@ -135,7 +135,7 @@ Another exposition of this corollary can be found in \cite[Section 8.4]{berglund
\section{Equivalence on rational spaces}
In this section we will prove that the adjunction in \CorollaryRef{minimal-model-adjucntion} is in fact an equivalence when restricted to certain subcategories. One of the restrictions is the following.
In this section we will prove that the adjunction in \CorollaryRef{minimal-model-adjunction} is in fact an equivalence when restricted to certain subcategories. One of the restrictions is the following.
\Definition{finite-type}{
A cdga $A$ is said to be of \Def{finite type} if $H(A)$ is finite dimensional in each degree. Similarly $X$ is of \Def{finite type} if $H^i(X; \Q)$ is finite dimensional for each $i$.
@ -149,13 +149,13 @@ 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-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.
and the second map is obtained by the map $A \to A(K(A))$ and using the bijection from \CorollaryRef{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-homotopy-groups} 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 (note that this is specific for $\Q$):
By \CorollaryRef{minimal-cdga-homotopy-groups} we know that $K(A)$ is an Eilenberg-MacLane space of type $K(\Q^\ast, n)$. The cohomology of an \linebreakEilenberg-MacLane space with coefficients in $\Q$ is known (note that this is specific for $\Q$):
$$ H^\ast(K(\Q^\ast, n); \Q)=\Q[x], $$
that is, the free commutative graded algebra with one generator $x$. This can be calculated, for example, with spectral sequences \cite{griffiths}.
@ -170,7 +170,7 @@ and the second map is obtained by the map $A \to A(K(A))$ and using the bijectio
\Lambda D(m) \ar[r]& B
}
\end{displaymath}
Then if $A \to A(K(A))$ is a weak equivalence, so is $B \to A(K(B))$
Then if $A \to A(K(A))$ is a weak equivalence, so is $B \to A(K(B))$.
}
\Proof{
Applying $K$ to the above diagram gives a pullback diagram of simplicial sets, where the induced vertical maps are fibrations (since $K$ is right Quillen). In other words, the induced square is a homotopy pullback.
@ -204,12 +204,12 @@ Note that by \RemarkRef{finited-dim-minimal-model} every cdga of finite type has
Let $(\Lambda V, d)$ be a $1$-connected minimal algebra with $V^i$ finite dimensional for all $i$. Then $(\Lambda V, d)\to A(K(\Lambda V, d))$ is a weak equivalence.
}
\Proof{
Note that if we want to prove the isomorphism $H^i(\Lambda V, d)\to H^i(A(K(\Lambda V, d)))$ it is enough to prove that $H^i(\Lambda V^{\leq i}, d)\to H^i(A(K(\Lambda V^{\leq i}, d)))$ is an isomorphism (as the elements of higher degree do not change the isomorphism). By the $1$-connectedness we can choose our filtration to respect the degree by \LemmaRef{1-reduced-minimal-model}.
Note that if we want to prove the isomorphism \linebreak$H^i(\Lambda V, d)\to H^i(A(K(\Lambda V, d)))$ it is enough to prove that\linebreak$H^i(\Lambda V^{\leq i}, d)\to H^i(A(K(\Lambda V^{\leq i}, d)))$ is an isomorphism (as the elements of higher degree do not change the isomorphism). By the $1$-connectedness we can choose our filtration to respect the degree by \LemmaRef{1-reduced-minimal-model}.
Now $V(n)$ is finitely generated for all $n$ by assumption. By the inductive procedure above we see that $(\Lambda V(n), d)\to A(K(\Lambda V(n), d))$ is a weak equivalence for all $n$. Hence $(\Lambda V, d)\to A(K(\Lambda V, d))$ is a weak equivalence.
Now $V(n)$ is finitely generated for all $n$ by assumption. By the inductive procedure above we see that $(\Lambda V(n), d)\to\linebreakA(K(\Lambda V(n), d))$ is a weak equivalence for all $n$. Hence\linebreak$(\Lambda V, d)\to A(K(\Lambda V, d))$ is a weak equivalence.
}
Now we want to prove that $X \to K(M(X))$ is a weak equivalence for a simply connected rational space $X$ of finite type. For this, we will use that $A$ preserves and detects such weak equivalences by the Serre-Whitehead theorem (\CorollaryRef{serre-whitehead}). To be precise: for a simply connected rational space $X$ the map $X \to K(M(X))$ is a weak equivalence if and only if $A(K(M(X)))\to A(X)$ is a weak equivalence.
Now we want to prove that $X \to K(M(X))$ is a weak equivalence for a simply connected rational space $X$ of finite type. For this, we will use that $A$ preserves and detects such weak equivalences by the Serre-Whitehead theorem (\CorollaryRef{rational-whitehead}). To be precise: for a simply connected rational space $X$ the map $X \to K(M(X))$ is a weak equivalence if and only if $A(K(M(X)))\to A(X)$ is a weak equivalence.
\Lemma{}{
The map $X \to K(M(X))$ is a weak equivalence for $1$-connected, rational spaces $X$ of finite type.
@ -223,7 +223,7 @@ Now we want to prove that $X \to K(M(X))$ is a weak equivalence for a simply con
The map on the right is a weak equivalence by \CorollaryRef{cdga-unit-we}. Then by the 2-out-of-3 property we see that the above composition is indeed a weak equivalence. Since $A$ detects weak equivalences, we conclude that $X \to K(M(X))$ is a weak equivalence.
}
We have proven the following theorem.
We have proven the following theorem.\newpage
\Theorem{main-theorem}{
The functors $A$ and $K$ induce an equivalence of homotopy categories, when restricted to rational, $1$-connected objects of finite type. More formally, we have:
One can check that $\Apl\in\simplicial{\CDGA_\k}$. We will denote the subspace of homogeneous elements of degree $k$ as $\Apl^k$, this is a simplicial $\k$-module as the maps $d_i$ and $s_i$ are graded maps of degree $0$.
@ -101,7 +101,8 @@ In this section we will prove that the rational cohomology of an H-space is free
An \Def{H-space} is a pointed topological space $x_0\in X$ with a map $\mu: X \times X \to X$, such that $\mu(x_0, -), \mu(-, x_0) : X \to X$ are homotopic to $\id_X$.
}
Let $X$ be an $0$-connected H-space of finite type, then we have the induced comultiplication map $\mu^\ast: H^\ast(X; \Q)\to H^\ast(X; \Q)\tensor H^\ast(X; \Q)$.
Let $X$ be an $0$-connected H-space of finite type, then we have the induced comultiplication map
Homotopic maps are sent to equal maps in cohomology, so we get $H^\ast(\mu(x_0, -))=\id_{H^\ast(X; \Q)}$. Now write $H^\ast(\mu(x_0, -))=(\counit\tensor\id)\circ H^\ast(\mu)$, where $\counit$ is the augmentation induced by $x_0$, to conclude that for any $h \in H^{+}(X; \Q)$ the image is of the form
$$ H^\ast(\mu)(h)= h \tensor1+1\tensor h +\psi, $$
@ -135,4 +136,4 @@ This allows us to directly relate the rational homotopy groups to the cohomology
The spheres $S^n$ are not H-spaces if $n$ is even.
}
In fact we have that $S^n_\Q$ is an H-space if and only if $n$ is odd. The only if part is precisely the corollary above, the if part follows from the fact that $S^n_\Q$ is the loop space $K(\Q^\ast, n)$ for odd $n$.
In fact we have that $S^n_\Q$ is an H-space if and only if $n$ is odd. The only if part is precisely the above corollary. The if part follows from the fact that $S^n_\Q$ is the Eilenberg-MacLane space $K(\Q^\ast, n)$ for odd $n$.
@ -37,3 +37,5 @@ Now that we have a bunch of localizations $X_\Q, X_2, X_3, X_5, \ldots$ we might
This theorem is known as \emph{the arithmetic square}, \emph{fracture theorem} or \emph{local-to-global theorem}.
As an example we find that if $X$ is an H-space, then so are its localizations. The converse also holds when certain compatibility requirements are satisfied \cite{sullivan}. In the previous section we were able to prove that $S^n_\Q$ is an H-space if and only if $n$ is odd. It turns out that the prime $p=2$ brings the key to Adams' theorem: for odd $n$ we have that $S^n_2$ is an H-space if and only if $n=1, 3$ or $7$. For the other primes $S^n_p$ is always an H-space for odd $n$. This observation leads to one approach to prove Adams' theorem.
@ -68,7 +68,7 @@ It is clear that induction will be an important technique when proving things ab
This finished the construction of $V$ and $m : \Lambda V \to A$. Now we will prove that $H(m)$ is an isomorphism. We will do so by proving surjectivity and injectivity by induction on $k$.
Start by noting that $H^i(m_2)$ is surjective for $i \leq2$. now assume$H^i(m_k)$ is surjective for $i \leq k$. Since $\im H(m_k)\subset\im H(m_{k+1})$ we see that $H^i(m_{k+1})$ is surjective for $i < k+1$. By construction it is also surjective in degree $k+1$. So $H^i(m_k)$ is surjective for all $i \leq k$ for all $k$.
Start by noting that $H^i(m_2)$ is surjective for $i \leq2$. Now assume by induction that$H^i(m_k)$ is surjective for $i \leq k$. Since $\im H(m_k)\subset\im H(m_{k+1})$ we see that $H^i(m_{k+1})$ is surjective for $i < k+1$. By construction it is also surjective in degree $k+1$. So $H^i(m_k)$ is surjective for all $i \leq k$ for all $k$.
For injectivity we note that $H^i(m_2)$ is injective for $i \leq3$, since $\Lambda V^{\leq2}$ has no elements of degree $3$. Assume $H^i(m_k)$ is injective for $i \leq k+1$ and let $[z]\in\ker H^i(m_{k+1})$. Now if $\deg{z}\leq k$ we get $[z]=0$ by induction and if $\deg{z}= k+2$ we get $[z]=0$ by construction. Finally if $\deg{z}= k+1$, then we write $z =\sum\lambda_\alpha v_\alpha+\sum\lambda'_\beta v'_\beta+ w$ where $v_\alpha, v'_\beta$ are the generators as above and $w \in\Lambda V^{\leq k}$. Now $d z =0$ and so $\sum\lambda'_\beta v'_\beta+ dw =0$, so that $\sum\lambda'_\beta[z_\beta]=0$. Since $\{[z_\beta]\}$ was a basis, we see that $\lambda'_\beta=0$ for all $\beta$. Now by applying $m_k$ we get $\sum\lambda_\alpha[b_\alpha]= H(m_k)[w]$, so that $\sum\lambda_\alpha[a_\alpha]=0$ in the cokernel, recall that $\{[a_\alpha]\}$ formed a basis and hence $\lambda_\alpha=0$ for all $\alpha$. Now $z = w$ and the statement follows by induction. Conclude that $H^i(m_{k+1})$ is injective for $i \leq k+2$.
@ -76,7 +76,7 @@ It is clear that induction will be an important technique when proving things ab
\end{proof}
\Remark{finited-dim-minimal-model}{
The above construction will construct an $r$-reduced minimal model for an $r$-connected cdga $A$.
The previous construction will construct an $r$-reduced minimal model for an $r$-connected cdga $A$.
Moreover if $H(A)$ is finite dimensional in each degree, then so is the minimal model $\Lambda V$. This follows inductively. First notice that $V^2$ is clearly finite dimensional. Now assume that $\Lambda V^{<k}$ is finite dimensional in each degree, then both the cokernel and kernel are, so we adjoin only finitely many elements in $V^k$.
@ -256,7 +256,7 @@ Note that whenever we have a full subcategory $\cat{C'} \subset \cat{C}$, where
\Example{ho-top}{
The category $\Ho(\Top)$ has as objects just topological spaces and homotopy classes between cofibrant replacements (note that every objects is already fibrant). Moreover, if we restrict to the full subcategory of CW complexes, maps are precisely homotopy classes between objects.
Homotopical invariants are often defined as functors on $\Top$. For example we have the $n$-th homotopy group functor $\pi_n: \Top\to\Grp$ and the $n$-th homology group functor $H_n: \Top\to\Ab$. But since they are homotopy invariant, they can be expressed as functors on $\Ho(\Top)$:
Homotopical invariants are often defined as functors on $\Top$. For example we have the $n$-th homotopy group functor \linebreak$\pi_n: \Top\to\Grp$ and the $n$-th homology group functor\linebreak$H_n: \Top\to\Ab$. But since they are homotopy invariant, they can be expressed as functors on $\Ho(\Top)$:
@ -6,12 +6,15 @@ Given a category $\cat{C}$ and a functor $F: \DELTA \to \cat{C}$, then define th
F_!(X) &= \colim_{\Delta[n]\to X} F[n] & X \in\sSet\\
F^\ast(C)_n &= \Hom_{\cat{C}}(F[n], Y) & C \in\cat{C}
\end{align*}
A simplicial map $X \to Y$ induces a map of the diagrams of which we take colimits. Applying $F$ on these diagrams, make it clear that $F_!$ is functorial. Secondly we see readily that $F^\ast$ is functorial. By using the definition of colimit and the Yoneda lemma (Y) we can prove that $F_!$ is left adjoint to $F^\ast$:
A simplicial map $X \to Y$ induces a map of the diagrams of which we take colimits. Applying $F$ on these diagrams, make it clear that $F_!$ is functorial. Secondly we see readily that $F^\ast$ is functorial. By using the definition of colimit and the Yoneda lemma (Y) we can prove that $F_!$ is left adjoint to $F^\ast$ by the following calculation:
@ -19,13 +22,12 @@ A simplicial map $X \to Y$ induces a map of the diagrams of which we take colimi
Furthermore we have $F_!\circ\Delta[-]\iso F$. In short we have the following:
\cdiagram{Kan_Extension}
In our case where $F =\Apl$ and $\cat{C}=\CDGA_\k$ we get:
\cdiagram{Apl_Extension}
\subsection{The cochain complex of polynomial forms}
In our case we take the opposite category, so the definition of $A$ is in terms of a limit instead of colimit. This allows us to give a nicer description:
In our case where $F =\Apl$ and $\cat{C}=\CDGA_\k$ we get:
\cdiagram{Apl_Extension}
Note that we have the opposite category of cdga's, so the definition of $A$ is in terms of a limit instead of colimit. This allows us to give a nicer description:
where $C^\ast(\Delta^n; \k)$ is the (normalized) singular cochain complex of $\Delta^n$ with coefficients in $\k$. 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}
This left adjoint functor $C^\ast : \sSet\to\opCat{\DGA_\k}$ is (just as above) defined as $C^\ast(X)=\Hom_\sSet(X, C^\ast(\Delta[-]; \k))$. To see that this is precisely the classical definition of the singular cochain complex, we make the following calculation.
This left adjoint functor $C^\ast : \sSet\to\opCat{\DGA_\k}$ is (just as above) defined as $C^\ast(X)=\Hom_\sSet(X, C^\ast(\Delta[-]; \k))$. To see that this is precisely the classical definition of the singular \linebreakcochain complex, we make the following calculation.
\begin{align*}
C^\ast(X) &= \Hom(X, C^\ast(\Delta[n])) \\
&= \Hom(X, \Hom(N \Z\Delta[n], \k)) \\
&\ison{1}\Hom(X, \Gamma(\k)) \\
&\iso\Hom(N \Z (X), \k)
\end{align*}
where $\Z$ is the free simplicial abelian group, $N$ is the normalized chain complex (this is the Dold-Kan equivalence) and $\Gamma$ its right adjoint. At \refison{1} we used the definition $\Gamma(C)=\Hom(N \Z\Delta[n], C)$. Now the conclusion of this calculation is that $C^\ast(X)$ is precisely the dual complex of $N \Z(X)$, which is the singular (normalized) chain complex.
where $\Z$ is the free simplicial abelian group, $N$ is the normalized chain complex (this is the Dold-Kan equivalence) and $\Gamma$ its right adjoint. At \refison{1} we use the definition of this right adjoint$\Gamma(C)=\Hom(N \Z\Delta[n], C)$. Now the conclusion of this calculation is that $C^\ast(X)$ is precisely the dual complex of $N \Z(X)$, which is the singular (normalized) chain complex.
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.
@ -137,7 +139,7 @@ We will now prove that the map $\oint: A(X) \to C^\ast(X)$ is a quasi isomorphis
Define a map $t: \prod_i B_i \to\prod_i B_i$ defined by $t(x_0, x_1, \dots)=(x_0+ b_0(x_1), x_1+ b_1(x_2), \dots)$. Note that $t$ is surjective and that $B \iso\ker(t)$. So we get the following natural short exact sequence and its associated natural long exact sequence in homology:
Define a map $t: \prod_i B_i \to\prod_i B_i$ defined by $t(x_0, x_1, \dots)=(x_0+ b_0(x_1), x_1+ b_1(x_2), \dots)$. Note that $t$ is surjective and that \linebreak$B \iso\ker(t)$. So we get the following natural short exact sequence and its associated natural long exact sequence in homology:
$$0\to B \tot{i}\prod_i B_i \tot{t}\prod_i B_i \to0, $$
Joshua Moerman
9 years ago