In this section $\k$ will be any commutative ring. We will recap some of the basic definitions of commutative algebra in a graded setting. By \emph{linear}, \emph{module}, \emph{tensor product}, etc\dots we always mean $\k$-linear, $\k$-module, tensor product over $\k$, etc\dots.
In this section $\k$ will be any commutative ring. We will recap some of the basic definitions of commutative algebra in a graded setting. By \emph{linear}, \emph{module}, \emph{tensor product}, etc\dots we always mean $\k$-linear, $\k$-module, tensor product over $\k$, etc\dots.
\subsection{Cochain models for the $n$-disk and $n$-sphere}
\section{Cochain models for the $n$-disk and $n$-sphere}
We will first define some basic cochain complexes which model the $n$-disk and $n$-sphere. $D(n)$ is the cochain complex generated by one element $b \in D(n)^n$ and its differential $c = d(b)\in D(n)^{n+1}$. $S(n)$ is the cochain complex generated by one element $a \in S(n)^n$ which differential vanishes (i.e. $da =0$). In other words:
So it is the free cdga with $n+1$ generators and their differentials such that $\sum_{i=0}^n x_i =1$ and in order to be well behaved $\sum_{i=0}^n dx_i =0$.
@ -7,7 +7,7 @@ In this section we will discuss the so called minimal models. These are cdga's w
An 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$V$ has a filtration $V(0)\subset V(1)\subset\cdots\subset\bigcup_{k \in\N} V(k)= V$ such that $d(V(k))\subset\Lambda V(k-1)$.
\end{itemize}
An cdga $(A, d)$ is a \emph{minimal (Sullivan) algebra} if in addition
@ -27,7 +27,7 @@ The requirement that there exists a filtration can be replaced by a stronger sta
Let $(A, d)$ be a cdga which is $1$-reduced, quasi-free and with a decomposable differential. Then $(A, d)$ is a minimal algebra.
\end{lemma}
\begin{proof}
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:
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 $\deg{x}, \deg{y}\geq2$ and so by the above $\deg{x}, \deg{y}\leq n-1$. Conclude that $d(V(k))\subset\Lambda(V(n-1))$.
\end{proof}
@ -67,7 +67,7 @@ Before we state the uniqueness theorem we need some more properties of minimal m
\cimage[scale=0.5]{Sullivan_Lifting}
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.
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.
\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
@ -101,7 +101,7 @@ Before we state the uniqueness theorem we need some more properties of minimal m
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 1.5.10]{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.
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}
@ -109,5 +109,5 @@ Before we state the uniqueness theorem we need some more properties of minimal m
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 $m' \circ\phi\eq m$.
\end{theorem}
\begin{proof}
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.
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.
Joshua Moerman
10 years ago