Machine-translated from Chinese.
Hodge-Laplacian
由内积$\left<{}e_x,e_y\right>=\delta_{xy}$,
我们可诱导链复形上$\partial$的对偶$\partial^\ast ,$ 使得
$$
\left<{}\partial u,v\right>=\left<{}u,\partial^\ast v\right>.
$$
我们可以定义$\Delta_p=\partial^\ast \partial+\partial\partial^\ast ,$
称为$p$-形式上的Hodge-Laplacian.
命题 1. $\Delta_p$是自伴半正定算子.
令$B$为$\partial:\Omega_p\rightarrow \Omega_{p-1}$的表示矩阵, 即
$$
B_{mi}=\left<{}\partial \alpha_i,\beta_m\right>,
$$
那么$\partial^\ast \partial$的表示矩阵是$B^TB.$ 类似的,
令$C$为$\partial^\ast :\Omega_p\rightarrow \Omega_{p+1}$的表示矩阵,
那么$\partial\partial^\ast $的表示矩阵为$C^TC.$ 对于$\Delta_p,$ 我们有
$$
(\Delta_p)_{ij}=\sum_m \left<{}\partial \alpha_i,\beta_m\right>\left<{}\partial\alpha_j,\beta_m\right>+\sum_n \left<{}\alpha_i,\partial \gamma_n\right>\left<{}\alpha_j,\partial\gamma_n\right>.
$$
特别的, 我们有
$$
(\Delta_0)_{ij}=\sum_{k\rightarrow l}\left<{}e_i,e_l-e_k\right>\left<{}e_j,e_l-e_k\right>=\sum_{k\rightarrow l}(\delta_{il}-\delta_{ik})(\delta_{jl}-\delta_{jk})=\sum_{k\rightarrow i}\delta_{ij}+\sum_{j\rightarrow l}\delta_{ij}-1_{i\rightarrow j}-1_{j\rightarrow i}.
$$
等式前两项构成度数矩阵, 后两项构成邻接矩阵.
类似的, 对于$\Delta_1,$ 下Laplacian $\partial^\ast \partial$有表示
$$
(\partial^\ast \partial)_{(ij)(i'j')}=\delta_{ii'}+\delta_{jj'}-\delta_{ij'}-\delta_{ji'}=:[ij,i'j'],
$$
衡量了两个边的连接情况.
问题: 当$\Delta_1=\operatorname{diag}\{\lambda\}$时, $G$是什么样的图?
文章最后更新于 2024-11-04 17:39:11
Hodge-Laplacian
From the inner product $\left<{}e_x,e_y\right>=\delta_{xy}$,
We can induce the dual $\partial^\ast ,$ of $\partial$ on the chain complex such that
$$
\left<{}\partial u,v\right>=\left<{}u,\partial^\ast v\right>.
$$
We can define $\Delta_p=\partial^\ast \partial+\partial\partial^\ast ,$
Called $p$-formally Hodge-Laplacian.
Proposition 1. $\Delta_p$ is a self-adjoint positive semidefinite operator.
Let $B$ be the representation matrix of $\partial:\Omega_p\rightarrow \Omega_{p-1}$, that is
$$
B_{mi}=\left<{}\partial \alpha_i,\beta_m\right>,
$$
Then the representation matrix of $\partial^\ast \partial$ is $B^TB.$. Similarly,
Let $C$ be the representation matrix of $\partial^\ast :\Omega_p\rightarrow \Omega_{p+1}$,
Then the representation matrix of $\partial\partial^\ast $ is $C^TC.$. For $\Delta_p,$ we have
$$
(\Delta_p)_{ij}=\sum_m \left<{}\partial \alpha_i,\beta_m\right>\left<{}\partial\alpha_j,\beta_m\right>+\sum_n \left<{}\alpha_i,\partial \gamma_n\right>\left<{}\alpha_j,\partial\gamma_n\right>.
$$
Specially, we have
$$
(\Delta_0)_{ij}=\sum_{k\rightarrow l}\left<{}e_i,e_l-e_k\right>\left<{}e_j,e_l-e_k\right>=\sum_{k\rightarrow l}(\delta_{il}-\delta_{ik})(\delta_{jl}-\delta_{jk})=\sum_{k\rightarrow i}\delta_{ij}+\sum_{j\rightarrow l}\delta_{ij}-1_{i\rightarrow j}-1_{j\rightarrow i}.
$$
The first two terms of the equation form the degree matrix, and the last two terms form the adjacency matrix.
Similarly, for $\Delta_1,$ the Laplacian $\partial^\ast \partial$ has the expression
$$
(\partial^\ast \partial)_{(ij)(i'j')}=\delta_{ii'}+\delta_{jj'}-\delta_{ij'}-\delta_{ji'}=:[ij,i'j'],
$$
Measures the connection between two edges.
Question: When $\Delta_1=\operatorname{diag}\{\lambda\}$, what kind of graph is $G$?
The article was last updated on 2024-11-04 17:39:11
