to a commutative bunch of elements of $U$, and you do the summation [pictures], I wish to visualize $\ell^{g}_{m,n}$ as a box with $m$ inputs and $n$ outputs [picture] and this will be the summation of all possible ways of how to insert $m$ of the commutative variables into the box, sum of all boxes and all possible insertions.  

I want $\Delta^2$ to be zero, I have these boxes, and then I apply $\Delta$ again, and then when I apply $\Delta$ again, I get $n$ boxes that are kind of moving, this must be zero. Everything is kind of commutative or skew-commutative. If it happens that these boxes are independent, then you'll get cancellation [pictures]. The only things that matter are the following compositions: [pictures].

Let's take some particular cases. If I assume that the only nontrivial members of this family are $\ell^0_{m,1}:S^{m}U\to U$, you get an $L_\infty$ algebra on $U$.

You can think of $IBL_\infty$ algebras as algebras over a resolution of a prop for involutive Lie bialgebras. What is this, from this point of view, this is a stongly homotopy version of the Lie bialgebra, along with the involutive condition that the comultiplication followed by the bracket is zero.

In practice we construct these in physics using structures on the surface, and we should be able to prove that this doesn't depend on structural choices. So we should have a notion of a map or isomorphism of these algebras. The notion of morphism given by Cieliebak and Latschev was much worse than the definition for the algebras, let me give a description of the definition.
\begin{defi}
Let $(S(U'),\Delta')$ and $(S(U''), \Delta'')$ be $IBL_\infty$ structures. Then a morphism $f$ is the same as a map $S(U')\to S(U'')$, a map of underlying spaces, such that $\exp(f)\circ \Delta'=\Delta''\circ\exp(f)$. 
\end{defi}
I'll explain what $\exp$ is in a minute. Since $f$ is not an endomorphism, you cannot say $\exp$ is just the ordinary exponential. I'll say that $\exp(f)$ is [pictures].

Now they needed to define composition, and they prove by very brutal force, the proof occupies many pages, that the morphisms compose, if $f:S(U')\to S(U'')$ and $g:S(U'')\to S(U''')$, there is, I can take $\exp(g)\circ \exp(f)$, and there exists a unique $h$ such that this composition is $\exp(h)$. Then they define $g \diamond f\coloneqq h$.

We realized that this exponential is actually an exponential in an appropriate commutative associative algebra, which I'll actually define very soon.

Then we can actually define $h$ as the logarithm of $\exp(g)\circ \exp(f)$, and I have to say that they didn't use the logarithm. If you do this for the particle case, you'll get precisely, I'll tell you what you get. [pictures].

Maybe I'll say what the results are, we realized that everything here lives in a bigger category which is trivial in a certain sense where these are real exponentiation in a commutative associative algebra. We were able to use this to get nice transfer for solutions to a Maurer--Cartan equation.

There are other interesting subcategories, like $L_\infty$ algebras, loop (quantum) $L_\infty$ algebras, other things.

Let's do some definitions. As Gabriel observed, I need assumptions so my power series converges.
\begin{defi}
Let $R$ be a complete local Noetherian ring with residue field $\fieldk$ of characteristic zero and $m$ the unique maximal ideal. In all examples, $R=\fieldk[[h]]$ or $R=\fieldk$ (or it could be $\fieldk[h]/(h^{n+1})$).

An $MV$-algebra is a $\fieldk$-vector space $V$ equipped with
\begin{enumerate}
\item a commutative associative multiplication $\mu$, 
\item a cocommutative conilpotent coassociative comultiplication $\delta$, 
\item a degree $1$ continuous map $\Delta:V\widehat{\otimes} R\to V\widehat{\otimes} R$, (where I define this as $\lim (V\otimes R)/m^k$) such that $\Delta^2=0$ and $\Delta(1)=0$, such that
\item The unit $\eta:\fieldk\to V$ is a coaugmentation for $\delta$ and the counit $\epsilon$ is an augmentation for $\mu$.
\end{enumerate}
\end{defi}
Notice there is no compatibility between product and coproduct except for the unit and counit compatibility.

What about examples. Any $\fieldk$-algebra $A$, augmented, is $MV$ over $\fieldk$ if you equip $A$ with a trivial coalgebra structure, and similarly a coaugmented coalgebra is an $MV$ algebra with a trivial product.

Should I repeat what a differential operator of order $\le k$ is? I probably don't need to repeat it. So a commutative $BV_\infty$ algebra is, I'm repeating what Sasha said, it's a commutative algebra, along with $\fieldk$-linear degree $1$ operator $\sum \Delta_k h^{k-1}$ that has $\Delta_k$ a differential operator of order $\le k$. Then an $IBL_\infty$ algebra is a $BV_\infty$ algebra with $A=S(U)$. Here I take the standard comultiplication of polynomial algebra. Then both of these are examples of $MV$-algebras.

Now I'm coming to morphisms to make it a category. Let $V'=(V',\mu',\delta',\Delta')$ and $(V'',\mu'',\delta'',\Delta'')$ be two $MV$ algebras over $R$. Then it is well-known, at least in some circles, that there is a commutative associative unital augmented algebra structure $*$ on linear maps $V'\to V''$, called convolution product, which is given by the following formula 
\[
f*g= \mu''(f\otimes g)\delta'.
\]
It can be proven that this is commutative associative unital augmented structure. The unit is $e=\eta''\circ \epsilon'$ and there is a similar form for augmentation that I'm not going to use. [pictures]

Very good, so everyone is happy so far? One may observe that this structure extends to the space of $R$-linear maps from $V'\widehat\otimes R$ to $V''\widehat\otimes R$ (of course I mean continuous maps), which I'll denote $\Lin_R(V',V'')$. Finally I'll denote
\[
\Lin^0_R(V',V'')=\{f\in \Lin_R(V',V'')| \im (f\circ \eta')\subset V''\widehat{\otimes}m\}
\]
\begin{prop}
All power series in $\Lin^0_R(V,V'')$ converge.
\end{prop}
To prove it, you need to combine the assumption with completion of $R$ and conilpotence and then you can do it.

In particular I know what $\exp(f)$ is, probably from kindergarten, and what $\log(e+g)$ is 
\[
g-\frac{g^{*2}}{2}+\frac{g^{*3}}+\cdots
\]
I'll give one precise formula that you can do by yourself, what does this do to $f\in \Lin^0_R$ from $S(U)$ into $(B,\mu,1)$? We say that 
\[
\exp(f)(u_1,\ldots, u_n)=
\sum\frac{1}{k!}\frac{epsilon(u)}{a_1!\cdots a_k!}f(u_{\sigma(1)},\ldots u_{\sigma(a_1)})\cdots f_{u_{\sigma(n-a_k+1)},\cdots u_{\sigma(n)}}
\]
where $a_1+\cdots +a_n=n$ and $\sigma$ runs over partitions. You may be annoyed by the appearance of the combinatorial factors, but they're inessential, we can just sum over shuffles to get rid of them.

So now $MV_R$ has objects $MV$-algebras over $R$ and morphisms linear maps in $\Lin_R(V',V'')$ such that $\Delta''\circ \exp(f)=\exp(f)\circ \Delta'$, and composition is given by the formula $g\diamond f=\log (\exp(g)\circ \exp(f))$.

Now convergence is taken care of, there is something slightly frustrating.
\begin{thm}
The category $MV_R$ is equivalent to the category of pointed cochain complexes.
\end{thm}
When we realized this, this equivalence is very easy, for algebras you forget everything except $\eta$ and $\Delta$ and at the level of morphisms you take the exponential. When we realized this Sasha got very depressed. I'm used to the fact that lots of things in life are depressing.

Why does this not destroy everything, because subcategories are not necessarily subcategories of chain complexes, for instance $L_\infty$ algebras, $IBL_\infty$ algebras, and so on.

We realized if we are supposed to increase the chances the paper will be published, we should try to prove something nontrivial, what is this nontrivial thing? 

I'll define a subcategory of $IBL_\infty$ algebras, a non-full subcategory of $MV_{k[[\hbar]]}$. I said what the objects are, $A=S(U)$. The morphisms are linear maps from $S(U')[[\hbar]]\to S(U'')[[\hbar]]$ and the following is required, that $f$ decomposes as $f^{(1)}+\hbar f^{(2)}+\cdots$ with the property that $\ker f^{(k)}\supset \bigoplus_{n>k}S^n(U')$, so that $f^{(k)}$ is trivial on everything of size greater than $k$. So this should have $k+1$ or fewer inputs. 

You need to prove that this is a subcategory, so the composition of maps with this property have this property. We were not able to find any statement in the literature, so we proved it, and the proof is nice, but let me not say it.

I may stop here or tell you in seven minutes something like the Quillen result, maybe seven and a half.

I'm sorry I'm with my favorite $R$, I have $MV_R$-algebra $V$, as before, and so the quantum master equation of $V$ over $R$ is a set of solutions of $\Delta e^S=0$ where $S$ is of degree $0$ in $V\widehat\otimes m$. Again convergence is guaranteed by completeness of $R$. This is an actual exponential. I have to say there are some other versions, where say you assume that $S$ is of the form $\tilde{S}/\hbar$, and you can interpret this in terms of the $\varphi^k$ that are popular here.
\begin{thm}
The set of solutions to the quantum master equation in $V$, the morphisms in the category of $MV$ algebras from $\fieldk$ to $V$ is isomorphic to quantum master equation solutions in $V$.
\end{thm}
The proof is simple, if you have $s:\fieldk\to V$, then $S\coloneqq s(1)$ is a solution to the master equation. Slightly more involved is to prove this does what you want.

The advantage is that this gives a transfer formula for solutions to the quantum master equation, which were not clear in Cieliebak--Latschev. If $S\in \QME_R(V')$ and $f\in MV_R(V',V'')$, then I can represent $\fieldk\xrightarrow{s}V'\xrightarrow{f}V''$ and composing in the diamond composition, I define $f_!(S)$ as $(f\diamond s)(1)$, so I can use the theorem again to translate this back and get $f_!(S)=\log(\exp(f)(e^S))$. They somehow constructed $f_!$ but gave no description of it, some element that satisfies a crazy condition.

\section{May 23: Sergei Merkulov: Props of ribbon graphs, involutive Lie bialgebras and moduli spaces of curves}

Thank you very much for the invitation, it's a great pleasure to be in this country and at this new institute. This project was started years ago and is based on joint work with Thomas Willwacher. This project was started after Thomas produced his famous paper of 2010 where he was able to compute some cohomology of the Kontsevich graph complex. One striking result was that $H^0(GC_2)$ (definitions later) is the Grothendieck Teichm\"uller Lie algebra. The group is a transcendental object, usually requires integrals, transcendental methods. To find out how this acts, you should develop quantum field theory, integrals of differential forms with singularities, so on. 

So we asked if it was possible to do semothing with the ribbon graph complex. We noticed at that time a connection to involutive Lie bialgebras. This result motivated why we studied Koszulness of this properad. These are popular in string topology. Our motivation came from the story I'll explain, viewing them as a map into something. Ribbon graphs turn out to be a prop, this is another surprise. There are operations that make it a properad. Vertices are inputs, boundaries are outputs. This lets us glue ribbon graphs.

Finally $\Mm_{g,n}$. We understood this prop, we saw some connections. We needed a rigorous homotopy theory. It took years to see this, we also needed a deformation theory for these. In fact, Kontsevich graph complexes can be defined as deformation complexes of even and odd Lie bialgebras.

We will able to get cohomology of graph complexes into $\Mm_{g,n}$, and we could show the cohomology of this thing is a differential graded Lie algebra with interesting differential. We also found more things which computes in the trivial case the cohomology of $\Mm_{g,n}$. There is a quantization from Lie bialgebras to Hopf algebras, and we don't know what the non-commutative analogue of our constructions compute.

So what is $\Mm_{g,n}$? A point, this is a complex orbifold, a point in it is a Riemann surface with $n$ marked different points. [picture] This is taken up to biholomorphism which keeps points on their places. This is what this moduli space classifies. Classes of such equivalences of compact Riemann surfaces. Penner found a nice combinatorial model for this theory. Essentially this model is based on the so-called Strebel differentials. Given such a surface $S$, a Strebel differential on $S$, choosing a collection of non-negative numbers, one for each puncture, is a meromorphic section $\omega$ in $(\Omega^1_S)^{\otimes 2}$ with poles only at the marked points such that the residue around every point is $-(\frac{c_i}{2\pi})^2(\frac{dz}{z})^2$.

When we work with a quadratic differential, tkae $S=\cC$ and $\omega=(dz)^2$, you recover horizontal and vertical lines by restricting to $y$-constant and $x$-constant lines, where we get $dx^2$ and $dy^2$. We define horizontal and vertical curves in terms of $z(t)$ by saying that $\omega$ restricted to a horizontal curve gives $f(t)dt^2$ with $f(t)>0$ (similarly vertical). Strebel showed that there is a unique such differential, which gives a ribbon graph immediately. Vertices are zeros of $\omega$ and edges are noncompact horizontal curves (ones which are not circles). [pictures].

We can consider now a complex, in the original Penner story because we have Strebel differentials, we have a metric, we have a positive definite metric on the ribbon graphs, we have lengths, so strictly speaking, Penner used metric ribbon graphs. What happens if we forget about this number? Still it's a nice complex.

Kontsevich observed, let $RGC^{\ge 3}$ be the vector space spanned by ribbon graphs. One can make it into a complex in a very simple way. You can contract an edge or split a vertex to make a new edge. We use the differential dual to edge contraction. If you like, it's a substitution, if I have a vertex, I substitute an edge in for each vertex, preserving the cyclic structure in all possible ways.

\begin{lemma}
\[H^*(RGC^{\ge 3})=\prod H^*(\Mm_{g,n},\qQ)\otimes_{\sS_n}\mathbf{1}_n\]
\end{lemma}

There is another one which Kontsevich missed but we found. There are two ribbon graph complexes, very different, I'll tell you about it. Now the new thing, I'll put the index $RGC^{\ge 3}_{\odd}$. Then we recover, this is the Kontsevich ribbon graph complex. The other one which emerged in our story, we first found the even one and then realized if you look at odd Lie bialgebras you get this one.

The new thing, there is $RGC^{\ge 3}_{\even}$, what is the difference? In this case, you start with a ribbon graph with all vertices numbered and boundaries numbered. To go to homology, you symmetrize boundary and skew-symmetrize vertices or the other way around. For Lie bialgebras, if they're both skew-symmetric, well, anyway you have an even ribbon graph complex whose cohomology computes
\[
\prod_{g,n} H^*(\Mm_{g,n,\qQ})\otimes_{S_n}.
\]
Taking only trivalent ribbon graphs is not really a good idea, we need at least bivalent, these are first of all easy to compute. I consider $\hH_{\odd;\even }\coloneqq H^*(RGC^{\ge 2}_{\odd;\even})$. First of all, this is not a complication of the story, this is $H^*(RGC^{\ge 3}_{\odd;\even})\oplus$ a piece $\sum_{k\ge 0} \qQ[-1-4k]$ in the even case and $\sum_{k\ge 0} [-2-4k]$ in the odd case. Why this is good and clever, both the odd and even, both the even and odd are Lie algebras when we complete it. Moreover, they have a differential, which, it makes sense only the completed thing, one of the differentials changes $n$ by any number of punctures. What it computes we do not know, but we know it's highly nontrivial and contains some cohomology classes from graph complexes.

\begin{thm}
There is a prop of ribbon graphs $\Rr\Gra_{\even; \odd}$ which is a collection $\rR\Gra(m,n)$ which is spanned by ribbon graphs with $m$ boundaries and $n$ vertices (plus orientation). 
\end{thm}
In the even case, orientation is an ordering of edges up to even permutation. You chose a direction on each edge in the odd case. The vertices and boundaries are numbered.

Now I have to go to Lie bialgebras. Let's have a break.

There is a purely combinatorial rule that explains how to compose boundaries of one ribbon graph with the vertices of another, this is a sum of ribbon graphs. What is important is that you can map Lie bialgebras into this prop. 
\begin{defi}
A \emph{Lie $(c,d)$-bialgebra} is
\begin{enumerate}
\item  a graded vector space $V$ with
\item a degree $d-1$ Lie bracket (skew-symmetric and
\item a degree $c-1$ Lie cobracket.
\end{enumerate}
[pictures]

These should satisfy graded skew-symmetry, graded Jacobi, and the (Drinfeld) Lie bialgebra relation.
\end{defi}
The standard case $d=1=c$ is due to Drinfeld. Its minimal resolution, it's highly nontrivial, the proof that this even case, when both $c$ and $d$ are odd we call it even. The minimal resolution was constructed first by Kontsevich in a letter to Martin Markl, and then Markl and Voronov wrote a paper which proved all of Maxim's claims. I have this letter, so Maxim kind of proved it. The basic idea is to introduce path filtration, Maxim counted the number of paths in graphs that connect inputs to outputs. The zeroth terms make this computable. There was already a candidate as in the involutive case.  This is the case when both generators are skew symmtric or both symmetric. The first odd case $d=0$, $c=1$ was done by me in 2005 or 2004, I was just training myself, I have the education of theoretical physics, I learned all this stuff and then switched to mathematics, and so when I looked at props and operads, algebra didn't kind of arouse me, this case was what was of interest to me because I could see geometry. The minimal resolution here when $c$ and $d$ have opposite parity, the proof that it's Koszul is elementary, what is less evident, if you look at representations, these are just Poisson structures, and formal Poisson structures all arise in this way. These are genes for Poisson structures.

Now we come back to ribbon graph complexes and this prop.
\begin{lemma}
There is a map of properads $\Lieb_{\even}\to\Rr\Gra_{\even}$ which sends the Lie generator to [interval] and the cobracket [loop]. The Drinfeld relation requires checking. In fact, this map factors through involutive Lie bialgebras.
\end{lemma}
One can notice, in other words, that this graph will be mapped, using our rules for composition, to the ribbon graph [picture]. As I told you because of orientations, this graph is identically zero, so in fact we have, our map factors thorugh a map $\Lieb_{\even}^{\diamondsuit}\to \Rr\Gra_{\even}$.

This is why we were interested in the theory of involutive things.

In the odd case it's a paper by myself, Thomas, and Khoroshkin, you might need to consider [picture] this graph. The conjecture is that in the odd case it factors through the version where this is zero.

Now I want to return to the story I started with, with ribbon graph complexes.

Consider a map $s^*:\Lieb_{\even;\odd}\to \rR\Gra_{\even;\odd}$ which is nontirvial only on the Lie generator. This is still a map of properads. The theorem is
\begin{thm}
The ribbon graph complex $RGC_{\even;\odd}$ is the deformation complex of $\Lieb_{\even;\odd}\to \Rr\Gra_{\even;\odd}$.
\end{thm}
The first (new) corollary is that these things have Lie algebra structures.
\begin{cor}
$RGC_{\even;\odd}$ are dg Lie algebras. The same is true (without differentials) for their cohomologies.
\end{cor}
The structure is nontrivial, related to our prop compositions, the composition of poles with zeroes of the structure.

Now I can say more, there will be $L_\infty$ in a moment. The standard Penner graph complex, I can put a parameter $\lambda$ on my loop for the cobracket. The standard ribbon graph complex is the standard case of a family of maps. What is the deformation complex of $s$?

\begin{thm}
The deformations of $\Lieb_{\even}\xrightarrow{s}\Rr\Gra_{\even}$ is 
\[
(RGC<_{\even},[,],\Delta_1)
\]
with differential induced by the loop.
\end{thm}
This goes over genus and $n$. The $\delta$ preserves genus and $n$. The $\Delta$ adds a boundary or kills it in the dual.

Now $n$ becomes a parameter.

This ribbon graph complex then becomes highly nontrivial. Then there is a map from the even graph complex to here, injective on infinitely many cohomology classes. 

I didn't define $GC_{\even}$, which is the deformations of Lie bialgebras to themselves. I do something on the input, this induces something, I can compose with $s$, this is why I can have this map.

One can write a spectral sequence where the first page gives Kontsevich's theorem (and ours). So the totality of moduli spaces has this differential as well and it's an open question what is the homology of this thing.

Another thing is that Lie bialgebras can be quantized, So we have a quantized version of this depending on an associator, and what this computes we have no idea. This is a highly non-trivial quantization. Therefore, it's another open direction in this story.

What happens in the odd case? In the odd case, the situation is the following.

I want a map $\Lieb_{\odd}\to \Rr\Gra_{\odd}$. Now the difference between the even and odd case, we don't see any way to make the cobracket nontrivial, for $d=1$, there is nothing of the right degree. Before we deformed into the loop and so on. Instead you can add these graphs [pictures] to the bracket.

I can study now what is the deformation complex of this map in the odd case. As a vector space this is the Penner space with the odd orientation. Now the thing increases the number of boundaries by an arbitrary number of terms. 

In the even case I got something that factors through involutive bialgebras. 
\begin{thm}
You can consider $\Def(\Lieb^\diamondsuit\to \Rr\Gra)$ and this gives $(RGC_{\even},d+\Delta_1+\Delta_2)$
\end{thm}
We couldn't prove it, but the conjecture is that this complex is acyclic. This $\Delta_2$ is very funny. What would this mean? It means that in fact the moduli space classes come in pairs which kill each other with respect to $\Delta_2$. It would be a great insight into this world, and we cut this out of the paper.

Anyway, what is $\Delta_3$ doing? There are many nontrivial classes, all the loops survive, and the conjecture is that everything survives. Ther's an involutive version for odd, the conjecture is that this factors through that minimal model. If it's, if the conjecture is true, then we have of course $\Delta_4$, which again makes these things (conjecturally) in pairs. As you see, in this classical story of algebraic geometry, we see all of this structure, new algebraic structure.

Let me formulate the final result.
\begin{thm}
The maps of graph complexes $\alpha^{\even;\odd}:GC_{\even;\odd}\to\Rr\Gra,\delta+\Delta_{1;3}$ are injective on infinitely many classes.
\end{thm}
The conjecture is that it's an injection.

Finally, let me give the description of $GC_{\even;\odd}$, this is spanned by ordinary graphs (with orientation). The differential is contraction (or splitting). It's a very hard theorem to prove, Kontsevich invented these things but never computed anything.

Both even and odd Lie bialgebras (this is true for odd as well), in one case you have ordering of edges but no direction and in the other you have directions but no ordering.

I have some time left but my talk is over.
\section{May 30: Jae-Suk Park: Non-commutative quantum field theory and Chen's iterated path integrals}

Iterated integrals are a natural generalization of line integrals that we learn in calculus class, and it has many applications, in homotopy theory, number theory (multiple zeta values), and in algebraic geometry, natural mixed Hodge structure, and diophantine geometry. There are also applications to string topology by our visitor, Professor Merkulov, a very nice updated summary, and I'll use the tool of $C_\infty$ algebras to deal with rational homotopy type.

I'll only spend like five minutes on iterated integrals. So $M$ is a manifold, and let's consider the path space, $P(M,*)$, I'll do the based space for simplicity, so these are closed loops, piecewise smooth, and $\omega$ is a real-valued one-form. Then if we have a path $\alpha$ from $[0,1]$ to our manifold, identified with its image, we can pull back the form $\omega$, so that $\alpha^*(\omega)=f(t)dt$, and we integrate $\int_0^1 f(t)dt$. This is $\int_\alpha \omega$. This is valued in $\rR$, and it doesn't depend on the parameterization of the path is the first thing we notice. If $d\omega=0$, then we know that this only depends on the homotopy type of $\alpha$. So this is a homotopy functional.

This is simple and very nice, and if I compose two paths, this is $\int_{\alpha\circ \beta}\omega=\int_\alpha \omega + \int_\beta \omega = \int_{\beta\circ \alpha} \omega$.

Chen generalized this, let $\omega_1,\ldots, \omega_n$ be in $\Omega^1(M)$, and $\gamma$ is a path and we define
\[
\int_\gamma \omega_1\cdots \omega_r=\int_{0\le t_1\le \cdots \le t_{r}\le 1} f_1(t_1)\cdots f_r(t_r)dt_1\cdots dt_r
\]
where
\[
\gamma^*\omega_i=f_i(t)dt.
\]
For instance 
\[
\int_\gamma \omega_1\omega_2=\int_0^1\left(\int_0^{t_2}f_1(t_1)dt_1\right)f_2(t_2)dt_2.
\]
When are iterated integrals homotopy functionals? We can easily see that in general, $\int_\gamma \omega_1\omega_2$ is not. The integrand should be the pullback of a closed $1$-form. If $\omega_1$ and $\omega_2$ are closed, and if $d\omega_{12}+\omega_1\wedge \omega_2=0$, then we can consider $\int_\gamma (\omega_1\omega_2+\omega_{12})$ and that turns out to be a homotopy functional.

What about with three terms? There $\int_\gamma(\omega_1\omega_2\omega_3+\omega_{12}\omega_3+\omega_1\omega_{23}+\omega_{123})$ is a homotopy functional, where
\begin{align*}
d\omega_1&=0\\
d\omega_2&=0\\
d\omega_3&=0\\
d\omega_{12}+\omega_1\wedge\omega_2&=0\\
d\omega_{23}+\omega_2\wedge \omega_3&=0\\
d\omega_{123}+\omega_1\wedge \omega_{23}+\omega_{12}+\omega_3&=0\\
\end{align*}
Generally, if $\omega_i,\ldots, \omega_m$ are in $\Omega^1(M)$ and $t^1,\ldots, t^m$ are \emph{dual variables}, then consider $\rR\langle\langle t^1,\ldots, t^n\rangle\rangle\otimes \Omega^1(M)$, and we want to find $\Gamma=\Gamma_1+\Gamma_2+\cdots$ where $\Gamma_n=t^{\alpha_1}\cdots t^{\alpha_n}\omega_{\alpha_1\cdots\alpha_n}$. Then we are looking for solutions to $d\Gamma+\Gamma\cdot\Gamma=0$ (this is not always possible). Then this decomposes into
\begin{align*}
d\Gamma_1&=0\\
d\Gamma_2+\Gamma_1\Gamma_1&=0\\
d\Gamma_3+\Gamma_2\Gamma_1+\Gamma_1\Gamma_2&=0
\end{align*}
So this tells you that the space $\rR\langle\langle t^1,\ldots, t^n\rangle\rangle\otimes \Omega^1(M)$ along with $d=\id\otimes d$ and the product, this is a differential graded algebra, and this is nothing but the Maurer--Cartan equation in this algebra.

Then $T_\alpha=1+\int_\alpha T+\int_\alpha T^2+\cdots$, and if you collect these in word lengths in $T$, this looks like
\begin{align*}
1+\int_\alpha&+\Gamma_1
\\&+ \int_\alpha \Gamma_1\Gamma_1+\Gamma_2
\\&+\int_\alpha \Gamma_1\Gamma_1\Gamma_1+\Gamma_2\Gamma_1+\Gamma_1\Gamma_2+\Gamma_3
\end{align*}
This is the generating function for the kind of things we had in the equations before. 

I prefer $\Pi_1=\Gamma_1$ and $\Pi_2=\Gamma_1\Gamma_1+\Gamma_2$ and then you get $\Pi_1=t^\alpha\int \omega_\alpha$, $\Pi_2=t^\alpha t^\beta \otimes \int \omega_\alpha\omega_\beta+\omega_{\alpha\beta}$. This is called transport, and $T_{\alpha\circ\beta}=T_\alpha T_\beta$.

You can check that this has the following properties:
\begin{align*}
\int_{\alpha,\beta}\omega_1\cdots \omega_m&=\int_\alpha \omega_1\omega_2\cdots \omega_n
\\&+\int_\alpha \omega_1\int_\beta \omega_2\cdots \omega_n
\\&+\cdots
\\&+\int_\alpha \omega_1\cdots\omega_{n-1}\int_\beta \omega_n
\\&+\int_\beta \omega_1\cdots \omega_n
\end{align*}

The iterated integral has the following property, discovered by the Korean mathematician Rimhak Ree:
\[
\int_\gamma \omega_1\cdots \omega_r \int_\gamma \omega_{r+1}\cdots \omega_{r+s}
=\sum_{\sigma\in \sh(r,s)}\int_\gamma  \omega_{\sigma(1)}\cdots \omega_{\sigma(r+s)}
\]
where $\sigma\in\sh(r,s)$ if $\sigma(1)<\sigma(2)<\cdots< \sigma(r)$ and $\sigma(r+1)<\sigma(r+2)<\cdots<\sigma(r+s)$.

So for $\sh(1,1)$ this looks like $\omega_1\omega_2+\omega_2\omega_1$; for $\sh(2,1)$ this looks like $\omega_1\omega_2\omega_3+\omega_1\omega_3\omega_2+\omega_3\omega_1\omega_2$.

There is an \emph{antipode}, 
\[
\int_{\alpha^{-1}}\omega_1\cdots \omega_n=(-1)^n\int_\alpha \omega_n\cdots \omega_1.
\]
I do not want to go further with iterated integrals, I'll come back to this guy. This is a natural generalization of the ordinary line integral, this is how it behaves with composition of paths, with path reversal, and so on.

One appearance of path integrals is in evaluation of perturbative Feynman diagrams for the Feynman path integral for quantum field theory. This was studied by Kreimer, Connes, Francis Bacon, others, Dmitri Doryn is working on this subject here. This path integral is unrelated to the iterated integrals described here. This is essentially the Picard method of solving an ordinary differential equation. I want to make iterated integrals and their generalization literally the path integral of quantum field theory.

So I want to show that these iterated path integrals can be modeled as the semi-classical part of a$(0+0)$-dimensional non-commutative quantum field theory of a very special kind. That is my purpose here.
\subsection{NC Binary Boolean QFT}
Let me try to review, you don't need to memorize this name, this is my personal classification, non-commutative binary Boolean QFT. So quantum field theory in general, I want to do it in an algebraic setting. The goal is to characterize quantum observables and quantum expectation and quantum correlations between quantum observables. 

Here $\Bb$ is a bigraded $\fieldk$-vector space. For safety the characteristic will be $0$. We'll say $B=\bigoplus B_{(j)}^i$, where $i\in \zZ$ and $j\in \qQ$. I call $i$ the \emph{ghost number} $\gh$ and $j$ the \emph{mass dimension} $j$. We also introduce a formal parameter $\hbar$, a sort of Planck constant, with $\gh(\hbar)=0$ and $\md(\hbar)=1$. Every physical quantity has a dimension. We're dealing with a $(0+0)$-dimensional theory so the only dimension left is mass dimension. You could say instead this is $\hbar$ dimension.

Let me consider $(B,1_B,\cdot)$, making $B$ an associative unital algebra and $B^{i}_{(j)}B^{i'}_{(j')}=B^{i+i'}_{(j+j')}$. Then I can form $B[[\hbar]]$ and extend the product linearly with $\hbar$-linear continuity, and so $B[[\hbar]]$ is an algebra over $\fieldk[[\hbar]]$. I call this binary because $\cdot$ is a binary multiplication.

\begin{defi}
A \emph{BQFT algebra} is $(B[[\hbar]],1_B,\cdot, K)$, with $K=K^{(0)}+\hbar K^{(1)}+\cdots$
with $K^{(i)}:B\to B$
and I demand that $\gh(K)=1$ and $\md(K)=0$. I call $K^{(0)}$ the \emph{classical limit} of $K$ and call it $Q$. I require that $K(1_B)=0$ and $K^2=0$ along with a divisibility condition. 

I define a family $\mu^K=\mu_1^K,\mu_2^K,\cdots$, such that
\begin{align*}
\mu_1^K(x)&=K(x)\\
\hbar\mu_2^K(x,y)&=K(x,y)-K(x)y-(-1)^{\gh(x)}xK(y)
\\(-\hbar)^{n-1}\mu_n^K(x_1,\ldots, x_k)
&=K(x_1\cdots x_n)
-\sum_{\substack{p\in I(n)\\|B_i|=n-|p|+1\\|p|\ne 1}}(-\hbar)^{n-|p|}x_{B_1}\cdots x_{B_{i-1}}\mu^K(x_{B_i})x_{B_{i+1}}\cdots x_{B_{|p|}}
\end{align*}
\end{defi}

In particular, $K^2=0$ so we have no condition. But for $\mu_2$, the failure is divisible by $\hbar$, and higher and higher things as you go on.

If I have a BQFT algbera $(B[[\hbar]],1_B,\cdot, K)$ and another $(B'[[\hbar]],1_{B'},\cdot',K')$. A morphism between the two is a $\fieldk[\hbar]$-linear map $B[[\hbar]]\to B'[[\hbar]]$ such that $\gh(f)=0$ and $\md(f)=0$, so $f=f^{(0)}+\hbar f^{(1)}+\cdots$, which preserves the unit and is a cochain map, $fK=K'f$, together with a $\hbar$ condition, which can be summarized into, $\phi^f=\phi_1^f,\phi_2^f,\cdots$, 
\begin{align*}
-\phi_1^f(x)&=f
\\
\hbar\phi_2^f(x,y)&=\phi_1^f(xy)-\phi_1^f(x)\phi_1^f(y)
\\
\hbar\phi_3^f(x,y,z)&=-\phi_1^f(x_1)\phi_1^f(x_2)\phi_1^f(x_3)-\hbar \phi_2^f(x_1,x_2)\phi_1(x_3)-\hbar\phi_1^f(x_1)\phi_2^f(x_2,x_3)+\phi_1(x_1x_2x_3)
\end{align*}
So we obtain $B[[\hbar]],1_B,\mu^K$ and $B'[[\hbar]],1_{B'},\mu^{K'}$ and the theorem is that this defines a functor from Boolean algebras to the categories of $sA_\infty$ algebras over $\fieldk[[\hbar]]$.

This always satisfies a shifted $A_\infty$ relation, and the family of multilinear maps defines a morphism between the two guys here. Secondly, you can show that this is a homotopy functor.

[some discussion]

\begin{defi}
a binary \emph{BQFT} is a homotopy class of BQFT algebra map $c:B[[\hbar]]_{\Bl}\to \fieldk[[\hbar]]$, called \emph{quantum expectation}, defined up to homotopy.
\end{defi}
It has the expansion $c=c^{(0)}+\hbar c^{(1)}+\cdots$ with the condition $c\circ K=0$.

I forgot to say, $B[[\hbar]]_{\Bl}\xrightarrow{c}\fieldk[[\hbar]]$, you can get the descendent $\phi^c$, and the binary QFT, you should think, is defined only in the homotopy category. The observables are simple. Consider a bigraded vector space, regarded as a $0$ $sA_\infty$ algebra, and then consider an $A_\infty$ morphism to $B[[\hbar]]$ with the descendant structure. Then we pick out $B[[\hbar]],K$, this is a cochain complex, and to here we can get a map $\Pi^{\varphi}$ from $(TV[[\hbar]],0)$, then I can cook up a cochain map from $TV[[\hbar]]$ to the original cochain complex, so that the cochain homotopy type depends only on the homotopy type of $\varphi$. Then I have the map $c$, and the composition $c\circ \Pi$ gives a map $TV[[\hbar]]\to c$, and this has $n$-point functions. I can also compose, say $c=circ \Pi^\varphi=\mu^\nu_1,\mu^\nu_2,\ldots$, and I can compose, $\phi^c\circ \varphi$, and this I can call $\chi^\nu_1,\chi^\nu_2,\ldots$, and then $1+\mu_1^\nu+\cdots=e^{\chi_1^\nu/\hbar+\cdots}$, so $Z=e^{F/\hbar}$.

[some discussion]
\begin{align*}
\Pi_1^\varphi(v_1)&=\varphi_1(v)\\
\Pi_2^\varphi(v_1, v_2)&=\varphi_1(v_1)\varphi_1(v_2)+\hbar \varphi_2(v_1,v_2)
\\
\Pi_3^\varphi(v_1,v_2,v_3)&=\varphi_1(v_1)\varphi_1(v_2)\varphi_1(v_3)+\hbar \varphi_2(v_1,v_2)\varphi_1(v_3)+\hbar \varphi_1(v_1)\varphi_2(v_2,v_3)+\hbar^2\varphi_3(v_1,v_2,v_3)
\end{align*}
and this looks like the iterated integrals.

So now look at this ``stupid'' example. I always thought this was stupid but I was the one who was stupid. Look at $V$ a bigraded $\fieldk$-vector space with $\md(V)=1$ and $\gh(V)$ integral. So $TV=V\oplus V^{\otimes 2}+\cdots$ and I'll define $B=\widehat{TV}$, the completion. Then I look at $(B[[\hbar]], 1_B,\cdot, K)$, and I assume that $V$ is an $sA_\infty$ algebra. For example, $V=\Omega^\cdot M[-1]$ with $d$ and $\mu_2=\wedge$.

Then I want I want to cook up $(TV[[\hbar]], e,\otimes,K)$ and let me define $K$. Remember, my $K=Q+\hbar K^{(1)}+\cdots$, this is the codifferential of the $A_\infty$ structure on this, viewed as a coalgebra.

Because $Q$ is a particular mass dimension, the only thing you can do, if $v_1,\ldots, v_n$ are in $V$, then 
\begin{align*}
K(v_1)&=dv_1\\
K(v_1\otimes v_2)&=dv_1\otimes v_2\pm v_1\otimes dv_2 + \hbar \mu_2(v_1,v_2)\\
K(v_1\otimes v_2\otimes v_3)&=dv_1\otimes v_2\otimes v_3 + v_1\otimes dv_2\otimes v_3+v_1\otimes v_2\otimes dv_3 + \hbar\mu_2(v_1,v_2)\otimes v_3+\hbar v_1\otimes \mu_2(v_2,v_3)+\hbar^2\mu_3(v_1,v_2,v_3).
\end{align*}
The condition of divisibility and so on force these to be true. I have a descendant, $\mu^K=K,\mu_2^K,\cdots$, and you can say that $\mu_n^K$ is an $sA_\infty$-algebra on $TV[[\hbar]]$ and I can take a classical limit. If you cut out just $V$ on this, it's just the original $\mu$. You start from $K$ and $\otimes$, you do the descendent, get $TV[[\hbar]], \mu^K$, and take the classical limit and get $(V,\mu)$. I can go back then, I can start from any of these, these are the same thing.

Now one more construction, if I have $(V,\mu)\xrightarrow{\varphi}(V',\mu')$, and out of these I create $K$ and $K'$ on $TV[[\hbar]]$ and $TV'[[\hbar]]$ with $\otimes$. Then out of $\varphi$ I can creat a QFT morphism $f$, which satisfies the divisibility condition that says 
\begin{align*}
f(v_1)&=\varphi_1(v_1)\\
f(v_1\otimes v_2)&=\varphi_1(v_1)\otimes \varphi_1(v_2)+\hbar \varphi_2(v_1,v_2)
\end{align*}
so that if this was an $A_\infty$ quasiisomorphism, I get an quasi-isomorphism of complexes. Again I can start wherever I want in this story and go back and forth, taking classical limits and so on. If I have an $A_\infty$ algebra and a morphism $\chi$ to $(\fieldk, 0)$, I can create $(TV[[\hbar]],K,\otimes)\to (\fieldk[[\hbar]],0)$.

This is not what we want, we want $C_\infty$ algebra. On $TV$ we have a shuffle product.
\[
v_1\otimes\cdots\otimes v_r \Sha v_{r+1}\otimes\cdots\otimes v_{r+s}=
\sum_{\sigma\in \sh(r,s)}v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(r+s)}
\]
For example, $v_1\Sha v_2= v_1\otimes v_2\pm v_2\otimes v_1$.

A $C_\infty$ algebra is an $sA_\infty$ algebra such that $\mu_n(v_1\otimes \cdots\otimes v_k\Sha v_{k+1}\otimes \cdots\otimes v_n)=0$ for all $k=1,\ldots, n-1$.

Similiarly, an $sC_\infty$ morphism is an $sA_\infty$ morphism killed by the shuffle product.

For example, the most famous example is the algebra of differential forms, the product is graded commutative and all higher things vanish.

Then if I constructed $K$ out of an $sC_\infty$ algebra, then $K(\alpha\Sha\beta)=K\alpha\Sha\beta?\alpha \Sha K\beta$. If I have the descendent morphism, $\varphi\leadsto f$, then $f$ is an algebra homomorphism for the shuffle product. Then the descendent is an $sC_\infty$ morphism on $TTV$ or whatever, and conversely, if I start with a Boolean QFT algebra with this condition, in the classical limit I get $sC_\infty$ algebra and morphism.

I know that if I have a group, the group ring forms a Hopf algebra. Similarly, if you have a Hopf algebra you can get a group. If I started with differential forms on a manifold, I should have access to the fundamental group, a quantized version, and hope it has more information.

I have $(TV[[\hbar]],\otimes, \Sha, K)\to (\fieldk[[\hbar]],0)$, and if $(V,\mu)$ was a $C_\infty$ algebra with a $C_\infty$ morphism to the ground field, I get the quantum expectation. If this wants to be iterated integrals, we shoud be able to compose these maps. If I have $\alpha$ and $\beta$, then I can define the composition as 
\[
\alpha\diamondsuit \beta(v_1\otimes \cdots \otimes v_m)
=\alpha(v_1\otimes\cdots\otimes v_m)+\alpha(v_1)\beta(v_2\otimes\cdots \otimes v_m)+\cdots+\beta(v_1\otimes v_m)
\]
and I have $\alpha^{-1}(v_1\otimes \cdots \otimes v_n)=(-1)^n \alpha(v_n\otimes\cdots\otimes v_1)$. We have to be sure that this defines a well-defined group structure in the homotopy category. You can do this explicitly, if $\alpha\sim\alpha'$ and $\beta\sim\beta'$ then $\alpha\beta\sim\alpha'\beta'$. Out of these you can construct $C_\infty$ morphisms. So in the space of $C_\infty$ morphisms, we can put a group structure. The target doesn't need to be a ground ring.

The space of morphisms between quantum algebras also has a group structure. The space of $C_\infty$ morphism, and homotopy type, there's a group. These all satisfy the relation $\alpha\circ K=0$. Assume that $V=\Omega^\bullet(M)[-1]$ with $d$ and $\wedge$. Then $\alpha$ is an algebra homomorphism, but I can compose the diamond guys, and the composition, is by definition exactly like the composition for iterated path integrals, and similarly for the inverse.

What is the quantum correlation? Some $sA_\infty$ morphism, now we restrict to the classical limit, then we have Chen's homotopy functional.

$K$ has a decomposition by $\hbar$, so $Q$ is a derivation of the tensor product, $Q(v_1\otimes v_2)$, it's a derivation, this is nothing but $d$, so if I have $(V,d,\mu_2,\cdots)$, you take the cohomology and get $(H,\mu^H)$. This has rational homotopy living in it. This tells you that the $Q$-cohomology is $TH$, along with zero differential. So we can always construct a quasi-isomorphism between $(TH,0)$ and $(TV,d)$ and so we want to get $(TV[[\hbar]], K=Q+\hbar K^{(1)}+\cdots)$ and we want $(TH[[\hbar]],\kappa=\hbar\kappa^{(1)}+\cdots)$

It turns out this $\kappa$ contains the information of the minimal structure, and $f=f^{(0)}+\cdots$ contains all the higher data of the $C_\infty$ morphism. Then of coruse, you have a tensor and shuffle. The shuffle will induce the shuffle, but tensor product will not descend. We have $\mu^{\{\kappa\}}$, which is an $sC_\infty$ algebra, descendant, and we use this to build some kind of universal correlators. If you restrict to the cut out classical piece you get all of rational homotopy theory.  We have a fundamental group with rational homotopy theory and a canonical quantization.

[Some discussion]

What I call BQFT is the binary one. This is a special example of homotopy BQFT, where the morphisms are not just cochain maps but have homotopies and higher homotopies and such, and you can combine these into a very big structure and what you get is some kind of, how can I say, an $sA_\infty$-equivariant $\hbar$-flat superconnection and flat supersection, in the sense, say, of Deligne (if you specialize to $\hbar=1$).
\section{July 11: Chang-Yeon Chough: Topological types of algebraic stacks}
I'm a new member here beginning in July. I'm a new member here still getting used to this place. The first half I'll briefly discuss \'etale homotopy type of schemes and Friedlander's homotopy type of simplicial schemes. Then I'll talk about the recent derived approach, and finally my generalization to algebraic stacks. The second half will be an application to a concrete problem in algebraic geometry.

The title is ``homotopy type of algebraic stacks.'' But let me say, the derived functor approach has to do with, imagine you study some kind of cohomology theory and eventually you want to put everything in triangulated or derived categories to have a systematic way of studying things. I want to use Quillen's model category theory in an analogous way. Why stacks? Some people like them, others don't. I like them. They naturally arise in moduli problems. That explains some of what I'm going to talk about.

One more thing, how did I get interested in this kind of thing? There's a concrete problem I'd like to solve, say the moduli problem of vector bundles. A few years ago, it was observed that all cohomologies vanish over a point. So maybe that stack is contractible. At that time I googled ``fundamental group of algebraic stacks'' and there were some things, but I want to generate a space containing all of the homotopical information of the stack. That's the motivation for this story.

Let me say some consequences of this approach. First I'll mention some generalizations of the Artin--Mazur comparison theorem. Say you have a multiplicative group scheme $\gG_m$ over $\cC$. If you analytify this scheme and look at the underlying space, this is literally a space, we can talk about its homotopy, look at the $\cC$-points, you can talk about the $\pi_1$ or the cohomology. The \'etale homotopy type of the scheme is this underlying space after some kind of completion. Now we can talk about the classifying stack $\Bb \gG_m$, and in this case you do the same thing, you look at $\cC$ points, $\Bb\gG_m(\cC)$, and you'll see (the original thing is $S^1$) and this is $\cC\pP^\infty$ and this is $K(\zZ,2)$. If you compute $H^*(\Bb\Gg_m,\qQ_\ell)$, this is $\qQ_\ell[c]$ with $c$ in degree $2$. So for example \wh computed this in his paper, but if you look at the more classical space, you can see it's $\cC\pP^\infty$ and get this result directly. So we bring this to the level of algebraic topology and get the result.

That's the first thing I can obtain out of my approach. We can also get, I won't talk about this today, but you can get a generalization of a theorem of Cox, which says if you have a real variety and you can base change to complex numbers, then the [missed]. With this kind of thing you can generalize to any base field. 

The last one is what I'll talk about at the last moment. I'll come back to this in the second half.

Let me go back to a long time ago. Suppose $X$ is a paracompact topological space. We learn about this thing in the first course in graduate school, I guess. Suppose $\Uu=\{U_i\}$ is a good cover, any intersection of $U_i$ is contractible. Let's look at the Cech nerve of this cover. So we take in degree $1$ the coproduct $\sqcup U_i$, and then $\sqcup U_ij$ and so on, this goes on forever. In each degree you have a topological space, and then you can take the components, and so I take $\pi_0$ of this and call this the nerve of $\Uu$. Then
\begin{thm}(Borsuk)
$X$ and $\pi_0(N\Uu)$ are homotopy equivalent.
\end{thm}
Then you recover all the homotopy information of this. Now I'll talk about how this was generalized to algebraic topology. Today every scheme will be locally Noetherian and every morphism will be finite type. Think about coverings $\Cov(X)$ of $X$, so \'etale coverings, you take the nerve and connected components and this gives some functor to $\SSet$. So $\Uu\mapsto \pi_0(N\Uu)$.

You try this kind of thing, and there are at least two problems, two issues. The first, you compute the cohomology of this guy, $H^*(\pi_0(N\Uu),A)$, and you see this is $\check{H}^*(X,\ul{A})$
, but we want \emph{derived} cohomology, not Cech cohomology. That's the first problem. The second problem, if you look at $\Cov(X)$, this is \emph{not} cofiltered (this is some technical thing) and you'll be in trouble working with this index category.

The idea is to use hypercovers, which I won't define, but let me give you a rough idea. You have a covering of a scheme and do the nerve, and once you know what you have in degree zero, that determines every following thing. So this is too easy. Refine this. You take an \'etale cover, \'etale surjection in each degree, and there you have the freedom of refining all these things. The intersections might not be so great even if everything is nice. So you want to refine each interection.

You think about the category of hypercoverings of a scheme $X$, and you let $HR(X)$ be the category of hypercoverings. You mod out by the simplicial homotopies to get something cofiltered. But then you do get something cofiltered and get a map $HR(X)^{\op}\to \Ho\SSet$ given by $\Uu_*\to \pi_0(\Uu_*)$. This is Artin--Mazur. I won't use their notation,so let me call this $h_{AM}(X)$.

One consequence of this definition is, maybe as a sanity check,
\begin{thm}
(Verdier's hypercovering theorem) 
$\lim_{\Uu} H^*(\pi_0(\Uu),A)$ makes sense because the indexing category is cofiltered and the limit is $H^*_{\text{\'et}}(X,\ul{A})$. 
\end{thm}
Once you do this, you can recover the cohomology of $X$, I forgot to say this at the beginning but from $X$ you want to get a space.

For example, $\pi_n$, you can do levelwise. We know something about $\pi_1$ of a scheme $X$. So you can take $\pi_1$ of $X$, and after profinite completion you can recover Grothendieck's $\pi_1$. 

There are a lot of results, like the proof of the Adams conjecture by various people. Next, I briefly introduced Artin--Mazur's homotopy type. There's an intermediate step, Friedlander's \'etale homotopy type of simplicial schemes. This extended from schemes to simplicial schemes. If you have a stack, you take an \'etale cover and then you can roughly define the homotopy type as the homotopy type of the cover. Second, the pro-object in the derived category should make you upset. You want an object in the derived category \wh subcategory. He was able to lift this to the homotopy category of pro-simplicial sets. I won't talk about his definition, but he rigidified \'etale covers and was able to do lots of things.

Now I'm moving toward the present. You'd like to put all these things in a certain derived setting. Here we'll use the model category, say $X$ is a local Noetherian scheme, look at the small \'etale topos on $X$, you have the connected component functor to $\Set$ since $X$ is locally Noetherian. Then you lift this to the category of simplicial objects. Then for some reason you look at pro-objects
\[
\pro-X_{\text{\'et}}^{\Delta^{\op}}\xrightarrow{\pi_0}\pro-\SSet.
\]
This is the result by Barnea and Schlank, and they put a nice model category structure here that captures this, so that the functor can be derived, and you can derive it to get $L\pi_0$. That's the main point but I won't talk about it at all, and you derive a final object in this topos, $L\pi_0(*)$, and you'll have something in the target category, a prosimplicial set. This is homotopy equivalent to the Artin--Mazur homotopy type of $X$.

Let's say $\Xx$ is an algebraic stack. Then there is one thing you can do, look at the big \'etale topos on the stack $\Xx$, let me write it as $LF\acute{E}(\Xx)$, this is the full subcategory, first look at schemes over $\Xx$ up to equivalence, locally of finite type over $\Xx$. This is a ``big \'etale'' topology on the stack $\Xx$. I'm not using small \'etale because this is not necessarily Deligne--Mumford. To study stacks, you only have a smooth cover from a scheme, not an \'etale cover. This is a site, this is your new topos. Now do the same thing. You have a well-defined connected component functor. Take the simplicial version, and pro versions, and you still have the connected component. But then when you go back to Barnea--Schlank, they do it for any topos. So you can apply their structure to this topos. Then you can do $L\pi_0(*)$, derive the final object along the left derived functor, defined to be the topos homotopy type of $\Xx$. So the question is whether this really contains homotopical information on $\Xx$.

So, it does. Why does this work well? In the scheme case, they use the small \'etale topology while I use the big \'etale topology. Say $S$ is a base scheme throughout the talk, Noetherian, think about the big \'etale topology on $S$ and you can do the same thing $\pro-(LF\acute{E}(S)^\sim)^{\Delta^{\op}}$, this goes by $\pi_0$ to $\pro-\SSet$. Look at the functor representing the scheme $X$, and then by descent theory the presheaf is a sheaf on this site, I really mean the representing object of this guy, some object here, which you can view as constant in the associated simplicial category, and then embed that into the pro-category. Then you have an object, $L\pi_0(X)$. We're living in the world, this then recovers the Artin--Mazur homotopy type. I use the big topos instead of the small topos but still get the same homotopy type.
 
You can also think of the \'etale topos as a simplicial scheme, now you know what I'm going to do, derive that simplicial scheme, $L\pi_0(X_*)$, and that's defined to be the topological type of this simplicial scheme over $S$. 

If you have an algebraic space, you can replace \wh. So then an object in here, well, this is a simplicial object, each scheme corresponds to a sheaf. Given a simplicial algebraic space, derive it along this machinery and you'll get a simplicial set.

This is the object we want, think about $h(X_n)$, I'm abusing my terminology, I'll omit the subscript because they're the same, we take $\hocolim$ over $\Delta^{\op}$ of $X_n$. and this is the homotopy type, $h(X_*)$. If you know what happens on each degree, you can get the homotopy type of the simplicial object. As a corollary, the immediate corollary, if you think about simplicial schemes, $h_F(X_*)$, you can think about this object and they're homotopy equivalent, everything is compatible with Artin--Mazur theory.

So this is the key technical result, and this justifies the definition.

Let me just mention a couple more things and then I'll stop and talk about my concrete application to make people happy. For example, say you have an algebraic stack $\Xx$ and take a smooth hypercover $X$, take the coskeleton of $X\to \Xx$ and then $h(\cosk_0(X/\Xx))\sim h(\Xx)$, and here you have $\Bb\gG_n$, you take a point, you get a simplicial scheme which is $\Bb\gG_n$, that's the moral.

I mentioned a comparison theorem, if you have an algebraic stack over complex numbers and look at $\cC$-points, then the topological \wh coincide, that's true. Then there is \wh for the Hurwitz space.

Next I'll talk about a more concrete approach to geometric problems.

In the second half I'll talk about a very concrete application of this theory to symmetric powers to make you happy. Let's start with the classical case. Say you have a pointed connected topological space $X$. Think about the symmetric power. This is a theorem by Dold and Kan. On the one hand you look at the symmetric powers of $X$, $\Sym^\infty(X)$, this is $X^n/\ S_n$, the usual quotient, and you take the colimit of these spaces, and you take $\pi_i$ of this gadget, and this is $\tilde{H}_i(X,\zZ)$, which tells you a relation between homology theories and homotopy theories via symmetric powers. Last year, \wh proved that this statement is still valid in the world of algebraic geometry.
\begin{thm}
Say $X$ is a geometrically normal and proper algebraic space over a separably closed field $k$. Then $h(X)$, the \'etale homotopy type of $X$, you can take $\Sym^n h(X)$, and on the other hand you can think about $\Sym^n(X)$, which is subtle because quotients are subtle in algebraic geometry, you take the $n$-fold fiber product and it has a canonical action from $X^n$, and this is the coarse moduli space of the quotient stack. You can take $h(\Sym^n(X))$. The theorem tells you that these two spaces are homotopy equivalent.
\end{thm}

If you look at this for a moment, you see that it's formal, it's about the commutativity of $\Sym^n$ and $h$. So this formality fits very nicely in the way we developed the homotopy theory of algebraic stacks.

How did he prove that? He proved that these two objects have isomorphic $\pi_1$ and isomorphic cohomology groups. For cohomology groups, he went back to Deligne's computation of cohomology groups of symmetric powers. I basically obtain this statement for free from my approach. Let me give you a diagram here.
\[
\begin{tikzcd}
S_n\times X^n\ar[r,shift right]\ar[r,shift left]&X^n\ar[r]&\Sym^n(X)
\end{tikzcd}
\]
Things aren't free so this isn't a coequalizer. Nonetheless, it's a ``qfh cover.'' I won't say what the qfh topology is, it's Voevodsky. That implies something. This implies that if you read off this diagram in the category of $qfh$ toposes, this becomes a coequalizer. In other words, the $n$th symmetric power is not an \'etale sheaf quotient but it \emph{is} a qfh sheaf quotient. Let me say $LFQ(k)$ is the qfh site over the base field $k$, and you have a canonical morphism of topoi
\[
LFQ(k)^\sim\to LF\acute{E}(k)^\sim
\]
The symmetric power is not a coequalizer in the codomain, but it is when you pull it back. So define symmetric powers like that. Consider $(LFQ(k)^\sim)^{\Delta^{\op}}$ and an object $F$ there. You define the symmetric power to be this coequalizer 
\[
\begin{tikzcd}
S_n\times F_\cdot^n\ar[r,shift right]\ar[r,shift left]&F_\cdot^n\to \Sym^n F_\cdot
\end{tikzcd}
\]
What's next? Let's look at this diagram. Let's look at the statement we want for the qfh topology. Look at
\[
\begin{tikzcd}
(LFQ^\sim)^{\Delta^{\op}}\ar[r,"\Sym^n"]\ar[d,"\pi"] 
&(LFQ^\sim)^{\Delta^{\op}}\ar[d,"\pi"]
\\
\SSet\ar[r,"\Sym^n"]&\SSet
\end{tikzcd}
\]
If you start with $X$ in the topos in the upper left, you get the symmetric power which preserves weak equivalences. In the other direction you obtain $\Sym^n( h_{\qfh}(X))$; in the other side you take the symmetric power, with respect to the qfh topology, in order to check weak equivalence you can just look at stalks. If you pull the diagram back to simplicial sets, this is a colimit, and when you take the stalks, this remains a coequalizer diagram, and you can check that the weak equivalence between simplicial sheaves at the level of stalks. We have the result thanks to this categorical statement. The conclusion is that the $n$th symmetric power functor preserves weak equivalence because this is true for stalks. So on the other two sides we take $h_{\qfh}(\Sym^n(X))$. By abstract nonsense, these two are homotopy equivalent. Then if you erase qfh topology and \'etale topology, then you're done.


In general you're in trouble, but you have a diagram
\[
\begin{tikzcd}
\Sym^n(h_{\qfh}(X))\ar[d]\ar[r]&h_{\qfh}(\Sym^n(X))\ar[d]\\
\Sym^n (h_{\text{\'et}}(X))\ar[r]&h_{\text{\'et}}(\Sym^n (X))
\end{tikzcd}
\]
The top map is an isomorphism in the homotopy category. We'd like to show the bottom is an isomorphism in the homotopy category.

To show this, we need to see that $\pi_1$ and all cohomology groups are isomorphic on the right, and $\pi_1$ uses the fact that this is a symmetric power; the cohomology follows from Voevodsky's work. Then on the left we can compare homotopy groups, using Dold--Kan.

This gives an alternative way of proving the theorem.




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