Revises the proof of existence of sullivan models
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@ -5,10 +5,12 @@
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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.
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\begin{definition}
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An cdga $(A, d)$ is a \emph{Sullivan algebra} if
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A cdga $(A, d)$ is a \emph{Sullivan algebra} if
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\begin{itemize}
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\item $(A, d)$ is quasi-free (or semi-free), i.e. $A = \Lambda V$ is free as a cdga, and
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\item $V$ has a filtration $V(0) \subset V(1) \subset \cdots \subset \bigcup_{k \in \N} V(k) = V$ such that $d(V(k)) \subset \Lambda V(k-1)$.
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\item $A = \Lambda V$ is free as a commutative graded algebra, and
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\item $V$ has a filtration
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$$ 0 = V(-1) \subset V(0) \subset V(1) \subset \cdots \subset \bigcup_{k \in \N} V(k) = V, $$
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such that $d(V(k)) \subset \Lambda V(k-1)$.
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\end{itemize}
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An cdga $(A, d)$ is a \emph{minimal (Sullivan) algebra} if in addition
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@ -37,22 +39,24 @@ The requirement that there exists a filtration can be replaced by a stronger sta
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\section{Existence}
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\begin{theorem}
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Let $(A, d)$ be an $1$-connected cdga, then it has a minimal model.
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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 $i \leq r$.
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\end{theorem}
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\begin{proof}
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We will construct a sequence of models $m_k: (M(k), d) \to (A, d)$ inductively.
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\begin{itemize}
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\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.
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\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 $b_\beta \in A$.
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Start by setting $V(0) = H^{\geq 1}(A)$ and $d = 0$. This extends to a morphism $m_0 : (\Lambda V(0), 0) \to (A, d)$.
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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 $m_k(v''_\beta) = b_\beta$.
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\end{itemize}
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This ends the construction. We will prove the following assertion for $k \geq 2$:
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$$ H^i(m_k) \text{ is } \begin{cases}
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\text{an isomorphism} &\text{ if } i \leq k \\
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\text{injective} &\text{ if } i = k + 1
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\end{cases}. $$
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\TODO{Finish proof: $m_k$ well behaved, above assertion.}
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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.
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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 $\deg{v_\alpha} = \deg{z_\alpha}-1$.
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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 $d^2 v_\alpha = 0$.
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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 $m_{k+1} d v_\alpha = m_{k+1} z_\alpha = d a_\alpha = d m_{k+1} v_\alpha$.
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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.
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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.
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\todo{minimality for $1$-connected}
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\end{proof}
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@ -38,7 +38,7 @@
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\newcommand{\Np}{{\mathbb{N}^{>0}}} % positive numbers
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\newcommand{\Z}{\mathbb{Z}} % integers
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\newcommand{\R}{\mathbb{R}} % reals
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\newcommand{\Q}{\mathbb{Q}} % rationals
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\DeclareRobustCommand{\Q}{\mathbb{Q}} % rationals
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\renewcommand{\k}{\mathds{k}} % default ground ring
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% Basic category stuff
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