\chapter{Minimal models}
\label{sec:minimal-models}

In this section we will discuss the so called minimal models. These cdga's enjoy the property that we can easily prove properties inductively. Moreover it will turn out that weakly equivalent minimal models are actually isomorphic.

\begin{definition}
	A cdga $(A, d)$ is a \Def{Sullivan algebra} if
	\begin{itemize}
		\item $A = \Lambda V$ is free as a commutative graded algebra, and
		\item $V$ has a filtration
		$$ 0 = V(-1) \subset V(0) \subset V(1) \subset \cdots \subset \bigcup_{k \in \N} V(k) = V, $$
		such that $d(V(k)) \subset \Lambda V(k-1)$.
	\end{itemize}

	An cdga $(A, d)$ is a \Def{minimal Sullivan algebra} if in addition
	\begin{itemize}
		\item $d$ is decomposable, i.e. $\im(d) \subset \Lambda^{\geq 2}V$.
	\end{itemize}
\end{definition}

\begin{definition}
	Let $(A, d)$ be any cdga. A \Def{(minimal) Sullivan model} is a (minimal) Sullivan algebra $(M, d)$ with a weak equivalence:
	$$ (M, d) \we (A, d). $$
\end{definition}

We will often say \Def{minimal model} or \Def{minimal algebra} to mean minimal Sullivan model or minimal Sullivan algebra. In many cases we can take the degree of the elements in $V$ to induce the filtration, as seen in the following lemma of which the proof is left out, as we are not going to use it.

\Lemma{1-reduced-minimal-model}{
	Let $(A, d)$ be a cdga which is $1$-reduced, such that $A$ is free as cga and $d$ is decomposable. Then $(A, d)$ is a minimal algebra.
}
\Proof{
	Let $V$ generate $A$. 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:
	$$ \deg{x} + \deg{y} = \deg{xy} = \deg{dv} = \deg{v} + 1 = n + 1. $$
	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 $d(V(k)) \subset \Lambda(V(n-1))$.
}

The above definition is the same as in \cite{felix} without assuming connectivity. We find some different definitions of (minimal) Sullivan algebras in the literature. For example we find a definition using well orderings in \cite{hess}. The decomposability of $d$ also admits a different characterization (at least in the connected case). The equivalence of the definitions is expressed in the following two lemmas.

\Lemma{}{
	A cdga $(\Lambda V, d)$ is a Sullivan algebra if and only if there exists a well order $J$ such that $V$ is generated by $v_j$ for $j \in J$ and $d v_j \in \Lambda V_{<j}$.
}

\Lemma{}{
	Let $(\Lambda V, d)$ be a Sullivan algebra with $V^0 = 0$, then $d$ is decomposable if and only if there is a well order $J$ as above such that $i < j$ implies $\deg{v_i} \leq \deg{v_j}$.
}

It is clear that induction will be an important technique when proving things about (minimal) Sullivan algebras. We will first prove that minimal models always exist for $1$-connected cdga's and afterwards prove uniqueness.


\section{Existence}

\begin{theorem}
	Let $(A, d)$ be an $0$-connected cdga, then it has a Sullivan model $(\Lambda V, d)$. Furthermore if $(A, d)$ is $r$-connected with $r \geq 1$ then $V^i = 0$ for all $i \leq r$ and in particular $(\Lambda V, d)$ is minimal.
\end{theorem}
\begin{proof}
	Start by setting $V(0) = H^{\geq 1}(A)$ and $d = 0$. This extends to a morphism $m_0 : (\Lambda V(0), 0) \to (A, d)$.
	Note that the freeness introduces products such that the map $H(m_0) : H(\Lambda V(0)) \to H(A)$ is \emph{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 $d^2 v_\alpha = 0$, so $d$ is still a differential.
	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 $m_{k+1} d v_\alpha = m_{k+1} z_\alpha = d a_\alpha = d m_{k+1} v_\alpha$, so $m_{k+1}$ is a cochain map.
	Now take $V(k+1) = V(k) \oplus V_{k+1}$.

	Complete the construction by taking the union: $V = \bigcup_k V(k)$. Clearly $H(m)$ is surjective, this was established 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.

	Now assume that $(A, d)$ is $r$-connected ($r \geq 1$), this means that $H^i(A) = 0$ for all $1 \leq i \leq r$, and so $V(0)^i = 0$ for all $i \leq r$. Now $H(m_0)$ is injective on $\Lambda^{\leq 1} V(0)$, and so the defects are in $\Lambda^{\geq 2} V(0)$ and have at least degree $2(r+1)$. This means two things in the first inductive step of the construction. First, the newly added elements have decomposable differential. Secondly, these elements are at least of degree $2(r+1) - 1$. After adding these elements, the new defects are in $\Lambda^{\geq 2} V(1)$ and have at least degree $2(2(r+1) - 1)$. We see that as the construction continues, the degrees of adjoined elements go up. Hence $V^i = 0$ for all $i \leq r$ and by the previous lemma $(\Lambda V, d)$ is minimal.
\end{proof}


\section{Uniqueness}

Before we state the uniqueness theorem we need some more properties of minimal models. In this section we will use some general facts about model categories.

\begin{lemma}
	Sullivan algebras are cofibrant and the inclusions induced by the filtration are cofibrations.
\end{lemma}
\begin{proof}
	Consider the following lifting problem, where $\Lambda V$ is a Sullivan algebra.
	\begin{displaymath}
		\xymatrix {
		\k \ar[r]^\unit \ar[d]^\unit & X \artfib[d]^p \\
		\Lambda V \ar[r]^g & Y
		}
	\end{displaymath}

	By the left adjointness of $\Lambda$ we only have to specify a map $\phi: V \to X$ such that $p \circ \phi = g$. We will do this by induction. Note that the induction step proves precisely that $(\Lambda V(k), d) \to (\Lambda V(k+1), d)$ is a cofibration.
	\begin{itemize}
		\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 $\phi(v_\alpha) = x_\alpha$.
		\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
		$$ d \phi d v_\alpha = \phi d^2 v_\alpha = 0 $$
		$$ p \phi d v_\alpha = g d v_\alpha = d g v_\alpha. $$
		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 $p \circ \phi = g$.
	\end{itemize}
\end{proof}

\begin{lemma}
	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
	$$ dx' = x \quad\text{ and }\quad px' = y'. $$
\end{lemma}
\begin{proof}
	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 $d x' = d \overline{x} - d d a = d \overline{x} = x$.
\end{proof}

In the following we will need to replace a map by a fibration. But the one given abstractly from the model structure will not fit our needs. So we will first consider the following factorization.

Let $A$ be any cochain complex (not an algebra) and define $C(A)^k = C^k \oplus C^{k-1}$. Then $C(A)$ is again a cochain complex when we define the differential to be $\delta(c_k, c_{k-1}) = (0, c_k)$. Note that this cochain complex is acyclic, furthermore there is an obvious surjection $C(A) \tot{\rho} A$. Now for a cochain algebra $A$, we can do the same construction (by forgetting the algebra structure) and apply $\Lambda$. This defines a cdga $\Lambda C(A)$ (which is still acyclic).

Now let $f: X \to Y$ be any map, then we can tensor $X$ with $\Lambda C(Y)$ to obtain:
$$ f: X \tot{x \mapsto x \tensor 1} X \tensor \Lambda C(Y) \tot{\psi: x \tensor y \mapsto f(x) \cdot \rho (y)} Y. $$
Where the second map is surjective. By the 2-out-of-3 property the second map is a weak equivalence if and only if $f$ is a weak equivalence. The remarkable thing is that the left map (which is a weak equivalence by the Künneth theorem) has a left inverse, given by $\phi: x \tensor y \mapsto x \cdot \counit (y)$, where $\counit$ is the augmentation.

Now if the map $f$ is a weak equivalence, both maps $\phi$ and $\psi$ are surjective and weak equivalences.

\Lemma{minimal-model-bijection}{
	Let $f: X \we Y$ be a weak equivalence between cdga's and $M$ a minimal algebra. Then $f$ induces an bijection:
	$$ f_\ast: [M, X] \tot{\iso} [M, Y]. $$
}
\begin{proof}
	If $f$ is surjective this follows from the fact that $M$ is cofibrant and $f$ being a trivial fibration, see \CorollaryRef{cdga_homotopy_properties}.

	By the factorization above, we can turn $f$ into two trivial fibrations (going in different directions). This induces
	$$ [M, X] \toti{\iso} [M, \Lambda C(Y)] \tot{\iso} [M, Y], $$
	compatible with $f_\ast$.
\end{proof}

\begin{lemma}
	Let $\phi: (M, d) \we (M', d')$ be a weak equivalence between minimal algebras. Then $\phi$ is an isomorphism.
\end{lemma}
\begin{proof}
	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 1.5.2]{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.
	Conclude that $\phi = \Lambda \phi_0$ is an isomorphism.
\end{proof}

\Theorem{unique-minimal-model}{
	Let $m: (M, d) \we (A, d)$ and $m': (M', d') \we (A, d)$ be two minimal models for $A$. Then there is an isomorphism $\phi (M, d) \tot{\iso} (M', d')$ such that $m' \circ \phi \eq m$.
}
\begin{proof}
	By the previous lemmas we have $[M', M] \iso [M', A]$. By going from right to left 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.
\end{proof}

The assignment of $X$ to its minimal model $M_X = (\Lambda V, d)$ can be extended to morphisms. Let $X$ and $Y$ be two cdga's and $f: X \to Y$ be a map. By considering their minimal models we get the following diagram.
\begin{displaymath}
	\xymatrix @C=1.5cm{
	X \ar[r]^f & Y \\
	M_X \arwe[u]^{m_X} \ar[ur]^{f m_X} & M_Y \arwe[u]^{m_Y}
	}
\end{displaymath}
Now by \LemmaRef{minimal-model-bijection} we get a bijection ${m_Y}_\ast^{-1} : [M_X, Y] \iso [M_X, M_Y]$. This gives a map $M(f) = {m_Y}_\ast^{-1} (f m_X)$ from $M_X$ to $M_Y$. Of course this does not define a functor of cdga's as it is only well defined on homotopy classes. However it is clear that it does define a functor on the homotopy category of cdga's.

\Corollary{}{
	The assignment $X \mapsto M_X$ defines a functor $M: \Ho(\CDGA^1_\Q) \to \Ho(\CDGA^1_\Q)$. Moreover, since the minimal model is weakly equivalent, $M$ gives an equivalence of categories:
	$$ M: \Ho(\CDGA^1_\Q) \iso \Ho(\text{Minimal algebras}^1), $$
	where weakly equivalent cdga's are sent to \emph{isomorphic} minimal models.
}

\section{The minimal model of the sphere}
We know from singular cohomology that the cohomology ring of a $n$-sphere is $\Z[X] / (X^2)$. This allows us to construct a minimal model for $S^n$.
\Definition{minimal-model-sphere}{
	Define $A(n)$ to be the cdga defined as
	$$ A(n) = \begin{cases}
		\Lambda(e) \quad \deg{e} = n \quad de = 0 \qquad &\text{ if $n$ is odd } \\
		\Lambda(e, f) \quad \deg{e} = n, \deg{f} = 2n-1 \quad df = e^2 \qquad &\text{ if $n$ is even }
	\end{cases}. $$
}

\todo{prove $HA(n) \tot{\iso} HA(S^n)$}