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\author{Gabriel C. Drummond-Cole}
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\title{Deformation Theory and Operads}
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Recall.  What you need for an operad is a sequence of spaces $\Oo(n)$ for $n\ge 1$.  You have a right $\Sigma_n$ action on $\Oo(n)$, a $\circ_i:\Oo(n)\otimes \Oo(m)\to \Oo(n+m-1)$ satisfying associativity and equivariance, and then a unit $\Oo(1)$ which satisfies a unit condition.  The two examples were the endomorphism operad of a vector space and of a topological space.  We also did $Com$ with $Com(n)=k$ for $n\ge 1$ with the trivial action.  I want to introduce a couple more operads today, associative and Lie, but before that I should talk about another example.

[What about the sign representation on Com?]

Tensoring with this $\widetilde{Com}$ gives an operad where $A$ is an algebra over $\Oo\otimes \widetilde{Com}$ if $\Pi A$ is an algebra over $\Oo$.  This is in a paper of Tamarkin's.

Let $\Oo$ be an operad in vector spaces such that $\Oo(n)=0$ for all $n\ge 2$.  So then $\Oo(1)=V$.  This means that we have a right $\Sigma_1$ action.  That's vacuous.  Then we have an operation $\circ_1:V\otimes V\to V$ satisfying associativity, along with a unit for $\circ_1$.  In other words, we get an operad if and only if $(V,\mu,\mathbf{1})$ is an associative algebras.  So associative algebras sit inside operads.

Let me make the definition of an algebra over an operad.  First, let $\Oo,\Pp$ be two operads in $\Cc$ where this is either $Top$ or $Vect$.  Then a morphism of operads $\Ff:\Oo\to \Pp$ consists of a sequence of morphisms $\Ff_n:\Oo(n)\to \Pp(n)$ such that
\begin{enumerate}
\item$\Ff_n(f)\circ_i \Ff_n(g)=\Ff_{n+m-1}(f\circ_i g)$. 
\item$\Ff_n(f\cdot \sigma)=\Ff_n(f)\cdot \sigma$
\item$\Ff_1(\mathbf{1}_\Oo)=\mathbf{1}_\Pp$
\end{enumerate}
We call $V$ an algebra over an operad $\Oo$ if we are given a map from $\Oo$ into the endomorphisms of $V$.

I guess the first interesting example is that a map $\Com\to End_V$ is really for each $n$ a map $Com(n)\to Hom(V^{\otimes n},V)$.

So $Com(n)$ is $k$ so all you do is choose, well, start with $Com(1)$.  This should go into $Hom(V,V)$ so it should be the identity, the unit maps to the unit.  $Com(2)$ maps into $Hom(V^{\otimes 2},V)$.  This is the only information, as we'll see in a moment.  Take the unit of the field and this maps to $\mu:V^{\otimes 2}\to V$.  There is a $\sigma_2$ action here, and it is trivial.  So the product $\mu$ is commutative.

All higher ones are generated by this; if you look at $1_3$ in $Com(3)$, this is $1_2\circ_1 1_2$, which is $\mu_3:V^{\otimes 3}\to 3$.  This means that $\mu_3(v_1,v_2,v_3)=\Ff_3(1_3)(v_1,v_2,v_3)$, which is $$\Ff_3(1_2\circ_1 1_2)(v_1,v_2,v_3)=\Ff_2(I_2)\circ_1 \Ff_2(I_2)(v_1,v_2,v_3)=\mu\circ_1\mu(v_1,v_2,v_3)=\mu(\mu(v_1,v_2),v_3)$$

The outcome is that $V$ is an algebra over $Com$ if and only if $(V,\mu)$ is a commutative associative algebra.

Let's get to a more interesting example.  The associative operad $Assoc(n)$ is the vector space spanned by non-commuting monomials in the variables $x_1,\ldots, x_n$, containing each $x_i$ exactly once.  This is the span of $\{ x_{\sigma(1)},\ldots,x_{\sigma(n)}|\sigma\in \Sigma_n\}$.  So maybe let's do the lowest one.  The lowest one is spanned by $x_1$.  $Assoc(2)$ is spanned by $x_1x_2$ and $x_2x_1$; $Assoc(3)$ has six permutations.  We need a unit, which will be $x_1$.  The $\Sigma_n$ action is that $(x_{\sigma(1)},\ldots, x_{\sigma(n)})\rho=x_{\sigma\rho(1)},\ldots,x_{\sigma\rho(n)}$.  Check that this is an action, that $(\bar{x}_\sigma\rho)\rho'=(\bar{x}_\sigma)\rho\rho'$.

Composition of $(x_{\sigma_1},\ldots, x_{\sigma_n})\circ_i (x_{\rho_1},\ldots, x_{\rho_m})$ is the substitution of $x_{\rho_1},\ldots, x_{\rho_m}$ at the $x_i$th spot.  So $(x_3x_1x_2x_4)\circ_2(x_3x_2x_1)$.  This should be in $Assoc(6)$.  It's $x_5x_1(x_4x_3x_2)x_6$.  I put the parentheses to acknowledge the placment.  They don't matter.

So what is an algebra over $Assoc$?  It's $\Ff_n:Assoc(n)\to Hom(V^{\otimes n},V)$.  So the unit goes to the identity again, and then $x_1x_2\in Assoc(2)$ goes to $\mu\in Hom(V^{\otimes 2},V)$.  Then $x_2x_1$ goes to $\tilde{\mu}\in Hom(V^{\otimes 2},V)$.  So $\mu(v_1,v_2)=\Ff_2(x_1x_2)=\Ff_2(x_2x_1.\tau)(v_1,v_2)=\Ff_2(x_2,x_1).\tau(v_1,v_2)=\tilde{\mu}.\tau(v_,v_2)=\tilde{\mu}(v_2,v_1)$ so $\tilde{\mu}=\mu^{op}$.  So the $\circ_i$ operation determine the higher operations in terms of the lower ones, and that the product is associative.

From the $\Sigma_n$ action we see that $$\Ff_n(x_{\sigma 1},\ldots, x_{\sigma n})(v_1,\ldots, v_n)=\Ff_n(x_1,\ldots, x_n .\sigma)(v_1,\ldots, v_n)=\Ff_n(x_1,\ldots, x_n).\sigma(v_1,\ldots, v_n)=\Ff_n(x_1,\ldots, x_n)(v_{\sigma^{-1} 1},\ldots, v_{\sigma^{-1} n})=(v_{\sigma^{-1} 1},\ldots, v_{\sigma^{-1} n})$$

So this is the general scheme.  If you have a structure and want to build the operad that governs it, we've done the associative and the commutative.  We could do the commutative differently.  The associative algebra was $x_1,\ldots, x_n$ and, well, let's see how I want to write this.  Note that we could have also written $\Com(n)$ to be the span of all monomials with exactly one $x_1$ up to $x_n$ living in the free symmetric algebra.

Let $Lie(n)$ be the span of all monomials with exactly one $x_1,\ldots, x_n$ in the free Lie algebra $L(x_1,\ldots,x_n)$.  Recall that $(L,[\,,\,])$ is a Lie algebra if $[\,,\,]:L\otimes L\to L$ satisfies $[\ell_1,\ell_2]=-[\ell_2,\ell_1]$ and $[[\ell_1,\ell_2],\ell_3]+[[\ell_2,\ell_3],\ell_1]+[[\ell_3,\ell_1],\ell_2]=0$ (Jacobi)  You think of the free Lie algebra as anything you can build out of brackets.  Any iterated brackets up to anticommutativity and Jacobi.

Recall that for a vector space $V$ we have the free tensor algebra $TV$, the sum $$\bigoplus_{n\ge 0} V^{\otimes n}$$ satisfying that it is free, namely for all vector space maps $V\to A$ there is a unique lift $TV\to A$ such that you have some commutative diagram.
$$
\xymatrix{
&TV\ar@{->}[d]_{\tilde{f}}\\
V\ar[ur]^i\ar[r]_f&A}
$$
This map just depends on the lowest component.  You eventually see that $\tilde{f}(v_1\otimes\cdots\otimes v_n)=f(v_1)\cdots f(v_n)$.  There is an induced Lie algebra structure on $TV$ given by $[x,y]=x\otimes y-y\otimes x$.  Then define $LV$ to be the Lie subalgebra of $TV,[\,,\,]$ generated by $V$.  This is explicitly the intersection of all Lie algebras inside $TV$ containing $V$.  This is $LV$ and one of the goals is to show that this is the free Lie algebra generated by $V$.  For each vector space morphism there is a unique lift to $LV$.  It's unfortunately hard to compute or calculate with $LV$.  

What I'd like to do is give a characterization of $LV$.  This is not done very often.  Note that $TV\otimes TV$ is an associative algebra: $(x\otimes y)\bullet (r\otimes s)=(x\cdot r)\otimes (y\cdot s)$ so using $\Delta:V\to TV\otimes TV$, which maps $V$ to $V\otimes 1+1\otimes V$, I just gave you a map of $V$ into some algebra.  Then using the universal property there is an induced algebra map $\tilde{\Delta}:TV\to TV\otimes TV$.  How does this act?  So $\tilde{\Delta}(v_1,\ldots, v_n)$ is determined by its lowest components:  It's $\prod \Delta(v_i)$  So this is $(v_1\otimes 1+1\otimes v_1)\cdots(v_n\otimes 1+1\otimes v_n)$, and multiplying this out you get, for each one, $v_i$ in one or the other factor.  This is called an unshuffle or a shuffle
$$
\sum_{p+q=n}\sum_{(p,q)-\text{shuffles}\sigma}\left(x_{\sigma 1}\otimes\cdots\otimes x_{\sigma p}\right)\otimes\left(x_{\sigma (p+1)}\otimes \cdots \otimes x_{\sigma n}\right)
$$
Where $\sigma$ is a $p,q$ shuffle if $\sigma(1)<\cdots<\sigma(p)$
and $\sigma(p+1)<\cdots<\sigma(n)$.

Let me state, well, this is commutative.  You can shuffle in the opposite order, so this is actually a commutative bialgebra.  To continue, this is a Hopf algebra with antipode given by $S(v_1,\ldots, v_n)$ is $(v_n\otimes\cdots \otimes v_1)$

Now let $\gamma:TV\to LV$ be given by $\gamma(1_{TV})=0$ and $\gamma(v)=v$.  Then $$\gamma(v_1\otimes\cdots\otimes v_n)=[v_1,[v_2,\cdots,[v_{n-1},v_n]\cdots]]$$
\begin{prop}
For $x\in V^{\otimes n}\subset TV$ the following are equivalent:
\begin{enumerate}
\item $x\in LV$
\item $\tilde{\Delta}(x)=x\otimes 1+1\otimes x$ (Friedrichs)
\item $\gamma(x)=nx$ (Dynkin-Specht-Weiner)
\end{enumerate}
\end{prop}
I'm out of time.

\end{document}
