Introduction to involution

The aim of a this blog:

Introduce the concept involution via the representation of $\mathbb{Z}/2\mathbb{Z}$$\mathbb Z/2\mathbb Z$

Talking about the examples in various branches.

Generalise the proposition that every function over $\mathbb{R}$$\R$ could be uniquely written by sum of odd function and even function.

Let $\mathcal{C}$$\mathcal C$ be a category, and $X\in \mathrm{Ob}(\mathcal{C})$$X\in\mathrm{Ob}({\mathcal C})$.

Definition. An involution on $X$$X$ is an endomorphism $\ast$$*$, satisfies the property $\ast \ast ={\mathrm{id}}_X$$**=\mathrm{id}_X$.

Easy to see that $\ast$$*$ is an automorphism. The inverse is itself.

If $\eta :(\mathbb{Z}/2\mathbb{Z},+)⟶{\mathrm{Aut}}_{\mathcal{C}}(X)$$\eta:(\mathbb Z/2\mathbb Z,+)\longrightarrow\mathrm{Aut}_{\mathcal C}(X)$ is a monomorphism (injection)

Then

$$\begin{array}{}\text{(1)}& \eta (1)\in {\mathrm{Aut}}_{\mathcal{C}}(X)\end{array}$$

Gives us a non-trivial involution.

Example. Boolean Algebra and Lattice with complement.

Let $X\in \mathrm{Ob}(\mathrm{Set})$$X\in\mathrm{Ob(Set)}$ and its power set the $P(X)$$P(X)$.

$$\begin{array}{}\text{(2)}& \begin{array}{c}\eta :(\mathbb{Z}/2\mathbb{Z},+)⟶{\mathrm{Aut}}_{\mathrm{Set}}(P(X))\\ 1\mapsto (-{)}^c\end{array}\end{array}$$

Here $(-{)}^c$$(-)^c$ is complement. It could be generalised to any lattice with a complement. You can find lots of examples in my blog.

This idea could connect with this blog.

Remark For people farmilar with Boolean Ring.

A more natural description is

Notice that $P(-)$$P(-)$ is a representable functor, it is natural isomorphic to $2^{(-)}:={\mathrm{Hom}}_{\mathrm{Set}}(-,\mathbb{Z}/2\mathbb{Z})$$2^{(-)}:=\mathrm{Hom}_{\mathrm{Set}}(-,\mathbb Z/2\mathbb Z)$.

Hence

$$\begin{array}{}\text{(3)}& \begin{array}{c}\eta :(\mathbb{Z}/2\mathbb{Z},+)⟶{\mathrm{Aut}}_{\mathrm{Set}}(2^X)\\ 1\mapsto 1_X\end{array}\end{array}$$

Moreover, we could define a local complement for $S\subset X$$S\subset X$ by

$$\begin{array}{}\text{(4)}& \begin{array}{c}\eta :(\mathbb{Z}/2\mathbb{Z},+)⟶{\mathrm{Aut}}_{\mathrm{Set}}(2^X)\\ 1\mapsto 1_S\end{array}\end{array}$$

Example. Group object.

Let $G\in \mathrm{Ob}(\mathcal{C})$$G\in\mathrm{Ob}(\mathcal C)$ be a group object (for example, topological group/ abelian group)

Then the inverse morphism $\mu :G\to G,\mu \in {\mathrm{Aut}}_{\mathcal{C}}(G)$$\mu:G\to G,\mu\in\mathrm{Aut}_{\mathcal C}(G)$ is an automorphism.

$$\begin{array}{}\text{(5)}& \begin{array}{c}\eta :(\mathbb{Z}/2\mathbb{Z},+)⟶{\mathrm{Aut}}_{\mathcal{C}}(G)\\ 1\mapsto \mu \end{array}\end{array}$$

Example. Galois Group.

Consider the field extension such as $\mathbb{R}\subset \mathbb{C},\mathbb{Q}\subseteq \mathbb{Q}(\sqrt{d})...$$\mathbb R\subset\mathbb C,\mathbb Q\subseteq\mathbb Q(\sqrt d)...$

$$\begin{array}{}\text{(6)}& \mathrm{Gal}(L/K)\cong \mathbb{Z}/2\mathbb{Z}\end{array}$$

The conjugate is a nontrivial involution.

$$\begin{array}{}\text{(7)}& a+b\sqrt{d}\mapsto a-\sqrt{d}\end{array}$$

Example in matrix.

Let $M\in GL(V)$$M\in GL(V)$

$$\begin{array}{}\text{(8)}& M^2=I\end{array}$$

Example. $C^{\ast }$$C^*$ Algebra in Linear Algebra.

Let $V$$V$ be an inner product vector space over $\mathbb{C}$$\mathbb C$.

For $T\in {\mathrm{End}}_{\mathbb{C}}(V)$$T\in\mathrm{End}_{\mathbb C}(V)$, define its adjoint be

$$\begin{array}{}\text{(9)}& ⟨Tv,w⟩=⟨v,T^{\ast }w⟩\end{array}$$

Easy to verify that $T^{\ast }{}^{\ast }=T$$T^*{^*}=T$ by the axiom of inner product.

Moreover

$$\begin{array}{}\text{(10)}& (T+S{)}^{\ast }=T^{\ast }+S^{\ast },(T\circ S{)}^{\ast }=T^{\ast }{\circ }_{\mathrm{op}}{S}^{\ast }=S^{\ast }\circ T^{\ast }\end{array}$$

Hence

$$\begin{array}{}\text{(11)}& \ast :{\mathrm{End}}_{\mathbb{C}}(V)\to {\mathrm{End}}_{\mathbb{C}}^{\mathrm{op}}(V)\end{array}$$

gives a ring isomorphism.

If we focus on the abelian group structure, $({\mathrm{End}}_{\mathbb{C}}(V),+)$$(\mathrm{End}_{\mathbb C}(V),+)$

Then

$$\begin{array}{}\text{(12)}& \begin{array}{c}\eta :(\mathbb{Z}/2\mathbb{Z},+)⟶{\mathrm{Aut}}_{\mathcal{C}}(V)\\ 1\mapsto \ast \end{array}\end{array}$$

If we consider the real vector space, then $\ast :A\mapsto A^T$$*:A\mapsto A^T$.

Example. Involution on function over $\mathbb{C}$$\mathbb C$.

Let $f\in C(\mathbb{C})$$f\in C(\mathbb C)$.

$$\begin{array}{}\text{(13)}& f^{\ast }(z)=f(-z)\end{array}$$

gives us an involution.

Let $f\in C[a,b]$$f\in C[a,b]$.

$$\begin{array}{}\text{(14)}& f^{\ast }(x)=f(a+b-x)\end{array}$$

Give us an involution.

Proposition.

Let $M$$M$ be a $R-$$R-$module, and $2\in R^{\ast }$$2\in R^*$

Consider the $R-$$R-$module ${\mathrm{Hom}}_{\mathrm{Set}}(S,M)$$\mathrm{Hom}_{\mathrm{Set}}(S,M)$ and the involution

$$\begin{array}{}\text{(15)}& \eta :(\mathbb{Z}/2\mathbb{Z},+)⟶{\mathrm{Aut}}_R({\mathrm{Hom}}_{\mathrm{Set}}(S,M))\end{array}$$

Definition.

Let $\mathrm{Odd}\subseteq {\mathrm{Hom}}_{\mathrm{Set}}(S,M)$$\mathrm{Odd}\subseteq \mathrm{Hom}_{\mathrm{Set}}(S,M)$ be the submodule satisfies that $f^{\ast }=-f$$f^*=-f$,

$\mathrm{Even}\subseteq {\mathrm{Hom}}_{\mathrm{Set}}(S,M)$$\mathrm{Even}\subseteq \mathrm{Hom}_{\mathrm{Set}}(S,M)$ be the submodule satisfies that $f^{\ast }=f$$f^*=f$.

The reason that $\mathrm{Odd}$$\mathrm{Odd}$ and $\mathrm{Even}$$\mathrm{Even}$ is submodule follows from $\ast$$*$ is a $R-$$R-$ module homomorphism.

The reader could think that $\mathrm{Odd}$$\mathrm{Odd}$ and $\mathrm{Even}$$\mathrm{Even}$ is a kind of eigenspace concerning $-1,1$$-1,1$. Especially when $R$$R$ is a field.

Proposition.

$$\begin{array}{}\text{(16)}& {\mathrm{Hom}}_{\mathrm{Set}}(S,M)=\mathrm{Odd}⨁\mathrm{Even}\end{array}$$

Proof.

Firstly we should prove that $\mathrm{Odd}\cap \mathrm{Even}=\{0\}$$\mathrm{Odd}\cap\mathrm{Even}=\{0\}$.

That is, $f=f^{\ast }=-f⟹f+f=0$$f=f^*=-f\implies f+f=0$. By the condition $2\in R^{\ast }$$2\in R^*$, $f=0$$f=0$.

For any $f\in {\mathrm{Hom}}_{\mathrm{Set}}(S,M)$$f\in \mathrm{Hom}_{\mathrm{Set}}(S,M)$, $g=f-f^{\ast }\in \mathrm{Odd},g^{\prime }f+f^{\ast }\in \mathrm{Even}$$g=f-f^*\in\mathrm{Odd}, g'f+f^*\in\mathrm{Even}$.

Hence $g+g^{\prime }=2f$$g+g'=2f$. By the condition $2\in R^{\ast }$$2\in R^*$, we get that $\begin{array}{c}f=\frac{g+g^{\prime }}{2}◻\end{array}$$f=\frac{g+g'}{2}\quad\square\\$

The reader can see that this is a generalisation of odd functions and even functions.

Corollary.

Every function on $\mathbb{C}$$\mathbb C$ could be uniquely written as the sum of an odd function and an even function.

Every matrix could be written as the sum of a self-adjoint matrix and a matrix satisfies $T^{\ast }=-T$$T^*=-T$.

For $C(\mathbb{C})$$C(\mathbb C)$ and the conjugate as involution,

the $\mathrm{Odd}$$\mathrm{Odd}$ is the module of the imaginary function, and the $\mathrm{Even}$$\mathrm{Even}$ is the module of the real function.

...

In the case

$$\begin{array}{}\text{(17)}& f^{\ast }(x)=f(a+b-x)\end{array}$$

If we let $\begin{array}{c}[a,b]=[0,\pi ]\end{array}$$[a,b]=[0,{\pi}]\\$

Then

$$\begin{array}{}\text{(18)}& (\cos x+i\sin x{)}^{\ast }=\cos (\pi -x)+i\sin (\pi -x)=i^2(\cos x-i\sin x)=-\cos x+i\sin x\end{array}$$

In this case, $\cos x\in \mathrm{Odd}$$\cos x\in\mathrm{Odd}$ and $\sin x\in \mathrm{Even}$$\sin x\in\mathrm{Even}$!

Proposition.

For $f\in \mathrm{Odd}$$f\in\mathrm{Odd}$,

$$\begin{array}{}\text{(19)}& f(\frac{a+b}{2})=0\end{array}$$

Proof.

$$\begin{array}{}\text{(20)}& f(a+b-x)=-f(x)⟹f(\frac{a+b}{2})=-f(\frac{a+b}{2})◻\end{array}$$

Hence

$$\begin{array}{}\text{(21)}& f\in \mathrm{Odd}⟹{\int }_a^{b}f(x)dx=0\end{array}$$

As you can see, this involution is just an generalization of odd function and even function.

Proposition.

$$\begin{array}{}\text{(22)}& {({\int }_a^{b}f(x)dx)}^{\ast }:={\int }_b^{a}f(a+b-x)d(a+b-x)={\int }_a^{b}f(x)dx\end{array}$$

Proof.

$$\begin{array}{}\text{(23)}& {({\int }_a^{b}f(x)dx)}^{\ast }={\int }_b^{a}f(a+b-x)d(a+b-x)=-{\int }_b^{a}f(x)dx={\int }_a^{b}f(x)dx◻\end{array}$$