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\author{Gabriel C. Drummond-Cole}
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\title{Deformation Theory and Operads}
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Let $\Aa$ be the category of local Artin $k$-algebras with residue field $k$.  Local means one maximal ideal.  This should be commutative, associative, and unital.  I have a map $k\to R$ and another map $R\to R/m$.  I assume that $R/m$ is isomorphic to $k$ and that the composition of the reduction with the unit is an isomorphism.  I don't need to say this if $k$ is algebraically closed.

Artin by definition means that, it's some finiteness condition on ideals which translates into the following:  Such rings $R$ are isomorphic to $k\oplus m_R$ where $m_R$ is finite dimensional and nilpotent.  In fact, there's a structure theorem giving each $R$ as a polynomial ring in a finite number of variables modulo an ideal which is nilpotent, where some power of $t$ is zero.  As an example, $k[t_1,t_2,t_3]/\langle t_1^4-t_2^2 t^3,t_2^10, t_3^2\rangle$.  The morphisms are morphisms of $k$-algebras.

Let $\hat{\Aa}$ be the category of complete local Noetherian $k$-algebras, let me say it this way, of Noetherian projective limits of local Artin algebras.  These are local rings with a maximal ideal $m$ such that $R/m$ is isomorphic to $k$ and $R/m^n$ is in $A$ for each $n$.  In fact $R$ is the limit of the diagrams $\cdots\to R/m^3\to R/m^2\to R/m=k$.  Every chain of ideals stabilizes, $I_1\subset I_2\subset\cdots$.  Every ascending chain of ideals is finite.  Artin means every descending chain is finite, and that is stronger and implies Noetherian.

Example: $k[t]/t^5$ is Artin.  The ideals are spanned by $t$, $t^2$, and so on.  There are only four nontrivial ideals, and so there is just one chain, which is finite.  Now, $k[[t]]$ is Noetherian.  You have the ideals generated by $t^n$.  But $k[t_1,t_2,\ldots]/\langle t_i^2\rangle$ is not Noetherian and you don't want to consider that.  This diagram does not relate well to topology.

\begin{defi}
A functor of Artin rings is a covariant functor $\Aa\to Sets$.  An element $\gamma\in F(A)$ is called a deformation of $\gamma_0\in F(A/m)$ where $\gamma_0$ is the image of $\gamma$ under the map $F(A)\mapsto F(A/m)$ where this is the induced map.

We define the tangent space to $F$ to be $t_F=F(k[\epsilon])$ where $k[\epsilon]$ means $k[\epsilon]/\epsilon^2$.
\end{defi}
Let's look at some examples.  Let $(V,\bullet)$ be an associative $k$-algebra.  Define $F(A)$ to be $F(A)=\text{Deformations of }(V,\bullet)\text{ over }A$ which is associative products from $(V\otimes m_A)^{\otimes_A 2}\to V\otimes m_A$.  Recall if $(A,\epsilon)$ is an augmented ring, where $\epsilon:A\to k$ then a deformation of $(V,\bullet)$ over $(A,\epsilon)$ is an associative product $\mu$ on $V\otimes A$ such that the map $V\otimes A\to V\otimes k\to V$ is a map of algebras.

Given an Artin local ring $A$ we have an augmentation $\epsilon:A\to A/m\cong k$.  So such a deformation is an associative product $\mu$ such that $V\otimes A\to V$ is a map of algebras.  Since $A\cong k\oplus m$ this is $V\oplus V\otimes m\to V$.  The part in the $V$ cannot change, and the map kills the $m$ side.

How do I define $f$ on morphisms?  Given a deformation over $A$ and a homomorphism over $B$ I need to define a homomorphism over $B$.  We can express $\mu(v,w)=vw+b(v,w)$ where $b(v,w)\in V\otimes m_A$.  So I can define a new product on $V\otimes B$ by $\mu_{\sigma}(v,w)=vw+\sigma(b(v,w))$.

The next example would have this be deformations modulo equivalence and I'd also get a functor.  But here's the example that I like.  Fix an $R$ in $\hat{A}$ and define $h:\Aa\to Sets$ by $h_R(A)=hom(R,A)$.  This defines a functor, and given a morphism $\sigma:A\to B$ then $h_R(\sigma)$ is defined as $hom_{A^\sim}:hom(R,A)\to hom(R,B)$ which takes $f$ to $\sigma f$.  
\begin{defi}
A functor $F$ of Artin rings is rpepresntable by $R\in \hat{A}$ if and only if $F$ is naturally equivalent to $h_{R}$ for $R\in \hat{A}$.  Remark: suppose $F=h_R$ for some $R\in \hat{A}$.  Then $F(A)\cong hom(R,A)$ for all $A\in \hat{A}$.  However, any $F:\Aa\to Set$ may be extended to $\hat{F}:\hat{A}\to Set$ by $\hat{F}(S)=lim\ F(S/m^n)$.  This is a sequence.  An element in $F(S)$ is a sequence of elements in $F(S/m^n)$
\end{defi}
You end up replacing the set of deformations with a set of ring homomorphisms.

The functors extend naturally to limits of Artin rings, so in particular, $F(R)\cong hom(R,R)$.

There exists a $\Gamma\in F(R)$ so that $\Gamma$ corresponds to the identity in $hom(R,R)$.

There's a particular deformation which satisfies a universal property.  Notice that $\Gamma$ satsifies a universal property:  Given any $A\in \Aa$ and $\gamma\in F(A)$, there exists a unique ring homomorphism $R\to A$ such that $F(\sigma)(\Gamma)=\gamma$.  This is the end of my remark and is a good stopping place.  You come up with this when you chase the diagrams around.  
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