In this section we aim to prove that the homotopy theory of rational spaces is the same as the homootopy theory of cdga's over $\Q$. Before we prove the equivalence, we will show that $A$ and $K$ form a Quillen pair. This already provides an adjunction between the homotopy categories. Besides the equivalence of the homotopy categories we will also investigate homotopy groups on a cdga directly. The homotopy groups of a space will be dual to the homotopy groups of the associated cdga.
In this section we aim to prove that the homotopy theory of rational spaces is the same as the homotopy theory of cdga's over $\Q$. Before we prove the equivalence, we will show that $A$ and $K$ form a Quillen pair. This already provides an adjunction between the homotopy categories. Besides the equivalence of the homotopy categories we will also investigate homotopy groups on a cdga directly. The homotopy groups of a space will be dual to the homotopy groups of the associated cdga.
We will prove that $A$ preserves cofibrations and trivial cofibrations. We only have to check this fact for the generating (trivial) cofibrations in $\sSet$. Note that the contravariance of $A$ means that a (trivial) cofibrations should be sent to a (trivial) fibration.
@ -109,13 +109,13 @@ For the equivalence of rational spaces and cdga's we need that the unit and coun
A &\to M(K(A))
\end{align*}
where the first of the two maps is given by the composition $X \to K(A(X))\tot{K(m_X)} K(M(X))$,
and the second map is obtained by the map $A \to A(K(A))$ and using the bijection from \LemmaRef{minimal-model-bijection}: $[A, A(K(A))]\iso[A, M(K(A))]$. By the 2-ouy-of-3 property the map $A \to M(K(A))$ is a weak equivalence if and only if the ordinary unit $A \to A(K(A))$ is a weak equivalence.
and the second map is obtained by the map $A \to A(K(A))$ and using the bijection from \LemmaRef{minimal-model-bijection}: $[A, A(K(A))]\iso[A, M(K(A))]$. By the 2-out-of-3 property the map $A \to M(K(A))$ is a weak equivalence if and only if the ordinary unit $A \to A(K(A))$ is a weak equivalence.
\Lemma{}{
(Base case) Let $A =(\Lambda(v), 0)$ be a minimal model with one generator of degree $\deg{v}= n \geq1$. Then $A \we A(K(A))$.
}
\Proof{
By \CorollaryRef{minimal-cdga-EM-space} we know that $K(A)$ is an Eilenberg-MacLane space of type $K(\Q^\ast, n)$. The cohomology of an Eilenberg-Maclane space with coefficients in $\Q$ is known:
By \CorollaryRef{minimal-cdga-EM-space} we know that $K(A)$ is an Eilenberg-MacLane space of type $K(\Q^\ast, n)$. The cohomology of an Eilenberg-MacLane space with coefficients in $\Q$ is known:
$$ H^\ast(K(\Q^\ast, n); \Q)=\Q[x], $$
that is, the free commutative graded algebra with one generator $x$. This can be calculated, for example, with spectral sequences \cite{griffiths}.
@ -144,7 +144,7 @@ and the second map is obtained by the map $A \to A(K(A))$ and using the bijectio
& A(K(T(m))) \ar[rr]&& A(K(B))
}
\end{displaymath}
Note that we have a weak equivalence in the top left corner, by the base case ($S(m+1)=(\Lambda(v), 0)$). The weak equivalence in the top right is by assumption. Finally the bottom left map is a weak equivalence because both cdga's are acylcic.
Note that we have a weak equivalence in the top left corner, by the base case ($S(m+1)=(\Lambda(v), 0)$). The weak equivalence in the top right is by assumption. Finally the bottom left map is a weak equivalence because both cdga's are acyclic.
To conclude that $B \to A(K(B))$ is a weak equivalence, we wish to prove that the front face of the cube is a homotopy pushout, as the back face clearly is one. This is a consequence of the Eilenberg-Moore spectral sequence \cite{mccleary}.
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.
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 \emph{Sullivan algebra} if
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
@ -13,18 +13,18 @@ In this section we will discuss the so called minimal models. These are cdga's w
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
An cdga $(A, d)$ is a \Def{minimal Sullivan algebra} if in addition
\begin{itemize}
\item$d$ is decomposable, i.e. $\im(d)\subset\Lambda^{\geq2}V$.
\end{itemize}
\end{definition}
\begin{definition}
Let $(A, d)$ be any cdga. A \emph{(minimal) Sullivan model} is a (minimal) Sullivan algebra $(M, d)$ with a weak equivalence:
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}
In the following lemma we see that the filtration is sometimes naturally there for $1$-reduced cdga's.
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.
\begin{lemma}
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.
@ -73,7 +73,7 @@ It is clear that induction will be an important technique when proving things ab
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 $(\Lambda V(k), d)\to(\Lambda V(k+1), d)$ are cofibrations.
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.
@ -84,7 +84,7 @@ Before we state the uniqueness theorem we need some more properties of minimal m
}
\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 cofibrations.
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
@ -102,6 +102,16 @@ Before we state the uniqueness theorem we need some more properties of minimal m
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:
$$ f: X \tot{x \mapsto x \tensor1} X \tensor\Lambda C(Y)\tot{\psi: x \tensor y \mapsto f(x)\cdot\rho(y)} Y. $$
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:
$$ f_\ast: [M, X]\tot{\iso}[M, Y]. $$
@ -109,10 +119,8 @@ Before we state the uniqueness theorem we need some more properties of minimal m
\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}.
\todo{Put this surjectivity trick in a lemma} In general we will reduce to the surjective case. Let $C$ be any cochain complex and define $\delta C^k = C^{k-1}$. Now $(C \oplus\delta C, \delta)$ is again a cochain complex and there is a surjective map to $C$. Define $E(Y)=\Lambda(Y \oplus\delta Y, \delta)$ (we consider $Y$ as a cochain complex). We obtain:
$$ f: X \tot{x \mapsto x \tensor1} X \tensor E(Y)\tot{x \tensor y \mapsto f(x)\cdot y} Y. $$
Now the first map has a left inverse. We have two trivial fibrations $E(Y)\to X$ and $E(Y)\to Y$. This induces
$$[M, X]\toti{\iso}[M, E(Y)]\tot{\iso}[M, Y], $$
By the factorization above, we can turn $f$ into two trivial fibrations (going in different directions). This induces
@ -125,7 +133,7 @@ Before we state the uniqueness theorem we need some more properties of minimal m
\end{proof}
\Theorem{unique-minimal-model}{
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 $m' \circ\phi\eq m$.
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.
@ -133,7 +141,7 @@ Before we state the uniqueness theorem we need some more properties of minimal m
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.
@ -62,7 +62,7 @@ where the left adjoint is precisely the functor $C^\ast$ as noted in \cite{felix
In this section we will prove that the singular cochain complex is quasi isomorphic to the cochain complex of polynomial forms. In order to do so we will define an integration map $\int_n : \Apl_n^n \to\k$, which will induce a map $\oint_n : \Apl_n \to C_n$. For the simplices $\Delta[n]$ we already showed the cohomology agrees by the acyclicity of $\Apl_n = A(\Delta[n])$ (\LemmaRef{apl-acyclic}).
Let$v \in\Apl_n^n$, then we can always write it as $v = p(x_1, \dots, x_n)dx_1\dots dx_n$ where $p$ is a polynomial in $n$ variables. If $\Q\subset\k\subset\mathbb{C}$ we can integrate geometrically on the $n$-simplex:
For any$v \in\Apl_n^n$, we can write $v$ as $v = p(x_1, \dots, x_n)dx_1\dots dx_n$ where $p$ is a polynomial in $n$ variables. If $\Q\subset\k\subset\mathbb{C}$ we can integrate geometrically on the $n$-simplex:
which defines a well-defined linear map $\int_n : \Apl_n^n \to\k$. For general fields of characteristic zero we can define it formally on the generators of $\Apl_n^n$ (as vector space):
For it to be a chain map, we need to prove $d \circ\oint_n =\oint_n \circ d$. This is very similar to\emph{Stokes' theorem}. \todo{proof this}
For it to be a chain map, we need to prove $d \circ\oint_n =\oint_n \circ d$. This is precisely the same calculation as\emph{Stokes' theorem}. \todo{prove this?}
We now proved that $\oint$ is indeed a simplicial chain map. Note that $\oint_n$ need not to preserve multiplication, so it fails to be a map of cochain algebras. However $\oint(1)=1$ and so the induced map on homology sends the class of $1$ in $H(\Apl_n)=\k\cdot[1]$ to the class of $1$ in $H(C_n)=\k\cdot[1]$. We have proven the following lemma.
Joshua Moerman
10 years ago