75 Each $i$-handle $C$ of $\jj$ has an $i$-dimensional tangential coordinate and, |
75 Each $i$-handle $C$ of $\jj$ has an $i$-dimensional tangential coordinate and, |
76 more importantly for our purposes, a $k{-}i$-dimensional normal coordinate. |
76 more importantly for our purposes, a $k{-}i$-dimensional normal coordinate. |
77 We will typically use the same notation for $i$-cells of $L$ and the |
77 We will typically use the same notation for $i$-cells of $L$ and the |
78 corresponding $i$-handles of $\jj$. |
78 corresponding $i$-handles of $\jj$. |
79 |
79 |
80 For each (top-dimensional) $k$-cell $C$ of each $K_\alpha$, choose a point $p_c \in C \sub P$. |
80 For each (top-dimensional) $k$-cell $C$ of each $K_\alpha$, choose a point $p(C) \in C \sub P$. |
81 Let $D$ be a $k$-handle of $\jj$. |
81 Let $D$ be a $k$-handle of $\jj$. |
82 To $D$ we associate the tuple $(c_\alpha)$ of $k$-cells of the $K_\alpha$'s |
82 For each $\alpha$ let $C(D, \alpha)$ be the $k$-cell of $K_\alpha$ which contains $D$ |
83 which contain $D$, and also the corresponding tuple $(p_{c_\alpha})$ of points in $P$. |
83 and let $p(D, \alpha) = p(C(D, \alpha))$. |
84 |
84 |
85 For $p \in D$ we define |
85 For $p \in D$ we define |
86 \eq{ |
86 \eq{ |
87 u(t, p, x) = (1-t)p + t \sum_\alpha r_\alpha(x) p_{c_\alpha} . |
87 u(t, p, x) = (1-t)p + t \sum_\alpha r_\alpha(x) p(D, \alpha) . |
88 } |
88 } |
89 (Recall that $P$ is a convex linear polyhedron, so the weighted average of points of $P$ |
89 (Recall that $P$ is a convex linear polyhedron, so the weighted average of points of $P$ |
90 makes sense.) |
90 makes sense.) |
91 |
91 |
92 So far we have defined $u(t, p, x)$ when $p$ lies in a $k$-handle of $\jj$. |
92 Thus far we have defined $u(t, p, x)$ when $p$ lies in a $k$-handle of $\jj$. |
93 For handles of $\jj$ of index less than $k$, we will define $u$ to |
93 We will now extend $u$ inductively to handles of index less than $k$. |
94 interpolate between the values on $k$-handles defined above. |
94 |
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95 Let $E$ be a $k{-}1$-handle. |
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96 $E$ is homeomorphic to $B^{k-1}\times [0,1]$, and meets |
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97 the $k$-handles at $B^{k-1}\times\{0\}$ and $B^{k-1}\times\{1\}$. |
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98 Let $\eta : E \to [0,1]$, $\eta(x, s) = s$ be the normal coordinate |
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99 of $E$. |
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100 Let $D_0$ and $D_1$ be the two $k$-handles of $\jj$ adjacent to $E$. |
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101 There is at most one index $\beta$ such that $C(D_0, \beta) \ne C(D_1, \beta)$. |
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102 (If there is no such index $\beta$, choose $\beta$ |
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103 arbitrarily.) |
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104 For $p \in E$, define |
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105 \eq{ |
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106 u(t, p, x) = (1-t)p + t \left( \sum_{\alpha \ne \beta} r_\alpha(x) p(D_0, \alpha) |
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107 + r_\beta(x) (\eta(p) p(D_0, p) + (1-\eta(p)) p(D_1, p)) \right) . |
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108 } |
95 |
109 |
96 \nn{*** resume revising here ***} |
110 \nn{*** resume revising here ***} |
97 |
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98 If $p$ lies in a $k{-}1$-handle $E$, let $\eta : E \to [0,1]$ be the normal coordinate |
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99 of $E$. |
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100 In particular, $\eta$ is equal to 0 or 1 only at the intersection of $E$ |
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101 with a $k$-handle. |
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102 Let $\beta$ be the index of the $K_\beta$ containing the $k{-}1$-cell |
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103 corresponding to $E$. |
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104 Let $q_0, q_1 \in P$ be the points associated to the two $k$-cells of $K_\beta$ |
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105 adjacent to the $k{-}1$-cell corresponding to $E$. |
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106 For $p \in E$, define |
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107 \eq{ |
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108 u(t, p, x) = (1-t)p + t \left( \sum_{\alpha \ne \beta} r_\alpha(x) p_{c_\alpha} |
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109 + r_\beta(x) (\eta(p) q_1 + (1-\eta(p)) q_0) \right) . |
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110 } |
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111 |
111 |
112 In general, for $E$ a $k{-}j$-handle, there is a normal coordinate |
112 In general, for $E$ a $k{-}j$-handle, there is a normal coordinate |
113 $\eta: E \to R$, where $R$ is some $j$-dimensional polyhedron. |
113 $\eta: E \to R$, where $R$ is some $j$-dimensional polyhedron. |
114 The vertices of $R$ are associated to $k$-cells of the $K_\alpha$, and thence to points of $P$. |
114 The vertices of $R$ are associated to $k$-cells of the $K_\alpha$, and thence to points of $P$. |
115 If we triangulate $R$ (without introducing new vertices), we can linearly extend |
115 If we triangulate $R$ (without introducing new vertices), we can linearly extend |