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$\varepsilon$-邻域定理
设 $Y^n\hookrightarrow \mathbb{R}^k$ 为 $n$ 维光滑嵌入子流形.
$\,\forall\,y\in Y,$ $n$ 维子空间
$T_yY\subset T_y\mathbb{R}^k\cong \mathbb{R}^k$ 决定了一个正交补空间
$$
N_y(Y):=\{v\in \mathbb{R}^k|v\perp T_yY\}.
$$
称$N(Y):=\{(y,v)\in Y\times \mathbb{R}^k|v\in N_yY\}\subset T\mathbb{R}^k$为$Y$在$\mathbb{R}^k$中的法丛.
命题 1. 设$Y^n\hookrightarrow \mathbb{R}^k$是嵌入子流形. 法丛$N(Y)$是$k$维流形, 且投影$\sigma:N(Y)\rightarrow Y$为淹没.
$\,\forall\,y\in Y,$ 存在$\mathbb{R}^k$中开集$\widetilde U,$
淹没$\phi:\widetilde U\rightarrow \mathbb{R}^l,$
使得$U=Y\cap \widetilde U=\phi^{-1}(0).$ 有$N(Y)$中开集
$$
N(U):=\{(x,v)\in U\times \mathbb{R}^k|v\in N_xY\}.
$$
由 $d\phi_x$ 为满射, $\ker d\phi_x=T_xY,$
$(d\phi_x)^T:\mathbb{R}^l\rightarrow N_xY$ 为线性同构
($N(A)^\perp=R(A^T)$). 取微分同胚
$$
\psi:U\times \mathbb{R}^l\rightarrow N(U),\quad (y,v)\mapsto (y,(d\phi_y)^T(v)).
$$
这就给出了$N(Y)$的局部参数化. $\sigma\circ \psi$为标准投影,
因此$\sigma$为淹没.
定理 2 ($\varepsilon$-邻域定理). $\mathbb{R}^k$ 任意光滑嵌入无边子流形$Y$有一个$\varepsilon$-邻域$Y^\varepsilon:=h(V)\subset \mathbb{R}^k,$ 其中$V=\{(y,v)\in N(Y)\mid |v|<\varepsilon(y)\}\subset N(Y)$为开子集, $\varepsilon:Y\rightarrow \mathbb{R}^+$为正连续函数, $h:(y,v)\mapsto y+v$为微分同胚, 且光滑映射$\pi:Y^\varepsilon\rightarrow Y$既是淹没, 又是收缩. 特别的, 当$Y$紧致时$\varepsilon$可取为常数.
$dh_{(y,0)}$为线性同构, 因此$Y$上每点有小邻域使得$h$在其上为微分同胚.
若$Y$紧, 可令小邻域充分小使得有限覆盖构成$\varepsilon$-邻域.
$\pi=\sigma\circ h^{-1}$既是淹没, 又是收缩. 若$Y$非紧, 取连续函数
$$
\rho(y):=\sup\{\delta\le 1|h_{V_\delta(y)}\text{为微分同胚}\},
$$
用$\varepsilon=\frac{\rho}{2}$控制邻域即可.
定理 3 (Whitney逼近定理). 设$X,Y$为光滑流形, $f:X\rightarrow Y$为连续映射. 则$f$同伦于光滑映射$g:X\rightarrow Y.$ 若$f|_A$在闭子集$A\subset X$上光滑, 则同伦可取得在$A$不变.
取$\mathbb{R}^k$上的光滑逼近, 利用管状邻域投影即可.
推论 4 (光滑映射的延拓引理). 设$X,Y$是光滑流形, $A\subset X$为闭子集, $f:A\rightarrow Y$为光滑映射. 则$f$有到$X$的光滑延拓当且仅当$f$有到$X$的连续延拓.
推论 5. 若$X^m$是$m$维光滑流形, $m<n,$ 则连续映射$f:X^m\rightarrow S^n$同伦于常值映射.
参数横截定理
定理 6. 设$F:X\rightarrow Y$为光滑映射, 只有$X$带边. $Z\subset Y$为任意无边嵌入子流形. 若$F\pitchfork Z,$ $\partial F\pitchfork Z,$ 则对几乎所有的$s\in S,$ $f_s\pitchfork Z,$ $\partial f_s\pitchfork Z.$
由横截原像定理, $W:=F^{-1}(Z)$为$X\times S$中的嵌入子流形, 且
$$
\partial W=F^{-1}(Z)\cap \partial (X\times S)=F^{-1}(Z)\cap (\partial X\times S).
$$
设$\pi:X\times S\rightarrow S$为投影映射,
只需证当$s$为$\pi|_W,\partial (\pi|_W)$正则值时,
$f_s,\partial f_s\pitchfork Z.$ 这样由Sard定理得到结论.
若$s$为$\pi|_W$正则值,
$$
d\pi_{(x,s)}(T_{(x,s)}W)=T_sS,\quad dF_{(x,s)}(T_{(x,s)}(X\times S))+T_zZ=T_zY.
$$
欲证
$$
df_s(T_xX)+T_zZ=T_zY.
$$
注意到
$$
T_{(x,s)}(X\times S)=T_xX\oplus T_sS,\quad dF_{(x,s)}|_{T_xX}=df_s.
$$
只需说明$\,\forall\,e\neq 0\in T_sS,$ $\,\exists\,w\in T_xX,$
使得$dF_{(x,s)}(w,e)\in T_zZ.$ 只需令$(w,e)\in T_{(x,s)}W$即可,
而这由正则性保证. 边界映射同理.
现设$f:X\rightarrow Y$为光滑映射,
$Y\hookrightarrow \mathbb{R}^k$有$\varepsilon$邻域.
取单位开球$S\subset \mathbb{R}^k,$ 定义
$$
\widetilde F:X\times S\rightarrow \mathbb{R}^k,\quad (x,s)\mapsto f(x)+\varepsilon(f(x))s
$$
为淹没, $\partial \widetilde F$也是.
进一步取$F:=\sigma \circ \widetilde F:X\times S\rightarrow Y$满足同样淹没条件.
这样对$Y$中无边嵌入子流形$Z,$ $F,\partial F\pitchfork Z.$
从而对于几乎所有的$s\in S,$ 有$f_s,\partial f_s\pitchfork Z.$
由于$f_0=f,$ 这就给出了横截同伦定理.
定理 7 (横截同伦定理). 对任意光滑映射$f:X\rightarrow Y$和嵌入子流形$Z\subset Y,$ $Y,Z$无边. 存在光滑映射$g:X\rightarrow Y,$ 使得$g\sim f,$ $g,\partial g\pitchfork Z.$
横截延拓定理
称$f:X\rightarrow Y$在$A$上与$Z\subset Y$横截, 若
$$
df_x(T_xX)+T_{f(x)}Z=T_{f(x)}Y,\quad \,\forall\,x\in A\cap f^{-1}(Z).
$$
记为$f\pitchfork_A Z.$
定理 8 (横截延拓定理). 设$Z\subset Y$是$Y$的无边闭子流形, $A\subset X$为闭子集. 设$f:X\rightarrow Y$为光滑映射, $f\pitchfork_A Z,$ $\partial f\pitchfork_{A\cap \partial X} Z.$ 则存在光滑映射$g:X\rightarrow Y,$ 使得$g\sim f,$ $g,\partial g\pitchfork Z,$ 且在$A$的一个邻域上$g=f.$
改造横截同伦定理中的$g,$ 使得在$A$附近保持不变即可.
方法是将$\widetilde F:X\times S\rightarrow \mathbb{R}^k$改造为
$$
(x,s)\mapsto f(x)+\varepsilon(f(x))\rho^2(x)s.
$$
$\rho(x)$为截断函数.
它的加入使得$A$的小邻域上$F(x,\rho^2(x)s)=F(x,0)=f(x).$
由于$\partial X$为闭子集, 我们有
推论 9. 若$f:X\rightarrow Y$为光滑映射, $Z\subset Y$为无边闭子流形, $\partial f\pitchfork Z,$ 则存在光滑映射$g:X\rightarrow Y$满足$g\sim f,$ $\partial g=\partial f,$ $g\pitchfork Z.$
文章最后更新于 2023-06-12 18:39:59
$\varepsilon$-Neighborhood Theorem
Let $Y^n\hookrightarrow \mathbb{R}^k$ be a $n$-dimensional smooth embedding submanifold.
$\,\forall\,y\in Y,$ $n$ dimensional subspace
$T_yY\subset T_y\mathbb{R}^k\cong \mathbb{R}^k$ determines an orthogonal complement space
$$
N_y(Y):=\{v\in \mathbb{R}^k|v\perp T_yY\}.
$$
Call $N(Y):=\{(y,v)\in Y\times \mathbb{R}^k|v\in N_yY\}\subset T\mathbb{R}^k$ the one where $Y$ is in $\mathbb{R}^k$ Fa Cong.
Proposition 1. Let $Y^n\hookrightarrow \mathbb{R}^k$ be the embedded submanifold. The normal bundle $N(Y)$ is the $k$-dimensional manifold, and the projection $\sigma:N(Y)\rightarrow Y$ is submerged.
$\,\forall\,y\in Y,$ There is an open set $\widetilde U,$ in $\mathbb{R}^k$
Submerged$\phi:\widetilde U\rightarrow \mathbb{R}^l,$
Let $U=Y\cap \widetilde U=\phi^{-1}(0).$ have an open set in $N(Y)$
$$
N(U):=\{(x,v)\in U\times \mathbb{R}^k|v\in N_xY\}.
$$
From $d\phi_x$ to surjection, $\ker d\phi_x=T_xY,$
$(d\phi_x)^T:\mathbb{R}^l\rightarrow N_xY$ is linear isomorphism
($N(A)^\perp=R(A^T)$). Take diffeomorphism
$$
\psi:U\times \mathbb{R}^l\rightarrow N(U),\quad (y,v)\mapsto (y,(d\phi_y)^T(v)).
$$
This gives the local parameterization of $N(Y)$. $\sigma\circ \psi$ is the standard projection,
Therefore $\sigma$ is submerged.
Theorem 2 ($\varepsilon$-neighborhood theorem). $\mathbb{R}^k$ Any smooth embedded edgeless submanifold $Y$ has a $\varepsilon$-neighborhood $Y^\varepsilon:=h(V)\subset \mathbb{R}^k,$ where $V=\{(y,v)\in N(Y)\mid |v|<\varepsilon(y)\}\subset N(Y)$ is an open subset, $\varepsilon:Y\rightarrow \mathbb{R}^+$ is a positive continuous function, $h:(y,v)\mapsto y+v$ is a diffeomorphism, and the smooth map $\pi:Y^\varepsilon\rightarrow Y$ is both submerged and contracted. In particular, $\varepsilon$ can be taken as a constant when $Y$ is compact.
$dh_{(y,0)}$ is a linear isomorphism, so each point on $Y$ has a small neighborhood so that $h$ is a diffeomorphism on it.
If $Y$ is compact, the small neighborhood can be made small enough to form a $\varepsilon$-neighborhood with limited coverage.
$\pi=\sigma\circ h^{-1}$ is both submerged and contracted. If $Y$ is not compact, take a continuous function
$$
\rho(y):=\sup\{\delta\le 1|h_{V_\delta(y)}\text{为微分同胚}\},
$$
Just use $\varepsilon=\frac{\rho}{2}$ to control the neighborhood.
Theorem 3 (Whitney's approximation theorem). Assume $X,Y$ is a smooth manifold and $f:X\rightarrow Y$ is a continuous map. Then $f$ is homotopic to the smooth map $g:X\rightarrow Y.$. If $f|_A$ is smooth on the closed subset $A\subset X$, then the homotopy can be obtained unchanged in $A$.
Just take a smooth approximation on $\mathbb{R}^k$ and use tubular neighborhood projection.
Corollary 4 (The continuation lemma of smooth mapping). Assume $X,Y$ is a smooth manifold, $A\subset X$ is a closed subset, and $f:A\rightarrow Y$ is a smooth map. Then $f$ has a smooth continuation to $X$ if and only if $f$ has a continuous continuation to $X$.
Corollary 5. If $X^m$ is a $m$-dimensional smooth manifold, $m<n,$ then the continuous map $f:X^m\rightarrow S^n$ is homotopic to the constant map.
Parametric transversality theorem
Theorem 6. Let $F:X\rightarrow Y$ be a smooth map, and only $X$ has edges. $Z\subset Y$ is any edgeless embedded submanifold. If $F\pitchfork Z,$ $\partial F\pitchfork Z,$, then for almost all $s\in S,$ $f_s\pitchfork Z,$ $\partial f_s\pitchfork Z.$
According to the transversal preimage theorem, $W:=F^{-1}(Z)$ is the embedded submanifold in $X\times S$, and
$$
\partial W=F^{-1}(Z)\cap \partial (X\times S)=F^{-1}(Z)\cap (\partial X\times S).
$$
Let $\pi:X\times S\rightarrow S$ be the projection mapping,
It only needs to be proved that when $s$ is the regular value of $\pi|_W,\partial (\pi|_W)$,
$f_s,\partial f_s\pitchfork Z.$ This is the conclusion obtained from Sard’s theorem.
If $s$ is $\pi|_W$ regular value,
$$
d\pi_{(x,s)}(T_{(x,s)}W)=T_sS,\quad dF_{(x,s)}(T_{(x,s)}(X\times S))+T_zZ=T_zY.
$$
Want to prove
$$
df_s(T_xX)+T_zZ=T_zY.
$$
noticed
$$
T_{(x,s)}(X\times S)=T_xX\oplus T_sS,\quad dF_{(x,s)}|_{T_xX}=df_s.
$$
Just state $\,\forall\,e\neq 0\in T_sS,$ $\,\exists\,w\in T_xX,$
To make $dF_{(x,s)}(w,e)\in T_zZ.$ just let $(w,e)\in T_{(x,s)}W$,
And this is guaranteed by regularity. The same goes for boundary mapping.
Now suppose $f:X\rightarrow Y$ is a smooth mapping,
$Y\hookrightarrow \mathbb{R}^k$ has $\varepsilon$ neighbors.
Take unit kickoff $S\subset \mathbb{R}^k,$ Definition
$$
\widetilde F:X\times S\rightarrow \mathbb{R}^k,\quad (x,s)\mapsto f(x)+\varepsilon(f(x))s
$$
is submerged, so is $\partial \widetilde F$.
Further take $F:=\sigma \circ \widetilde F:X\times S\rightarrow Y$ to satisfy the same flooding condition.
In this way, the edgeless embedded submanifold $Z,$ $F,\partial F\pitchfork Z.$ in $Y$
Thus for almost all $s\in S,$ there is $f_s,\partial f_s\pitchfork Z.$
Since $f_0=f,$ this gives the transverse homotopy theorem.
Theorem 7 (Transverse homotopy theorem). For any smooth map $f:X\rightarrow Y$ and embedded submanifold $Z\subset Y,$ $Y,Z$, there is no edge. There is a smooth map $g:X\rightarrow Y,$ such that $g\sim f,$ $g,\partial g\pitchfork Z.$
transversal continuation theorem
It is said that $f:X\rightarrow Y$ is transverse to $Z\subset Y$ on $A$, if
$$
df_x(T_xX)+T_{f(x)}Z=T_{f(x)}Y,\quad \,\forall\,x\in A\cap f^{-1}(Z).
$$
Recorded as $f\pitchfork_A Z.$
Theorem 8 (Transversal continuation theorem). Let $Z\subset Y$ be the edgeless closed submanifold of $Y$, $A\subset X$ be a closed subset. Let $f:X\rightarrow Y$ be a smooth map, $f\pitchfork_A Z,$ $\partial f\pitchfork_{A\cap \partial X} Z.$, then there is a smooth map $g:X\rightarrow Y,$ such that $g\sim f,$ $g,\partial g\pitchfork Z,$ and on a neighborhood of $A$ $g=f.$
Just modify $g,$ in the transverse homotopy theorem so that it remains unchanged near $A$.
The method is to transform $\widetilde F:X\times S\rightarrow \mathbb{R}^k$ into
$$
(x,s)\mapsto f(x)+\varepsilon(f(x))\rho^2(x)s.
$$
$\rho(x)$ is the truncation function.
Its addition makes the small neighborhood of $A$ on $F(x,\rho^2(x)s)=F(x,0)=f(x).$
Since $\partial X$ is a closed subset, we have
Corollary 9. If $f:X\rightarrow Y$ is a smooth map, $Z\subset Y$ is an edgeless closed submanifold, $\partial f\pitchfork Z,$ then there is a smooth map $g:X\rightarrow Y$ satisfying $g\sim f,$ $\partial g=\partial f,$ $g\pitchfork Z.$
The article was last updated on 2023-06-12 18:39:59
