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模$2$相交数
设$Y$的两个子流形$X,Z\subset Y$维数互补, 即$\dim X+\dim Z=\dim Y.$
此时由横截原像定理,
$$
\operatorname{codim}(X\cap Z)=\operatorname{codim}X+\operatorname{codim}Z=\dim Z+\dim X=\dim Y,
$$
从而$X\cap Z$为零维嵌入子流形, 在紧流形中为有限点集.
设 $X$ 紧致, $f:X\rightarrow Y$ 为光滑映射, $f\pitchfork Z,$
$Z\subset Y$ 为与 $X$ 维数互补的闭子流形.
那么$f^{-1}(Z)$为$X$的零维闭子流形, 是有限点集.
定义$f$和$Z$的模$2$相交数为
$$
I_2(f,Z):=\# f^{-1}(Z)\mod 2.
$$
对一般的光滑映射$g:X\rightarrow Y,$ 由横截同伦定理定义即可.
合理性由如下定理保证.
定理 1. 如上假设, 若$f_0\sim f_1$且$f_0,f_1\pitchfork Z,$ 那么$I_2(f_0,Z)=I_2(f_1,Z).$
取光滑同伦 $F:X\times I\rightarrow Y.$ 那么 $\partial F\pitchfork Z.$
由横截延拓定理, 存在 $G:X\times I\rightarrow Y,$ $G\pitchfork Z$ 且
$\partial G=\partial F.$ 此时 $G^{-1}(Z)$ 为 $1$ 维紧致带边子流形,
$\partial G^{-1}(Z)=f_0^{-1}(Z)\cup f_1^{-1}(Z)$ 含偶数个点, 从而
$\#_2 f_0^{-1}(Z)=\#_2f_1^{-1}(Z).$
推论 2. 模$2$相交数为同伦不变量.
定理 3 (边界定理). 设$W$为紧致流形, $X=\partial W,$ $g:X\rightarrow Y$为光滑映射. 若$g$可以延拓到$W,$ 则$I_2(g,Z)=0,$ $\,\forall\,Z\subset Y$为与$X$维数互补闭子流形.
设 $G$ 为 $g$ 的延拓. 存在 $F\sim G,$ 使得 $F,\partial F\pitchfork Z.$
$F^{-1}(Z)$ 为一维紧致带边子流形, $\partial F^{-1}(Z)$ 含偶数个点, 从而
$$
I_2(g,Z)=I_2(\partial F,Z)=0.
$$
模$2$映射度
设 $f:X\rightarrow Y$ 为紧流形到连通流形上的光滑映射, $\dim X=\dim Y,$
那么任取 $y\in Y,$ 定义模 $2$ 映射度为
$$
\deg_2f:=I_2(f,\{y\}).
$$
它与$y$选取无关. 因为任取$Y$上正则值$y$, 由唱片引理,
存在$\bigsqcup_{i=1}^n U_i= f^{-1}(V),$ $f:U_i\rightarrow V$为微分同胚,
因此对于$y$周围的点,
$$
I_2(f,\{z\})\equiv n\mod 2,\quad \,\forall\,z\in V.
$$
即模$2$映射度是局部常值的. 由连通性得到良定性.
模$2$映射度自然也是同伦不变量, 且若$X=\partial W,$
$f:X\rightarrow Y$可延拓到$W$上, 那么$\deg_2 f=0.$ 特别地,
$\deg_2f\neq 0$给出了不可延拓的障碍, 内部存在奇点.
这可以用来证明奇数次复多项式必有根.
模$2$环绕数
设 $X^{n-1}$ 为 $n-1$ 维紧致连通流形, $f:X\rightarrow \mathbb{R}^n$
为光滑映射. $\,\forall\,z\notin f(X),$ 定义映射 $u:X\rightarrow S^{n-1}$
为单位向量 $u(x)=\frac{f(x)-z}{|f(x)-z|}.$ 定义 $f$ 围绕 $z$ 的模 $2$
环绕数为
$$
W_2(f,z):=\deg_2 u.
$$
定理 4. 设 $D^n$ 为紧致带边流形, $X=\partial D^n$ 为紧致连通流形, $f:X^{n-1}\rightarrow \mathbb{R}^n$ 为光滑映射. 设 $F:D^n\rightarrow \mathbb{R}^n$ 为 $f$ 的光滑延拓. 若 $z\notin \operatorname{Im}f$ 为 $F$ 的正则值, 则 $F^{-1}(z)$ 为有限点集, 且 $W_2(f,z)=\# F^{-1}(z)\mod 2.$
由于$z$为$F$的正则值, $F^{-1}(z)$为零维嵌入子流形, 是有限点集.
若$z\notin \operatorname{Im}F,$ $u$可延拓, 立即得证;
若$z\in \operatorname{Im}F,$ 由唱片引理得证.
Jordan-Brouwer分离定理
定理 5. 设$X\subset \mathbb{R}^n$为$\mathbb{R}^n$中的紧致光滑连通嵌入超曲面, 则$\mathbb{R}^n\setminus X$由外部$D_0$和内部$D_1$两个连通开集组成, 并且$\overline D_1$为紧致流形, 满足$\partial\overline D_1=X.$
首先证明$\mathbb{R}^n\setminus X$至多有两个连通分支, 接下来利用射线说明
$$
D_i:=\{z\mid W_2(\iota_X,z)=i\},\quad i=0,1
$$
分别决定了外部和内部即可. 特别的,
构造射线可以轻易看出点在区域内部还是外部.
Borsuk-Ulam定理
定理 6. 设$f:S^k\rightarrow \mathbb{R}^{k+1}\setminus\{0\}$为光滑映射, 且$f(-x)=-f(x),$ $\,\forall\,x\in S^k.$ 则$W_2(f,0)=1.$
该定理可以推出Brouwer不动点定理.
证明将球面上的环绕数与赤道上的环绕数等同起来, 由归纳法得到.
方法是将映射限制在上半球面上, 再对像空间做投影(以使用归纳假设).
推论 7. 若$f:S^k\rightarrow \mathbb{R}^{k+1}\setminus\{0\}$关于原点对称, 则$f$与任意过原点的直线至少相交一次.
定理 8. 设$f_1,\cdots,f_k$为$S^k$上的$k$个关于原点对称的光滑函数, 它们必有公共零点.
考虑$f(x)=(f_1(x),\cdots,f_k(x),0),$ 应用推论即可,
定理 9. 对$S^k$上任意$k$个光滑函数$g_1,\cdots,g_k,$ 必存在$p\in S^k,$ 使得$g_i(p)=g_i(-p).$
人为构造对称函数$f_i(x)=g_i(x)-g_i(-x).$
文章最后更新于 2023-06-13 12:34:42
Modulo $2$ intersection number
Assume that the dimensions of the two submanifolds $X,Z\subset Y$ of $Y$ are complementary, that is, $\dim X+\dim Z=\dim Y.$
At this time, according to the transverse original image theorem,
$$
\operatorname{codim}(X\cap Z)=\operatorname{codim}X+\operatorname{codim}Z=\dim Z+\dim X=\dim Y,
$$
Therefore $X\cap Z$ is a zero-dimensional embedded submanifold, which is a finite set of points in a compact manifold.
Let $X$ be compact, $f:X\rightarrow Y$ be smooth mapping, $f\pitchfork Z,$
$Z\subset Y$ is a closed submanifold with dimensions complementary to $X$.
Then $f^{-1}(Z)$ is the zero-dimensional closed submanifold of $X$, which is a finite point set.
Define the intersection number of $f$ and $Z$ modulo $2$ as
$$
I_2(f,Z):=\# f^{-1}(Z)\mod 2.
$$
For general smooth mapping $g:X\rightarrow Y,$, it can be defined by the transverse homotopy theorem.
Rationality is guaranteed by the following theorem.
Theorem 1. Assuming above, if $f_0\sim f_1$ and $f_0,f_1\pitchfork Z,$ then $I_2(f_0,Z)=I_2(f_1,Z).$
Take smooth homotopy $F:X\times I\rightarrow Y.$ then $\partial F\pitchfork Z.$
According to the transversal continuation theorem, there exists $G:X\times I\rightarrow Y,$ $G\pitchfork Z$ and
$\partial G=\partial F.$ At this time $G^{-1}(Z)$ is a $1$-dimensional compact submanifold with edges,
$\partial G^{-1}(Z)=f_0^{-1}(Z)\cup f_1^{-1}(Z)$ contains an even number of points, so
$\#_2 f_0^{-1}(Z)=\#_2f_1^{-1}(Z).$
Corollary 2. The intersection number modulo $2$ is a homotopy invariant.
Theorem 3 (Boundary theorem). Let $W$ be a compact manifold, $X=\partial W,$ $g:X\rightarrow Y$ be a smooth map. If $g$ can be extended to $W,$, then $I_2(g,Z)=0,$ $\,\forall\,Z\subset Y$ is a complementary closed submanifold with the dimension $X$.
Let $G$ be the continuation of $g$. There exists $F\sim G,$ such that $F,\partial F\pitchfork Z.$
$F^{-1}(Z)$ is a one-dimensional compact submanifold with edges, $\partial F^{-1}(Z)$ contains an even number of points, so
$$
I_2(g,Z)=I_2(\partial F,Z)=0.
$$
Modulo $2$ mapping degree
Let $f:X\rightarrow Y$ be a smooth mapping from compact manifold to connected manifold, $\dim X=\dim Y,$
Then take $y\in Y,$ and define the mapping degree modulo $2$ as
$$
\deg_2f:=I_2(f,\{y\}).
$$
It has nothing to do with the selection of $y$. Because any regular value $y$ on $Y$ is chosen, according to the record lemma,
Existence $\bigsqcup_{i=1}^n U_i= f^{-1}(V),$ $f:U_i\rightarrow V$ is diffeomorphism,
So for the points around $y$,
$$
I_2(f,\{z\})\equiv n\mod 2,\quad \,\forall\,z\in V.
$$
That is, the modulus $2$ mapping degree is locally constant. Good character is obtained from connectivity.
The mapping degree modulo $2$ is naturally a homotopy invariant, and if $X=\partial W,$
$f:X\rightarrow Y$ can be extended to $W$, then $\deg_2 f=0.$ In particular,
$\deg_2f\neq 0$ gives a non-extendible obstacle, and there is a singularity inside.
This can be used to prove that complex polynomials of odd degree must have roots.
Modulo $2$ wrap number
Let $X^{n-1}$ be a $n-1$-dimensional compact connected manifold, $f:X\rightarrow \mathbb{R}^n$
is a smooth mapping. $\,\forall\,z\notin f(X),$ defines the mapping $u:X\rightarrow S^{n-1}$
Define $f$ modulo $2$ around $z$ for the unit vector $u(x)=\frac{f(x)-z}{|f(x)-z|}.$
The number of surrounds is
$$
W_2(f,z):=\deg_2 u.
$$
Theorem 4. Let $D^n$ be a compact edge manifold, $X=\partial D^n$ be a compact connected manifold, and $f:X^{n-1}\rightarrow \mathbb{R}^n$ be a smooth map. Let $F:D^n\rightarrow \mathbb{R}^n$ be the smooth continuation of $f$. If $z\notin \operatorname{Im}f$ is the regular value of $F$, then $F^{-1}(z)$ is a finite point set, and $W_2(f,z)=\# F^{-1}(z)\mod 2.$
Since $z$ is the regular value of $F$, $F^{-1}(z)$ is a zero-dimensional embedded submanifold, which is a finite point set.
If $z\notin \operatorname{Im}F,$ $u$ can be extended, it will be proved immediately;
If $z\in \operatorname{Im}F,$ is proved by the record lemma.
Jordan-Brouwer separation theorem
Theorem 5. Assume $X\subset \mathbb{R}^n$ is a compact and smooth connected embedded hypersurface in $\mathbb{R}^n$, then $\mathbb{R}^n\setminus X$ consists of two connected open sets, the outer $D_0$ and the inner $D_1$, and $\overline D_1$ is a compact manifold, satisfying $\partial\overline D_1=X.$
First prove that $\mathbb{R}^n\setminus X$ has at most two connected components, and then use rays to illustrate
$$
D_i:=\{z\mid W_2(\iota_X,z)=i\},\quad i=0,1
$$
Just determine the exterior and interior respectively. In particular,
Constructing rays makes it easy to see whether a point is inside or outside a region.
Borsuk-Ulam theorem
Theorem 6. Let $f:S^k\rightarrow \mathbb{R}^{k+1}\setminus\{0\}$ be a smooth mapping, and $f(-x)=-f(x),$ $\,\forall\,x\in S^k.$ then $W_2(f,0)=1.$
This theorem can lead to Brouwer's fixed point theorem.
Proof equates the number of orbits on the sphere with the number of orbits on the equator, obtained by induction.
The method is to restrict the mapping to the upper hemisphere and then project it into the image space (to use the inductive hypothesis).
Corollary 7. If $f:S^k\rightarrow \mathbb{R}^{k+1}\setminus\{0\}$ is symmetric about the origin, then $f$ intersects any straight line passing through the origin at least once.
Theorem 8. Let $f_1,\cdots,f_k$ be $k$ smooth functions on $S^k$ that are symmetric about the origin, and they must have common zeros.
Just consider $f(x)=(f_1(x),\cdots,f_k(x),0),$ and apply the inference,
Theorem 9. For any $k$ smooth function $g_1,\cdots,g_k,$ on $S^k$, there must be $p\in S^k,$ such that $g_i(p)=g_i(-p).$
Artificially constructed symmetry function $f_i(x)=g_i(x)-g_i(-x).$
The article was last updated on 2023-06-13 12:34:42
