Machine-translated from Chinese.
谱分析
我们关心$\operatorname{Spec}\Delta_p,$ 特别是其是否有零特征值:
$\Delta_pu=0.$ 这样的$u$被称为调和的.
引理 1. $u\in \Omega_p$是调和的当且仅当$\partial u,\partial^\ast u=0.$
只需注意到有
$$
0=\left<{}\Delta_p,u\right>=\Vert\partial u\Vert^2+\Vert\partial^\ast u\Vert^2.
$$
定理 2 (Hodge decomposition). 我们有$\Omega_p=\partial \Omega_{p+1}\oplus \partial^\ast \Omega_{p-1}\oplus \mathcal{H}_p,$ 这里$\mathcal{H}_p$是全体$p$调和形式, 直和为正交和.
只需注意到$\,\forall\,w\in \partial^\ast \Omega_{p-1}^\perp,$
$\partial w=0;$ 另一边类似, 即可.
推论 3. 我们有$\mathcal{H}_p\cong H_p.$ 特别的, $0\in\operatorname{Spec}\Delta_p\Leftrightarrow \beta_p>0,$ 且零特征值的重数为$\beta_p.$
迹分析
$\operatorname{tr}\Delta_0=2E,$ $E$为有向边的个数.
定理 4. *记$T,S,D$分别为三角形, 正方形(线性无关), 双箭头的个数, 那么
$$
\operatorname{tr}\Delta_1=2E+3T+2S+4D.
$$
对于一个$m$-正方形, 正交化后的基形如
$$
\begin{aligned}
\omega_1&=e_{ab_0c}-e_{ab_1c},\\
\omega_2&=e_{ab_0c}+e_{ab_1c}-2e_{ab_2c},\\
\cdots.
\end{aligned}
$$
我们关心$\Delta_1$的谱, 特别是其没有零特征值的情形.
命题 5. $\lambda_{\max}(\Delta_0)\le 2\max_{i\in V}\deg(i).$
利用
$\lambda_{\max}(\Delta_0)=\max \frac{\Vert\partial^\ast u\Vert^2}{\Vert u\Vert^2}$
即可.
定理 6. *若$G$不含多重正方形, 无双箭头, 那么
$$
\lambda_{\max}(\Delta_1)\le 2\max_i\deg(i)+3\max_{i\rightarrow j} \deg_\Delta(ij)+2\max_{i\rightarrow j}\deg_{\square}(ij).
$$
Spectral Analysis
We care about $\operatorname{Spec}\Delta_p,$ specifically whether it has zero eigenvalues:
$\Delta_pu=0.$ Such $u$ is called harmonic.
Lemma 1. $u\in \Omega_p$ is harmonic if and only if $\partial u,\partial^\ast u=0.$
Just note that there
$$
0=\left<{}\Delta_p,u\right>=\Vert\partial u\Vert^2+\Vert\partial^\ast u\Vert^2.
$$
Theorem 2 (Hodge decomposition). We have $\Omega_p=\partial \Omega_{p+1}\oplus \partial^\ast \Omega_{p-1}\oplus \mathcal{H}_p,$ where $\mathcal{H}_p$ is the harmonic form of the whole $p$, and the direct sum is the orthogonal sum.
Just note that $\,\forall\,w\in \partial^\ast \Omega_{p-1}^\perp,$
$\partial w=0;$ The other side is similar, that’s all.
Corollary 3. We have $\mathcal{H}_p\cong H_p.$ in particular, $0\in\operatorname{Spec}\Delta_p\Leftrightarrow \beta_p>0,$ and the multiplicity of zero eigenvalues is $\beta_p.$
trace analysis
$\operatorname{tr}\Delta_0=2E,$ $E$ is the number of directed edges.
Theorem 4. *Remember $T,S,D$ as the number of triangles, squares (linearly independent), and double arrows respectively, then
$$
\operatorname{tr}\Delta_1=2E+3T+2S+4D.
$$
For a $m$-square, the basis shape after orthogonalization is as follows
$$
\begin{aligned}
\omega_1&=e_{ab_0c}-e_{ab_1c},\\
\omega_2&=e_{ab_0c}+e_{ab_1c}-2e_{ab_2c},\\
\cdots.
\end{aligned}
$$
We care about the spectrum of $\Delta_1$, especially the case where it has no zero eigenvalues.
Proposition 5. $\lambda_{\max}(\Delta_0)\le 2\max_{i\in V}\deg(i).$
take advantage of
$\lambda_{\max}(\Delta_0)=\max \frac{\Vert\partial^\ast u\Vert^2}{\Vert u\Vert^2}$
That’s it.
Theorem 6. *If $G$ does not contain multiple squares or double arrows, then
$$
\lambda_{\max}(\Delta_1)\le 2\max_i\deg(i)+3\max_{i\rightarrow j} \deg_\Delta(ij)+2\max_{i\rightarrow j}\deg_{\square}(ij).
$$
