Machine-translated from Chinese.
$\Omega_2$结构
给定有向图$G=(V,E),$
$\mathcal{A}_p=\left<{}e_{i_0\cdots i_p}:i_0\rightarrow\cdots\rightarrow i_p\right>_\mathbb{K},$
为由准许基本$p$-道路张成的空间.
记$\Omega_p=\left<{}v\in \mathcal{A}_p:\partial v\in \mathcal{A}_{p-1}\right>,$
那么$(\Omega_\ast ,\partial)$构成$G$的道路链复形.
我们知道有$\Omega_0=\mathcal{A}_0,$ $\Omega_1=\mathcal{A}_1,$
但之后只能保证$\Omega_2\subset \mathcal{A}_2,$ $\cdots.$
称$(x,y)$为半有向边若其不是有向边且$x\neq y$,
但存在中转点$z$使得$x\rightarrow z\rightarrow y.$
记为$x\rightharpoonup y.$
定理 1. $|\Omega_2|=|\mathcal{A}_2|-s,$ $s$为半有向边个数. 事实上$\Omega_2$由全体三角形, 正方形, 双箭头生成.
设$v=\sum_{a\rightarrow b\rightarrow c}v^{abc}e_{abc}\in \mathcal{A}_2,$
那么
$$
\partial v=\sum_{a\rightarrow b\rightarrow c} v^{abc}(e_{bc}-e_{ac}+e_{ab})\equiv-\sum_{a\rightarrow b\rightarrow c} v^{abc}e_{ac}\mod \mathcal{A}_1.
$$
若$a\rightarrow c$那么显然$\partial v\in \mathcal{A}_1,$ 此时为三角形;
若$a=c$更显然, 此时为双箭头; 若$a\rightharpoonup c,$
那么只需$\sum_{b:a\rightarrow b\rightarrow c} v^{abc}=0,$
即有$\partial v\in \mathcal{A}_1,$ 这对应于若干正方形($v^{ab_1c}=1,$
$v^{ab_2c}=-1,$ 其它为零)的线性组合. 共有$s$个这样的约束,
因此$|\Omega_2|=|\mathcal{A}_2|-s.$
若半有向边$a\rightharpoonup c$有$m+1$个中转点$b_i,$
称它们构成一个$m$-正方形. 此时这部分构成的有向图满足$|\Omega_2|=m,$
$|\mathcal{A}_2|=m+1,$ 半有向边只有一条$a\rightharpoonup c.$
有向图同态
对于有向图$X,Y$, 称$f:X\rightarrow Y$为有向图同态(映射),
若对于任意$X$中的$a\rightarrow b,$ 在$Y$中要么有$f(a)\rightarrow f(b),$
要么有$f(a)=f(b).$
给定这样的同态, 有诱导映射$f_\ast :\Lambda_n(X)\rightarrow \Lambda_n(Y),$
$f_\ast (e_{i_0\cdots i_n}):=e_{f(i_0)\cdots f(i_n)}.$
命题 2. 令$f:X\rightarrow Y$为有向图间的同态, 那么$f_\ast $延拓为链映射$f_\ast :\Omega_n(X)\rightarrow \Omega_n(Y),$ 进而$f_\ast :H_n(X)\rightarrow H_n(Y).$
只需注意到$f_\ast $将非正则基本道路映为非正则基本道路,
取模后将准许基本道路映为准许基本道路(可能被模掉为零).
同时显然有$\partial f_\ast =f_\ast \partial.$
正方形通过有向图同态可以变成三角形或双箭头.
有向多边形
设$G$为嵌入在$S^1$上的一个$n$边形,
每个有向边$\xi$定义$\sigma^\xi=\pm 1,$ 取正若有向边为逆时针方向,
反之为负. $\sigma=\sum_\xi \sigma^\xi e_\xi \in \mathcal{A}_1.$
引理 3. 我们有$\partial \sigma=0,$ 进而$\sigma\in \Omega_1$ 进一步, $\ker \sigma|_{\Omega_1}=\left<{}\sigma\right>.$
$\,\forall\,v\in \mathcal{A}_1,$ $v=\sum_\xi v^\xi e_\xi.$
$\partial v=\sum_{k=0}^{n-1} c_ke_k.$
设$(k-1)\xrightarrow{\xi} k\xrightarrow{\eta} (k+1),$
那么$c_k=v^{\xi}\sigma^\xi - v^\eta \sigma^\eta.$
于是$c_k=0$当且仅当$v^\xi \sigma^\xi=v^\eta \sigma^\eta,$
而$v^\xi=\sigma^\xi$构成解的基底.
命题 4. 令$P$为$n$边形, 若$P$为三角形或正方形, 那么$|\Omega_2|=1,$ $|\Omega_p|=0,$ $p\ge 3,$ $|H_p|=0,$ $p\ge 1;$ 不然, $|\Omega_p|=0,$ $\,\forall\,p\ge 2,$ $|H_1|=1,$ $|H_p|=0,$ $p\ge 2.$
文章最后更新于 2024-02-28 16:15:26
$\Omega_2$ Structure
Given a directed graph $G=(V,E),$
$\mathcal{A}_p=\left<{}e_{i_0\cdots i_p}:i_0\rightarrow\cdots\rightarrow i_p\right>_\mathbb{K},$
The ground allows the basic $p$ - the space formed by the road.
Note$\Omega_p=\left<{}v\in \mathcal{A}_p:\partial v\in \mathcal{A}_{p-1}\right>,$
Then $(\Omega_\ast ,\partial)$ constitutes the complex of the road link of $G$.
We know that there are $\Omega_0=\mathcal{A}_0,$ $\Omega_1=\mathcal{A}_1,$
But then only $\Omega_2\subset \mathcal{A}_2,$ $\cdots.$ is guaranteed
$(x,y)$ is called a semi-directed edge if it is not a directed edge and $x\neq y$,
But there is a transfer point $z$ such that $x\rightarrow z\rightarrow y.$
Recorded as $x\rightharpoonup y.$
Theorem 1. $|\Omega_2|=|\mathcal{A}_2|-s,$ $s$ is the number of semi-directed edges. In fact, $\Omega_2$ is generated from all triangles, squares, and double arrows.
Let $v=\sum_{a\rightarrow b\rightarrow c}v^{abc}e_{abc}\in \mathcal{A}_2,$
Then
$$
\partial v=\sum_{a\rightarrow b\rightarrow c} v^{abc}(e_{bc}-e_{ac}+e_{ab})\equiv-\sum_{a\rightarrow b\rightarrow c} v^{abc}e_{ac}\mod \mathcal{A}_1.
$$
If $a\rightarrow c$ then obviously $\partial v\in \mathcal{A}_1,$ is a triangle at this time;
If $a=c$ is more obvious, it is a double arrow; if $a\rightharpoonup c,$
Then just $\sum_{b:a\rightarrow b\rightarrow c} v^{abc}=0,$
That is, there is $\partial v\in \mathcal{A}_1,$ which corresponds to a number of squares ($v^{ab_1c}=1,$
$v^{ab_2c}=-1,$ linear combinations (others are zero). There are $s$ such constraints in total,
Therefore $|\Omega_2|=|\mathcal{A}_2|-s.$
If the semi-directed edge $a\rightharpoonup c$ has $m+1$ intermediate points $b_i,$
They are said to form a $m$-square. At this time, the directed graph formed by this part satisfies $|\Omega_2|=m,$
$|\mathcal{A}_2|=m+1,$ There is only one semi-directed edge $a\rightharpoonup c.$
directed graph homomorphism
For the directed graph $X,Y$, we call $f:X\rightarrow Y$ a directed graph homomorphism (mapping),
If for any $a\rightarrow b,$ in $X$, there is either $f(a)\rightarrow f(b),$ in $Y$
Either there $f(a)=f(b).$
Given such a homomorphism, there is an induced map $f_\ast :\Lambda_n(X)\rightarrow \Lambda_n(Y),$
$f_\ast (e_{i_0\cdots i_n}):=e_{f(i_0)\cdots f(i_n)}.$
Proposition 2. Let $f:X\rightarrow Y$ be a homomorphism between directed graphs, then $f_\ast $ extends to chain mapping $f_\ast :\Omega_n(X)\rightarrow \Omega_n(Y),$ and then $f_\ast :H_n(X)\rightarrow H_n(Y).$
Just note that $f_\ast $ maps non-regular basic roads to non-regular basic roads,
After taking the modulus, the allowed basic roads are mapped to the allowed basic roads (which may be moduloed to zero).
At the same time, there is obviously $\partial f_\ast =f_\ast \partial.$
A square can be transformed into a triangle or a double arrow through a directed graph homomorphism.
directed polygon
Let $G$ be an $n$ polygon embedded on $S^1$,
Each directed edge $\xi$ defines $\sigma^\xi=\pm 1,$ if the directed edge is counterclockwise,
Otherwise it is negative. $\sigma=\sum_\xi \sigma^\xi e_\xi \in \mathcal{A}_1.$
Lemma 3. We have $\partial \sigma=0,$ and then $\sigma\in \Omega_1$ and further, $\ker \sigma|_{\Omega_1}=\left<{}\sigma\right>.$
$\,\forall\,v\in \mathcal{A}_1,$ $v=\sum_\xi v^\xi e_\xi.$
$\partial v=\sum_{k=0}^{n-1} c_ke_k.$
Let $(k-1)\xrightarrow{\xi} k\xrightarrow{\eta} (k+1),$
Then $c_k=v^{\xi}\sigma^\xi - v^\eta \sigma^\eta.$
Then $c_k=0$ if and only if $v^\xi \sigma^\xi=v^\eta \sigma^\eta,$
And $v^\xi=\sigma^\xi$ forms the basis of the solution.
Proposition 4. Let $P$ be $n$ polygon, if $P$ is a triangle or square, then $|\Omega_2|=1,$ $|\Omega_p|=0,$ $p\ge 3,$ $|H_p|=0,$ $p\ge 1;$ Otherwise, $|\Omega_p|=0,$ $\,\forall\,p\ge 2,$ $|H_1|=1,$ $|H_p|=0,$ $p\ge 2.$
The article was last updated on 2024-02-28 16:15:26
