Joshua Moerman
11 years ago
11 changed files with 226 additions and 10 deletions
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\subsection{Cochain models for the $n$-disk and $n$-sphere} |
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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 $c = d(b) \in D(n)^{n+1}$. $S(n)$ is the cochain complex generated by one element $a \in S(n)^n$ which differential vanishes (i.e. $da = 0$). In other words: |
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$$ D(n) = ... \to 0 \to \k \to \k \to 0 \to ... $$ |
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$$ S(n) = ... \to 0 \to \k \to 0 \to 0 \to ... $$ |
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Note that $D(n)$ is acyclic for all $n$, or put in different words: $j_n : 0 \to D(n)$ is a quasi isomorphism. The sphere $S(n)$ has exactly one non-trivial cohomology group $H^n(S(n)) = \k \cdot [a]$. There is an injective function $i_n : S(n+1) \to D(n)$, sending $a$ to $c$. The maps $j_n$ and $i_n$ play the following important role in the model structure of cochain complexes: |
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\begin{claim} |
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The set $I = \{i_n : S(n+1) \to D(n) \I n \in \N\}$ generates all cofibrations and the set $J = \{j_n : 0 \to D(n) \I n \in \N\}$ generates all trivial cofibrations. |
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\end{claim} |
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The proof is omitted but can be found in different sources \todo{Cite sources}. In the next section we will prove a similar result for cdga's, so the reader can also refer to that proof. |
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$S(n)$ 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: $\Hom(S(n), X) / ~ \iso H^n(X)$. So the cohomology groups can be considered as honest homotopy groups. |
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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 $b^n$ and $c b^n$ when $n$ is even, and $c^n b$ and $c^n$ when $n$ is odd. In both cases we can compute the differentials using the Leibniz rule: |
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$$ d(b^n) = n \cdot c b^{n-1} $$ |
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$$ d(c b^n) = 0 $$ |
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$$ d(c^n b) = c^{n+1} $$ |
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$$ d(c^n) = 0 $$ |
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Those cocycles are in fact coboundaries (remember that $\k$ is a field of characteristic $0$): |
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$$ c b^n = \frac{1}{n} d(b^{n+1}) $$ |
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$$ c^n = d(b c^{n-1}) $$ |
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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. |
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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 (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. |
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\begin{theorem} |
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The sets $\Lambda(I)$ and $\Lambda(J)$ generate a model structure on $\CDGA_\k$ where: |
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\begin{itemize} |
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\item weak equivalences are quasi isomorphisms, |
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\item fibrations are (degree wise) surjective maps and |
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\item cofibrations are maps with the left lifting property against trivial fibrations. |
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\end{itemize} |
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\end{theorem} |
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We will prove this theorem in the next section. Note that the functors $\Lambda$ and $U$ thus form a Quillen pair with this model structure. |
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\subsection{Why we need $\Char{\k} = 0$ for algebras} |
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The above Quillen pair $(\Lambda, U)$ fails to be a Quillen pair if $\Char{\k} = p \neq 0$. We will show this by proving that the maps $\Lambda(j_n)$ are not weak equivalences for even $n$. Consider $b^p \in D(n)$, then by the Leibniz rule: |
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$$ d(b^p) = p \cdot c b^{p-1} = 0. $$ |
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So $b^p$ is a cocycle. Now assume $b^p = dx$ for some $x$ of degree $pn - 1$, then $x$ contains a factor $c$ for degree reasons. By the calculations above we see that any element containing $c$ has a trivial differential or has a factor $c$ in its differential, contradicting $b^p = dx$. So this cocycle is not a coboundary and $\Lambda D(n)$ is not acyclic. |
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\subsection{CDGA of Polynomials} |
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\newcommand{\Apl}[0]{{A_{PL}}} |
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We will now give a cdga model for the $n$-simplex $\Delta^n$. This then allows for simplicial methods. In the following definition one should be reminded of the topological $n$-simplex defined as convex span. |
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\begin{definition} |
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For all $n \in \N$ define the following cdga: |
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$$ (\Apl)_n = \frac{\Lambda(x_0, \ldots, x_n, dx_0, \ldots, dx_n)}{(\sum_{i=0}^n) x_i - 1, \sum_{i=0}^n dx_i)} $$ |
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So it is the free cdga with $n+1$ generators and their differentials such that $\sum_{i=0}^n x_i = 1$ and in order to be well behaved $\sum_{i=0}^n dx_i = 0$. |
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\end{definition} |
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Note that the inclusion $\Lambda(x_1, \ldots, x_n, dx_1, \ldots, dx_n) \to \Apl_n$ is an isomorphism of cdga's. So $\Apl_n$ is free and (algebra) maps from it are determined by their images on $x_i$ for $i = 1, \ldots, n$ (also note that this determines the images for $dx_i$). This fact will be used throughout. |
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These cdga's will assemble into a simplicial cdga when we define the face and degeneracy maps as follows ($j = 1, \ldots, n$): |
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$$ d_i(x_j) = \begin{cases} |
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x_{j-1}, &\text{ if } i < j \\ |
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0, &\text{ if } i = j \\ |
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x_j, &\text{ if } i > j |
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\end{cases} \qquad d_i : \Apl_n \to \Apl_{n-1} $$ |
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$$ s_i(x_j) = \begin{cases} |
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x_{j+1}, &\text{ if } i < j \\ |
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x_j + x_{j+1}, &\text{ if } i = j \\ |
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x_j, &\text{ if } i > j |
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\end{cases} \qquad s_i : \Apl_n \to \Apl_{n+1} $$ |
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One can check that $\Apl \in \simplicial{\CDGA_\k}$. We will denote the subspace of homogeneous elements of degree $k$ as $\Apl^k \in \simplicial{\Mod{\k}}$, this is indeed a simplicial $\k$-module as the maps $d_i$ and $s_i$ are graded maps of degree $0$. |
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\begin{lemma} |
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$\Apl^k$ is contractible. |
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\end{lemma} |
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\begin{proof} |
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We will prove this by defining an extra degeneracy $s: \Apl_n \to \Apl_{n+1}$. Define for $i = 1, \ldots, n$: |
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\begin{align*} |
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s(1) &= (1-x_0)^2 \\ |
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s(x_i) &= (1-x_0) \cdot x_{i+1} |
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\end{align*} |
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Extend on the differentials and multiplicatively on $\Apl_n$. As $s(1) \neq 1$ this map is not an algebra map, however it well-defined as a map of cochain complexes. In particular when restricted to degree $k$ we get a linear map: |
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$$ s: \Apl^k_n \to \Apl^k_{n+1}. $$ |
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Proving the necessary properties of an extra degeneracy is fairly easy. For $n \geq 1$ we get (on generators): |
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\begin{align*} |
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d_0 s(1) &= d_0 (1 - x_0)^2 = (1 - 0) \cdot (1 - 0) = 1 \\ |
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d_0 s(x_i) &= d_0((1-x_0)x_{i+1}) = d_0(1-x_0) \cdot x_i \\ |
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&= (1-0) \cdot x_i = x_i |
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\end{align*} |
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So $d_0 s = \id$. |
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\begin{align*} |
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d_{i+1} s(1) &= d_{i+1} (1 - x_0)^2 = d_{i+1} (\sum_{j=1}^n x_j)^2 \\ |
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&= (\sum_{j=1}^{n-1} x_j)^2 = (1-x_0)^2 = s d_i(1) \\ |
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d_{i+1} s(x_j) &= d_{i+1}(1-x_0) d_{i+1}(x_j) = (1-x_0) d_i(x_{j+1}) = s d_i (x_j) |
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\end{align*} |
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So $d_{i+1} s = s d_i$. Similarly $s_{i+1} s = s s_i$. And finally for $n=0$ we have $d_1 s = 0$. |
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So we have an extra degeneracy $s: \Apl^k \to \Apl^k$, and hence (see for example \cite{goerss}) we have that $\Apl^k$ is contractible. As a consequence $\Apl \to \ast$ is a weak equivalence. |
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\end{proof} |
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\begin{lemma} |
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$\Apl_n^k$ is a Kan complex. |
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\end{lemma} |
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\begin{proof} |
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By the simple fact that $\Apl_n^k$ is a simplicial group, it is a Kan complex \cite{goerss}. |
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\end{proof} |
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\begin{corollary} |
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$\Apl^k \to \ast$ is a trivial fibration in the standard model structure on $\sSet$. |
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\end{corollary} |
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\subsection{The free cdga} |
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Just as in ordinary linear algebra we can form an algebra from any graded module. Furthermore we will see that a differential induces a derivation. |
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\begin{definition} |
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The \emph{tensor algebra} of a graded module $M$ is defined as |
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$$ T(M) = \bigoplus_{n\in\N} M^{\tensor n}, $$ |
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where $M^{\tensor 0} = \k$. An element $m = m_1 \tensor \ldots \tensor m_n$ has a \emph{word length} of $n$ and its degree is $\deg{m} = \sum_{i=i}^n \deg{m_i}$. The multiplication is given by the tensor product (note that the bilinearity follows immediately). |
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\end{definition} |
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Note that this construction is functorial and it is free in the following sense. |
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\begin{lemma} |
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Let $M$ be a graded module and $A$ a graded algebra. |
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\begin{itemize} |
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\item A graded map $f: M \to A$ of degree $0$ extends uniquely to an algebra map $\overline{f} : TM \to A$. |
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\item A differential $d: M \to M$ extends uniquely to a derivation $d: TM \to TM$. |
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\end{itemize} |
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\end{lemma} |
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\begin{corollary} |
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Let $U$ be the forgetful functor from graded algebras to graded modules, then $T$ and $U$ form an adjoint pair: |
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$$ T: \grMod{\k} \leftadj \grAlg{\k} $$ |
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Moreover it extends and restricts to |
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$$ T: \dgMod{\k} \leftadj \dgAlg{\k} $$ |
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$$ T: \CoCh{\k} \leftadj \DGA{\k} $$ |
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\end{corollary} |
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As with the symmetric algebra and exterior algebra of a vector space, we can turn this graded tensor algebra in a commutative graded algebra. |
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\begin{definition} |
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Let $A$ be a graded algebra and define |
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$$ I = < ab - (-1)^{\deg{a}\deg{b}}ba \I a,b \in A >. $$ |
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Then $A / I$ is a commutative graded algebra. |
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For a graded module $M$ we define the \emph{free commutative graded algebra} as |
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$$ \Lambda(M) = TM / I $$ |
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\end{definition} |
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Again this extends to differential graded modules (i.e. the ideal is preserved by the derivative) and restricts to cochain complexes. |
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\begin{lemma} |
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We have the following adjunctions: |
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$$ \Lambda: \grMod{\k} \leftadj \grAlg{\k}^{comm} $$ |
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$$ \Lambda: \dgMod{\k} \leftadj \dgAlg{\k}^{comm} $$ |
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$$ \Lambda: \CoCh{\k} \leftadj \CDGA_\k $$ |
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\end{lemma} |
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We can now easily construct cdga's by specifying generators and their differentials. |
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\subsection{Polynomial Forms} |
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There is a general way to construct functors from $\sSet$ whenever we have some simplicial object. In our case we have the simplicial cdga $\Apl$ (which is nothing more than a functor $\opCat{\DELTA} \to \CDGA$) and we want to extend to a contravariant functor $\sSet \to \CDGA_\k$. This will be done via Kan extensions. |
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Given a category $\cat{C}$ and a functor $F: \DELTA \to \cat{C}$, then define the following on objects: |
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\begin{align*} |
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F_!(X) &= \colim_{\Delta[n] \to X} F[n] &\quad X \in \sSet \\ |
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F^\ast(C)_n &= \Hom_{\cat{C}}(F[n], Y) &\quad C \in \cat{C} |
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\end{align*} |
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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. |
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