Machine-translated from Chinese.
回顾
对于对称多项式$S=\mathbb{Z}[t_1,\cdots,t_n]^{S_n},$
$S^k$为$k$次对称多项式, 它有两种基的选取, $\{\sigma_I\}$或$\{m_I\},$
$I$为$k$-划分. $m_I=S_I(\sigma_1,\cdots,\sigma_n).$
拓扑上, $H^\ast (BU(1)^{\times n})\cong \mathbb{Z}[t_1,\cdots,t_n],$
$|t_i|=2.$
$H^\ast (BU(n))\cong \mathbb{Z}[t_1,\cdots,t_n]^{S_n}=\mathbb{Z}[c_1,\cdots,c_n],$
$c_i=\sigma_i.$
定义$S_I(c(\omega)):=S_I(c_1(\omega),\cdots,c_n(\omega)).$
引理 1 (Thom). $S_I(c(\omega\oplus\omega'))=\sum_{JK=I}S_J(c(\omega))S_K(c(\omega')).$
有由之定义的示性类.
新示性类
推论 2. 若$m,n\neq 0,$ 那么$S_{m+n}[K^m\times L^n]=0.$
对于$\tau=\tau_{\mathbb{C}\mathrm{P}^n},$ $c(\tau)=(1+a)^{n+1},$
$c_k(\tau)=\sigma_k(a,\cdots,a),$
$$
S_k(c_1,\cdots,c_k)=m_k(a,\cdots,a)=(n+1)a^k.
$$
取$n=k,$
那么$S_n[\mathbb{C}\mathrm{P}^n]=\left<{}S_n(\tau_{\mathbb{C}\mathrm{P}^n}),[\mathbb{C}\mathrm{P}^n]\right>=n+1\neq 0.$
因此$\mathbb{C}\mathrm{P}^n\neq K\times L,$
除非$K$或$L$为$\mathbb{C}\mathrm{P}^n.$
类似地, 对于Pontrjagin示性类,
$$
S_I(p(\xi)):=S_I(p_1(\xi),\cdots,p_n(\xi))\in H^{4n}(B;\mathbb{Z}).
$$
同样的, 我们有
命题 3. $S_I(p(\xi\oplus\xi'))=\sum_{JK=I}S_J(p(\xi))S_k(p(\xi'))+2\:torsions.$
推论 4. $S_I(p)[M\times N]=\sum_{JK=I}S_J(p)[M]\cdot S_K(p)[N].$ 两侧都是整数, 不涉及挠部.
$S_I(p)[M\times N]=0$ 除非 $4|\dim_\mathbb{R}M, \dim_\mathbb{R}N.$
Chern数和Pontrjagin数线性无关.
定理 5 (Thom). 若$K^1,\cdots,K^n$为复流形, 使得 $S_k(c)[K^k]\neq 0,$ 那么$p(m)\times p(n)$维矩阵 $[c_{i_1}\cdots c_{i_r}[K^{j_1}\times\cdots\times K^{j_s}]]_{I,J}$ 非奇异.
这给出$\{c_I\}$作为复流形不变量是独立的. 取$K^k=\mathbb{C}\mathrm{P}^k,$
那么这满足要求, 因此存在这样的复流形列.
对于$2$维时的情况, 唯一的划分为$1+1,2.$ 那么矩阵为
$$
\begin{bmatrix}
c_1c_1[\mathbb{C}\mathrm{P}^1\times \mathbb{C}\mathrm{P}^1]&c_1c_1[\mathbb{C}\mathrm{P}^2]\\
c_2[\mathbb{C}\mathrm{P}^1\times \mathbb{C}\mathrm{P}^1]& c_2[\mathbb{C}\mathrm{P}^2]
\end{bmatrix}
$$
类似的结论对Pontrjagin示性类也成立.
定理 6 (Thom). 若$M^{4k}$为定向流形, $S_k(p)[M^{4k}]\neq 0,$ $\,\forall\,k,$ 那么矩阵$[p_{i_1}\cdots p_{i_r}[M^{4j_1}\times\cdots\times M^{4j_s}]]_{I,J}$非奇异.
此时取得流形例子为$M^{4k}=\mathbb{C}\mathrm{P}^{2k},$
$p(\mathbb{C}\mathrm{P}^{2k})=(1+a^2)^{2k+1},$
那么$S_k(p)[\mathbb{C}\mathrm{P}^{2k}]=2k+1\neq 0.$
注意$\{c_I\},$ $\{S_I(c)\}$都构成$H^{2n}(BU(n);\mathbb{Z})$的一组基,
那么基改变后, 只需证明
$$
[S_I(K^{j_1}\times\cdots\times K^{j_s})]_{I,J}
$$
非奇异.
$$
(\ast )=\sum_{I_1\cdots I_s=I}S_{I_1}[K^{j_1}]\cdots S_{I_s}[K^{j_s}].
$$
上式只在$I$加细$J$的时候非零. 因此$[S_I[K^J]]_{I,J}$为上三角阵,
若合适排序$I,J.$
因此,
$$
\det=\prod_I S_I[K^I]=\prod_I S_{i_1}[K^{i_1}]\cdots S_{i_r}[K^{i_r}]
$$
定向配边环$\Omega_\ast $
设$(X^{n+1},\partial X)$为光滑流形,
$TX$上的定向诱导了$T(\partial X)$上的定向. 称$v_1,\cdots, v_n$正定向,
若$\,\exists\,v_0\in T_xX\setminus T_x(\partial X),$
$v_0,\cdots,v_n$正定向.
记$M$为定向闭流形, $-M$为其反定向, $M+M':=M\sqcup M'.$
定义两个$n$维光滑闭流形$M,M'$具正定向配边, 若存在紧定向流形, 具边界$X,$
使得$\partial X\cong M+(-M'),$ 同胚是保定向的. 容易验证这是一个等价关系.
等价类称为配边类.
$$
\Omega_n:=\{[M^n]|M^n\text{光滑,闭, 定向}\}.
$$
这构成了加法下的Abel群.
记
$$
\Omega_\ast :=\bigoplus_{n=0}^\infty \Omega_n,
$$
构成了分级交换环,
$[M]\times [N]:=[M\times N].$
注意$[M^m\times N^n]=(-1)^{mn}[N^n\times M^m].$
回忆$M^{4k}=\partial V^{4k+1},$ 那么它所有的$p_I[M]=0,$
$\,\forall\,I$划分$k.$
推论 7. $\,\forall\,I$划分$k,$ 我们有群同态$p_I:\Omega_{4k}\rightarrow \mathbb{Z}$良定, $[M^{4k}]\mapsto p_I[M^{4k}]=0.$
推论 8. $\{\mathbb{C}\mathrm{P}^{2i_1}\times\cdots\times \mathbb{C}\mathrm{P}^{2i_r}|I\text{为$k$划分}\}\subset \Omega_{4k}.$ 是$\mathbb{Z}$-线性无关的.
因此, $r(\Omega_{4k})\ge p(k).$
$[P_I(\mathbb{C}\mathrm{P}^J)]_{I,J}$非奇异.
之后我们说明$r(\Omega_{4k})=p(k).$ 我们有如下事实:
$\Omega_0\cong \mathbb{Z}=\mathbb{Z}\{[\ast ]\},$ $\Omega_1=0,$
$\Omega_2=0,$ $\Omega_3=0,$
$\Omega_4\cong \mathbb{Z}=\mathbb{Z}\{[\mathbb{C}\mathrm{P}^2]\},$
$\Omega_5\cong \mathbb{Z}_2=\left<{}Y\right>,$ $\Omega_6=0,$
$\Omega_7=0,$ $\Omega_8=\mathbb{Z}\oplus\mathbb{Z},$
$\Omega_9=\mathbb{Z}_2\oplus\mathbb{Z}_2,$ $\Omega_{10}=\mathbb{Z}_2,$
$\cdots.$
文章最后更新于 2022-12-08 10:21:47
review
For symmetric polynomial $S=\mathbb{Z}[t_1,\cdots,t_n]^{S_n},$
$S^k$ is a symmetric polynomial of degree $k$, which has two base options, $\{\sigma_I\}$ or $\{m_I\},$
$I$ is divided into $k$-divided. $m_I=S_I(\sigma_1,\cdots,\sigma_n).$
Topologically, $H^\ast (BU(1)^{\times n})\cong \mathbb{Z}[t_1,\cdots,t_n],$
$|t_i|=2.$
$H^\ast (BU(n))\cong \mathbb{Z}[t_1,\cdots,t_n]^{S_n}=\mathbb{Z}[c_1,\cdots,c_n],$
$c_i=\sigma_i.$
Definition$S_I(c(\omega)):=S_I(c_1(\omega),\cdots,c_n(\omega)).$
Lemma 1 (Thom). $S_I(c(\omega\oplus\omega'))=\sum_{JK=I}S_J(c(\omega))S_K(c(\omega')).$
There are characteristic classes defined by them.
Novelty
Corollary 2. If $m,n\neq 0,$ then $S_{m+n}[K^m\times L^n]=0.$
For $\tau=\tau_{\mathbb{C}\mathrm{P}^n},$ $c(\tau)=(1+a)^{n+1},$
$c_k(\tau)=\sigma_k(a,\cdots,a),$
$$
S_k(c_1,\cdots,c_k)=m_k(a,\cdots,a)=(n+1)a^k.
$$
Take $n=k,$
Then $S_n[\mathbb{C}\mathrm{P}^n]=\left<{}S_n(\tau_{\mathbb{C}\mathrm{P}^n}),[\mathbb{C}\mathrm{P}^n]\right>=n+1\neq 0.$
Therefore $\mathbb{C}\mathrm{P}^n\neq K\times L,$
Unless $K$ or $L$ is $\mathbb{C}\mathrm{P}^n.$
Similarly, for the Pontrjagin characteristic class,
$$
S_I(p(\xi)):=S_I(p_1(\xi),\cdots,p_n(\xi))\in H^{4n}(B;\mathbb{Z}).
$$
Likewise, we have
Proposition 3. $S_I(p(\xi\oplus\xi'))=\sum_{JK=I}S_J(p(\xi))S_k(p(\xi'))+2\:torsions.$
Corollary 4. $S_I(p)[M\times N]=\sum_{JK=I}S_J(p)[M]\cdot S_K(p)[N].$ Both sides are integers, no deflection is involved.
$S_I(p)[M\times N]=0$ unless $4|\dim_\mathbb{R}M, \dim_\mathbb{R}N.$
The Chern number and the Pontrjagin number are linearly independent.
Theorem 5 (Thom). If $K^1,\cdots,K^n$ is a complex manifold, such that $S_k(c)[K^k]\neq 0,$ then the $p(m)\times p(n)$-dimensional matrix $[c_{i_1}\cdots c_{i_r}[K^{j_1}\times\cdots\times K^{j_s}]]_{I,J}$ is non-singular.
This gives $\{c_I\}$ as complex manifold invariants that are independent. Take $K^k=\mathbb{C}\mathrm{P}^k,$
Then this satisfies the requirement, so there exists such a complex manifold sequence.
For the case of $2$ dimension, the only division is $1+1,2.$, then the matrix is
$$
\begin{bmatrix}
c_1c_1[\mathbb{C}\mathrm{P}^1\times \mathbb{C}\mathrm{P}^1]&c_1c_1[\mathbb{C}\mathrm{P}^2]\\
c_2[\mathbb{C}\mathrm{P}^1\times \mathbb{C}\mathrm{P}^1]& c_2[\mathbb{C}\mathrm{P}^2]
\end{bmatrix}
$$
A similar conclusion holds for Pontrjagin characteristic classes.
Theorem 6 (Thom). If $M^{4k}$ is a directional manifold, $S_k(p)[M^{4k}]\neq 0,$ $\,\forall\,k,$ then the matrix $[p_{i_1}\cdots p_{i_r}[M^{4j_1}\times\cdots\times M^{4j_s}]]_{I,J}$ is non-singular.
At this time, the example of obtaining the manifold is $M^{4k}=\mathbb{C}\mathrm{P}^{2k},$
$p(\mathbb{C}\mathrm{P}^{2k})=(1+a^2)^{2k+1},$
Then $S_k(p)[\mathbb{C}\mathrm{P}^{2k}]=2k+1\neq 0.$
Note that $\{c_I\},$ $\{S_I(c)\}$ all constitute a set of bases of $H^{2n}(BU(n);\mathbb{Z})$,
Then after the basis is changed, we only need to prove
$$
[S_I(K^{j_1}\times\cdots\times K^{j_s})]_{I,J}
$$
Not strange.
$$
(\ast )=\sum_{I_1\cdots I_s=I}S_{I_1}[K^{j_1}]\cdots S_{I_s}[K^{j_s}].
$$
The above formula is non-zero only when $I$ is added to $J$. Therefore, $[S_I[K^J]]_{I,J}$ is an upper triangular matrix,
Sort if appropriate $I,J.$
Therefore,
$$
\det=\prod_I S_I[K^I]=\prod_I S_{i_1}[K^{i_1}]\cdots S_{i_r}[K^{i_r}]
$$
Oriented edge ring$\Omega_\ast $
Let $(X^{n+1},\partial X)$ be a smooth manifold,
The orientation on $TX$ induces the orientation on $T(\partial X)$. It is called $v_1,\cdots, v_n$ positive orientation,
If$\,\exists\,v_0\in T_xX\setminus T_x(\partial X),$
$v_0,\cdots,v_n$ Forward direction.
Let $M$ be the oriented closed manifold, $-M$ be its reverse orientation, $M+M':=M\sqcup M'.$
Define two $n$-dimensional smooth closed manifolds $M,M'$ with positively oriented edges, and if there is a tightly oriented manifold, with boundaries $X,$
This makes $\partial X\cong M+(-M'),$ homeomorphisms orientation-preserving. It is easy to verify that this is an equivalence relationship.
Equivalence classes are called coordination classes.
$$
\Omega_n:=\{[M^n]|M^n\text{光滑,闭, 定向}\}.
$$
This forms the Abelian group under addition.
remember
$$
\Omega_\ast :=\bigoplus_{n=0}^\infty \Omega_n,
$$
A hierarchical exchange ring is formed,
$[M]\times [N]:=[M\times N].$
Note$[M^m\times N^n]=(-1)^{mn}[N^n\times M^m].$
Memories$M^{4k}=\partial V^{4k+1},$ then all of it$p_I[M]=0,$
$\,\forall\,I$ Division $k.$
Corollary 7. $\,\forall\,I$ Partition $k,$ We have group homomorphism $p_I:\Omega_{4k}\rightarrow \mathbb{Z}$ Well-determined, $[M^{4k}]\mapsto p_I[M^{4k}]=0.$
Corollary 8. $\{\mathbb{C}\mathrm{P}^{2i_1}\times\cdots\times \mathbb{C}\mathrm{P}^{2i_r}|I\text{为$k$划分}\}\subset \Omega_{4k}.$ is $\mathbb{Z}$-linearly independent.
Therefore, $r(\Omega_{4k})\ge p(k).$
$[P_I(\mathbb{C}\mathrm{P}^J)]_{I,J}$ Not singular.
Then we show $r(\Omega_{4k})=p(k).$ that we have the following facts:
$\Omega_0\cong \mathbb{Z}=\mathbb{Z}\{[\ast ]\},$ $\Omega_1=0,$
$\Omega_2=0,$ $\Omega_3=0,$
$\Omega_4\cong \mathbb{Z}=\mathbb{Z}\{[\mathbb{C}\mathrm{P}^2]\},$
$\Omega_5\cong \mathbb{Z}_2=\left<{}Y\right>,$ $\Omega_6=0,$
$\Omega_7=0,$ $\Omega_8=\mathbb{Z}\oplus\mathbb{Z},$
$\Omega_9=\mathbb{Z}_2\oplus\mathbb{Z}_2,$ $\Omega_{10}=\mathbb{Z}_2,$
$\cdots.$
The article was last updated on 2022-12-08 10:21:47
