Recall the following facts about cdga's over a ring $\k$:
Recall that a cdga $A$ is a commutative differential graded algebra, meaning that
\begin{itemize}
\item it has a grading: $A =\bigoplus_{n\in\N} A^n$,
\item it has a differential: $d: A \to A$ with $d^2=0$,
\item it has an associative and unital multiplication: $\mu: A \tensor A \to A$ and
\item it is commutative: $x y =(-1)^{\deg{x}\cdot\deg{y}} y x$.
\end{itemize}
And all of the above structure is compatible with eachother (e.g. the differential is a derivative, the maps are graded, \dots). We have a left adjoint $\Lambda$ to the forgetful functor $U$ which assigns the free graded commutative algebras $\Lambda V$ to a graded module $V$. This extends to an adjunction (also called $\Lambda$ and $U$) between commutative differential graded algebras and differential graded modules.
In homological algebra we are especially interested in \emph{quasi isomorphisms}, i.e. the maps $f: A \to B$ inducing an isomorphism on homology: $H(f): HA \iso HB$. This notions makes sense for any object with a differential.
We furthermore have the following categorical properties of cdga's:
\begin{itemize}
\begin{itemize}
\item A map $f: A \to B$ in $\CDGA_\k$ is a \emph{quasi isomorphism} if it induces isomorphisms in cohomology.
\item The finite coproduct in $\CDGA_\k$ is the (graded) tensor product.
\item The finite coproduct in $\CDGA_\k$ is the (graded) tensor product.
\item The finite product in $\CDGA_\k$ is the cartesian product (with pointwise operations).
\item The finite product in $\CDGA_\k$ is the cartesian product (with pointwise operations).
\item The equalizer (resp. coequalizer) of $f$ and $g$ is given by the kernel (resp. cokernel) of $f - g$. Together with the (co)products this defines pullbacks and pushouts.
\item The equalizer (resp. coequalizer) of $f$ and $g$ is given by the kernel (resp. cokernel) of $f - g$. Together with the (co)products this defines pullbacks and pushouts.
\item$\k$ and $0$ are the initial and final object.
\item$\k$ and $0$ are the initial and final object.
\end{itemize}
\end{itemize}
In this chapter the ring $\k$ is assumed to be a field of characteristic zero.
In this chapter the ring $\k$ is assumed to be a field of characteristic zero. In particular the modules are vector spaces.
\section{Cochain models for the $n$-disk and $n$-sphere}
\section{Cochain models for the $n$-disk and $n$-sphere}
@ -16,7 +16,7 @@ In the following definition \emph{space} is to be understood as a topological sp
$$\pi_i(X)\tensor\Q\quad\forall i > 0.$$
$$\pi_i(X)\tensor\Q\quad\forall i > 0.$$
}
}
In order to define the tensor product $\pi_1(X)\tensor\Q$ we need that the fundamental group is abelian, that is the rational homotopy groups are only defined for simple spaces. There is a more general approach using \Def{nilpotent groups}, which admit $\Q$-completions \cite{bousfield}. Since this is rather technical we will often restrict ourselves to simple spaces or even simply connected spaces.
In order to define the tensor product $\pi_1(X)\tensor\Q$ we need that the fundamental group is abelian, that is why the definition requires simple spaces. There is a more general approach using \Def{nilpotent groups}, which admit $\Q$-completions \cite{bousfield}. Since this is rather technical we will often restrict ourselves to simple spaces or even simply connected spaces.
Note that for a rational space $X$, the ordinary homotopy groups are isomorphic to the rational homotopy groups, i.e. $\pi_i(X)\tensor\Q\iso\pi_i(X)$.
Note that for a rational space $X$, the ordinary homotopy groups are isomorphic to the rational homotopy groups, i.e. $\pi_i(X)\tensor\Q\iso\pi_i(X)$.
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 $c = d(b)\in D(n)^{n+1}$. \todo{Herschrijf}$S(n)$ is the cochain complex generated by one element $a \in S(n)^n$ with trivial differential (i.e. $d a =0$). In other words:
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 $c = d(b)\in D(n)^{n+1}$. On the other hand we define $S(n)$ to be the cochain complex generated by one element $a \in S(n)^n$ with trivial differential (i.e. $d a =0$). In other words:
$$ D(n)= ... \to0\to\k\to\k\to0\to ... $$
$$ D(n)= ... \to0\to\k\to\k\to0\to ... $$
$$ S(n)= ... \to0\to\k\to0\to0\to ... $$
$$ S(n)= ... \to0\to\k\to0\to0\to ... $$
@ -15,20 +14,20 @@ As we do not directly need this claim, we omit the proof. However, in the next s
$S(n)$ plays a another special role: maps from $S(n)$ to some cochain complex $X$ correspond directly to elements in the kernel of $\restr{d}{X^n}$. Any such map is null-homotopic precisely when the corresponding elements in the kernel is a coboundary. So there is a natural isomorphism: $\Hom(S(n), X)/{\simeq}\iso H^n(X)$.
$S(n)$ plays a another special role: maps from $S(n)$ to some cochain complex $X$ correspond directly to elements in the kernel of $\restr{d}{X^n}$. Any such map is null-homotopic precisely when the corresponding elements in the kernel is a coboundary. So there is a natural isomorphism: $\Hom(S(n), X)/{\simeq}\iso H^n(X)$.
By using the free cdga functor we can turn these cochain complexes into cdga's $\Lambda(D(n))$ and $\Lambda(S(n))$. So $\Lambda(D(n))$ consists of linear combinations of $b^n$\todo{gebruik niet weer $n$} and $c b^n$ when $n$ is even, and it consists of linear combinations of $c^n b$ and $c^n$ when $n$ is odd. In both cases we can compute the differentials using the Leibniz rule:
By using the free cdga functor we can turn these cochain complexes into cdga's $\Lambda(D(n))$ and $\Lambda(S(n))$. So $\Lambda(D(n))$ consists of linear combinations of $b^k$ and $c b^k$ when $n$ is even, and it consists of linear combinations of $c^k b$ and $c^k$ when $n$ is odd. In both cases we can compute the differentials using the Leibniz rule:
$$ d(b^n)= n\cdot c b^{n-1}$$
$$ d(b^k)= k\cdot c b^{k-1}$$
$$ d(c b^n)=0$$
$$ d(c b^k)=0$$
$$ d(c^n b)= c^{n+1}$$
$$ d(c^k b)= c^{k+1}$$
$$ d(c^n)=0$$
$$ d(c^k)=0$$
Those cocycles are in fact coboundaries (using that $\k$ is a field of characteristic $0$):
Those cocycles are in fact coboundaries (using that $\k$ is a field of characteristic $0$):
$$ c b^n=\frac{1}{n} d(b^{n+1})$$
$$ c b^k=\frac{1}{k} d(b^{k+1})$$
$$ c^n= d(b c^{n-1})$$
$$ c^k= d(b c^{k-1})$$
There are no additional cocycles in $\Lambda(D(n))$ besides the constants and $c$. So we conclude that $\Lambda(D(n))$ is acyclic as an algebra. In other words $\Lambda(j_n): \k\to\Lambda D(n)$ is a quasi isomorphism.
There are no additional cocycles in $\Lambda(D(n))$ besides the constants and $c$. So we conclude that $\Lambda(D(n))$ is acyclic as an algebra. In other words $\Lambda(j_n): \k\to\Lambda D(n)$ is a quasi isomorphism.
The situation for $\Lambda S(n)$ is easier: when $n$ is even it is given by polynomials in $a$, if $n$ is odd it is an exterior algebra \todo{?} (i.e.$a^2=0$). Again the sets $\Lambda(I)=\{\Lambda(i_n) : \Lambda S(n+1)\to\Lambda D(n)\I n \in\N\}$ and $\Lambda(J)=\{\Lambda(j_n) : \k\to\Lambda D(n)\I n \in\N\}$ play an important role.
The situation for $\Lambda S(n)$ is easier as it has only one generator (as algebra). For even $n$ this means it is given by polynomials in $a$. For odd $n$ it is an exterior algebra, meaning$a^2=0$. Again the sets $\Lambda(I)=\{\Lambda(i_n) : \Lambda S(n+1)\to\Lambda D(n)\I n \in\N\}$ and $\Lambda(J)=\{\Lambda(j_n) : \k\to\Lambda D(n)\I n \in\N\}$ play an important role.
\begin{theorem}
\begin{theorem}
The sets $\Lambda(I)$ and $\Lambda(J)$ generate a model structure on $\CDGA_\k$ where:
The sets $\Lambda(I)$ and $\Lambda(J)$ generate a model structure on $\CDGA_\k$ where:
In this section we will prove the existence of rationalizations $X \to X_\Q$. We will do this in a cellular way. The $n$-spheres play an important role here, so their rationalizations will be discussed first. Again spaces (except for $S^1$) are assumed to be $1$-connected.
In this section we will prove the existence of rationalizations $X \to X_\Q$. We will do this in a cellular way. The $n$-spheres play an important role here, so their rationalizations will be discussed first. In this section $1$-connectedness of spaces will play an important role.
\section{Rationalization of \texorpdfstring{$S^n$}{Sn}}
\section{Rationalization of \texorpdfstring{$S^n$}{Sn}}
In this section we fix $n>0$. We will construct $S^n_\Q$ in stages $S^n(1), S^n(2), \ldots$, where at each stage we wedge a sphere and then glue a $n+1$-cell to ``invert'' some element in the $n$th homotopy group.
In this section we fix $n>0$. We will construct $S^n_\Q$ in stages $S^n(1), S^n(2), \ldots$, where at each stage we wedge a sphere and then glue a $n+1$-cell to ``invert'' some element in the $n$th homotopy group.
@ -12,7 +12,6 @@ In this section we fix $n>0$. We will construct $S^n_\Q$ in stages $S^n(1), S^n(
We start the construction with $S^n(1)= S^n$. Assume we constructed $S^n(r)=\bigvee_{i=1}^{r} S^{n}\cup_{h}\coprod_{i=1}^{r-1} D^{n+1}$, where $h$ is a specific attaching map. Assume furthermore the following two properties. Firstly, the inclusion $i_r : S^n \to S^n(r)$ of the terminal sphere is a weak equivalence. Secondly, the inclusion $i_1 : S^n \to S^n(r)$ of the initial sphere induces the multiplication $\pi_n(S^n)\tot{\times r!}\pi_n(S^n(r))$ under the identification of $\pi_n(S^n)=\pi_n(S^n(r))=\Z$.
We start the construction with $S^n(1)= S^n$. Assume we constructed $S^n(r)=\bigvee_{i=1}^{r} S^{n}\cup_{h}\coprod_{i=1}^{r-1} D^{n+1}$, where $h$ is a specific attaching map. Assume furthermore the following two properties. Firstly, the inclusion $i_r : S^n \to S^n(r)$ of the terminal sphere is a weak equivalence. Secondly, the inclusion $i_1 : S^n \to S^n(r)$ of the initial sphere induces the multiplication $\pi_n(S^n)\tot{\times r!}\pi_n(S^n(r))$ under the identification of $\pi_n(S^n)=\pi_n(S^n(r))=\Z$.
We will construct $S^n(r+1)$ with similar properties as follows. Let $f: S^n \to S^n(r)$ be a representative for $1\in\Z\iso\pi_n(S^n(r))$ and $g: S^n \to S^n$ be a representative for $r+1\in\Z\iso\pi_n(S^n)$. These maps combine into $\phi: S^n \to S^n \vee S^n \tot{f \vee g} S^n(r)\vee S^n$. We define $S^n(r+1)$ as the pushout in the following diagram.
We will construct $S^n(r+1)$ with similar properties as follows. Let $f: S^n \to S^n(r)$ be a representative for $1\in\Z\iso\pi_n(S^n(r))$ and $g: S^n \to S^n$ be a representative for $r+1\in\Z\iso\pi_n(S^n)$. These maps combine into $\phi: S^n \to S^n \vee S^n \tot{f \vee g} S^n(r)\vee S^n$. We define $S^n(r+1)$ as the pushout in the following diagram.
@ -158,7 +158,6 @@ For the main theorem we need the following construction. \todo{Geef de construct
}
}
\Proof{
\Proof{
Consider the mapping cylinder $B_f$ of $f$, i.e. factor the map $f$ as a cofibration followed by a trivial fibration $f: A \cof B_f \fib B$. The inclusion $A \subset B_f$ gives a long exact sequence of homotopy groups and homology groups:
Consider the mapping cylinder $B_f$ of $f$, i.e. factor the map $f$ as a cofibration followed by a trivial fibration $f: A \cof B_f \fib B$. The inclusion $A \subset B_f$ gives a long exact sequence of homotopy groups and homology groups:
\cdiagram{Serre_Whitehead_LES}
\cdiagram{Serre_Whitehead_LES}
We now have the equivalence of the following statements:
We now have the equivalence of the following statements:
Joshua Moerman
10 years ago