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Grassmannian上同调计算
我们希望得到$H^\ast (G_n(\mathbb{R}^\infty);\mathbb{Z}_2)=\mathbb{Z}_2[w_1,\cdots,w_n].$
回忆我们希望给出$G_n(\mathbb{R}^\infty)$上的CW结构. 固定flag:
$$
\mathbb{R}^0\subset \mathbb{R}^1\subset \cdots \subset \mathbb{R}^m.
$$
任意$X\in G_n(\mathbb{R}^m)$给出一个序列:
$$
0\le \dim (X\cap \mathbb{R}^1)\le\cdots\le \dim(X\cap \mathbb{R}^m)=n.
$$
定义Schubert符号$\sigma=(\sigma_1,\cdots,\sigma_n)$记录了序列中跳跃的部分,
$1\le \sigma_1<\cdots<\sigma_n\le m.$
$$
e(\sigma):=\{X\in G_n(\mathbb{R}^m)|\dim (X\cap \mathbb{R}^{\sigma_i})=i,\:\dim(X\cap \mathbb{R}^{\sigma_i-1})=i-1\}.
$$
断言$\{e(\sigma)\}$组成了$G_n(\mathbb{R}^m)$的CW结构.
令$H^k\subset \mathbb{R}^k$为$x_k>0$的上半平面,
引理 1. 每个$X\in e(\sigma)$有唯一一个正交基$(x_1,\cdots,x_n)\in H^{\sigma_1}\times\cdots\times H^{\sigma_n}.$
只需注意到下面的序列, 由线性空间的基本知识即得.
$$
\dim(X\cap \mathbb{R}^{\sigma_1})=1,\quad \dim(X\cap \mathbb{R}^{\sigma_2})=2,\quad \cdots.
$$
$$
e'(\sigma):=\{(x_1,\cdots,x_n)|\{x_i\}\text{正交, }x_i\in H^{\sigma_i}\}.
$$
对$n$做归纳,
即可得到$\overline e'(\sigma)\cong$维数为$d(\sigma):=\sum_i(\sigma_i-i)$的闭球.
那么映射
$$
\overline e'(\sigma)\ni (x_1,\cdots,x_n)\mapsto \operatorname{span}\{x_1,\cdots,x_n\} \in G_n(\mathbb{R}^m).
$$
将$e'(\sigma)$同胚地映到$e(\sigma)$上. 这就实现了贴胞腔,
给出了$G_n(\mathbb{R}^m)$上的CW结构.
注 2. $G_n(\mathbb{R}^m)$是一个有限CW复形, 有$\binom{m}{n}$个胞腔. 令$m\rightarrow \infty,$ $G_n(\mathbb{R}^\infty)$为一个CW复形.
一个问题是$G_n(\mathbb{R}^m)$中有多少个$r$胞腔?
即$\dim e(\sigma)=\sum_i (\sigma_i-i)=r$的Schubert数.
记$\sigma_i-i=\mu_i,$ 那么
$$
0\le \mu_1\le \cdots\le \mu_n\le m-n,
$$
满足$\sum_i\mu_i=r.$
推论 3. $G_n(\mathbb{R}^m)$的$r$胞腔数为将$r$划分为至多$n$个小于等于$m-n$的正整数的方案数. 也称划分数.
定理 4. $H^\ast (G_n;\mathbb{Z}_2)\cong \mathbb{Z}_2[w_1,\cdots,w_n].$
我们希望构造一个同构:
$H^\ast (G_n)\xrightarrow{f^\ast }\mathbb{Z}_2[s_1,\cdots,s_n],$
$w_k(\gamma^n)\mapsto s_k.$
考虑$f:(\mathbb{R}\mathrm{P}^\infty)^{\times n}\rightarrow G_n,$
使得$(\gamma^1)^{\times n}=f^\ast \gamma^n.$
我们有$f^\ast :H^\ast (G_n)\rightarrow H^\ast ((\mathbb{R}\mathrm{P}^\infty)^{\times n}).$
注意到$\,\forall\,\tau\in S^n$作用在$(\mathbb{R}\mathrm{P}^\infty)^{\times n}$上,
$\tau^\ast (\gamma^1)^{\times n}\cong (\gamma^1)^{\times n}.$
因此$f\circ \tau$也能够将万有丛拉回. 那么$f\circ \tau\simeq f,$
$\tau^\ast \circ f^\ast =f^\ast .$
即$S_n$在$\operatorname{Im}(f^\ast )\subset H^\ast ((\mathbb{R}\mathrm{P}^\infty)^{\times n})$上作用平凡.
回忆我们有
$$
H^\ast ((\mathbb{R}\mathrm{P}^\infty)^{\times n})\cong H^\ast (\mathbb{R}\mathrm{P}^\infty)^{\otimes n}\cong \mathbb{Z}_2[a_1]\otimes \cdots\otimes \mathbb{Z}_2[a_n]\cong \mathbb{Z}_2[a_1,\cdots,a_n],\quad \deg a_i=1.
$$
我们有对称多项式基本定理,
$$
\mathbb{Z}_2[a_1,\cdots,a_n]=\mathbb{Z}_2[s_1,\cdots,s_n],\quad s_k=s_k(a_1,\cdots,a_n).
$$
断言这诱导了$w_k(\gamma^n)\mapsto s_k(a_1,\cdots,a_n).$ 回忆
$$
w(\gamma^1)=1+a,\quad w((\gamma^1)^{\times n})=(1+a_1)\times\cdots\times (1+a_n)\in H^\ast (\mathbb{R}\mathrm{P}^\infty)^{\otimes n},
$$
$$
w((\gamma^1)^{\times n})=(1+a_1)\cdots(1+a_n)=\sum_{k=0}^n s_k(a_1,\cdots,a_n).
$$
这就给出了$f^\ast w_k(\gamma^n)=w_k(f^\ast \gamma^n)=w_k((\gamma^1)^{\times n})=s_k.$
只需再验证$f^\ast $为单射. 希望证明$\,\forall\,r,$
$\dim H^r(G_n)\le \dim f^\ast H^r(G_n).$ 左侧小于等于$G_n$的胞腔数,
等同于度数为$r$的单项$s_1^{r_1}\cdots s_n^{r_n}$的个数,
又小于等于右侧的维数. 因此的确不等式成立.
特别地, 我们证明完了$f^\ast $是同构, 因此其实是取等.
那么这就说明了Grassmannian流形上的微分都是零.
注 5. 每个$e(\sigma)$都表示了$H_\ast (G_n)$中的一个非平凡元. $\dim H_r(G_n)$等于$r$维胞腔数, 与划分数联系. 每个$\mathbb{Z}_2$示性类都是$SW$示性类的乘积. 这也说明了$SW$示性类是唯一的.
SW示性类存在性
定理 6. SW示性类存在.
我们有至少四种方式来定义SW示性类.
第一种方式是通过谱序列直接计算$H^\ast (G_n;\mathbb{Z}_2),$
得到$\mathbb{Z}_2[w_1,\cdots,w_n].$ 将生成元拉回就给出了SW类.
第二种方式是通过障碍性理论. 第三种方式是通过Steenrod操作.
第四种方式也是要介绍的方式: 通过Leray-Hirsch定理. 回忆它是这么表述的:
定理 7 (Leray-Hirsch). 设$F\xrightarrow{i} E\xrightarrow{p} B$为纤维丛. 取定某个交换环$R,$ 若$H^n(F;R)$为自由有限生成的$R$模, 存在$c_j\in H^\ast (E;R)$使得$\{i^\ast c_j\}$组成了$H^\ast (F;R)\cong R\{c_j\}_j$的一组$R$-基. 那么有如下$H^\ast (B;R)$模同构:
$$
H^\ast (B;R)\bigotimes H^\ast (F;R)\cong H^\ast (E;R), \quad \sum_{i,j}b_i\otimes i^\ast (c_j)\mapsto \sum_{i,j}p^\ast (b_i)\cup c_j.
$$
文章最后更新于 2022-10-24 19:03:42
Grassmannian cohomology calculation
We want to get $H^\ast (G_n(\mathbb{R}^\infty);\mathbb{Z}_2)=\mathbb{Z}_2[w_1,\cdots,w_n].$
Recall that we want to give the CW structure on $G_n(\mathbb{R}^\infty)$. Fixed flag:
$$
\mathbb{R}^0\subset \mathbb{R}^1\subset \cdots \subset \mathbb{R}^m.
$$
Any $X\in G_n(\mathbb{R}^m)$ gives a sequence:
$$
0\le \dim (X\cap \mathbb{R}^1)\le\cdots\le \dim(X\cap \mathbb{R}^m)=n.
$$
Define Schubert symbol $\sigma=(\sigma_1,\cdots,\sigma_n)$ to record the jump part in the sequence,
$1\le \sigma_1<\cdots<\sigma_n\le m.$
$$
e(\sigma):=\{X\in G_n(\mathbb{R}^m)|\dim (X\cap \mathbb{R}^{\sigma_i})=i,\:\dim(X\cap \mathbb{R}^{\sigma_i-1})=i-1\}.
$$
Assert that $\{e(\sigma)\}$ forms the CW structure of $G_n(\mathbb{R}^m)$.
Let $H^k\subset \mathbb{R}^k$ be the upper half plane of $x_k>0$,
Lemma 1. Each $X\in e(\sigma)$ has a unique orthonormal basis $(x_1,\cdots,x_n)\in H^{\sigma_1}\times\cdots\times H^{\sigma_n}.$
Just note the following sequence, which is obtained from the basic knowledge of linear spaces.
$$
\dim(X\cap \mathbb{R}^{\sigma_1})=1,\quad \dim(X\cap \mathbb{R}^{\sigma_2})=2,\quad \cdots.
$$
$$
e'(\sigma):=\{(x_1,\cdots,x_n)|\{x_i\}\text{正交, }x_i\in H^{\sigma_i}\}.
$$
Make a generalization about $n$,
You can get a closed ball with $\overline e'(\sigma)\cong$ dimension and $d(\sigma):=\sum_i(\sigma_i-i)$ dimension.
Then map
$$
\overline e'(\sigma)\ni (x_1,\cdots,x_n)\mapsto \operatorname{span}\{x_1,\cdots,x_n\} \in G_n(\mathbb{R}^m).
$$
Map $e'(\sigma)$ homeomorphically onto $e(\sigma)$. This achieves the adhesion to the cell,
The CW structure on $G_n(\mathbb{R}^m)$ is given.
Note 2. $G_n(\mathbb{R}^m)$ is a finite CW complex with $\binom{m}{n}$ cells. Let $m\rightarrow \infty,$ $G_n(\mathbb{R}^\infty)$ be a CW complex.
One question is how many $r$ cells are there in $G_n(\mathbb{R}^m)$?
That is the Schubert number of $\dim e(\sigma)=\sum_i (\sigma_i-i)=r$.
Remember $\sigma_i-i=\mu_i,$ Then
$$
0\le \mu_1\le \cdots\le \mu_n\le m-n,
$$
Satisfy$\sum_i\mu_i=r.$
Corollary 3. The number of $r$ cells of $G_n(\mathbb{R}^m)$ is the number of options for dividing $r$ into at most $n$ positive integers less than or equal to $m-n$. It is also called the partition number.
Theorem 4. $H^\ast (G_n;\mathbb{Z}_2)\cong \mathbb{Z}_2[w_1,\cdots,w_n].$
We wish to construct an isomorphism:
$H^\ast (G_n)\xrightarrow{f^\ast }\mathbb{Z}_2[s_1,\cdots,s_n],$
$w_k(\gamma^n)\mapsto s_k.$
Consider $f:(\mathbb{R}\mathrm{P}^\infty)^{\times n}\rightarrow G_n,$
Make$(\gamma^1)^{\times n}=f^\ast \gamma^n.$
We have $f^\ast :H^\ast (G_n)\rightarrow H^\ast ((\mathbb{R}\mathrm{P}^\infty)^{\times n}).$
Notice that $\,\forall\,\tau\in S^n$ acts on $(\mathbb{R}\mathrm{P}^\infty)^{\times n}$,
$\tau^\ast (\gamma^1)^{\times n}\cong (\gamma^1)^{\times n}.$
Therefore $f\circ \tau$ can also pull back the universal bundle. Then $f\circ \tau\simeq f,$
$\tau^\ast \circ f^\ast =f^\ast .$
That is, $S_n$ has a trivial effect on $\operatorname{Im}(f^\ast )\subset H^\ast ((\mathbb{R}\mathrm{P}^\infty)^{\times n})$.
memories we have
$$
H^\ast ((\mathbb{R}\mathrm{P}^\infty)^{\times n})\cong H^\ast (\mathbb{R}\mathrm{P}^\infty)^{\otimes n}\cong \mathbb{Z}_2[a_1]\otimes \cdots\otimes \mathbb{Z}_2[a_n]\cong \mathbb{Z}_2[a_1,\cdots,a_n],\quad \deg a_i=1.
$$
We have the fundamental theorem of symmetric polynomials,
$$
\mathbb{Z}_2[a_1,\cdots,a_n]=\mathbb{Z}_2[s_1,\cdots,s_n],\quad s_k=s_k(a_1,\cdots,a_n).
$$
Assert that this induced $w_k(\gamma^n)\mapsto s_k(a_1,\cdots,a_n).$ recall
$$
w(\gamma^1)=1+a,\quad w((\gamma^1)^{\times n})=(1+a_1)\times\cdots\times (1+a_n)\in H^\ast (\mathbb{R}\mathrm{P}^\infty)^{\otimes n},
$$
$$
w((\gamma^1)^{\times n})=(1+a_1)\cdots(1+a_n)=\sum_{k=0}^n s_k(a_1,\cdots,a_n).
$$
This gives $f^\ast w_k(\gamma^n)=w_k(f^\ast \gamma^n)=w_k((\gamma^1)^{\times n})=s_k.$
Just need to verify that $f^\ast $ is injective. Hope to prove $\,\forall\,r,$
$\dim H^r(G_n)\le \dim f^\ast H^r(G_n).$ The number of cells on the left side that is less than or equal to $G_n$,
Equivalent to the number of single terms $s_1^{r_1}\cdots s_n^{r_n}$ with degree $r$,
It is less than or equal to the dimension on the right side. Therefore, the inequality is indeed established.
In particular, we have proved that $f^\ast $ is isomorphic, so it is actually equal.
Then this means that the differentials on the Grassmannian manifold are all zero.
Note 5. Each $e(\sigma)$ represents a non-trivial element in $H_\ast (G_n)$. $\dim H_r(G_n)$ is equal to the number of $r$ dimensional cells, which is related to the partition number. Each $\mathbb{Z}_2$ characteristic class is the product of the $SW$ characteristic class. This also shows that the $SW$ characteristic class is unique.
SW representational existence
Theorem 6. SW indicates the existence of sex type.
We have at least four ways to define SW characteristic classes.
The first way is to directly calculate $H^\ast (G_n;\mathbb{Z}_2),$ through the spectral sequence
Obtaining $\mathbb{Z}_2[w_1,\cdots,w_n].$ and pulling back the generator gives the SW class.
The second way is through the obstacle theory. The third way is through the Steenrod operation.
The fourth way is also the way to be introduced: through the Leray-Hirsch theorem. Recall that it is stated like this:
Theorem 7 (Leray-Hirsch). Let $F\xrightarrow{i} E\xrightarrow{p} B$ be a fiber bundle. Let a certain commutative ring $R,$ be determined. If $H^n(F;R)$ is a free finitely generated $R$ module, $c_j\in H^\ast (E;R)$ exists such that $\{i^\ast c_j\}$ forms a set of $R$-basis of $H^\ast (F;R)\cong R\{c_j\}_j$. Then there is the following $H^\ast (B;R)$ module isomorphism:
$$
H^\ast (B;R)\bigotimes H^\ast (F;R)\cong H^\ast (E;R), \quad \sum_{i,j}b_i\otimes i^\ast (c_j)\mapsto \sum_{i,j}p^\ast (b_i)\cup c_j.
$$
The article was last updated on 2022-10-24 19:03:42
