Degree sequence: an algebra approach

In graph theory, we know an important graph invariant is degree sequence.

In this essay, I will make this invariant much more interesting.

I will consider a function

$$\begin{array}{}\text{(1)}& F:\mathrm{FGraph}⟶\mathbb{Z}\mathbb{-}\mathrm{A}\mathrm{l}\mathrm{g}\end{array}$$

The idea is simple.

If you consider any finite connected graph $G$$G$, you could consider all the functions from vertex to $\mathbb{Z}$$\mathbb Z$.

This will form a Ring, but it is pretty boring. Since $F{G}_1\cong F{G}_2⟺|G_1|=|G_2|$$FG_1\cong F G_2\iff|G_1|=|G_2|$.

i.e.

$$\begin{array}{}\text{(3)}& FG={\mathbb{Z}}^{\oplus |G|}\end{array}$$

Glue two graph together correspond to $\oplus$$\oplus$, cut some part of graph correspond to quotient.

To improve this algebra invariant, we could consider those functions taking value at ${\mathbb{Z}}_{\mathrm{deg}(v_i)}$$\mathbb Z_{\deg(v_i)}$ at each vertex.

Where ${\mathbb{Z}}_{\mathrm{deg}(v_i)}$$\mathbb Z_{\deg(v_i)}$ is $\mathbb{Z}/\mathrm{deg}(v_i)\mathbb{Z}$$\mathbb Z/{\deg(v_i)}\mathbb Z$.

Let $I$$I$ be the function (not a functor).

i.e.

$$\begin{array}{}\text{(4)}& I(G)=\underset{i=1}{\overset{n}{⨁}}{\mathbb{Z}}_{\mathrm{deg}(v_i)}\end{array}$$

Remark

$†.G$$\dagger.$ The degree function on a graph is $0_G$$0_G$ in this algebra!

$†.G$$\dagger.$ is a loop or a line $⟺$$\iff$ $I(G)\cong {\mathbb{Z}}_2^{\oplus n}$$I(G)\cong\mathbb Z_{2}^{\oplus n}$ i.e. Boolean Ring!

$†.G$$\dagger.$ have not singleton point $⟺$$\iff$ $|I(G)|<\mathrm{\infty }$$|I(G)|<\infin$.

 

 

 

Connecting Differential Operators and Lie Brackets in ODEs

In ODE An Algebraic Approach 1, we observe a beautiful identity:

$$\begin{array}{}\text{(1)}& (D-\lambda )=e^{\lambda x}\circ D\circ e^{-\lambda x}\end{array}$$

In this essay, I would like to explain the connection between $(1)$$(1)$ and Lie bracket.

Firstly, we can prove this identity from the Lie bracket.

i.e.

$$\begin{array}{}\text{(2)}& [D,e^{\lambda x}]=\lambda e^{\lambda x}=D\circ e^{\lambda x}-e^{\lambda x}\circ D\end{array}$$

Thus

$$\begin{array}{}\text{(3)}& e^{\lambda x}\circ D+e^{\lambda x}\lambda =D\circ e^{\lambda x}⟹e^{\lambda x}(D+\lambda )=D\circ e^{\lambda x}\end{array}$$

Then take the inverse of $e^{\lambda x}$$e^{\lambda x}$

$$\begin{array}{}\text{(4)}& \begin{array}{c}\\ ⟹(D+\lambda )=e^{-\lambda x}\circ D\circ e^{\lambda x}⟹e^{\lambda x}\circ D\circ e^{-x}=(D-\lambda )\end{array}\end{array}$$

Conversely,

$$\begin{array}{}\text{(5)}& \frac{d}{dt}{|}_{t=0}(e^{tX}\circ Y\circ e^{-tX})=(X{e}^{tX}\circ Y\circ e^{-tX}+e^{tX}\circ Y\circ -X){|}_{t=0}=XY-YX=[X,Y]\end{array}$$