From Physics to Mathematics: Exploring Directional Resistance in Conductors through Involution

The initial idea of this blog comes from physics.

Some conductors exhibit changes in resistance when oriented in different directions, primarily due to the anisotropy of the materials. Anisotropy means that the physical properties of a material vary in different directions, including resistance and thermal conductivity, etc. For example, certain crystalline materials, composite materials, and some specially processed metallic materials may display this property.

In such materials, the arrangement of atoms or molecules might be more tightly or regularly aligned in certain directions, making it easier for electrons to move in those directions, thereby resulting in lower resistance; whereas in other directions, the less favorable alignment can hinder electron movement, resulting in higher resistance.

Hence we can consider a set of some conductors with two directions. Let $x$$x$ be a conductor, define $x^{†}$$x^{\dagger}$ to be the opposite direction of $x$$x$. Using some algebra to study this, the following essay generalizes this initial idea.

This blog（ Math Essays: Introduction to involution (marco-yuze-zheng.blogspot.com) will help you understand what is involution.

Let $(X,†,\mu )$$(X,\dagger,\mu)$ be a set with an involution operator $†$$\dagger$ and a valuation $\mu :X\to R$$\mu: X \to R$, where $R$$R$ is a ring.

Example. Let $X$$X$ be the set of conductors, $x^{†}$$x^\dagger$ means the opposite direction of $x$$x$. The valuation function $\mu :X\to \mathbb{R}$$\mu: X \to \mathbb{R}$ gives you the resistance of $x$$x$.

Consider the free $\mathbb{Z}$$\mathbb{Z}$-module of $(X,†,\mu )$$(X,\dagger,\mu)$ denoted as $[F(X),†,\mu ]$$[F(X),\dagger,\mu]$. The $†$$\dagger$ will be uniquely extended to the $\mathbb{Z}$$\mathbb{Z}$-linear map. i.e., $(x+y{)}^{†}=x^{†}+y^{†}$$(x+y)^\dagger = x^{\dagger} + y^{\dagger}$. We can define a map $d:F(X)\to F(X),d(x)=x-x^{†}$$d: F(X) \to F(X), d(x) = x - x^{\dagger}$.

Remark. According to the previous blog $d(x)\in \mathrm{Odd}$$d(x) \in \mathrm{Odd}$. Since $d(x{)}^{†}=-d(x)$$d(x)^{\dagger} = -d(x)$.

Notice that $d$$d$ is linear since $d=\mathrm{id}-(-{)}^{†}$$d = \mathrm{id} - (-)^\dagger$. The kernel of $d$$d$ is those "self-adjoint" elements, i.e., $x=x^{†}$$x = x^{\dagger}$.

Notice that $\dim (\mathrm{Ker}d)$$\mathrm{dim}(\mathrm{Ker} d)$ is equal to the cardinality of those "self-adjoint" elements.

Then we could consider $\mu \circ d(x)$$\mu \circ d(x)$ to measure the difference between $x$$x$ and $x^{†}$$x^{\dagger}$.

Proposition. $d(x^{†})=d(x{)}^{†}$$d(x^\dagger) = d(x)^\dagger$.

Proof. $d(x^{†})=x^{†}-(x^{†}{)}^{†}=x^{†}-x=(x-x^{†}{)}^{†}=d(x{)}^{†}$$d(x^\dagger) = x^{\dagger} - (x^{\dagger})^{\dagger} = x^{\dagger} - x = (x - x^{\dagger})^\dagger = d(x)^{\dagger}$.

Proposition. $d^2(x)=2d(x)$$d^2(x) = 2d(x)$.

Proof. $d^2(x)=d(d(x))=d(x-x^{†})=d(x)-d(x{)}^{†}$$d^2(x) = d(d(x)) = d(x - x^\dagger) = d(x) - d(x)^{\dagger}$. But $d(x{)}^{†}=-d(x)$$d(x)^{\dagger} = -d(x)$, hence $d^2(x)=2d(x)$$d^2(x) = 2d(x)$.

Hence it is a $\mathbb{Z}[T]/(T^2-2)\cong \mathbb{Z}[\sqrt{2}]$$\mathbb{Z}[T]/(T^2-2) \cong \mathbb{Z}[\sqrt{2}]$ module. Hence to count $d^n(x)$$d^n(x)$, we only need to consider $(\sqrt{2}{)}^n$$(\sqrt{2})^n$.

In general, we do not need $\mu$$\mu$. Let us consider the category of dagger sets.

Definition. ${\mathrm{Set}}^{†}$$\mathrm{Set}^\dagger$. The objects of ${\mathrm{Set}}^{†}$$\mathrm{Set}^{\dagger}$ are sets with involution. The morphism of two dagger sets $(X,†)$$(X,\dagger)$ and $(Y,\ast )$$(Y,*)$ should preserve the involution.

$$\begin{array}{}\text{(1)}& f:(X,†)\to (Y,†),f(x^{†})=f(x{)}^{†}\end{array}$$

Hence ${\mathrm{Set}}^{†}$$\mathrm{Set}^{\dagger}$ is just the category of $\mathbb{Z}/2\mathbb{Z}$$\mathbb{Z}/2\mathbb{Z}$-Set.

Example. Let $X={\mathrm{End}}_{\mathbb{C}-\mathrm{Vect}}(V)$$X = \mathrm{End}_{\mathrm{\mathbb{C}-Vect}}(V)$ or ${\mathrm{End}}_{\mathbb{R}-\mathrm{Vect}}(V)$$\mathrm{End}_{\mathrm{\mathbb{R}-Vect}}(V)$. $T^{†}=T^{\ast }$$T^\dagger = T^*$ is the adjoint of $T$$T$, $Y$$Y$ be the $M_{n,n}(\mathbb{F})$$M_{n,n}(\mathbb{F})$.

Let $f$$f$ be the matrix representation of $X$$X$. Then

$$\begin{array}{}\text{(2)}& f(S^{†})=f(S{)}^T\end{array}$$

Here $f(S{)}^T$$f(S)^{T}$ is the transpose or conjugate transpose.

The kernel of $d(T)$$d(T)$ is the self-adjoint map. Notice that if $X={\mathrm{End}}_{\mathbb{R}-\mathrm{Vect}}(V)$$X = \mathrm{End}_{\mathrm{\mathbb{R}-Vect}}(V)$ then $†$$\dagger$ is linear. Hence $d$$d$ will be linear as well.

For the dagger set $(M_{n,n}(\mathbb{C}),†)$$(M_{n,n}(\mathbb C),\dagger)$, we could equip it with the norm to make it becomes $(X,†,\mu )$$(X,\dagger,\mu)$.

Then $\mu \circ d(A)\le ϵ$$\mu\circ d(A)\le\epsilon$ may tell us some interesting thing.