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\author{Gabriel C. Drummond-Cole}
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\title{Deformation Theory Workshop}
\begin{document}\maketitle
\section{Ginot}
Higher order Hochschild, I will start with this, (co)homology was coined by Pirashvidi.  Everyone knows that Hochschild homology has to do with circle actions.  The higher order idea is that we want a homology theory which is modeled on a ``space'' $HH^X$ such that $HH^{S^1}$ is usual Hochschild homology.  It should be functorial on the space.  It will give you a lot of possibility to construct things.  The surface product, many people know that Chas and Sullivan defined string topology, a product on the homology of the free loop space $H_.(LM)$ with the dimension shifted by the dimension of $M$.  This is a theorem that in fact there is an isomorphism of algebras in between this thing and $HH^*(\Omega_M,\Omega_M)$.  This is due to Cohen-Jones, Merkulov, and others.  This is defined for $S^1$ but you can define the analogue where you replace $S^1$ by genus $g$ surfaces, oriented and compact.  The goal is to find a well-defined Hochschild model for which this theorem holds. 

So I will define higher order Hochschild cohomology and hopefully it will be very elementary.  A bit of terminology.  All algebras are commutative (differential graded) over a fixed field $k$, and I will assume characteristic zero.  I don't need that here but will need it later on.

Before defining this over a space, I will speak obout a set.  Let $S$ be a pointed finite set.  From now on, all sets will be pointed, unless I say otherwise.  For instance, $S$ can be $\{0,\ldots, n\}$ and $0$ will be my base point.  Then $CH^S(A,M)$ for $A$ a commutative algebra and $M$ an $A$-bimodule will be, if $n+1$ is the order of $S$, will be $(M\otimes \underbrace{A\otimes \cdots \otimes A}_{n \text{times}})$.  Let $f$ be a map of pointed finite sets.  Then I define the map $f_*:CH^S(A,M)\to CH^T(A,M)$.  It sends $(m\otimes a_1\otimes\cdots a_n)\mapsto n\otimes b_1\otimes \cdots b_q$ where $b_i=\prod_{j\in f^{-1}(i)}a_j$, with an empty product equal to one.  These algebras should have been unital.  So $n=m\prod_{j\ne 0, f(j)=0}a_j$  We have a functor $CH^?(A,M)$ from finite pointed sets to $k$-modules.

Okay, so well, now I can speak about spaces and semisimplicial sets.  If $Y$ is a simplicial pointed finite set, we have face and degeneracy maps $Y_2\leftrightarrow Y_1\leftrightarrow Y_0$, then we can do the same construction, and define $CH^{Y_.}(A,M)$ is the simplicial $k$-vector space formed by applying the functor $CH^?(A,M)$ to the sets $Y_i$.

Okay, so we can define $b=\sum (-1)^i (d_i)_*$ which is a differential.  We call $(CH^{Y_.}(A,M),b)$ the (higher order) Hochschild chain copmelex modelled on $Y$.  I should give a few remarks.  Remark one.  I can assume that $A$ and $M$ are graded, differential graded, I can do the same construction, and add the differential.

Remark two is, what if my simplicial set is not finite?  For pointed sets, not finite, or not necessarily finite, you define $CH^S(A,\Gamma)$ is the limit over all finite sets $T\to S$ of $CH^T(A,M)$.  You can define that, it might be totally crazy.

\begin{lemma}How much of this structure depends on the simplicial structure?
\\
If $f:Y_.\to X_.$ is a map of pointed simplicial sets which is an isomorphism on homology, then $CH^{Y_.}(A,M)\to CH^{X_.}(A,M)$ is a quasiisomorphism.  In particular it does not depend on the choice of simplicial sets.  In particular, if you are given any space, any topological space, then we can always look at $CH^{\Delta X}(A,M)$ where $\Delta_k(X)=Map(\Delta^k\to X)$.  Of course, that's a horrible thing that you should never look at.

So far I have only spoken about cohomology but of course you can dualize to get homology.  $CH_{Y_.}(A,M)=Hom_A(CH^{Y_.}(A,A),M)$.

SO now I will give you a lot of examples.  If $Y$ is a point, then $(CH^{pt}(A,M),b)$ is $\cdots M\stackrel{0}{\to}M\stackrel{id}{\to}M\stackrel{0}{\to}M$.  The next example is with $Y$ the interval, with the regular simplicial model.  Then $I_n$ will be $\{0,\ldots n+1\}$.  What does the complex look like?  $CH^{Y_n}(A,M)=M\otimes A^{\otimes n+1}$, and the simplicial structure, $d_i$ is the map which identifies $i$ and $i+1$ so we get that $(d_i)_*(a_1\otimes \cdots\otimes a_n)=a_1\otimes \cdots\otimes a_ia_{i+1}\otimes\cdots\otimes a_n$.  What is this thing in standard terms?  It's just the bar resolution of $M$, $B(M,A,A)$.  So can I do $S^1$?  There are many models, but I can consider the interval where I have collapsed the two endpoints.  What is the simplicial way, I just take away the last guy.  So $CH^{S^1_n}(A,M)\cong M\otimes A^{\otimes n}$, and the last differential also has an $mn_0$ term.  This is the standard Hochschild complex, hopefully.

I'm going to do a few other examples.  What about $I^2$?  or $S^2$ or the torus?  For $I^2$ I take the Cartesian product of $I$, and so $I^2_n=I_n\times I_n$ has $(n+2)^2$ terms.  So what does $CH^{I^2_n}$ look like?  It is labeled by $a_{0,0}$ through $a_{n+1,n+1}$.  You can write down this thing as a matrix.  I have a big matrix, and the simplicial differential is to apply diagonally.  $(d_i)_*$ multiplies the line $i$ and $i+1$, and the same for the rows, and the guy in the $i,i$ place gets four things.  Okay, so then what about $CH^{S^2_n}$?  That's a square where you collapse the boundary to a basepoint?  It means that all the things from the zero lines collapse to a point  So it is the same square where you just have the inner part, an $n\times n$ matrix.

So then $CH^T_n(A,M)$ is a square where the left and right, the top and bottom are identified.  The slight differences in the matrices lead to big differences in the answers.

So I'm going to state something.  Let $X$ be a manifold.

There is a map $I_t^Y:HH^{Y_.}(\Omega_X,\Omega_X)\to H^*(X^{|Y_.|})$ which is a quasiisomorphism when $Y$ is $n$dimensional and $X$ is $n$-connected.

So, you can see the relationship with string topology.  Now let me say what is the surface product.  This should be $H_.(M^{S^1})\otimes H_.(M^{S^1})\to H_.(M^{S^1})$.  This is the string product.  You have 
$$\xymatrix{
M^{S^1}\times M^{S^1}\ar[d]&M^{S^1\vee S^1}\ar[l]_\iota\ar[d]\\
M\times M& M\ar[l]_\Delta
}$$

So you have these families of loops.  If you look at the intersection at their basepoints, you get something of dimension $p+q-dim(M)$, a family of loop pairs.  Now you can use the map $M^{S^1\vee S^1}\stackrel{(p)^*}\to M^{S^1}$ where you pinch $p:S^1\to S^1\vee S^1$.  So the product, you use the usual things, and you get a product $H_.(M^{S^1}\times M^{S^1})\to H_{.dim(M)}(M^{S^1\vee S^1})\to H(M^{S^1})$ using an umkehr map.

So now we will do the same thing with $\Sigma^g$ a surface of genus $g$.  You end up with some kind of thing, and you get exactly the same, you get your map

$H_.(M^{\Sigma^g}\times M^{\Sigma^h})\to H_{.dim(M)}(M^{\Sigma^g\vee \Sigma^h})\to H(M^{\Sigma^{g+h}})$
So this defines an algebra, noncommutative on $\bigoplus_g H_.(M^{\Sigma^g})$.

I want to do the same for higher Hochschild, so I need a nice simplicial model for $\Sigma^g$ and a pinching map.  I will stick to $g\ge 1$ because it's easier.  For something of genus $g$ I can divide up a square into a number of smaller squares and identify some outside edges, and I can give a model for $\Sigma^g$ by dividing into $g^2$ squares.  I want a pinching map that will send this to two squares touching at their corners.  So to pinch I want to move across diagonally so I need to subdivide all of the off-diagonal squares along their diagonals.  To do this I need a model for triangles, which I didn't give you.  It's two squares identifying opposite edges to a point, that's two triangles back to back.  You can compute that $\Sigma^g_k$ has $(2g^2-g)k^2+(3g^2-g)k+1$ simplices.

So, the cup product, if $X_.$ and $Y_.$ are finite pointed sets, I can do $CH_{X_p}(A,A)\otimes CH_{Y_q}(A,A)$ and I can use Alexander Whitney to get a diagonal simplicial thing $CH_{X_{pq}}(A,A)\otimes CH_{Y_{pq}}(A,A)$ and you compute something that he wrote down and said too fast because his time was up and he knew it.  Oh!  I was wrong!  He's about to keep going!  I just saw Imma look at the clock.  I looked at the clock too.  This room gets really hot in the afternoons.  I can define the following.  $CH_{\Sigma^g}(A,A)\otimes CH_{\Sigma^h}(A,A)\to CH_{\Sigma^g\wedge \Sigma^h}(A,A)\to CH_{\Sigma^{g+h}}(A,A)$ and the theorem is that $\bigoplus HH_{\Sigma^g}(A,A)$ with this $\cup$ is an associative algebra but has the sphere in the center, and if $A=\Omega_X$ for $X$ an oriented manifold, the morphism is a morphism of algebras when $X$ is $2$-connected.

\end{document}
 