\section{Degree by Degree Constructions} \label{sec:dbd}
\parshape 1 \parindent \quotewidth
{\small\noindent{\bf Summary. }
  Using standard deformation theory techniques, we show that if
  $H^2(G,\bbQ)=0$ then $G$ has a Taylor expansion that can be constructed
  inductively.
}
\vskip 1mm

An ``expansion to degree $d$'' (sometimes, ``partial
expansion''), often denoted $Z_d$, is the same as an expansion
(Definition~\ref{def:Expansion}), except that it takes values in the
part $\calA_{\leq d}(G)$ of degrees at most than $d$ of $\calA(G)$
(equivalently, in $\calA(G)/\calA_{>d}(G)$), and that the condition for an
expansion, Equation~\eqref{eq:ZExpansion}, is imposed only for $n\leq d$.
It similarly makes sense to speak of ``multiplicative'' and ``Taylor''
expansions, and of $A$-expansions to degree $d$.

Note that by Proposition~\ref{prop:Deg1}, for any group $G$ and any $g\in
G$ we must set $Z_1(g)=1+\bar{g}$;
this definition is Taylor to degree $1$.

We say that a multiplicative (or Taylor or $A$-) expansion to degree $d$
named $Z_d$ is ``extendible'' if we can find a $Z_{d+1}$, a multiplicative
(or Taylor or \text{$A$-)} expansion to degree $d+1$, whose restriction
to degrees $\leq d$ is $Z_d$. (By an argument similar to the proof of
Proposition~\ref{prop:ExpansionsExist}, all partial expansions are
extendible as ``plain expansions'', without multiplicativity or a Taylor
property).

We say that a group $G$ has the mutiplicative extension property (or the
Taylor extension property) is for every $d\geq 1$ every multiplicative
(or Taylor) expansion to degree $d$ is extendible.

\begin{example} Free groups have the multiplicative / Taylor extension
property. For the multiplicative extension property, simply extend any
expansion to degree $d$ by choosing degree $d+1$ values for the extension on
the generators in an arbitrary manner. For the Taylor extension property
use the fact cited below in Capsule~\ref{cap:GroupLike} that group-like
elements to degree $d$ always extend to group-like elements to degree
$d+1$.
\end{example}

We give a very brief definition of the second cohomology
$H^2 = H^2(G;V)$ of a group $G$, with coefficients in some
vector space $V$ over some field $\bbF$ (which we will later fix to be
$\bbQ$). Further information can be found e.g.\ in
Weibel's~\cite{Weibel:HomologicalAlgebra}.

Given a group $G$, let $C^k\coloneqq C^k(G;V) \coloneqq \{\varphi\colon
G^k\to V\}$ be the set of $k$-ary $V$-valued functions on $G$, for
$k=1,2,3$. Define $d^1\colon C^1\to C^2$ and $d^2\colon C^2\to C^3$ on
$\phi\in C^1$ and $\epsilon\in C^2$ as
follows:
\[ (d^1\phi)(g_1,g_2) \coloneqq
  \phi(g_1) - \phi(g_1g_2) + \phi(g_2),
\]
\begin{multline*}
  (d^2\epsilon)(g_1,g_2,g_3) \coloneqq \epsilon(g_2,g_3) 
    - \epsilon(g_1g_2,g_3) \\
    + \epsilon(g_1,g_2g_3) - \epsilon(g_1,g_2).
\end{multline*}
It is easy to verify that $d^1\act d^2=0$, and hence it makes sense to set
$H^2 \coloneqq H^2(G;V) \coloneqq \ker(d^2)/\im(d^1)$.

Our plan is to extend expansions. As a first step, we prove a general
``extension'' lemma. Let $(B,\eta)$ be an augmented $\bbF$-algebra (an
$\bbF$-algebra $B$ along with a multiplicative $\eta\colon B\to\bbF$), and
let $C$ be an ideal within $B$ which is trivial as a $B$-bimodule; namely,
such that $bc=cb=\eta(b)c$ for every $b\in B$ and $c\in C$. Let $\pi\colon
B\to B/C$ be the projection. It is a morphism of augmented algebras.

\parpic[r]{$\xymatrix{
 & B \ar[d]^\pi \\
 G \ar[r]^{Z_0} \ar@{-->}[ru]^{\exists Z} & B/C
}$}
\begin{lemma} \label{lem:H2} If $H^2(G;C)=0$, then every multiplicative
map $Z_0\colon G\to B/C$ for which $Z_0\act\eta=1$ can be lifted to a
multiplicative $Z\colon G\to B$ for which $Z\act\eta=1$.
\end{lemma}

\begin{proof} Pick an arbitrary, possibly not-multiplicative, lift $Z'\colon
G\to B$. Its failure to be multiplicative is measured by a functional
$\epsilon'\coloneqq\epsilon(Z')\in C^2(G;C)$, where
\begin{equation} \label{eq:epsilon}
  \epsilon(Z')(g_1,g_2) \coloneqq Z'(g_1g_2) - Z'(g_1)Z'(g_2)
\end{equation}
(the multiplicativity of $\pi$ implies that $\epsilon'\act\pi=0$, so
$\epsilon'$ takes values in $C$). We claim that $d^2\epsilon'=0$. Indeed,
in the diamond
\[ \xymatrix@C=-6pt{
  & Z'(g_1g_2)Z'(g_3) \ar@{-}[rd]^{\epsilon'(g_1,g_2)} & \\
  Z'(g_1g_2g_3)
      \ar@{-}[ru]^{\epsilon'(g_1g_2,g_3)}
      \ar@{-}[rd]^{\epsilon'(g_1,g_2g_3)}
    & & Z'(g_1)Z'(g_2)Z'(g_3) \\
  & Z'(g_1)Z'(g_2g_3) \ar@{-}[ru]^{\epsilon'(g_2,g_3)} &
} \]
the quantity indicated on each edge is the difference of the quantity on
the vertex to its left with the quantity on the vertex to its right (for
the two edges on the right, we use the triviallity of the action of $B$ on
$C$ and the condition $Z_0\act\eta=1$). Hence by telescoping the sum of
the upper two edge-quantities is equal to the sum of the bottom two. That
is, $d^2\epsilon'=0$.

As $H^2(G;C)=0$, we can find $\phi\in C^1(G;C)$ such that
$d_1\phi = \epsilon'$. Set $Z\coloneqq
Z'+\phi$. Studying~\eqref{eq:epsilon} (and again using the triviallity
of the action of $B$ on $C$ and the condition $Z_0\act\eta=1$) we find that
\[ \epsilon(Z) = \epsilon(Z') - d^1\phi = \epsilon'-\epsilon' = 0, \]
which means that $Z$ is multiplicative.\qed
\end{proof}

The road is now clear.

\begin{theorem} \label{thm:H2} If $H^2(G;\bbQ)=0$ then $G$ has the
multiplicative extension property: every multiplicative partial
expansion $Z_d$, for $d\geq 1$, can be extended. In particular, $G$
has a multiplicative (full) expansion $Z$.
\end{theorem}

\begin{proof} If $H^2(G;\bbQ)=0$ then $H^2(G;V)=0$ for any $\bbQ$-vector
space $V$. For the first statement of the theorem, take $B=\calA_{\leq
d+1}(G)$, $C=\calA_{d+1}(G)$, $B/C=\calA_{\leq d}(G)$, and use the
lemma. The second statement is proven by induction starting from the
fact that $Z_1$ always exists.\qed
\end{proof}

{\red MORE. It would be nice to relate the above with standard group
cohomology techniques: $\operatorname{Ext}$ groups, group extensions,
Schur multipliers.}

Recall that a multiplicative expansion $Z$ is ``Taylor'' if $Z(g)$ is
group-like for every $g\in G$ (Definition~\ref{def:ZCoMultiplicative}.
Before we can prove the ``Taylor'' version of Theorem~\ref{thm:H2},
we need to recall a few well-known facts about group-like elements in
a bi-algebra.

\begin{capsule} \label{cap:GroupLike} {\red (consider moving to an earlier
location)} (Following many sources starting
with~\cite{MilnorMoore:Hopf}. A quick introduction is
at~\cite[Appendix~A.2]{ChmutovDuzhinMostovoy:VassilievBook}). In a graded
degree-completed connected co-commutative bialgebra $A$, $\zeta$ is called
``group-like'' if it satisfies $\Box\zeta=\zeta\otimes\zeta$, and $\phi$
is ``primitive'' if $\Box\phi = \phi\otimes 1+1\otimes\phi$. Group-like
elements form a group (denoted $A_{\exp}$) under multiplication, and
primitive elements form a graded Lie algebra (denoted $A_{\text{prim}})$
under the commutator bracket. There is a bijection between group-like
and primitive elements: if $\phi$ is primitive then $\exp(\phi)$ is
group-like, and if $\zeta$ is group-like, $\log\phi$ makes sense and is
primitive. Both notions make sense ``up to degree $d$'' (or ``modulo
degrees higher than $d$''), and the bijection persists. A primitive up
to degree $d$ element always extends to a primitive up to degree $d+1$
element: simply extend by $0$ at degree $d+1$. The same extension
property is true for group-like elements: take the logarithm, extend by
$0$, and exponentiate back again. Finally, if $\zeta$ is group-like to
degree $d$ and $\phi$ is of degree $d$, then $\zeta+\phi$ is group-like
to degree $d$ iff $\phi$ is primitive.\endpar{cap:GroupLike}
\end{capsule}

\begin{theorem} \label{thm:H2T} If $H^2(G;\bbQ)=0$ then $G$ has the
Taylor extension property: every partial Taylor
expansion $Z_d$, for $d\geq 1$, can be extended. In particular, $G$
has a (full) Taylor expansion $Z$.
\end{theorem}

\begin{proof} Use the same procedures as in the proof of Theorem~\ref{thm:H2}
and Lemma~\ref{lem:H2}, yet note that at each stage the lift $Z'$ can be
chosen to be group-like (by the extension property for group-like elements
above), and then $\epsilon'$ is primitive as the least-degree difference of
two group-like elements. As $H^2(G;\calA_{\text prim}(G))=0$, we can choose
$\phi$ to be primitive, and then $Z=Z'+\phi$ remains group-like.\qed
\end{proof}

In rare ocassions, the existence of Taylor expansions for a
group $G$ implies the same for a quotient of $G$:

\begin{definition} We say that a normal subgroup $R$ of a group $G$
is ``robust'' if it is normally generated within $G$ by elements $r_i$
whose images in $\bbQ\otimes G^{ab}$ are linearly independent.
\end{definition}

\begin{theorem} If $G$ has the multiplicative (or Taylor) extension
property and $R$ is a robust normal subgroup of $G$, then $G/R$ also
has the multiplicative (or Taylor) extension property.
\end{theorem}

\begin{proof} As every group has expansions to degree $1$, it is enough to
prove an extension lemma 
\end{proof}

{\red MORE: Finish this!

Are there examples beyond LOT groups?

Is it true that Lemma~\ref{lem:H2} is an iff and that if $R$ is robust and
$H^2(G)=0$ then $H^2(G/R)=0$?
}