From ff5ade6894a790a92693716148964c8200bd0e5b Mon Sep 17 00:00:00 2001 From: Joshua Moerman Date: Fri, 9 Jan 2015 17:33:07 +0100 Subject: [PATCH] Fixes a proof (to some extend) --- thesis/notes/Minimal_Models.tex | 15 ++++++++------- 1 file changed, 8 insertions(+), 7 deletions(-) diff --git a/thesis/notes/Minimal_Models.tex b/thesis/notes/Minimal_Models.tex index 23b7423..33dfe2e 100644 --- a/thesis/notes/Minimal_Models.tex +++ b/thesis/notes/Minimal_Models.tex @@ -47,6 +47,8 @@ The above definition is the same as in \cite{felix} without assuming connectivit 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. +\todo{at the moment this is just cut n pasted. Rewrite to make sense in this context} +Minimal models admit very nice homotopy groups. Note that for a minimal algebra $\Lambda V$ there is a natural augmentation and the the differential is decomposable. Hence $Q \Lambda V$ is naturally isomorphic to $(V, 0)$. In particular the homotopy groups are simply given by $\pi^n(\Lambda V) = V^n$. \section{Existence} @@ -128,8 +130,12 @@ Now if the map $f$ is a weak equivalence, both maps $\phi$ and $\psi$ are surjec Let $\phi: (M, d) \we (M', d')$ be a weak equivalence between minimal algebras. Then $\phi$ is an isomorphism. \end{lemma} \begin{proof} - \todo{introduce homotopy groups before this point. Prove it using that} 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. + Since both $M$ and $M'$ are minimal, they are cofibrant and so the weak equivalence is a strong homotopy equivalence (\CorollaryRef{cdga_homotopy_properties}). And so the induced map $\pi^n(\phi) : \pi^n(M) \to \pi^n(M')$ is an isomorphism (\LemmaRef{cdga-homotopic-maps-equal-pin}). + + Since $M$ (resp. $M'$) is free as a cga's, it is generated by some graded vector space $V$ (resp. $V'$). By an earlier remark \todo{where?} the homotopy groups were eassy to calculate and we conclude that $\phi$ induces an isomorphism from $V$ to $V'$: + \[ \pi^\ast(\phi) : V \tot{\iso} V'. \] + + Conclude that $\phi = \Lambda \phi_0$ \todo{why?} is an isomorphism. \end{proof} \Theorem{unique-minimal-model}{ @@ -155,11 +161,6 @@ Now by \LemmaRef{minimal-model-bijection} we get a bijection ${m_Y}_\ast^{-1} : } -\section{Homotopy groups of minimal models} -\todo{at the moment this is just cut n pasted. Rewrite to make sense in this context} -Minimal models admit very nice homotopy groups. Note that for a minimal algebra $\Lambda V$ there is a natural augmentation and the the differential is decomposable. Hence $Q \Lambda V$ is naturally isomorphic to $(V, 0)$. In particular the homotopy groups are simply given by $\pi^n(\Lambda V) = V^n$. - - \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}{