32 |
32 |
33 \newcommand{\bdy}{\partial} |
33 \newcommand{\bdy}{\partial} |
34 \newcommand{\iso}{\cong} |
34 \newcommand{\iso}{\cong} |
35 \newcommand{\tensor}{\otimes} |
35 \newcommand{\tensor}{\otimes} |
36 \newcommand{\Tensor}{\bigotimes} |
36 \newcommand{\Tensor}{\bigotimes} |
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37 \newcommand{\directSum}{\oplus} |
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38 \newcommand{\DirectSum}{\bigoplus} |
37 |
39 |
38 \newcommand{\into}{\hookrightarrow} |
40 \newcommand{\into}{\hookrightarrow} |
39 |
41 |
40 \newcommand{\restrict}[2]{#1{}_{\mid #2}{}} |
42 \newcommand{\restrict}[2]{#1{}_{\mid #2}{}} |
41 \newcommand{\set}[1]{\left\{#1\right\}} |
43 \newcommand{\set}[1]{\left\{#1\right\}} |
42 \newcommand{\setc}[2]{\setcl{#1}{#2}} |
44 \newcommand{\setc}[2]{\setcl{#1}{#2}} |
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45 \newcommand{\setcl}[2]{\left\{ \left. #1 \;\right| \; #2 \right\}} |
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46 \newcommand{\setcr}[2]{\left\{ #1 \;\left| \; #2 \right\}\right.} |
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47 |
43 |
48 |
44 % tricky way to iterate macros over a list |
49 % tricky way to iterate macros over a list |
45 \def\semicolon{;} |
50 \def\semicolon{;} |
46 \def\applytolist#1{ |
51 \def\applytolist#1{ |
47 \expandafter\def\csname multi#1\endcsname##1{ |
52 \expandafter\def\csname multi#1\endcsname##1{ |
248 |
253 |
249 Suppose an $n$-manifold $X$ contains a copy of $Y$, an $n-1$ manifold, as a codimension $0$ submanifold of its boundary. Fix a boundary condition $c \in \cF(\bdy X \setminus Y)$. Then the collection $A(X; c \bullet d)$, as $d$ varies over $\cF(Y)$, forms a module over the $1$-category $A(Y)$. The action is via gluing a collar onto $Y$, then applying a `collaring homeomorphism' $X \cup_Y Y \times I \to X$. |
254 Suppose an $n$-manifold $X$ contains a copy of $Y$, an $n-1$ manifold, as a codimension $0$ submanifold of its boundary. Fix a boundary condition $c \in \cF(\bdy X \setminus Y)$. Then the collection $A(X; c \bullet d)$, as $d$ varies over $\cF(Y)$, forms a module over the $1$-category $A(Y)$. The action is via gluing a collar onto $Y$, then applying a `collaring homeomorphism' $X \cup_Y Y \times I \to X$. |
250 |
255 |
251 If $X$ contains two copies of $Y$, $A(X)$ is then a bimodule over $A(Y)$. Below, we'll compute the invariant of the `glued up' manifold $X \bigcup_Y \selfarrow$ as the self-tensor product of this bimodule. |
256 If $X$ contains two copies of $Y$, $A(X)$ is then a bimodule over $A(Y)$. Below, we'll compute the invariant of the `glued up' manifold $X \bigcup_Y \selfarrow$ as the self-tensor product of this bimodule. |
252 |
257 |
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258 What is the self-tensor product of a bimodule over a category? First, what is the tensor product of a left-module and a right-module? Recall a module $\cM$ over a category $\cC$ is a collection of vector spaces $\cM_c$ indexed by an object $c \in \cC$, along with a linear map $f: \cM_a \to \cM_b$ for each morphism $f:a \mapsto b$ of $\cC$. The tensor product of a left- and a right-module is defined to be the vector space |
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259 $$\cM \Tensor_\cC \cN = \left(\DirectSum_{c \in \cC} \cM_c \tensor {}_c\cN\right) \Big/ \setc{mf \tensor n - m \tensor fn}{f: c \mapsto c'}$$ |
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260 just as you'd expect. The self-tensor product of a bimodule ${}_{\cC}\cX_{\cC}$ is then |
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261 $$\cX \Tensor_{\cC} \selfarrow = \DirectSum_{c \in \cC} {}_c \cX_c / (xf - fx).$$ |
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262 |
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263 |
253 \begin{lem} |
264 \begin{lem} |
254 Any isotopy of $X \bigcup_Y \selfarrow$ is homotopic to a composition of `collar shift' isotopies and isotopies that are constant on $Y$ (i.e. the image of an isotopy of $X$ itself). |
265 Any isotopy of $X \bigcup_Y \selfarrow$ is homotopic to a composition of `collar shift' isotopies and isotopies that are constant on $Y$ (i.e. the image of an isotopy of $X$ itself). |
255 \end{lem} |
266 \end{lem} |
256 \begin{proof} |
267 \begin{proof} |
257 First make the isotopy act locally. When it's acting in a small ball overlapping $Y$, conjugate by a collar shift to move it off. |
268 First make the isotopy act locally. When it's acting in a small ball overlapping $Y$, conjugate by a collar shift to move it off. |
259 |
270 |
260 \begin{thm} |
271 \begin{thm} |
261 $$A(X \bigcup_Y \selfarrow) \iso A(X) \Tensor_{A(Y)} \selfarrow$$ |
272 $$A(X \bigcup_Y \selfarrow) \iso A(X) \Tensor_{A(Y)} \selfarrow$$ |
262 \end{thm} |
273 \end{thm} |
263 \begin{proof} |
274 \begin{proof} |
264 Certainly there is a map $A(X) \to A(X \bigcup_Y \selfarrow)$. We send an element of $A(X)$ to the corresponding `glued up' element of $A(X \bigcup_Y \selfarrow)$. This is well-defined since $\cU(X)$ maps into $\cU(X \bigcup_Y \selfarrow)$. This map descends down to a map |
275 Certainly there is a map $\DirectSum_{y \in \cF(Y)} A(X; y \bullet y) \to A(X \bigcup_Y \selfarrow)$. We send an element of $A(X)$ to the corresponding `glued up' element of $A(X \bigcup_Y \selfarrow)$. This is well-defined since $\cU(X)$ maps into $\cU(X \bigcup_Y \selfarrow)$. This map descends down to a map |
265 $$A(X) \Tensor_{A(Y)} \selfarrow \to A(X \bigcup_Y \selfarrow)$$ |
276 $$A(X) \Tensor_{A(Y)} \selfarrow \to A(X \bigcup_Y \selfarrow)$$ |
266 since the fields $ev$ and $ve$ (here $e \in A(Y), v \in A(X)$) are isotopic on $X \bigcup_Y \selfarrow$ (see Figure \ref{fig:ev-ve}). |
277 since the fields $ev$ and $ve$ (here $e \in A(Y), v \in A(X)$) are isotopic on $X \bigcup_Y \selfarrow$ (see Figure \ref{fig:ev-ve}). |
267 |
278 |
268 \begin{figure}[!ht] |
279 \begin{figure}[!ht] |
269 $$ |
280 $$ |