Machine-translated from Chinese.
准备工作
引理 1. 若$u\in\mathcal{R}_p(X),$ $\varphi\in \mathcal{R}_{p'}(X),$ $v\in \mathcal{R}_q(Y),$ $\psi\in\mathcal{R}_{q'}(Y),$ 则 $\left<{}u\times v,\varphi\times \psi\right>=\binom{p+q}{p}\left<{}u,\varphi\right>\left<{}v,\psi\right>.$
注意到$(u\times v)^z=(-1)^{L(z)}u^xv^y,$ 其中$x,y$为$z$的投影,
$z\in S(z),$ 所有阶梯型道路构成的集合. 接下来考虑内积即可.
引理 2. $\,\forall\,w\in \Omega_\ast (Z),$ $w=\sum_{\begin{subarray}{c} x\in A(X)\\ y\in A(Y) \end{subarray} }c^{xy}e_x\times e_y.$
注意对于道路的连接, 连接空间里的准许道路都是准许道路的连接,
但乘积空间中不一定, 需要$\partial$-不变才可以.
注意到对于两个只差一个拐角的道路, 为了求边界后仍然准许,
它们的系数应该是相反数.
因此所有$\Omega_\ast (Z)$中的道路都是形如$e_x\times e_y$的组合. 特别的,
可以直接写出$c^{xy}=(-1)^{L(z)}w^z,$ 这与$z$选取无关.
引理 3. $\,\forall\,w\in \Omega_\ast (z),$ 存在表示$w=\sum_{x\in A(X)}e_x\times a^x,$ $a^x\in \Omega_\ast (Y),$ 以及$w=\sum_{y\in A(Y)}\ell^y\times e_y,$ $\ell^y\in \Omega_\ast (X).$
引理 4. $u\in \Omega_p^\perp(X),$ $v\in \mathcal{A}_q(Y),$ $u\times v\in \Omega_r^\perp(Z).$
定理 5. $\Omega_\ast (X\square Y)\cong \Omega_\ast (X)\otimes \Omega_\ast (Y),$ $\otimes \mapsto \times.$ 同调群也对.
文章最后更新于 2024-11-04 17:39:01
Preparation
Lemma 1. If $u\in\mathcal{R}_p(X),$ $\varphi\in \mathcal{R}_{p'}(X),$ $v\in \mathcal{R}_q(Y),$ $\psi\in\mathcal{R}_{q'}(Y),$ then $\left<{}u\times v,\varphi\times \psi\right>=\binom{p+q}{p}\left<{}u,\varphi\right>\left<{}v,\psi\right>.$
Note that $(u\times v)^z=(-1)^{L(z)}u^xv^y,$ where $x,y$ is the projection of $z$,
$z\in S(z),$ The set of all stepped roads. Next, just consider the inner product.
Lemma 2. $\,\forall\,w\in \Omega_\ast (Z),$ $w=\sum_{\begin{subarray}{c} x\in A(X)\\ y\in A(Y) \end{subarray} }c^{xy}e_x\times e_y.$
Note that for road connections, the allowed roads in the connection space are all allowed road connections.
But it is not necessarily the case in the product space, it needs to be $\partial$-invariant.
Note that for two roads that only differ by one corner, in order to find the boundary, it is still allowed,
Their coefficients should be opposite numbers.
Therefore, all roads in $\Omega_\ast (Z)$ are combinations of the shape $e_x\times e_y$. In particular,
You can write $c^{xy}=(-1)^{L(z)}w^z,$ directly, which has nothing to do with the selection of $z$.
Lemma 3. $\,\forall\,w\in \Omega_\ast (z),$ exists to represent $w=\sum_{x\in A(X)}e_x\times a^x,$ $a^x\in \Omega_\ast (Y),$ and $w=\sum_{y\in A(Y)}\ell^y\times e_y,$ $\ell^y\in \Omega_\ast (X).$
Lemma 4. $u\in \Omega_p^\perp(X),$ $v\in \mathcal{A}_q(Y),$ $u\times v\in \Omega_r^\perp(Z).$
Theorem 5. $\Omega_\ast (X\square Y)\cong \Omega_\ast (X)\otimes \Omega_\ast (Y),$ $\otimes \mapsto \times.$ The homology group is also correct.
The article was last updated on 2024-11-04 17:39:01
