author | kevin@6e1638ff-ae45-0410-89bd-df963105f760 |
Sun, 18 Oct 2009 23:54:43 +0000 | |
changeset 128 | 0387384c9acd |
parent 123 | a5e863658e74 |
child 133 | 7a880cdaac70 |
permissions | -rw-r--r-- |
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%!TEX root = ../blob1.tex |
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\section{The blob complex for $A_\infty$ $n$-categories} |
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\label{sec:ainfblob} |
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Given an $A_\infty$ $n$-category $\cC$ and an $n$-manifold $M$, we define the blob |
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complex $\bc_*(M)$ to the be the colimit $\cC(M)$ of Section \ref{sec:ncats}. |
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\nn{say something about this being anticlimatically tautological?} |
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We will show below |
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\nn{give ref} |
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that this agrees (up to homotopy) with our original definition of the blob complex |
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in the case of plain $n$-categories. |
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When we need to distinguish between the new and old definitions, we will refer to the |
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new-fangled and old-fashioned blob complex. |
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\medskip |
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Let $M^n = Y^k\times F^{n-k}$. |
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Let $C$ be a plain $n$-category. |
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Let $\cF$ be the $A_\infty$ $k$-category which assigns to a $k$-ball |
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$X$ the old-fashioned blob complex $\bc_*(X\times F)$. |
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\begin{thm} \label{product_thm} |
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The old-fashioned blob complex $\bc_*^C(Y\times F)$ is homotopy equivalent to the |
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new-fangled blob complex $\bc_*^\cF(Y)$. |
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\end{thm} |
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\begin{proof} |
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We will use the concrete description of the colimit from Subsection \ref{ss:ncat_fields}. |
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First we define a map |
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\[ |
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\psi: \bc_*^\cF(Y) \to \bc_*^C(Y\times F) . |
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\] |
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In filtration degree 0 we just glue together the various blob diagrams on $X\times F$ |
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(where $X$ is a component of a permissible decomposition of $Y$) to get a blob diagram on |
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$Y\times F$. |
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In filtration degrees 1 and higher we define the map to be zero. |
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It is easy to check that this is a chain map. |
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Next we define a map |
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\[ |
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\phi: \bc_*^C(Y\times F) \to \bc_*^\cF(Y) . |
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\] |
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Actually, we will define it on the homotopy equivalent subcomplex |
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$\cS_* \sub \bc_*^C(Y\times F)$ generated by blob diagrams which are small with |
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respect to some open cover |
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of $Y\times F$. |
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\nn{need reference to small blob lemma} |
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We will have to show eventually that this is independent (up to homotopy) of the choice of cover. |
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Also, for a fixed choice of cover we will only be able to define the map for blob degree less than |
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some bound, but this bound goes to infinity as the cover become finer. |
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Given a decomposition $K$ of $Y$ into $k$-balls $X_i$, let $K\times F$ denote the corresponding |
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decomposition of $Y\times F$ into the pieces $X_i\times F$. |
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%We will define $\phi$ inductively, starting at blob degree 0. |
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%Given a 0-blob diagram $x$ on $Y\times F$, we can choose a decomposition $K$ of $Y$ |
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%such that $x$ is splittable with respect to $K\times F$. |
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%This defines a filtration degree 0 element of $\bc_*^\cF(Y)$ |
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We will define $\phi$ using a variant of the method of acyclic models. |
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Let $a\in \cS_m$ be a blob diagram on $Y\times F$. |
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For $m$ sufficiently small there exists a decomposition $K$ of $Y$ into $k$-balls such that the |
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codimension 1 cells of $K\times F$ miss the blobs of $a$, and more generally such that $a$ is splittable along (the codimension-1 part of) $K\times F$. |
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Let $D(a)$ denote the subcomplex of $\bc_*^\cF(Y)$ generated by all $(a, \bar{K})$ |
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such that each $K_i$ has the aforementioned splittable property |
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(see Subsection \ref{ss:ncat_fields}). |
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\nn{need to define $D(a)$ more clearly; also includes $(b_j, \bar{K})$ where |
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$\bd(a) = \sum b_j$.} |
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(By $(a, \bar{K})$ we really mean $(a^\sharp, \bar{K})$, where $a^\sharp$ is |
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$a$ split according to $K_0\times F$. |
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To simplify notation we will just write plain $a$ instead of $a^\sharp$.) |
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Roughly speaking, $D(a)$ consists of filtration degree 0 stuff which glues up to give |
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$a$, filtration degree 1 stuff which makes all of the filtration degree 0 stuff homologous, |
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filtration degree 2 stuff which kills the homology created by the |
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filtration degree 1 stuff, and so on. |
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More formally, |
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\begin{lemma} |
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$D(a)$ is acyclic. |
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\end{lemma} |
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\begin{proof} |
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We will prove acyclicity in the first couple of degrees, and \nn{in this draft, at least} |
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leave the general case to the reader. |
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Let $K$ and $K'$ be two decompositions of $Y$ compatible with $a$. |
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We want to show that $(a, K)$ and $(a, K')$ are homologous via filtration degree 1 stuff. |
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\nn{need to say this better; these two chains don't have the same boundary.} |
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We might hope that $K$ and $K'$ have a common refinement, but this is not necessarily |
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the case. |
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(Consider the $x$-axis and the graph of $y = x^2\sin(1/x)$ in $\r^2$.) |
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However, we {\it can} find another decomposition $L$ such that $L$ shares common |
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refinements with both $K$ and $K'$. |
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Let $KL$ and $K'L$ denote these two refinements. |
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Then filtration degree 1 chains associated to the four anti-refinemnts |
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$KL\to K$, $KL\to L$, $K'L\to L$ and $K'L\to K'$ |
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give the desired chain connecting $(a, K)$ and $(a, K')$ |
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(see Figure \ref{zzz4}). |
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\begin{figure}[!ht] |
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\begin{equation*} |
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\mathfig{.63}{tempkw/zz4} |
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\end{equation*} |
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\caption{Connecting $K$ and $K'$ via $L$} |
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\label{zzz4} |
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\end{figure} |
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Consider a different choice of decomposition $L'$ in place of $L$ above. |
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This leads to a cycle consisting of filtration degree 1 stuff. |
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We want to show that this cycle bounds a chain of filtration degree 2 stuff. |
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Choose a decomposition $M$ which has common refinements with each of |
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$K$, $KL$, $L$, $K'L$, $K'$, $K'L'$, $L'$ and $KL'$. |
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b62214646c4f
preparing for semi-public version soon
kevin@6e1638ff-ae45-0410-89bd-df963105f760
parents:
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diff
changeset
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\nn{need to also require that $KLM$ antirefines to $KM$, etc.} |
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Then we have a filtration degree 2 chain, as shown in Figure \ref{zzz5}, which does the trick. |
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(Each small triangle in Figure \ref{zzz5} can be filled with a filtration degree 2 chain.) |
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\begin{figure}[!ht] |
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\begin{equation*} |
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\mathfig{1.0}{tempkw/zz5} |
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\end{equation*} |
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\caption{Filling in $K$-$KL$-$L$-$K'L$-$K'$-$K'L'$-$L'$-$KL'$-$K$} |
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\label{zzz5} |
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\end{figure} |
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Continuing in this way we see that $D(a)$ is acyclic. |
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\end{proof} |
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We are now in a position to apply the method of acyclic models to get a map |
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$\phi:\cS_* \to \bc_*^\cF(Y)$. |
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This map is defined in sufficiently low degrees, sends a blob diagram $a$ to $D(a)$, |
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and is well-defined up to (iterated) homotopy. |
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The subcomplex $\cS_* \subset \bc_*^C(Y\times F)$ depends on choice of cover of $Y\times F$. |
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If we refine that cover, we get a complex $\cS'_* \subset \cS_*$ |
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and a map $\phi':\cS'_* \to \bc_*^\cF(Y)$. |
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$\phi'$ is defined only on homological degrees below some bound, but this bound is higher than |
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the corresponding bound for $\phi$. |
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We must show that $\phi$ and $\phi'$ agree, up to homotopy, |
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on the intersection of the subcomplexes on which they are defined. |
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This is clear, since the acyclic subcomplexes $D(a)$ above used in the definition of |
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$\phi$ and $\phi'$ do not depend on the choice of cover. |
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\nn{need to say (and justify) that we now have a map $\phi$ indep of choice of cover} |
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We now show that $\phi\circ\psi$ and $\psi\circ\phi$ are homotopic to the identity. |
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$\psi\circ\phi$ is the identity. $\phi$ takes a blob diagram $a$ and chops it into pieces |
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according to some decomposition $K$ of $Y$. |
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$\psi$ glues those pieces back together, yielding the same $a$ we started with. |
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$\phi\circ\psi$ is the identity up to homotopy by another MoAM argument... |
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This concludes the proof of Theorem \ref{product_thm}. |
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\nn{at least I think it does; it's pretty rough at this point.} |
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\end{proof} |
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\nn{need to say something about dim $< n$ above} |
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\medskip |
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\begin{cor} |
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The new-fangled and old-fashioned blob complexes are homotopic. |
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\end{cor} |
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\begin{proof} |
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Apply Theorem \ref{product_thm} with the fiber $F$ equal to a point. |
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\end{proof} |
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\medskip |
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\hrule |
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\medskip |
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\nn{to be continued...} |
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\medskip |
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