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定向
向量空间的定向由有序基决定. 由此可定义矩阵的保(反)定向.
称流形可定向, 若$\,\forall\,x\in X,$
有坐标卡$(U,\phi)$使得$\,\forall\,y\in U,$
$d\phi_y: T_yX\rightarrow \mathbb{R}^k$保定向.
命题 1. 连通可定向流形有且仅有两个定向.
$X\times Y$的乘积定向由$T_xX+T_yY$的直和定向决定. 具体的,
决定定向的有序基为$\{(\alpha,0),(0,\beta)\},$
$\alpha,\beta$分别为$T_xX$和$T_yY$的有序基.
带边流形在边界处的定向可由流形诱导. 具体的,
令边界处有序基$\beta$的符号由$\{n,\beta\}$决定, $n$为外法向量.
设$f:X\rightarrow Y$为光滑映射, $f,\partial f\pitchfork Z.$
$X,Y,Z$为定向流形, $Y,Z$无边. 记$S:=f^{-1}(Z)$为嵌入子流形,
欲取定逆像定向.
由$T_xS^\perp\oplus T_xS=T_xX,$ 可通过直和定向决定其定向.
先取$T_xS^\perp$定向. 由横截性,
$$
df_x(T_xX)+T_zZ=T_zY,\quad df_x(T_xS^\perp) \oplus T_zZ=T_zY.
$$
设$df_x$保定向,
即$T_xS^\perp$定向通过$df_x(T_xS^\perp)$由直和定向决定的定向诱导.
此时$\partial S=\partial (f^{-1}(Z))=(\partial f)^{-1}(Z)$有两种定向方式.
一种是逆像定向, 考虑
$$
T_x \partial S^\perp\oplus T_x\partial S=T_x\partial X,\quad d\partial f_x(T_x\partial S^\perp)\oplus T_zZ=T_zY.
$$
注意到$T_xS^\perp=T_x\partial S^\perp,$ 这两个定向是一致的.
取其正定向有序基$\alpha,$ 那么
$$
\operatorname{sgn}(\beta)=\operatorname{sgn}(\alpha,\beta)=\operatorname{sgn}(n,\alpha,\beta).
$$
$n$为$\partial S$在$S$中的外法向量. 另一种是边界定向.
取定$T_x\partial S$上的有序基$\beta,$ 那么
$\operatorname{sgn}(\beta)=\operatorname{sgn}(n,\beta)$
在$T_xS$中的符号决定, 即
$$
\operatorname{sgn}(\beta)=\operatorname{sgn}(n,\beta)=\operatorname{sgn}(\alpha,n,\beta).
$$
由于$\alpha$是$T_xS^\perp$的有序基,
维数为$\operatorname{codim}_XS=\operatorname{codim}_YZ.$
因此这两种定向也差了$(-1)^{\operatorname{codim}_YZ}.$
定向相交数
基本假设同模$2$相交数. 额外假设所有流形都是可定向的.
$df_x(T_xX)$的定向由$X$赋予.
若$df_x(T_xX)\oplus T_zZ=T_zY$两侧定向定向一致,
则记$x$处的定向数为$\operatorname{sgn}_x(f,Z)=1,$ 不然为$-1.$
称定向数的和为$f$与$Z$的定向相交数, 记为
$$
I(f,Z):=\sum_{x\in f^{-1}(Z)}\operatorname{sgn}_x(f,Z).
$$
由于任意紧致定向$1$维带边流形的边界点定向数之和为$0,$
同模$2$相交数可证明定向相交数为同伦不变量.
因此对一般的映射可通过横截同伦定理给出定义. 同理我们还有
命题 2. 若$X=\partial W,$ $W$紧致, $f:X\rightarrow Y$可延拓到$W,$ $Z\subset Y$为闭子流形, 则$I(f,Z)=0.$
若$\dim X=\dim Y,$ 则可定义$\deg f:=I(f,\{y\}),$ $y$在$Y$中任意选取.
良定性由横截同伦定理与唱片引理得证. 此时若$y$为$f$的正则值,
有$\deg f=\sum_{x\in f^{-1}(y)}\operatorname{sgn}_x(f,y),$
$\operatorname{sgn}_x(f,y)=1,$ 若$df_x$保定向. 反之为$-1.$
映射度也是同伦不变量, 且对于可从边界延拓到内部的函数, 映射度为零.
用映射度就可以给出代数学基本定理的证明了, 和模$2$相交数一致.
同时在紧区域边界处可用$\deg \frac{p}{|p|}$来数内部零点个数.
接下来讨论映射间的定向相交数. 设$f:X\rightarrow Y,$ $g:Z\rightarrow Y.$
若$f\pitchfork g,$ 可定义$I_{(x,z)}(f,g),$
正负号由$df_x(T_xX)\oplus dg_z(T_zZ)=T_yY$左右两侧定向决定.
命题 3. $f\pitchfork g$当且仅当$f\times g\pitchfork \Delta,$ 且$I(f,g)=(-1)^{\dim Z}I(f\times g,\Delta).$
这主要就是如下的引理
引理 4. 设$U,W$为$V$的定向子空间, 那么$U\oplus W=V$当且仅当$(U\times W)\oplus \Delta=(-1)^{\dim W}V\times V.$
这样就可以用横截同伦定理对一般的函数给出定义了.
此时可说明$I(f,Z)=I(f,\iota_Z),$
特别的$\deg f=I(f,\{y\})$与$y$选取无关可给出新的证明. 比较直和定向有:
命题 5. $I(f,g)=(-1)^{\dim X\dim Z}I(g,f),$ $I(X,Z)=(-1)^{\dim X\dim Z}I(Z,X).$
特别的, 若$Z=X,$ 奇数维流形自相交数$I(X,X)=0.$ 若$X$为紧致定向流形,
定义其欧拉示性数为$\chi(Y):=I(\Delta,\Delta).$ 因此它是同伦不变量.
推论 6. 奇数维紧致可定向流形欧拉示性数为$0.$
文章最后更新于 2023-06-13 12:33:48
Orientation
The orientation of the vector space is determined by the ordered basis. From this, the preserving (inverse) orientation of the matrix can be defined.
manifold Orientable, if$\,\forall\,x\in X,$
There is a coordinate card $(U,\phi)$ that makes $\,\forall\,y\in U,$
$d\phi_y: T_yX\rightarrow \mathbb{R}^k$ Bao direction.
Proposition 1. A connected orientable manifold has and has only two orientations.
The product orientation of $X\times Y$ is determined by the direct sum orientation of $T_xX+T_yY$. Specifically,
The ordered basis that determines the orientation is $\{(\alpha,0),(0,\beta)\},$
$\alpha,\beta$ are the ordered bases of $T_xX$ and $T_yY$ respectively.
The orientation of a manifold with edges at the boundary can be induced by the manifold. Specifically,
Let the sign of the ordered basis $\beta$ at the boundary be determined by $\{n,\beta\}$, and $n$ be the external normal vector.
Let $f:X\rightarrow Y$ be a smooth mapping, $f,\partial f\pitchfork Z.$
$X,Y,Z$ is a directional manifold, $Y,Z$ is edgeless. Let $S:=f^{-1}(Z)$ be an embedded submanifold,
Want to get the reverse image orientation.
From $T_xS^\perp\oplus T_xS=T_xX,$, its orientation can be determined by direct sum orientation.
First take the $T_xS^\perp$ orientation. From the transversality,
$$
df_x(T_xX)+T_zZ=T_zY,\quad df_x(T_xS^\perp) \oplus T_zZ=T_zY.
$$
Assume $df_x$ is oriented,
That is, the $T_xS^\perp$ orientation is induced by the $df_x(T_xS^\perp)$ orientation determined by the straight sum orientation.
At this time $\partial S=\partial (f^{-1}(Z))=(\partial f)^{-1}(Z)$ has two orientation modes.
One is inverse image orientation, considering
$$
T_x \partial S^\perp\oplus T_x\partial S=T_x\partial X,\quad d\partial f_x(T_x\partial S^\perp)\oplus T_zZ=T_zY.
$$
Note that the two orientations $T_xS^\perp=T_x\partial S^\perp,$ are consistent.
Take its positive directional ordered basis $\alpha,$, then
$$
\operatorname{sgn}(\beta)=\operatorname{sgn}(\alpha,\beta)=\operatorname{sgn}(n,\alpha,\beta).
$$
$n$ is the external normal vector of $\partial S$ in $S$. The other is boundary orientation.
Taking the ordered basis $\beta,$ on $T_x\partial S$, then
$\operatorname{sgn}(\beta)=\operatorname{sgn}(n,\beta)$
The symbol decision in $T_xS$ is
$$
\operatorname{sgn}(\beta)=\operatorname{sgn}(n,\beta)=\operatorname{sgn}(\alpha,n,\beta).
$$
Since $\alpha$ is the ordered basis of $T_xS^\perp$,
The dimension is $\operatorname{codim}_XS=\operatorname{codim}_YZ.$
Therefore, the two orientations are also different $(-1)^{\operatorname{codim}_YZ}.$
Number of directional intersections
The basic assumption is that the same module $2$ intersection number is used. The additional assumption is that all manifolds are orientable.
The orientation of $df_x(T_xX)$ is given by $X$.
If $df_x(T_xX)\oplus T_zZ=T_zY$ has the same orientation on both sides,
Then record the orientation number at $x$ as $\operatorname{sgn}_x(f,Z)=1,$, otherwise it is $-1.$
The sum of the orientation numbers is called the orientation intersection number of $f$ and $Z$, recorded as
$$
I(f,Z):=\sum_{x\in f^{-1}(Z)}\operatorname{sgn}_x(f,Z).
$$
Since the sum of the orientation numbers of boundary points of any compactly oriented $1$-dimensional manifold with edges is $0,$
The intersection number of the same module $2$ can prove that the directional intersection number is a homotopy invariant.
Therefore, the general mapping can be defined by the transverse homotopy theorem. In the same way, we also have
Proposition 2. If $X=\partial W,$ $W$ is compact, $f:X\rightarrow Y$ can be extended to $W,$ $Z\subset Y$ is a closed submanifold, then $I(f,Z)=0.$
If $\dim X=\dim Y,$, you can define $\deg f:=I(f,\{y\}),$ $y$ and choose arbitrarily among $Y$.
Wellness is proved by the transverse homotopy theorem and the record lemma. At this time, if $y$ is the regular value of $f$,
Yes $\deg f=\sum_{x\in f^{-1}(y)}\operatorname{sgn}_x(f,y),$
$\operatorname{sgn}_x(f,y)=1,$ If $df_x$ maintains the direction. Otherwise, it is $-1.$
The mapping degree is also a homotopy invariant and is zero for functions that extend from the boundary to the interior.
The proof of the basic theorem of algebra can be given using the mapping degree, which is consistent with the intersection number modulo $2$.
At the same time, $\deg \frac{p}{|p|}$ can be used to count the number of internal zero points at the boundary of the tight region.
Next, we discuss the directional intersection number between mappings. Let $f:X\rightarrow Y,$ $g:Z\rightarrow Y.$
If $f\pitchfork g,$ can define $I_{(x,z)}(f,g),$
The sign is determined by the orientation of the left and right sides of $df_x(T_xX)\oplus dg_z(T_zZ)=T_yY$.
Proposition 3. $f\pitchfork g$ if and only if $f\times g\pitchfork \Delta,$ and $I(f,g)=(-1)^{\dim Z}I(f\times g,\Delta).$
This is mainly the following lemma
Lemma 4. Let $U,W$ be the directed subspace of $V$, then $U\oplus W=V$ if and only if $(U\times W)\oplus \Delta=(-1)^{\dim W}V\times V.$
In this way, general functions can be defined using the transverse homotopy theorem.
At this point it can be explained $I(f,Z)=I(f,\iota_Z),$
In particular, a new proof can be given that $\deg f=I(f,\{y\})$ has nothing to do with the selection of $y$. Comparing straightness and orientation are:
Proposition 5. $I(f,g)=(-1)^{\dim X\dim Z}I(g,f),$ $I(X,Z)=(-1)^{\dim X\dim Z}I(Z,X).$
In particular, if $Z=X,$ is the self-intersection number of an odd-dimensional manifold $I(X,X)=0.$ and if $X$ is a compact directional manifold,
Define its Euler characteristic number as $\chi(Y):=I(\Delta,\Delta).$, so it is a homotopy invariant.
Corollary 6. The Euler characteristic number of an odd-dimensional compact orientable manifold is $0.$
The article was last updated on 2023-06-13 12:33:48
