diff --git a/thesis/notes/A_K_Quillen_Pair.tex b/thesis/notes/A_K_Quillen_Pair.tex index 3a241c7..c91a777 100644 --- a/thesis/notes/A_K_Quillen_Pair.tex +++ b/thesis/notes/A_K_Quillen_Pair.tex @@ -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 $\deg{v} = n \geq 1$. 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 \linebreak Eilenberg-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 $(\Lambda V(n), d) \to \linebreak A(K(\Lambda V(n), d))$ 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: $$ \Ho(\sSet_{\Q,1,f}) \iso \Ho(\CDGA_{\Q,1,f}). $$ diff --git a/thesis/notes/CDGA_Of_Polynomials.tex b/thesis/notes/CDGA_Of_Polynomials.tex index 583f5cb..2a663d3 100644 --- a/thesis/notes/CDGA_Of_Polynomials.tex +++ b/thesis/notes/CDGA_Of_Polynomials.tex @@ -24,6 +24,7 @@ $$ s_i(x_j) = \begin{cases} One can check that $\Apl \in \simplicial{\CDGA_\k}$. We will denote the subspace of homogeneous elements of degree $k$ as $\Apl^k$, this is a simplicial $\k$-module as the maps $d_i$ and $s_i$ are graded maps of degree $0$. +\pagebreak \Lemma{apl-contractible}{ $\Apl^k$ is contractible. } diff --git a/thesis/notes/Calculations.tex b/thesis/notes/Calculations.tex index 2e5ff5a..f56e131 100644 --- a/thesis/notes/Calculations.tex +++ b/thesis/notes/Calculations.tex @@ -101,7 +101,8 @@ In this section we will prove that the rational cohomology of an H-space is free An \Def{H-space} is a pointed topological space $x_0 \in X$ with a map $\mu: X \times X \to X$, such that $\mu(x_0, -), \mu(-, x_0) : X \to X$ are homotopic to $\id_X$. } -Let $X$ be an $0$-connected H-space of finite type, then we have the induced comultiplication map $\mu^\ast: H^\ast(X; \Q) \to H^\ast(X; \Q) \tensor H^\ast(X; \Q)$. +Let $X$ be an $0$-connected H-space of finite type, then we have the induced comultiplication map +$$\mu^\ast: H^\ast(X; \Q) \to H^\ast(X; \Q) \tensor H^\ast(X; \Q).$$ Homotopic maps are sent to equal maps in cohomology, so we get $H^\ast(\mu(x_0, -)) = \id_{H^\ast(X; \Q)}$. Now write $H^\ast(\mu(x_0, -)) = (\counit \tensor \id) \circ H^\ast(\mu)$, where $\counit$ is the augmentation induced by $x_0$, to conclude that for any $h \in H^{+}(X; \Q)$ the image is of the form $$ H^\ast(\mu)(h) = h \tensor 1 + 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$. diff --git a/thesis/notes/Further_Topics.tex b/thesis/notes/Further_Topics.tex index a89384a..82f83a3 100644 --- a/thesis/notes/Further_Topics.tex +++ b/thesis/notes/Further_Topics.tex @@ -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. + +\blankpage diff --git a/thesis/notes/Minimal_Models.tex b/thesis/notes/Minimal_Models.tex index 7d3d15f..f14bdf6 100644 --- a/thesis/notes/Minimal_Models.tex +++ b/thesis/notes/Minimal_Models.tex @@ -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 \leq 2$. 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 \leq 2$. 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 \leq 3$, since $\Lambda V^{\leq 2}$ 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^{>}[rr] & & B_2 \arfib[r]^-{b_1} & B_1 \arfib[r]^-{b_0} & B_0 } \end{displaymath} - 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 \to 0, $$ $$ \cdots \tot{\Delta} H^n(B) \tot{i_\ast} H^n(\prod_i B_i) \tot{t_\ast} H^n(\prod_i B_i) \tot{\Delta} \cdots $$