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\author{Gabriel C. Drummond-Cole}
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\title{Deformation Theory and Operads}
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I ended last time with two homework problems.  I defined $A_t$ as $\rR[x]/x^3-tx$.  The first question is, is $A_1$ isomorphic to $A_2$.  The other question is whether $A_0$ is isomorphic to $A_t$.  Maybe I won't spoil the answer for anyone.  It's a slightly interesting problem, actually.  
Let me make a definition.  Let $(A,\cdot)$ be an associative algebra over a field $\fieldk$.  Then a formal one-parameter deformation of $(A,\cdot)$ is an associative algebra structure $\mu:A[[t]]\otimes_{\fieldk[[t]]}A[[t]]\to A[[t]]$ such that the map such that the map $A[[t]]\to A$ (evaluation at zero) is a map of algebras over $\fieldk$. You could assemble this from what I said last time except that I didn't say $A[[t]]$.  This is actually $A\hat{\otimes} k[[t]]$.  Generally you want to deal with the completed tensor product.  Just to point out the difference:  If I do $A\otimes k[[t]]$, this would be finite sums of $a_i\otimes p_i(t)$.  If I take $A[[t]]$, this is infinite sums, power series in $t$ with coefficients in $a$.  The first one only allows finitely many distinct things from $A$.  If $\fieldk$ is topological, you can ask that this converge but in general you won't ask.  A formal one parameter deformation of $(A,\cdot)$ is determined by a collection $\{\mu_i\}_{i=1}^\infty$ of bilinear maps $A\times A\to A$ as follows, $\mu(x,y)=x\cdot y+\mu_1(x,y)t+\mu_2(x,y) t^2+\cdots$  I wanted to frame the second homework question in this language.  If $\mu$ is a one-parameter formal deformation of $(A,\cdot)$ then $\mu$ is associative is equivalent to an infinite collection of relations among the $\{\mu_i\}$.  In particular, $\mu_1$, well, $\mu(x,\mu(y,z))-\mu(\mu(x,y),z)=0$.  This is $x\cdot \mu(y,z)+t\mu_1(x,\mu(y,z)+\cdots-(
\mu(x,y)\cdot z+t\mu_1(\mu(x,y),z)$

I can expand the other $\mu$ and eventually get some equations that I can write down.  If these will be zero over $A[[t]]$ then each coefficient of $t^n$ is zero.  The constant coefficient is $0$ so $(xy)z-x(yz)=0$ and $x\mu_1(y,z)+\mu_1(x,yz)-\mu_1(x,y)z-\mu_1(xy,z)$
I wanted to isolate the $t^1$ condition.  Let me write this in an easier way to remember.

That is, $x\mu_1(y,z)-\mu_1(xy,z)+\mu_1(x,yz)-\mu_1(x,y)\cdot z=0$.

Let $(A,\cdot)$ be a $\fieldk$ algebra.  Two deformations $\mu$ and $\mu'$ of $(A,\cdot)$ over $R$ are equivalent if and only if there exists an isomorphism of $R$-algebras $\varphi:A\otimes R\to A\otimes R$ so that the diagram commutes:
$$
\xymatrix{
A\otimes R,\mu\ar[rr]\ar[dr]&& A\otimes R,\mu'\ar[dl]\\
&(A,\cdot)}
$$
In the case of two formal one parameter deformations $\mu$ and $\mu'$ they are equivalent if and only if there exists a sequence of linear maps $\varphi_1,\cdots$ so that for all $x$ and $y$, $\mu(x,y)=\varphi_t^{-1}\mu(\varphi_t(x),\varphi_t(y))$ where $\varphi_t(x)=x+t\varphi_1(x)+\cdots$

Given an associative algebra $(A,\cdot)$ over a field $\fieldk$, one always has the very trivial defined by $\mu_t$ is trivial if and only if it is equivalent to the veryt trivial one.

[\wh]

Now let me erase everything on the board.  I want to define the Hochschild complex.  

Everyone can make up their own example of a trivial deformation.  Pick any old maps.

\begin{defi}$C^n(A):=Hom_\fieldk(\underbrace{A\otimes\cdots\otimes A}_{n+1},A)$.  Then let $C:=\bigoplus{n\ge 0} C^n$.  This is a graded vector space.  I want to define a differential $C^{n-1}\to C^{n}$ by $(\delta f)(a_1,\ldots, a_{n+1}):=a_1f(a_2,\ldots, a_{n+1})-f(a_1a_2,a_3,\ldots, a_{n+1})+\cdots +(-1)^n f(a_1,\ldots, a_na_{n+1})+(-1)^{n+1}f(a_1,\ldots, a_n)\cdot a_{n+1}$.
\end{defi}

You might ask about the grading.  Let me finish the construction.  Fact, exercise, $\delta^2=0$.

So if $f:A\otimes A\to A$ then $\delta f(x,y,z)=xf(y,z)-f(xy,z)+f(x,yz)-f(x,y)z$.  The formula that was there before shows up, and we see that if $\mu$ is a formal deformation of $\cdot$ then $\delta\mu_1=0$.  How shall I summarize this?  
This leads in Kodaira-Spencer terms to the ``question of existence'': Given $(A,\cdot)$ and a Hochschild $1$-cocycle $\mu_1$, does there exist a formal 1 parameter deformation $\mu$ of $(A,\cdot)$ so that $\mu(x,y)=x\cdot y + t\mu_1(x,y)+\cdots$

Let me make this an exercise.  Show that if $\mu$ is a trivial $1$-parameter formal deformation then $\mu_1$ is a Hochschild coboundary.  That is, there exists a $\phi_1:A\to A$ so that $\delta\phi_1=\mu_1$.

This is a special case.  If two deformations are equivalent, then, if $\mu$ and $\mu'$ are equivalent then $\mu_1-\mu_1'$ is a Hochschild coboundary.  So the infinitessimal deformations, modulo equivalence, are the Hochschild $1$-cohomology.

The question of existence is typically hard.

I can make the following definition
\begin{defi}
Given an associative algebra $(A,\cdot)$ and a formal one parameter deformation $\mu_t$, we define $\frac{d\mu_t}{dt}|_{t=0}$ to be $[\mu_1]\in H(C^1,\delta)$.  The idea is that Hochschild $1$-cohomology classes are infinitessimal deformations of $(A,\cdot)$.
\end{defi}

Let's go to graded algebras.  Well, Gerstenhaber's circle product.  Given $f\in C^r,g\in C^s$, define $f\circ g\in C^{r+s}$ by $f\circ g(a_1,a_2,\ldots, r+s+1)=f(g(a_1,\ldots, s+1),a_{s+2},\ldots)+(-1)^s f(a_1,g(a_2,\ldots, a_{s+2}),\ldots, a_{r+s+1})+(-1)^{2s}\cdots$  So these signs will either all be plus or will alternate, depending on the sign of $s$.

I like to think of this pictorially.  [Pictures]  This is $f$ with some inputs and an output.  I let $g$ act on some number of inputs.  So I evaluate $g$ inside of $f$ like this, plus or minus, this, plus or minus this, and so on.

Here's a fact.  The associative product $\cdot$ on $A$ is an element of $C^1(A)$ since it's a map from $A\otimes A\to A$.  I'm already saying it's associative.  I want to say that $\cdot$ is associative is equivalent to saying $\cdot \circ \cdot=0$.

The condition that $\delta^2=0$, well, you have to use the associativity of $\cdot$ for that.  The differential $\delta$ can also be defined by $\delta(f)=\cdot \circ f-(-1)^{|f|} f\circ \cdot$.

Let's see that this is the same.  $\cdot \circ f(a_1,\ldots, a_{n+1})$, I evaluate $f$, so it's $f(a_1,\ldots, a_n)$

I skipped one part of trying to get the signs and I see I paid for it.  Elements that $\circ$ with themselves and give zero, those are interesting.  If you're in $C^1$, and do that, it's associative.  The interesting $1$ to $1$ things are differentials, but those are only in graded algebras.  The first thing I'd like to say, is how can I tell that $C^n=Hom(A^{\otimes n+1},A)$ should be degree $n$.  Let me give a practical reason.  Given a graded vector space $V$, define the shift $V[k]$ for an integer $k$ to be the graded vector space where $(V[k])^j = V^{k+j}$.  If I shift by one, for example, then $V[1]^j = V^{j+1}$.  If $V$ were in degree zero, then $V[1]$, well, everything will be concentrated in degree $-1$.

If $\varphi$ is in $Hom(V,V)$ has degree $j$, then $\varphi:V^i\to V^{i+j}$.  Or a map $\varphi:V^i\to V^k$ has degree in the homomorphism complex (forget complex, this is a vector space), $k-i$.  So if $A$ is a vector space that's not graded, you can view it as graded and concentrated in degree zero.  Then any map $\varphi:A^{\otimes n+1}\to A$ can be viewed as a map $\varphi:(A[1]^{\otimes n+1}\to A[1])$ of degree $n$. 

The formula for the circle product will be a little different for a graded vector space, but the Hochschild complex is robust enough to give you things where square zero elements are interesting.

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