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\author{Gabriel C. Drummond-Cole}
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\title{Deformation Theory and Operads}
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Recall that we defined a formal 1 parameter deformation of an algebra $(A,\mu)$ as a $\mu\in C^1(A,A)$.  Then $[\mu,\mu]=0$ if and only if $\mu$ is associative.  Then let $(A,\mu)$ be associative let $\mu'=\mu+t\tilde{\mu}$.  Then $\mu'$ is a deformation of $\mu$ if and only if $\mu'$ is associative if and only if $[\mu',\mu']=0$ if and only if $2[\mu,t\tilde{\mu}]=2\delta(t\mu)+[t\tilde{\mu},t\tilde{\mu}]=0$ so the Maurer Cartan equation for $t\mu$ or the master equation is $$\delta(t\tilde{\mu})+\frac{1}{2}[t\tilde{\mu},t\tilde{\mu}]=0$$

Let's do the Ext-Tor interpretation of Hochschild.  Let $R$ be a ring, let $P$ be a left-module over $R$.  Then $P$ is called projective if it is a direct summand of a free $R$-module $F$ so that there exists $Q$ so that $F=P\oplus Q$.  Let $A$ be a free $R$-module; then a projective resolution of $A$ is a chain complex $P_.=\to P_3\to P_2\to P_1\to P_0$ so that $\epsilon P_0\to A$, each $P_i$ is projective, and so that $\ldots\to P_1\to P_0\to A\to \{0\}$ is an exact sequence.

For example, if $A$ is projective, then $P_.=\ldots\to \{0\}\to \{0\}\to A=P_0$ is a projective resolution of $A$.

If $A$ is an (left) $R$-module, and $P_.$ is a projective resolution of $A$, define the following:
\begin{defi}
If $B$ is a right $R$ module, let $Tor_.^R(A,B)$ be the homology of $P_.\otimes_R B$
\end{defi}
One point is that these are independent of choices of $B$.

As an alternative, you look at the functor $(\ \otimes_R B):R-mod\to Ab$, and say that this is a covariant right-exact functor and then define $Tor^R$ is the left-derived of that functor.

The other one, Ext, is gotten by taking
\begin{defi} If $B$ is a left $R$-module, then let $Ext_R(A,B)$ is the homology of $Hom(P_n,B)$ with the reversed differential.  Alternatively, you could say $Hom_R(\ , B)$ is a contravariant left-exact functor from the category of $R$-modules into the category of Abelian groups, and $Ext_R$ is the left derived of this.
\end{defi}
The main point is that $Tor$ and $Ext$ are independent of the choice of resolution.

I want to specialize to the case of Hochschild.  Let's say $(A,\mu)$ be an associative algebra over some commutative ring $S$.  What is the opposite algebra?  Let $(A^{op},\mu^{op})$ be the opposite algebra.  (This is ungraded).  This means $A^{op}$ is $A$ and $\mu^{op}(a,b)=\mu(b,a)$.  So switch the order of inputs; it's the opposite way of multiplying.  Let $(A^e,\mu^e)$ be the enveloping algebra, so $A^e=(A\otimes A^{op},\mu_e=\mu\otimes \mu^{op})$.  We want to take $R=A^e$, so this will be our ground ring.  What is $Tor^{A^e}$ and $Ext_{A^e}$?  So first, what is an $A^e$-module.  If $M$ is a left $A^e$-module, then we have maps $A^e\otimes M\to M$, so maps $A\otimes A^{op}\otimes M\to M$, so that's like $a\otimes b\otimes m\mapsto a.m.b$.  This said, I think it's fairly easy to see that this has the module property.  This is equivalent to $M$ being a bimodule over $A$.  Well, $A$ is always a bimodule (right-$A$ bimodule or left $A$-bimodule).  So what is $Tor^{A^e}(A,M)$ or $Ext_{A^e}(A,M)$?  We need a projective $A^e$-resolution of $A$.  We can take $\cdots\to A^{\otimes 5}\to A^{\otimes 4}\to A^{\otimes 3}\to A^{\otimes 2}$ and you stop there.

So $A^{\otimes n\ge 2}$ is an $A$-bimodule, where you multiply on the left and on the right.  It's projective, and a resolution of $A$ as an $A^e$-module.  Here the projection, on $(a_1\to a_n)$, multiplies successive pairs togother, so is very similar to the Hochschild differential.  So this is $(a_1a_2)\otimes a_3\otimes \cdots \otimes a_n+\cdots + (-1)^n a_1\otimes\cdots\otimes \cdots(a_{n-1}\otimes a_n)$
This is a differential.  We haven't discussed the Hochschild differential.  $\epsilon$ is the product in $A$.

Now to prove this, why is $P$ projective?  If $A$ is projetive over $S$ and $A$ has a left unit, well, [discussion].  Why is the sequence exact?  What you do is define $s:A^{\otimes n}\to A^{\otimes n+1}$ which takes $a_1\otimes\cdots \otimes a_n\mapsto 1\otimes a_1\otimes\cdots\otimes a_n$.  Then $b'\circ s+s\circ b'=id_{A^{\otimes n}}$, where $b'$ is the projection map (including augmentation).  To calculate we see
$$b'\circ s(a_1\otimes \cdots\otimes a_n)=
a_1\otimes\cdots a_n-1\otimes (a_1a_2)\otimes\cdots a_n+\cdots \pm 1\otimes a_1\otimes\cdots \otimes (a_{n-1})(a_n)$$
On the other hand, $s\circ b'$ is the same thing without the first term.  So some of the terms cancel and you just get the leading term of the first.   This is a typical calculation.

Now we can just plug this in.  Let $M$ be a bimodule over $A$.  Then $Tor^{A^e}(A,M)=H_.(P_.*\otimes_{A^e} M)$ is the homology of $A^{\otimes 4}\otimes_M\to A^{\otimes 3}\otimes M\to A^{\otimes 2}\otimes M\cong M$

What is the differential?  We get for example $b'(a_1\otimes a_2\otimes m),$ which can be written $(1\otimes a_2\otimes a_3\otimes 1)\otimes_{A^e} m$ which is $a_2\otimes a_3\otimes s\otimes m - (1\otimes a_2a_3\otimes 1)\otimes m + (1\otimes a_2\otimes m_3)\otimes m$.  So under the identification this gives $a_3\otimes m a_2 - (a_2a_3)\otimes m +a_2\otimes (a_3m)$.

In general I get the Hochschild differential.  You multiply things together and get $a_2\otimes\cdots\otimes (ma_1)-a_1a_2\otimes\cdots\otimes m$ and so on.
$Ext_{A^e}(A,M)$ has exactly the Hochschild differential.  

There are a few easy things.  So let $A=TV=\bigoplus V^{\otimes}$ be the tensor algebra with product the tensor product.  Then by this lemma, we can choose any projective resolution.  This one has a small resolution, $P_.^{sm}$ is $TV\otimes V\otimes TV\to TV\otimes TV$.  This is a resolution, with the only map $b'$ equal to
$$
b'((v_1,\ldots, v_n)\otimes v \otimes (w_1,\ldots, w_m)):=(v_1,\ldots, v_n,v)\otimes (w_1,\ldots, w_m)-(v_1\ldots, v_n)\otimes v,w_1,\ldots, w_n)
$$
I claim this is a projective resolution of $TV$.  What you get is that $Tor^{TV^e}(TV,TV)=H_.((TV\otimes V)\to TV)$ which is $TV_\tau$ (as a homework), the coinvariants under the cyclic rotation, so $TV/1-\tau$, in $i=0$ and for $i=1$ the elements $v\in TV$ so that $\tau(v)=v$.

One has to sit down and do these calculations, but you can just sit down and do them, it's totally spelled out, it's doable.  The second example using this is $A=SV$.  We can do the same thing, and again we need a resolution.  Note that if $V$ is $1$-dimensional, then $TV=SV$, so wo can use the first case.  If this is not one-dimensional, assume it is finite dimensional.  You want to build the resolution out of these for , well, we get $P^{SV}-SV\otimes \wedge V\otimes SV$.  So $P_.^{sm}$ has $SV\otimes \wedge^d V\otimes SV\to SV\otimes \wedge^{d+1}V\otimes SV\to \cdots
$  Now you do the same thing as before and you get $(u_1,\ldots, u_n)\otimes (v_1\wedge \cdots\wedge v_m)\otimes (w_1,\ldots, w_k)$ is the sum of bringing one $v_i$ into the left and right factor in all possible ways with the appropriate signs.

I will say some more next time, some resolution, and then some examples.  The homology is $HH_i(SV,SV)$ is $SV\otimes \wedge^iV$ for $0\le i\le d$.  All the differentials become zero.  I think I'll do more about this next time and the coderivation.  This example where the algebra is $SV$ is like functions on the manifold.  So the Hochschild on functions on a manifold is the smooth forms.











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