Joshua Moerman
11 years ago
5 changed files with 152 additions and 12 deletions
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\section{Minimal models} |
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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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\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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\end{itemize} |
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An cdga $(A, d)$ is a \emph{minimal (Sullivan) algebra} if in addition |
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\begin{itemize} |
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\item $d$ is decomposable, i.e. $\im(d) \subset \Lambda^{\geq 2}V$. |
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\end{itemize} |
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\end{definition} |
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\begin{definition} |
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Let $(A, d)$ be any cdga. A \emph{(minimal) Sullivan model} is a (minimal) Sullivan algebra $(M, d)$ with a weak equivalence: |
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$$ (M, d) \we (A, d). $$ |
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\end{definition} |
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The requirement that there exists a filtration can be replaced by a stronger statement. |
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\begin{lemma} |
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Let $(A, d)$ be a cdga which is $1$-reduced, quasi-free and with a decomposable differential. Then $(A, d)$ is a minimal algebra. |
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\end{lemma} |
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\begin{proof} |
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Take $V(n) = \bigoplus_{k=0}^n V^k$ (note that $V^0 = v^1 = 0$). Since $d$ is decomposable we see that for $v \in V^n$: $d(v) = x \cdot y$ for some $x, y \in A$. Assuming $dv$ to be non-zero we can compute the degrees: |
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$$ \deg{x} + \deg{y} = \deg{xy} = \deg{dv} = \deg{v} + 1 = n + 1. $$ |
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As $A$ is $1$-reduced we have $\deg{x}, \deg{y} \geq 2$ and so by the above $\deg{x}, \deg{y} \leq n-1$. Conclude that $d(V(k)) \subset \Lambda(V(n-1))$. |
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\end{proof} |
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\subsection{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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\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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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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\end{proof} |
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\subsection{Uniqueness} |
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Before we state the uniqueness theorem we need some more properties of minimal models. |
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\begin{lemma} |
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Sullivan algebras are cofibrant. |
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\end{lemma} |
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\begin{proof} |
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Consider the following lifting problem, where $\Lambda V$ is a Sullivan algebra. |
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\cimage[scale=0.5]{Sullivan_Lifting} |
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By the left adjointness of $\Lambda$ we only have to specify a map $\phi: V \to X$ sucht that $p \circ \phi = g$. We will do this by induction. |
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\begin{itemize} |
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\item Suppose $\{v_\alpha\}$ is a basis for $V(0)$. Define $V(0) \to X$ by choosing preimages $x_\alpha$ such that $p(x_\alpha) = g(v_\alpha)$ ($p$ is surjective). Define $\phi(v_\alpha) = x_\alpha$. |
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\item Suppose $\phi$ has been defined on $V(n)$. Write $V(n+1) = V(n) \oplus V'$ and let $\{v_\alpha\}$ be a basis for $V'$. Then $dv_\alpha \in \Lambda V(n)$, hence $\phi(dv_\alpha)$ is defined and |
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$$ d \phi d v_\alpha = \phi d^2 v_\alpha = 0 $$ |
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$$ p \phi d v_\alpha = g d v_\alpha = d g v_\alpha. $$ |
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Now $\phi d v_\alpha$ is a cycle and $p \phi d v_\alpha$ is a boundary of $g v_\alpha$. By the following lemma there is a $x_\alpha \in X$ such that $d x_\alpha = \phi d v_\alpha$ and $p x_\alpha = g v_\alpha$. The former property proves that $\phi$ is a chain map, the latter proves the needed commutativity $p \circ \phi = g$. |
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\end{itemize} |
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\end{proof} |
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\begin{lemma} |
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Let $p: X \to Y$ be a trivial fibration, $x \in X$ a cycle, $p(x) \in Y$ a boundary of $y' \in Y$. Then there is a $x' \in X$ such that |
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$$ dx' = x \quad\text{ and }\quad px' = y'. $$ |
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\end{lemma} |
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\begin{proof} |
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We have $p^\ast [x] = [px] = 0$, since $p^\ast$ is injective we have $x = d \overline{x}$ for some $\overline{x} \in X$. Now $p \overline{x} = y' + db$ for some $b \in Y$. Choose $a \in X$ with $p a = b$, then define $x' = \overline{x} - da$. Now check the requirements: $p x' = p \overline{x} - p a = y'$ and $d x' = d \overline{x} - d d a = d \overline{x} = x$. |
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\end{proof} |
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\begin{lemma} |
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Let $f: X \we Y$ be a weak equivalence between cdga's and $M$ a minimal model for $X$. Then $f$ induces an bijection: |
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$$ f_\ast: [M, X] \tot{\iso} [M, Y]. $$ |
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\end{lemma} |
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\begin{proof} |
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If $f$ is surjective this follows from the fact that $M$ is cofibrant and $f$ being a trivial fibration (see \cite[lemma 4.9]{dwyer}). |
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In general we can construct a cdga $Z$ and trivial fibrations $X \to Z$ and $Y \to Z$ inducing bijections: |
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$$ [M, X] \tot{\iso} [M, Z] \toti{\iso} [M, Y], $$ |
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compatible with $f_\ast$. \cite[Proposition 12.9]{felix}. |
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\end{proof} |
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\begin{lemma} |
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Let $\phi: (M, d) \we (M', d')$ be a weak equivalence between minimal algebras. Then $\phi$ is an isomorphism. |
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\end{lemma} |
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\begin{proof} |
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Let $M$ and $M'$ be generated by $V$ and $V'$. Then $\phi$ induces a weak equivalence on the linear part $\phi_0: V \we V'$ \cite[Theorem 1.5.10]{loday}. Since the differentials are decomposable, their linear part vanishes. So we see that $\phi_0: (V, 0) \tot{\iso} (V', 0)$ is an isomorphism. |
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Conclude that $\phi = \Lambda \phi_0$ is an isomorphism. |
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\end{proof} |
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\begin{theorem} |
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Let $m: (M, d) \we (A, d)$ and $m': (M', d') \we (A, d)$ be two minimal models. Then there is an isomorphism $\phi (M, d) \tot{\iso} (M', d')$ such that $m' \circ \phi \eq m$. |
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\end{theorem} |
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\begin{proof} |
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By the previous lemmas we have $[M', M] \iso [M', A]$. By going from right to elft we get a map $\phi: M' \to M$ such that $m' \circ \phi \eq m$. On homology we get $H(m') \circ H(\phi) = H(m)$, proving that (2-out-of-3) $\phi$ is a weak equivalence. The previous lemma states that $\phi$ is then an isomorphism. |
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\end{proof} |
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