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\author{Gabriel C. Drummond-Cole}
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\title{Deformation Theory and Operads}
\begin{document}\maketitle

Recall that for $V$ a vector space over $k$, we had $TV=\bigoplus V^{\otimes n}$ and $LV=\bigcap_{V\subset g\subset TV} g$.  Then 
$$\tilde{\Delta}(v_1,\ldots, v_n)=\sum_{p,q\text{-shuffles, }p+q=n} (v_{\sigma(1)},\ldots, v_{\sigma(p)})\otimes (v_{\sigma(p+1)},\ldots, v_{\sigma(n)})$$
So $\gamma:TV\to LV$ takes $1$ to $0$ and $v\in V$ to itself.  It takes $v_1,\ldots, v_n$ to $[v_1,[v_2,\ldots,[v_{n-1},v_n]\ldots]]$
\begin{prop}
For $x\in V^{\otimes n}\subset TV$ the following are equivalent:
\begin{enumerate}
\item $x\in LV$
\item $\tilde{\Delta}(x)=x\otimes 1 + 1\otimes x$
\item $\gamma(x)=n.x$
\end{enumerate}
\end{prop}
The easiest implication is the third to the first.  Assume that $\gamma(x)=n.x$.  Then $x=\frac{\gamma(x)}{n}$ so it's in the free Lie algebra, since $\gamma$ is a map to the free Lie algebra.  All iterated brackets are in the free Lie algebra.

Let me show the first implies the second.  Let $\tilde{g}$ be those $x$ in $TV$ such that $\tilde{\Delta}(x)=x\otimes 1 + 1\otimes x$.  I claim that $\tilde{g}$ is a Lie subalgebra of $TV$.  If I have $x,y\in \tilde{g}$, and I take their bracket, I take $\tilde{\Delta}([x,y])$, what is that?  Let's see.  It's $\tilde{\Delta}(x y - yx)$.  Recall that $\tilde{\Delta}$ is an algebra map, so this is
$$\tilde{\Delta}(x)\tilde{\Delta}(y)-\tilde{\Delta}(y)\tilde{\Delta}(x)
=
(x\otimes 1+1\otimes x)(y\otimes 1+1\otimes y) - (y\otimes 1 + 1\otimes y)(x\otimes 1+1\otimes x)$$
If you foil this, you get
$$[x,y]\otimes 1 + 1 \otimes [x,y]$$.
So this is a subalgebra of $TX$ as a Lie algebra.  So $LV\subset \tilde{g}$.

The last implication assumes that $\tilde{\Delta}(x)=x\otimes 1 + 1\otimes x$.  We will see in a moment that $n.x=\bullet\circ(\gamma\otimes id_TV)\circ \tilde{\Delta}$.  This is a map from $TV$ to $TV\otimes TV$ to $LV\otimes TV$ to $TV$.  Then this is
$\bullet(\gamma\otimes id)(x\otimes 1 + 1\otimes x)$.  Since $\gamma(1)=0$ we have
$\bullet(\gamma(x)\otimes 1)=\gamma(x)$.  The only thing I owe you is to show that $nx=\bullet(\gamma\otimes id)\tilde{\Delta}$.  We show this by induction on $n$.  The first couple of pieces are pretty trivial.  If $n=0$ this is a scalar and will eventually go to $0$.  If you start with $v\in V$ you get $\bullet(\gamma(v)\otimes 1)=\gamma(v)$.

Let $n\ge 2$; we'll do a little induction here.  Assume that, I'm going to show it for homogeneous elements.  Let $x=(v_1,\ldots, v_n)$.  Write $\tilde{\Delta}$, apply it to $(v_2,\ldots, v_n)$, so this is $1\otimes (v_2\ldots v_n )+\sum_i a_i\otimes b_i$.  The only thing I do is that $a_i\in V^{\otimes \ge 1}$.  

By induction, $(n-1)(v_2,\ldots, v_n)$ is $\bullet(\gamma\otimes id)\tilde{\Delta}(v_2,\ldots, v_n)=\bullet(\gamma\otimes id) (1\otimes (v_2,\ldots, v_n)+\sum a_i\otimes b_i]
=\sum\gamma_i(a_i), b_i)$.

Then $$
\bullet(\gamma \otimes id)\tilde{\Delta}(v_1,\ldots, v_n)=\bullet(\gamma\otimes id)[v_1\otimes 1
+1\otimes v_1)(1\otimes v_2\ldots v_n)]
=v_1\otimes v_2\ldots, v_n+\sum v_ia_i\otimes b_i
+1\ldots v_1\ldots v_n +\sum a_i\otimes vb_i
$$
$$
=\bullet(v_1\otimes v_2,\ldots, v_n)+\sum \gamma(v_1a_i)b_i+\sum(\gamma(a_i))v_i b_i
$$
The last piece of this is $[v_1,\gamma(a_i)]-v_1\gamma(a_i)-\gamma(a_i)v_1$
so we get
$$
v_1v_2\ldots, v_n+\sum v_1\gamma(a_i)b_i=v_1\ldots v_n +(n-1)v_1\ldots v_n=n(v_1\ldots v_n)
$$
\begin{cor}
for all $\theta:V\to g$ there exists a unique map $\tilde{\theta}:LV\to g$ a Lie algebra map which extends:
$$
\xymatrix{
&LV\ar@{->}[d]^{\tilde{\theta}}\\
V\ar[ur]^{inc}\ar[r]_\theta&g
}$$
\end{cor}
{\it Proof.} Define $\tilde{\theta}([v_1,[v_2,\ldots, [v_{n-1},v_n]\ldots ]])$ to be $\{\theta v_1,\{\theta v_2,\ldots\{\theta v_{n-1},\theta v_n\}\ldots\}\}$ be the unique map induced by the Lie relation.  We need to show that for $x,y$ in $LV$, we have $\tilde{\theta}([x,y])=\{\tilde{\theta}x,\tilde{\theta}y\}$
By induction on $x\in L^nV$ we see for $n=1$ that $x\in V$.  For $n>1$ let $x=[v,\bar{x}]$ Then $$\tilde{\theta}([x,y])=\tilde{\theta}([[v,\bar{x}],y])\underbrace{=}_{\text{Jacobi}}\tilde{\theta}(\pm [v,[\bar{x},y]]\pm [\bar{x},[v,y]])
\underbrace{=}_{\text{induction}}\pm\{\tilde{\theta}v,\tilde{\theta}[\bar{x},y]\}\pm \{\tilde{\theta}\bar{x},\tilde{\theta}[v,y]\}
\underbrace{=}_{\text{induction}}\pm\{\tilde{\theta}v,\tilde{\theta}\{\tilde{\theta}\bar{x},\tilde{\theta}y\}\}\pm \{\tilde{\theta}\bar{x},\tilde{\theta}\{\tilde{\theta}v,\tilde{\theta}y\}\}
\underbrace{=}_{\text{Jacobi}}\{\{\tilde{\theta}v,\tilde{\theta}\bar{x}\},\tilde{\theta}y\}\underbrace{=}_{\text{induction}}\{\tilde{\theta}[v,\bar{x}],\tilde{\theta} y\}=\{\tilde{\theta}x,\tilde{\theta}y\}$$
Back to the Lie operad.  $Lie(n)$ is the span of monomials in $LV$ with exactly one $v_1,\ldots, v_n$
\begin{lemma}The brackets $[v_{\sigma(1)},[v_{\sigma(2)},\ldots,[v_{\sigma(n-1)},v_n],\ldots,]]$ for $\sigma\in \Sigma_{n-1}$ form a basis for $Lie(n)$.  In particular, the dimension is $(n-1)!$
\end{lemma}
You just use anticommutativity to put one thing at the end, and then Jacobi to move the other brackets to this form.

[Discussion of $Lie\to Ass\to Com$.  We agree that the maps don't go in the other direction.]

Let's to another example, the little disks operad.  Let $\Cc$ be the topological category, and let $D^k$ be $\{x\in \rR^k|\,|x|\le 1\}$ be the $k$-ball.  We define a little disk to be a map $f:D^k\to D^k$ of the form $f(x_1,\ldots, x_k)= y_1+cx_1,\ldots, y_k+cx_k$ so that $c\le 1$ and $f(D^k)\subset D^k$.  The little $k$-disk operad consists of $D_k(n)=\{f_1,\ldots, f_n\}$ where each $f_i$ is a little disk such that their images can touch but not intersect in the interiors.

[What if they can't touch?] I think something breaks but I don't know what.  

Let me draw a picture. [picture]

This sits in $\rR^{n(k+1)}$.
Now I need to tell you the structures.  The identity in $D_k(1)$ is the identity of $D^k$.  I'm using the cartesian product, so $(f_1,\ldots, f_n).\sigma= (f_\sigma(1),\ldots, f_{\sigma(n)})$.

You can do this for cubes too, but it's very hard to write a map between the little disks to little cubes.

Let me finish this example.  The composition is as follows.  $\circ_i:D_k(n)\times D_k(m)\to D_k(n+m-1)$ takes $(f_1,\ldots, f_n)\circ_i (g_1,\ldots, g_n)\mapsto (f_1,\ldots, f_{i-1},f_i\circ g_1,\ldots f_i\circ g_m,f_{i+1},\ldots, f_n)$.  So here's an example [picture].

Let $(X,*)$ be a based topological space.  Let $\Omega^k X=\{r:D^k\to X|r(\boundary D^k)=*\}$.  This is the $k$th iterated loop space.  The way I've defined it is already nice.  I can do an algebra map.  We claim that $\Omega^k X$ is an algebra over $D_k$, so there is an operad map from $\Ff:D_k\to End_{\Omega^k X}$ which means that for all $n$ there are maps $\Ff_n:D_k(n)\to Hom(\Omega^k X)^n,\Omega^k X)$.

[Picture describing the map].

So for $t\in D^k$, given $r_1,\ldots, r_n$ from $D^k\to X$, $t$ maps to $r_j\circ f_j^{-1}(t)$ if $t\in f_j(D^k)$ for some $j$ and otherwise to the basepoint.





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