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\author{Gabriel C. Drummond-Cole}
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\title{Deformation Theory and Operads}
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Remember that we were talking about functors $F:\Aa\to Set$ where $\Aa$ was the category of Artin local rings.  We had a set of conditions:
\begin{enumerate}
\item[H0]$F(k)$ is a point
\item[H2']$\alpha:F(A\times_C B)\o F(A)\times_{F(C)}F(B)$ is a bijection for $A=B=\kk[\epsilon],C=\kk$
\item[H2]$\alpha$ is a bijection when $A=\kk[\epsilon],C=\kk$
\item[H3]$F(k[\epsilon])$ which is the definition of the tangent space $T_F$ is finite dimensional.
\item[H4]$\alpha$ is a bijection for $A\to C$ small with $B=\kk$, so that $F(A')\to F(A)$ is a bijection 
\end{enumerate}
The first two of these say that $T_F$ is a vector space; addition comes from $\kk[\epsilon]\times_\kk \kk[\epsilon]\to \kk[\epsilon]$

[Discussion broken by H4]

\section{Differential Graded Lie algebras}
A differential graded Lie algebra, or dgLa, is a triple $(V,d,[\,,\,])$ here $V$ is a $\zZ$-graded vector space, $d:V^i\to V^{i+1}$ is a square zero degree one operator, $[\,,\,]$ is a degree zero bilinear operator on $V$ which is graded skew symmetric and satisfies graded Jacobi:
$$[x,y]+(-1)^{|x||y|}[y,x]=0$$
$$
[x,[y,z]]=[[x,y],z]+(-1)^{|x||y|}[y,[x,z]]
$$
The compatibility between differential and bracket is
$$d[x,y]=[dx,y]+(-1)^{|x|}[x,dy]$$
I think of this as saying that bracketing with $x$ is a derivation of degree $|x|$ of the bracket, and $d$ is a derivation of the bracket.

If $x$ is even, then $[x,x]=0$.  If $x$ is odd, then $[x,[x,x]]=0$.  Also, $[x,[x,y]]=[[x,x],y]+(-1)^{|x||x|}[x,[x,y]]$ so that $2[x,[x,y]]=[[x,x],y]$.

If $C=(W,d,\bullet)$ is a graded commutative associative algebra and $L=(V,d,[\,,\,])$ is a dgLa, then $L\otimes C$ which is defined as $(V\otimes W,d\otimes 1+1\otimes d,[\,,\,])$ is a dgLa.

This could have been discussed in terms of chain complexes.  The commutative operad has just one operad in each arity; tensoring with the commutative operad is trivial, and so tensoring with a commutative algebra preserves the type of algebra you started with.

The formula for the bracket in $L\otimes C$ says that $[v\otimes a,w\otimes c]=(-1)^{|a||w|}[v,w]\otimes ab$.

Now I want to define a functor of Artin rings, fixing a dgLa.  One may deform $d$ of a dgLa in the direction of an inner derivation.  You want the new operator $d_a:=d+[a,-]$ to be a square zero degree one derivation of the bracket.  It will always be a derivation; for the grading to be correct you want $a$ to be in degree one.  If $d_a+\frac{1}{2}[a,a]=0$ and $a\in V^1$ then $d_a$ squares to zero.

$$d_a^2(x)=d^2x+d[a,x]+[a,dx]+[a,[dx]]+[a,[a,x]]
=[da,x]-[a,dx]+[a,dx]+[a,[a,x]]
$$
If $a$ is odd then this means $[da+\frac{1}{2}[a,a],x]=0$.
This leads to the equation for an element of degree one satisfying $da+\frac{1}{2}[a,a]=0$, which is called the Maurer Cartan equation or the classical master equation, or even the deformation equation.

An observation: given a graded Lie algebra and an element $a\in V^1$ satisfying the equation $[a,a]=0$, one has a diferential graded Lie algebra $(V,d,[\,,\,])$ where $dx:=[a,x]$

Let's look the other way around.  Given a dgLa $L$ define a new $dgLa$ by $L'=(V\oplus kd,[\,,\,]_{new})$ where $[v+\alpha d,w+\beta d]=[v,w]+\alpha dw +(-1)^{|v|}\beta dv$ which lives in $V\subset V\oplus kd$
Then $[d,d]_{new}=0$ so we have a differential on $L'$ defined by $d'=[d,-]$.  One thing I didn't check was the Jacobi identity here.  Let me go on.  Now $d'(v+\alpha d):=[d,v+\alpha d]_{new}=dv$.  I should have said that the degree of my new piece is $1$.  So now I get a new differential graded Lie algebra of the form $L'=(V\oplus kd, d'=[d,-],[\,,\,]_{new})$.  The idea is that by extending the Lie algebra, I can think of the differential as being bracketing with a degree one element.  So it's an extension of $L$, which is a subalgebra.

In this context, we have a map $L\to L'$, and I'll give a new map, $a\mapsto a+d$.  Notice that, I would like this to be a map of the degree one part of $L$ into the degree one part of $L'$; well, let me say it like this: if $a\in V^1$, then $[d+a,d+a]_{new}=0$ in $L'$ if and only if $da+\frac{1}{2}[a,a]=0$ in $L$.

Recall that in the Hochschild complex, well, if we have $V$ we have $HC^.(V,V):=\bigoplus Hom(V^{\otimes n},V)$.  There is a bracket on this complex, making it into a graded Lie algebra.  If $\bullet:V^{\otimes 2}\to V$ is a degree one element, and $[\bullet,\bullet]=0$, then $\bullet$ is an associative product.  That was if and only if.  Then we obtain a differential graded Lie algebra $(HC^.(V,V),d=[\bullet,-],[\,,\,])$.  I sort of think of the fundamental equation being $[a,a]=0$.

Now, let's define, well, let $L$ be a differential graded Lie algebra, and define a functor of Artin rings $MC_L:\Aa\to Set$ by $MC_L(A)=\{\gamma\in (L\otimes m_A)^1|d\gamma+\frac{1}{2}[\gamma,\gamma]=0\}$, which are equal to deformations of the differential $d$ in the direction $\gamma$, including parameters from $A$.

Given an associative algebra $(B,\bullet)$, define a $F_B:\Aa\to Sets$ by $F_B(A)$ are deformations of $(B,\bullet)$ over $A$.  Let $L=(HC(B,B),d=[\bullet,\,],[\,,\,])$.  Then $MC_L\cong F_B$ as functors of Artin rings.  I believe that this statement doesn't need an argument.

Let's look at, let's illustrate the bijection.  Start with $\gamma\in MC_L(A)\subset HC(B,B)\otimes m_A$.  So $\gamma:B\otimes B\to B$.  So $\gamma\in Hom(B\otimes B,B)\otimes m_A$, which is equal to, I claim, $Hom(B\otimes B,B\otimes m_A)$.  Then $\gamma(b_1,b_2)\in B\otimes m_A$.  Define a product $\bullet_\gamma:(B\otimes A)^{\otimes 2}\to B\otimes A$ by $\bullet_\gamma(b_1,b_2)=b_1\bullet b_2 + \gamma(b_1,b_2)$ where these pieces are in $B\otimes \kk$ and $B\otimes m_A$ so their sum is in $B\otimes A$.

Maurer-Cartan equation for $\gamma$ implies associativity for $\bullet_\gamma$ and the new product when you map down to $A$ gives you the old product.  I claim that's a bijection.  Remember that a deformation of $B$ over an augmented ring.  Modding out by the maximal ideal, you should get the old thing.  So it's exactly something of this form.  The associativity gets you Maurer Cartan for $\gamma$.

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