@ -77,7 +77,7 @@ Note that for a minimal algebra $\Lambda V$ there is a natural augmentation and
This means that $f_\ast={d_1}_\ast h_\ast={d_0}_\ast h_\ast= g_\ast$. \todo{detail}
}
Consider the augmented cdga $V(n)=D(n)\oplus\k$, with trivial multiplication and where the term $\k$ is used for the unit and augmentation. There is a weak equivalence $A(n)\to V(n)$ (recall \DefinitionRef{minimal-model-sphere}). This augmented cdga can be thought of as a specific model of the sphere. In particular the homotopy groups can be expressed as follows.
Consider the augmented cdga $V(n)=S(n)\oplus\k$, with trivial multiplication and where the term $\k$ is used for the unit and augmentation. There is a weak equivalence $A(n)\to V(n)$ (recall \DefinitionRef{minimal-model-sphere}). This augmented cdga can be thought of as a specific model of the sphere. In particular the homotopy groups can be expressed as follows.
\Lemma{cdga-dual-homotopy-groups}{
There is a natural bijection for any augmented cdga $A$
@ -24,7 +24,7 @@ In this section we will discuss the so called minimal models. These cdga's enjoy
$$(M, d)\we(A, d). $$
\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.
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.
\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.
@ -35,7 +35,7 @@ We will often say \Def{minimal model} or \Def{minimal algebra} to mean minimal S
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 $d(V(k))\subset\Lambda(V(n-1))$.
}
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.
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.\todo{to prove or not to prove}
\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 $d v_j \in\Lambda V_{<j}$.
@ -57,14 +57,14 @@ It is clear that induction will be an important technique when proving things ab
Start by setting $V(0)= H^{\geq1}(A)$ 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 $\deg{v_\alpha}=\deg{z_\alpha}-1$.
Suppose $V(k)$ and $m_k$have been 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 ($r \geq1$), 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 $\Lambda^{\leq1} V(0)$, and so the defects are in $\Lambda^{\geq2} V(0)$ 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 $\Lambda^{\geq2} V(1)$ 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.
\todo{Rewrite this section}Now assume that $(A, d)$ is $r$-connected ($r \geq1$), 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 $\Lambda^{\leq1} V(0)$, and so the defects are in $\Lambda^{\geq2} V(0)$ 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 $\Lambda^{\geq2} V(1)$ 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 \LemmaRef{1-reduced-minimal-model}$(\Lambda V, d)$ is minimal.
\end{proof}
@ -102,7 +102,7 @@ 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.
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.\todo{This actually is the same as in chapter 4}
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).
@ -128,7 +128,7 @@ 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}
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.
\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.
\end{proof}
@ -139,7 +139,7 @@ Now if the map $f$ is a weak equivalence, both maps $\phi$ and $\psi$ are surjec
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.
The assignment to $X$ of 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.
@ -46,8 +46,7 @@ Another way to model the $n$-simplex is by the singular cochain complex associat
$$ C_n = C^\ast(\Delta^n; \k). $$
The inclusion maps $d^i : \Delta^n \to\Delta^{n+1}$ and the maps $s^i: \Delta^n \to\Delta^{n-1}$ induce face and degeneracy maps on the dga's $C_n$, turning $C$ into a simplicial dga. Again we can extend this to functors by Kan extensions
\cdiagram{C_Extension}
where the left adjoint is precisely the functor $C^\ast$ as noted in \cite{felix}. We will relate $\Apl$ and $C$ in order to obtain a natural quasi isomorphism $A(X)\we C^\ast(X)$ for every $X \in\sSet$. Furthermore this map preserves multiplication on the homology algebras.
\todo{show that $C^\ast$ really is sing cohom} where the left adjoint is precisely the functor $C^\ast$ as noted in \cite{felix}. We will relate $\Apl$ and $C$ in order to obtain a natural quasi isomorphism $A(X)\we C^\ast(X)$ for every $X \in\sSet$. Furthermore this map preserves multiplication on the homology algebras.
\subsection{Integration and Stokes' theorem for polynomial forms}
Joshua Moerman
10 years ago