Machine-translated from Chinese.
连接图上$\Delta_p$的谱
考虑增广链复形
$$
0\leftarrow \mathbb{K}\leftarrow \Omega_0\leftarrow \Omega_1\leftarrow \cdots
$$
上的增广Laplacian$\widetilde\Delta_p.$
显然$\widetilde\Delta_p=\Delta_p,$ $p\ge 1.$
而$\widetilde\Delta_{-1}=|V|$(标量变换),
$\widetilde\Delta_0=\Delta_0+E,$ $E$是各处全为$1$的矩阵.
引理 1. $\widetilde\Delta_r(u\ast v)=(\widetilde\Delta_pu)\ast v+u\ast (\widetilde\Delta_qv),$ $r=p+q+1.$
引理 2. $\partial^\ast (u\ast v)=(\partial^\ast u)\ast v+(-1)^{p+1}u\ast (\partial^\ast v).$
利用内积, 转移即可.
注意我们有$\Omega_r=\bigoplus_{r=p+q+1} \Omega_p \ast \Omega_q.$
定理 3. $\operatorname{Spec}\widetilde\Delta_r(X\ast Y)=\bigsqcup_{r=p+q+1}\left(\operatorname{Spec}\widetilde\Delta_p(X)+\operatorname{Spec}\widetilde\Delta_q(Y)\right)$
设$D_m$是$m$个独立点, $D_m^n=D_m\ast \cdots \ast D_m,$ 是$n$个$D_m$的连接.
定理 4. *$\,\forall\,n,m\ge 1,$ $r\ge 2,$
$$
\operatorname{Spec}\Delta_{r-1}(D_m^n)=\left\{((n-k)m)_{\binom{r}{k}\binom{n}{r}(m-1)^k}\right\}_{k=0}^r
$$
考虑$\left<{}e_x,e_y\right>_a=\frac{1}{a_p}\delta_x^y$为带权内积,
此时$\partial^\ast _a=\frac{a_{p+1} }{a_p}\partial^\ast .$
命题 5. 令$a_p=p!,$ 我们有$\Delta_r^{(a)}(u\times v)=\Delta_p^{(a)}u\times v+u\times \Delta_q^{(a)}v.$
定理 6. $\operatorname{Spec}\Delta_r^{(a)}(X\square Y)=\bigsqcup_{p+q=r}(\operatorname{Spec}\Delta_p^{(a)}(X)+\operatorname{Spec}\Delta_q^{(a)}(Y)).$
文章最后更新于 2024-11-04 17:39:28
Spectrum of $\Delta_p$ on a Graph Join
Consider the augmenting chain complex
$$
0\leftarrow \mathbb{K}\leftarrow \Omega_0\leftarrow \Omega_1\leftarrow \cdots
$$
Augmented Laplacian$\widetilde\Delta_p.$
Obviously $\widetilde\Delta_p=\Delta_p,$ $p\ge 1.$
And $\widetilde\Delta_{-1}=|V|$ (scalar transformation),
$\widetilde\Delta_0=\Delta_0+E,$ $E$ is a matrix with $1$ everywhere.
Lemma 1. $\widetilde\Delta_r(u\ast v)=(\widetilde\Delta_pu)\ast v+u\ast (\widetilde\Delta_qv),$ $r=p+q+1.$
Lemma 2. $\partial^\ast (u\ast v)=(\partial^\ast u)\ast v+(-1)^{p+1}u\ast (\partial^\ast v).$
Just use the inner product and transfer.
Note that we have $\Omega_r=\bigoplus_{r=p+q+1} \Omega_p \ast \Omega_q.$
Theorem 3. $\operatorname{Spec}\widetilde\Delta_r(X\ast Y)=\bigsqcup_{r=p+q+1}\left(\operatorname{Spec}\widetilde\Delta_p(X)+\operatorname{Spec}\widetilde\Delta_q(Y)\right)$
Let $D_m$ be $m$ independent points, $D_m^n=D_m\ast \cdots \ast D_m,$ be $n$ $D_m$ connections.
Theorem 4. *$\,\forall\,n,m\ge 1,$ $r\ge 2,$
$$
\operatorname{Spec}\Delta_{r-1}(D_m^n)=\left\{((n-k)m)_{\binom{r}{k}\binom{n}{r}(m-1)^k}\right\}_{k=0}^r
$$
Consider $\left<{}e_x,e_y\right>_a=\frac{1}{a_p}\delta_x^y$ as the weighted inner product,
At this time $\partial^\ast _a=\frac{a_{p+1} }{a_p}\partial^\ast .$
Proposition 5. Let $a_p=p!,$ we have $\Delta_r^{(a)}(u\times v)=\Delta_p^{(a)}u\times v+u\times \Delta_q^{(a)}v.$
Theorem 6. $\operatorname{Spec}\Delta_r^{(a)}(X\square Y)=\bigsqcup_{p+q=r}(\operatorname{Spec}\Delta_p^{(a)}(X)+\operatorname{Spec}\Delta_q^{(a)}(Y)).$
The article was last updated on 2024-11-04 17:39:28
