In this section we will discuss the so called minimal models. These are cdga's with the property that a quasi isomorphism between them is an actual isomorphism.
\begin{definition}
An cdga $(A, d)$ is a \emph{Sullivan algebra} if
A cdga $(A, d)$ is a \emph{Sullivan algebra} if
\begin{itemize}
\item$(A, d)$ is quasi-free (or semi-free), i.e. $A =\Lambda V$ is free as a cdga, and
\item$V$ has a filtration $V(0)\subset V(1)\subset\cdots\subset\bigcup_{k \in\N} V(k)= V$ such that $d(V(k))\subset\Lambda V(k-1)$.
\item$A =\Lambda V$ is free as a commutative graded algebra, and
An cdga $(A, d)$ is a \emph{minimal (Sullivan) algebra} if in addition
@ -37,22 +39,24 @@ The requirement that there exists a filtration can be replaced by a stronger sta
\section{Existence}
\begin{theorem}
Let $(A, d)$ be an $1$-connected cdga, then it has a minimal model.
Let $(A, d)$ be an $0$-connected cdga, then it has a Sullivan model $(\Lambda V, d)$. Furthermore if $(A, d)$ is $r$-connected, $V^i =0$ for all $i \leq r$.
\end{theorem}
\begin{proof}
We will construct a sequence of models $m_k: (M(k), d)\to(A, d)$ inductively.
\begin{itemize}
\item First define $V(0)= V(1)=0$ and $m_0= m_1=0$. Then set $V(2)= H^2(A)$ and define a map $m_2: V(2)\to A$ by picking representatives.
\item Suppose $m_k: (\Lambda V(k), d)\to(A, d)$ is constructed. Choose cocycles $a_\alpha\in A^{k+1}$ and $z_\beta\in(\Lambda V(k))^{k+2}$ such that $H^{k+1}(A)=\im(H^{k+1}(m_k))\oplus\bigoplus_\alpha\k\cdot[a_\alpha]$ (so $m_k$ together with $a_\alpha$ span $H^{k+1}(A)$) and $\ker(H^{k+2}(m_k))=\bigoplus_\beta\k\cdot[z_\beta]$. Note that $m_k z_\beta= db_\beta$ for some $b_\beta\in A$.
Start by setting $V(0)= H^{\geq1}(A)$ and $d =0$. This extends to a morphism $m_0 : (\Lambda V(0), 0)\to(A, d)$.
Define $V(k+1)=\bigoplus_\alpha\k\cdot v'_\alpha\oplus\bigoplus\k\cdot v''_\beta$ and set $dv'_\alpha=0$, $dv''_\beta= z_\beta$, $m_k(v'_\alpha)= a_\alpha$ and $m_k(v''_\beta)= b_\beta$.
\end{itemize}
This ends the construction. We will prove the following assertion for $k \geq2$:
$$ H^i(m_k)\text{ is }\begin{cases}
\text{an isomorphism}&\text{ if } i \leq k \\
\text{injective}&\text{ if } i = k + 1
\end{cases}. $$
\TODO{Finish proof: $m_k$ well behaved, above assertion.}
Note that the freeness introduces products such that the map $H(m_0) : H(\Lambda V(0))\to H(A)$ is not an isomorphism. We will ``kill'' these defects inductively.
Suppose $V(k)$ and $m_k$ are constructed. Consider the defect $\ker H(m_k)$ and let $\{[z_\alpha]\}_{\alpha\in A}$ be a basis for it. Define $V_{k+1}=\bigoplus_{\alpha\in A}\k\cdot v_\alpha$ with the degrees $\deg{v_\alpha}=\deg{z_\alpha}-1$.
Now extend the differential by defining $d(v_\alpha)= z_\alpha$. This step kills the defect, but also introduces new defects which will be killed later. Notice that $z_\alpha$ is a cocycle and hence $d^2 v_\alpha=0$.
Since $[z_\alpha]$ is in the kernel of $H(m_k)$ we see that $m_k z_\alpha= d a_\alpha$ for some $a_\alpha$. Extend $m_k$ to $m_{k+1}$ by defining $m_{k+1}(v_\alpha)= a_\alpha$. Notice that $m_{k+1} d v_\alpha= m_{k+1} z_\alpha= d a_\alpha= d m_{k+1} v_\alpha$.
Now take $V(k+1)= V(k)\oplus V_{k+1}$. We already proved that $d$ is indeed a differential, and that $m_{k+1}$ is indeed a chain map.
Complete the construction by taking the union: $V =\bigcup_k V(k)$. Clearly $H(m)$ is surjective, this was establsihed in the first step. Now if $H(m)[z]=0$, then we know $z \in\Lambda V(k)$ for some stage $k$ and hence by construction is was killed, i.e. $[z]=0$. So we see that $m$ is a quasi isomorphism and by construction $(\Lambda V, d)$ is a sullivan algebra.
Joshua Moerman
10 years ago