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\author{Gabriel C. Drummond-Cole}
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\title{Deformation Theory and Operads}
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I want to do at least one more example.  The other example I did was \begin{enumerate}\item$A=TV$.
\item Let $A=SV=TV/(v_iv_j=v_jv_i)$ be the symmetric algebra on $V$ over $k$.  Denote by $\wedge^.V$ the exterior algebra, $TV/(v_iv_j=-v_jv_i)$.  Denote $u\odot v$ the product in $SV$ and $u\wedge v$ the product in $\wedge V$, and $u\in SV$ is $u_1\odot u_2\odot\cdots$ and in $\wedge V$ we have $u=u_1\wedge u_2\wedge\cdots$  I'm assuming my vector spaces are finite dimensional.  Define the following complex for $V$, $d$-dimensional:
$$
\to SV\otimes \wedge^d V\otimes SV\to\cdots\to SV\otimes V\otimes SV\to SV\otimes SV\to 0
$$
with an augmentation to $SV$.

First of all I need to give you these maps $b'$.  So 
$$
b'(u\otimes v_1\wedge\cdots\wedge v_r\otimes w)=\sum_{i=1}^r (-1)^{i+1}(u\odot v_i\otimes v_1\wedge\cdots\wedge \hat{v_i}\wedge \cdots\wedge v_r\otimes w -
u \otimes v_1\wedge\cdots\wedge \hat{v_i}\wedge \cdots\wedge v_r\otimes v_i\odot w)
$$
The augmentation is $\odot$.  Note that if $V$ is one dimensional then $SV$ equals $TV$ and this resolution is the same resolution we had before.  If $U$ and $X$ are finite dimensional vector spaces then $P(U)\otimes P(X)\cong P(U\oplus X)$.  So this is an isomorphism of dgas.  It respects, you can go backwards.  [Calculation]  So we have an isomorphism of chain complexes and then we can put these together to get that this is a resolution for $X$ of any dimension.
\end{enumerate}
So now we can find out what our Hochschild is, tensoring over $SV\otimes SV^{op}$ with $SV$, and we get $wedge V\otimes SV$.
Let me do this more precisely:  the maps $b'\otimes id$ transfer to $\wedge^*V\otimes SV$ as $0$.  Let me look at the diagram and show that it commutes.  If we have $u\otimes v\otimes w$ and we tensor this with $x$.  This maps to $v\otimes w\odot x\odot u$ which goes to zero.  The other direction this will go to $\sum \pm(((u\odot v_i )\otimes \hat{v}^i\otimes w)\otimes x - (u\otimes \hat{v}^i\otimes(v_i\odot w))\otimes x)$ which clearly goes to zero under this map.

So $HH_.(SV,SV)=H_.(P\otimes SV)=\wedge V\otimes SV$.  As a homework, calculate $HH^.(SV,SV)$.

I at least want to state the third example that one can do.  The third example is to take $A=k[x]/x^{n+1}=0$.  The claim is that the following is a projective $A^e$ resolution: $\cdots\to A^e\to A^e\to\cdots\to A^e$, where the augmentation is the product and you multiply by $u$ and $v$ alternatingly, where $u=x\otimes 1-1\otimes x$ and $v=x^n\otimes 1 +x^{n-1}\otimes x+\cdots+x^n\otimes A\otimes A^{op}$.  Then you can show that for every bimodule $M$ over $A$, the $HH_.(A,M)$ and $HH^.(A,M)$ are two periodic and if $\frac{1}{n+1}\in k$ then $HH_i(A,A)\cong HH^j(A,A)\cong A/x^nA$ 

I want to continue with the coderivation point of view.

Let me start by saying what a derivation is.  Let $(A,\mu)$ be an associative algebra.  A derivation $D:A\to A$ is a linear map satisfying $D(a\cdot b)=Da\cdot b + (-1)^{|a||D|}a\cdot Db$.  Here $A$ is graded $A=\oplus A^i$ and the degree of $a$ is $|a|$.  (Assume $\mu$ is of degree zero).

The diagram I just wrote down was 
$$\xymatrix{
A\otimes A\ar[r]^\mu\ar[d]_{D\otimes id + id\otimes D}&A\ar[d]^D\\
A\otimes A\ar[r]_\mu&A
}$$
The sign looks weird, but $(id\otimes D)(a\otimes b)$ is $(-1)^{|D||a|}a\otimes D(b)$.

A coalgebra is a pair $(C,\Delta)$ where $C$ is a graded $S$-module and $\Delta:C\to C\otimes C$ such that we have coassociativity, degree zero.  Coassociativity means that 
$$\xymatrix{
C\ar[r]^\Delta\ar[d]_\Delta&C\otimes C\ar[d]^{\Delta\otimes id}\\
C\otimes C\ar[r]_{id\otimes \Delta}&C\otimes C\otimes C
}$$
commutes.

A coderivation is a map $f:C\to C$ to make the following commute:
$$\xymatrix{
C\ar[d]_f\ar[r]^\Delta&C\otimes C\ar[d]^{f\otimes id+id\otimes f}
\\C\ar[r]_\Delta&C\otimes C
}$$
The main example we'll be using is the tensor coalgebra $T^cV$ which is $TV$ with the structure
$$\Delta(v_1,\ldots, v_n)=1\otimes (v_1,\ldots, v_n)+v_1\otimes (v_2,\ldots, v_n)+\cdots +(v_1,\ldots, v_n)\otimes 1$$ where $1\in k=V^0$.  Coassociativity corresponds to breaking things up in two ways, in either order.

You can characterize the coderivations on the tensor coalgebra nicely as follows.  Every coderivation $TV\to TV$ is uniquely determined by its component $f^1:TV\to TV\to V$.  There should be enough time to prove this.  I don't want to write out the formula but in the end you will see it come out.

First of all, let $f^k:TV\to TV\to V^{\otimes k}$.  We need to show that $f^k$ is determined by $f^1$.  We have the following, we know that $f$ is a coderivation, so we have the following diagram:
$$
\xymatrix{TV\ar[r]^\Delta\ar[d]_f&TV\otimes TV\ar[d]_{f\otimes id+id\otimes f}\\
TV\ar[r]_\Delta&TV\otimes TV}
$$
If there was an $f^0$ then we'd have
$$
\xymatrix{
(v_1,\ldots, v_m)\ar[r]^\Delta\ar[d]&1\otimes (v_1,\ldots, v_m)+\cdots +(v_1,\ldots, v_m)\otimes 1\ar[dr] 
\\
f^0(v_1,\ldots, v_m)1_{TV}+\ldots\ar[r]_\Delta&f^0(v_1,\ldots, v_m)1\otimes 1&2f^0(v_1,\ldots, v_m)1\otimes 1
}$$
These are different unless $0=1$ in $k$.  If we assume that $m$ is minimal in the sense that $f^0(v_1,\ldots, v_k)=0$ for $k\le m$, then we get a contradiction that $1=2$.

So in the general case, $(v_1,\ldots, v_m)$ goes to $\cdots+f^k(v_1,\ldots, v_m)+\cdots$ where this part will be called $(w_1,\ldots, w_k)$ which splits in all possible ways, and lands in $V^{\otimes i}\otimes V^{\otimes k-i}$.  For a given component $V^i\otimes V^{k-i}$, we can reconstruct $f^k$ if we know what the value is on the other side.  So by applying $\Delta$ we get, well, the only things that will land in $V^i\otimes V^{k-i}$ will be the things that have either $i$ factors in the first part or $k-i$ in the second, and then what maps to use are determined by the factors they must land in.

So I get the statement that $f^k(v_1,\ldots, v_m)=f^i(v_1,\ldots, v_{m-k+i},v_{m-k+i+1},\ldots v_m)+(-1)^{|f|(|v_1|+\ldots +|v_i|}(v_1,\ldots, v_i, f^{k-i}(v_{i+1},\ldots, v_m)$.
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