text/explicit.tex
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     1 %!TEX root = ../blob1.tex
       
     2 
       
     3 \nn{Here's the ``explicit'' version.}
       
     4 
       
     5 Fix a finite open cover of $X$, say $(U_l)_{l=1}^L$, along with an
       
     6 associated partition of unity $(r_l)$.
       
     7 
       
     8 We'll define the homotopy $H:I \times P \times X \to X$ via a function
       
     9 $u:I \times P \times X \to P$, with
       
    10 \begin{equation*}
       
    11 H(t,p,x) = F(u(t,p,x),x).
       
    12 \end{equation*}
       
    13 
       
    14 To begin, we'll define a function $u'' : I \times P \times X \to P$, and
       
    15 a corresponding homotopy $H''$. This homotopy will just be a homotopy of
       
    16 $F$ through families of maps, not through families of diffeomorphisms. On
       
    17 the other hand, it will be quite simple to describe, and we'll later
       
    18 explain how to build the desired function $u$ out of it.
       
    19 
       
    20 For each $l = 1, \ldots, L$, pick some $C^\infty$ function $f_l : I \to
       
    21 I$ which is identically $0$ on a neighborhood of the closed interval $[0,\frac{l-1}{L}]$
       
    22 and identically $1$ on a neighborhood of the closed interval $[\frac{l}{L},1]$. (Monotonic?
       
    23 Fix a bound for the derivative?) We'll extend it to a function on
       
    24 $k$-tuples $f_l : I^k \to I^k$ pointwise.
       
    25 
       
    26 Define $$u''(t,p,x) = \sum_{l=1}^L r_l(x) u_l(t,p),$$ with
       
    27 $$u_l(t,p) = t f_l(p) + (1-t)p.$$ Notice that the $i$-th component of $u''(t,p,x)$ depends only on the $i$-th component of $p$.
       
    28 
       
    29 Let's now establish some properties of $u''$ and $H''$. First,
       
    30 \begin{align*}
       
    31 H''(0,p,x) & = F(u''(0,p,x),x) \\
       
    32            & = F(\sum_{l=1}^L r_l(x) p, x) \\
       
    33            & = F(p,x).
       
    34 \end{align*}
       
    35 Next, calculate the derivatives
       
    36 \begin{align*}
       
    37 \partial_{p_i} H''(1,p,x) & = \partial_{p_i}u''(1,p,x) \partial_1 F(u(1,p,x),x) \\
       
    38 \intertext{and}
       
    39 \partial_{p_i}u''(1,p,x) & = \sum_{l=1}^L r_l(x) \partial_{p_i} f_l(p).
       
    40 \end{align*}
       
    41 Now $\partial_{p_i} f_l(p) = 0$ unless $\frac{l-1}{L} < p_i < \frac{l}{L}$, and $r_l(x) = 0$ unless $x \in U_l$,
       
    42 so we conclude that for a fixed $p$, $\partial_p H''(1,p,x) = 0$ for all $x$ outside the union of $k$ open sets from the open cover, namely
       
    43 $\bigcup_{i=1}^k U_{l_i}$ where for each $i$, we choose $l_i$ so $\frac{l_i -1}{L} \leq p_i \leq \frac{l_i}{L}$. It may be helpful to refer to Figure \ref{fig:supports}.
       
    44 
       
    45 \begin{figure}[!ht]
       
    46 \begin{equation*}
       
    47 \mathfig{0.5}{explicit/supports}
       
    48 \end{equation*}
       
    49 \caption{The supports of the derivatives {\color{green}$\partial_p f_1$}, {\color{blue}$\partial_p f_2$} and {\color{red}$\partial_p f_3$}, illustrating the case $k=2$, $L=3$. Notice that any
       
    50 point $p$ lies in the intersection of at most $k$ supports. The support of $\partial_p u''(1,p,x)$ is contained in the union of these supports.}
       
    51 \label{fig:supports}
       
    52 \end{figure}
       
    53 
       
    54 Unfortunately, $H''$ does not have the desired property that it's a homotopy through diffeomorphisms. To achieve this, we'll paste together several copies
       
    55 of the map $u''$. First, glue together $2^k$ copies, defining $u':I \times P \times X$ by
       
    56 \begin{align*}
       
    57 u'(t,p,x)_i & =
       
    58 \begin{cases}
       
    59 \frac{1}{2} u''(t, 2p_i, x)_i & \text{if $0 \leq p_i \leq \frac{1}{2}$} \\
       
    60 1-\frac{1}{2} u''(t, 2-2p_i, x)_i & \text{if $\frac{1}{2} \leq p_i \leq 1$}.
       
    61 \end{cases}
       
    62 \end{align*}
       
    63 (Note that we're abusing notation somewhat, using the fact that $u''(t,p,x)_i$ depends on $p$ only through $p_i$.)
       
    64 To see what's going on here, it may be helpful to look at Figure \ref{fig:supports_4}, which shows the support of $\partial_p u'(1,p,x)$.
       
    65 \begin{figure}[!ht]
       
    66 \begin{equation*}
       
    67 \mathfig{0.4}{explicit/supports_4} \qquad \qquad \mathfig{0.4}{explicit/supports_36}
       
    68 \end{equation*}
       
    69 \caption{The supports of $\partial_p u'(1,p,x)$ and of $\partial_p u(1,p,x)$ (with $K=3$) are subsets of the indicated region.}
       
    70 \label{fig:supports_4}
       
    71 \end{figure}
       
    72 
       
    73 Second, pick some $K$, and define
       
    74 \begin{align*}
       
    75 u(t,p,x) & = \frac{\floor{K p}}{K} + \frac{1}{K} u'\left(t, K \left(p - \frac{\floor{K p}}{K}\right), x\right).
       
    76 \end{align*}
       
    77 
       
    78 \todo{Explain that the localisation property survives for $u'$ and $u$.}
       
    79 
       
    80 We now check that by making $K$ large enough, $H$ becomes a homotopy through diffeomorphisms. We start with
       
    81 $$\partial_x H(t,p,x) = \partial_x u(t,p,x) \partial_1 F(u(t,p,x), x) + \partial_2 F(u(t,p,x), x)$$
       
    82 and observe that since $F(p, -)$ is a diffeomorphism, the second term $\partial_2 F(u(t,p,x), x)$ is bounded away from $0$. Thus if we can control the
       
    83 size of the first term $\partial_x u(t,p,x) \partial_1 F(u(t,p,x), x)$ we're done. The factor $\partial_1 F(u(t,p,x), x)$ is bounded, and we
       
    84 calculate \todo{err... this is a mess, and probably wrong.}
       
    85 \begin{align*}
       
    86 \partial_x u(t,p,x)_i & = \partial_x \frac{1}{K} u'\left(t, K\left(p - \frac{\floor{K p}}{K}\right), x\right)_i \\
       
    87                       & = \pm \frac{1}{2 K} \partial_x u''\left(t, (1\mp1)\pm 2K\left(p_i-\frac{\floor{K p}}{K}\right), x\right)_i \\
       
    88                       & = \pm \frac{1}{2 K} \sum_{l=1}^L (\partial_x r_l(x)) u_l\left(t, (1\mp1)\pm 2K\left(p_i-\frac{\floor{K p}}{K}\right)\right)_i. \\
       
    89 \intertext{Since the target of $u_l$ is just the unit cube $I^k$, we can make the estimate}
       
    90 \norm{\partial_x u(t,p,x)_i} & \leq \frac{1}{2 K} \sum_{l=1}^L \norm{\partial_x r_l(x)}.
       
    91 \end{align*}
       
    92 The sum here is bounded, so for large enough $K$ this is small enough that $\partial_x H(t,p,x)$ is never zero.