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准备工作
定义$\left<{}e_{i_0\cdots i_p},e_{j_0\cdots j_q}\right>=\delta_{J}^I,$
由此扩展到全体正则道路$\mathcal{R}_\ast =\bigoplus_{p\ge -1} \mathcal{R}_p$上的配对.
我们记大写字母表示基本道路, 花体字母表示任意道路,
即由基本道路张成的各类空间.
引理 1. $\,\forall\,u,\varphi\in \mathcal{R}_\ast (X),$ $v,\psi\in \mathcal{R}_\ast (Y),$ $\left<{}u\ast v,\varphi\ast \psi\right>=\left<{}u,\varphi\right>\left<{}v,\psi\right>.$
这是因为$u\ast v$和$\varphi\ast \psi$相同当且仅当它们分别相同.
引理 2. $\,\forall\,w\in \Omega_\ast (Z),$ $w=\sum_{x\in A(X)}e_x\ast a^x=\sum_{y\in A(Y)}\ell^y\ast e_y,$ $a^x\in \Omega_\ast (Y),$ $\ell^y\in \Omega_\ast (X).$
注意到$\,\forall\,w\in \mathcal{A}_\ast (Z),$
$w=\sum\limits_{\begin{subarray}{c}
x\in A(X)\\
y\in A(Y)
\end{subarray} } c^{xy}e_x*e_y,$
只需说明当$w\in \Omega_\ast (Z)$时,
$a^x=\sum_{y\in A(Y)}c^{xy}e_y\in \Omega_\ast (Y).$
记$\partial e_x=\sum_{x'\in R(X)}\varepsilon^x_{x'}e_{x'},$ 那么
$$
\begin{aligned}
\partial w&=\sum\limits_{\begin{subarray}{c}
x\in A(X)\\
x'\in R(X)
\end{subarray} } \varepsilon^x_{x'} e_{x'} *a^x+\sigma_x e^x*\partial a^x\\
&=\sum_{x,x'\in A(X)}e_x*(\varepsilon_{x}^{x'}a^{x'}+\sigma_x\partial a^x)+\sum_{x\in R(X)\setminus A(X)}\varepsilon_x^{x'} e_x*a^{x'}.
\end{aligned}
$$
当$\partial w\in \mathcal{A}_*(Z)$时,
后一项和消失.
记$\widetilde a^x=\sum_{x'\in A(X)}\varepsilon_x^{x'}a^x+\sigma_x\partial a^x,$
它必须是准许的. 那么做差即可得到$\partial a^x\in \mathcal{A}_\ast (Y).$
记$\Omega_p^{\perp}=\{u\in \mathcal{A}_p:\left<{}u,v\right>=0,\,\forall\,v\in \Omega_p\}.$
引理 3. $\Omega_p^\perp(X)\ast A_q(Y),A_p(X)\ast \Omega_q^\perp(Y)\subset \Omega_r^\perp(X\ast Y),$ $r=p+q+1.$
利用引理$1$即可.
主要定理
定理 4. $\,\forall\,w\in \Omega_r(Z),$ $w=\sum_j u_j\ast v_j,$ $u_j\in \Omega_{p_j}(X),$ $v_j\in \Omega_{q_j}(Y),$ $p_j+q_j+1=r.$
已知$A_r(Z)=\sum_{p+q+1=r}A_p(X)\ast A_q(Y).$
记$\Omega_r'(Z)=\sum_{p+q+1=r} \Omega_p(X)\ast \Omega_q(Y),$
希望证明$\Omega_r'(Z)=\Omega_r(Z).$
显然$\Omega_r'(Z)\subset \Omega_r(Z).$
$\,\forall\,u\in A_\ast (Z),$ 有分解$u=u_\Omega+u_\perp.$
$u\ast v=u_\Omega\ast v_\Omega+(\cdots),$ 括号内的部分都在$\Omega_r^\perp(Z)$中,
前一部分在$\Omega_r'(Z)$中.
于是若$u\ast v\in\Omega_r(Z)$中,
立即有$u\ast v=u_\Omega\ast v_\Omega\in \Omega_r'(Z).$ 这就证明了结论.
赋予合适的代数结构, 我们就得到了
定理 5. $\Omega_\ast (Z)\cong \bigoplus \Omega_\ast (X)\otimes \Omega_\ast (Y).$
文章最后更新于 2024-03-08 10:52:13
Preparation
Definition$\left<{}e_{i_0\cdots i_p},e_{j_0\cdots j_q}\right>=\delta_{J}^I,$
This is extended to pairings on all regular roads $\mathcal{R}_\ast =\bigoplus_{p\ge -1} \mathcal{R}_p$.
We remember capital letters to represent basic roads, and curly letters to represent arbitrary roads.
That is, various spaces formed by basic roads.
Lemma 1. $\,\forall\,u,\varphi\in \mathcal{R}_\ast (X),$ $v,\psi\in \mathcal{R}_\ast (Y),$ $\left<{}u\ast v,\varphi\ast \psi\right>=\left<{}u,\varphi\right>\left<{}v,\psi\right>.$
This is because $u\ast v$ and $\varphi\ast \psi$ are the same if and only if they are respectively the same.
Lemma 2. $\,\forall\,w\in \Omega_\ast (Z),$ $w=\sum_{x\in A(X)}e_x\ast a^x=\sum_{y\in A(Y)}\ell^y\ast e_y,$ $a^x\in \Omega_\ast (Y),$ $\ell^y\in \Omega_\ast (X).$
Notice$\,\forall\,w\in \mathcal{A}_\ast (Z),$
$w=\sum\limits_{\begin{subarray}{c}
x\in A(X)\\
y\in A(Y)
\end{subarray} } c^{xy}e_x*e_y,$
It only needs to be stated that when $w\in \Omega_\ast (Z)$,
$a^x=\sum_{y\in A(Y)}c^{xy}e_y\in \Omega_\ast (Y).$
Remember $\partial e_x=\sum_{x'\in R(X)}\varepsilon^x_{x'}e_{x'},$ then
$$
\begin{aligned}
\partial w&=\sum\limits_{\begin{subarray}{c}
x\in A(X)\\
x'\in R(X)
\end{subarray} } \varepsilon^x_{x'} e_{x'} *a^x+\sigma_x e^x*\partial a^x\\
&=\sum_{x,x'\in A(X)}e_x*(\varepsilon_{x}^{x'}a^{x'}+\sigma_x\partial a^x)+\sum_{x\in R(X)\setminus A(X)}\varepsilon_x^{x'} e_x*a^{x'}.
\end{aligned}
$$
When $\partial w\in \mathcal{A}_*(Z)$,
The latter sum disappears.
Note$\widetilde a^x=\sum_{x'\in A(X)}\varepsilon_x^{x'}a^x+\sigma_x\partial a^x,$
It must be allowed. Then make a difference and get $\partial a^x\in \mathcal{A}_\ast (Y).$
Remember$\Omega_p^{\perp}=\{u\in \mathcal{A}_p:\left<{}u,v\right>=0,\,\forall\,v\in \Omega_p\}.$
Lemma 3. $\Omega_p^\perp(X)\ast A_q(Y),A_p(X)\ast \Omega_q^\perp(Y)\subset \Omega_r^\perp(X\ast Y),$ $r=p+q+1.$
Just use Lemma $1$.
main theorem
Theorem 4. $\,\forall\,w\in \Omega_r(Z),$ $w=\sum_j u_j\ast v_j,$ $u_j\in \Omega_{p_j}(X),$ $v_j\in \Omega_{q_j}(Y),$ $p_j+q_j+1=r.$
Known $A_r(Z)=\sum_{p+q+1=r}A_p(X)\ast A_q(Y).$
Note$\Omega_r'(Z)=\sum_{p+q+1=r} \Omega_p(X)\ast \Omega_q(Y),$
Hope to prove$\Omega_r'(Z)=\Omega_r(Z).$
Obviously $\Omega_r'(Z)\subset \Omega_r(Z).$
$\,\forall\,u\in A_\ast (Z),$ With decomposition $u=u_\Omega+u_\perp.$
$u\ast v=u_\Omega\ast v_\Omega+(\cdots),$ The parts in brackets are in $\Omega_r^\perp(Z)$,
The previous part is in $\Omega_r'(Z)$.
So if $u\ast v\in\Omega_r(Z)$ is in,
Immediately there is $u\ast v=u_\Omega\ast v_\Omega\in \Omega_r'(Z).$ which proves the conclusion.
Given the appropriate algebraic structure, we get
Theorem 5. $\Omega_\ast (Z)\cong \bigoplus \Omega_\ast (X)\otimes \Omega_\ast (Y).$
The article was last updated on 2024-03-08 10:52:13
