Universal property of periodic function and an interesting question

On C^1

Consider a group action $\iota :\mathbb{R}\times C(\mathbb{R})\to C(\mathbb{R})$$\iota:\mathbb R\times C(\mathbb R)\to C(\mathbb R)$,

$$\begin{array}{}\text{(1)}& \iota (\tau ,f):=f(x+\tau )\end{array}$$

Let $g$$g$ be a non-constant periodic function, and define

$$\begin{array}{}\text{(2)}& T_g=inf\{T^{\prime }\in {\mathbb{R}}^{+}|g(x+T^{\prime })=g(x)\}\end{array}$$

We have

$$\begin{array}{}\text{(3)}& {\mathrm{Stab}}_{\mathbb{R}}(g):=\{\tau \in \mathbb{R}|\iota (\tau ,g)=g\}=T\mathbb{Z}\end{array}$$

Then

$$\begin{array}{}\text{(4)}& \mathbb{R}/{\mathrm{Stab}}_{\mathbb{R}}(g)=\{x+nT|x\in [0,T),n\in \mathbb{Z}\}\cong S^1\end{array}$$

Thus $\mathrm{\forall }$$\forall$ non-constant periodic continuous function, we have following universal property. (Click the link!)

Where $\pi :\mathbb{R}\to \mathbb{R}/T\mathbb{Z}\cong S^1$$\pi:\mathbb R\to\mathbb R/T\mathbb Z\cong S^1$.

But I would like to consider a more interesting idea here.

Let $N(\mathbb{R})$$N(\mathbb R)$ be the algebra of all the continuous periodic functions with a period $T^{\prime }\in {\mathbb{N}}^{+}$$T'\in\mathbb N^+$.

Similarly, define

$$\begin{array}{}\text{(5)}& T_g=inf\{T^{\prime }\in \mathbb{N}|g(x+T^{\prime })=g(x)\}\end{array}$$

be the least period of $g$$g$.

In addition, if we have two function $f,h$$f,h$, with $T_f=p,T_h=q$$T_f=p,T_h=q$ is a prime number.

We have $T_{(f+g)}=T_{fg}=pq$$T_{(f+g)}=T_{fg}=pq$ (? Maybe there exists some counter-example.)

So for safety, we could say that if $p$$p$ is a period of $f$$f$ and $q$$q$ is a period of $g$$g$, then $pq$$pq$ is a period of $f+g,fg$$f+g,fg$.

I would like to define that ''ideal''( it is not ideal of the ring, but it correspondence to $n\mathbb{Z}$$n\mathbb Z$) of $N(\mathbb{R})$$N(\mathbb R)$ to be

$$\begin{array}{}\text{(6)}& (n):=\{f\in N(\mathbb{R})|f(x+n)=f(x)\}\end{array}$$

It is not an ideal but it is a subring!

Then similarly we could have a prime ideal, then $\mathrm{Spec}N(\mathbb{R})$$\mathrm{Spec}\,N(\mathbb R)$.

A natural question is, let $T_f$$T_f$ be a square-free number, do we have $(T_f)=(p_{i_1},...,p_{i_j})$$(T_f)=(p_{i_1},...,p_{i_j})$?

The reason I consider square-free number is, for example, let $T_f=4$$T_f=4$.

For example, let $T_f=6$$T_f=6$, could we have $f=\sum a_i{b}_j$$f=\sum a_{i}b_j$, where $a_i\in (2),b_j\in (3)$$a_i\in(2),b_j\in(3)$.

Like

$$\begin{array}{}\text{(7)}& \sin (2\pi x/6)=\sin (2\pi x(\frac{1}{3}-\frac{1}{2}))=\sin (\frac{2\pi x}{3})\cos (\frac{2\pi x}{2})-\cos (\frac{2\pi x}{3})\sin (\frac{2\pi x}{2})\end{array}$$

I do not know!

Another interesting thing is if we consider the periodic function $g$$g$ defined on $\mathbb{Z}$$\mathbb Z$.

It is not hard to see that we could induce the group

$$\begin{array}{}\text{(8)}& \mathbb{Z}/T_g\mathbb{Z}\end{array}$$

So, let us consider the function space $\mathbb{T}p=\{f\in {\mathrm{Hom}}_{\mathrm{Set}}(\mathbb{Z},\mathbb{C}),f(z+p)=f(z)\}$$\mathbb Tp=\{f\in\mathrm{Hom}_{\mathrm{Set}}(\mathbb Z,\mathbb C),f(z+p)=f(z)\}$, where $p$$p$ is a prime number.

By the universal property, it is equivalence to say we are studying ${\mathrm{Hom}}_{\mathrm{Set}}(\mathbb{Z}/p\mathbb{Z},\mathbb{C})$$\mathrm{Hom}_{\mathrm{Set}}(\mathbb Z/p\mathbb Z,\mathbb C)$.

But then we could consider ${\mathrm{Hom}}_{\mathrm{Set}}(\mathbb{Z}/p^2\mathbb{Z},\mathbb{C})$$\mathrm{Hom}_{\mathrm{Set}}(\mathbb Z/p^2\mathbb Z,\mathbb C)$...

If $f=g$$f=g$ in ${\mathrm{Hom}}_{\mathrm{Set}}(\mathbb{Z}/p\mathbb{Z},\mathbb{C})$$\mathrm{Hom}_{\mathrm{Set}}(\mathbb Z/p\mathbb Z,\mathbb C)$, it does not imply that $f=g$$f=g$ in ${\mathrm{Hom}}_{\mathrm{Set}}(\mathbb{Z}/p^2\mathbb{Z},\mathbb{C})$$\mathrm{Hom}_{\mathrm{Set}}(\mathbb Z/p^2\mathbb Z,\mathbb C)$

We could consider this interesting thing

$$\begin{array}{}\text{(9)}& f\equiv g\mathrm{mod}{p}^n\end{array}$$

That means $f=g$$f=g$ in ${\mathrm{Hom}}_{\mathrm{Set}}(\mathbb{Z}/p^n\mathbb{Z},\mathbb{C})$$\mathrm{Hom}_{\mathrm{Set}}(\mathbb Z/p^n\mathbb Z,\mathbb C)$, and it will related to the p-adic number.

On smooth function

Let $D$$D$ be $\begin{array}{c}\frac{d}{dx}\end{array}$$\frac{d}{dx}\\$, we could rewrite the group action $\iota (\tau ,f)$$\iota(\tau,f)$ on real analytic function space as follows.

$$\begin{array}{}\text{(10)}& \iota (\tau ,f)=e^{\lambda D}f(x)=f(x+\lambda )\end{array}$$

Proof

by Taylor series,

$$\begin{array}{}\text{(11)}& f(x+\lambda )=f(x)+\lambda f^{\prime }(x)+\frac{{\lambda }^2{f}^{″}(x)}{2}+...\end{array}$$

i.e.

$$\begin{array}{}\text{(12)}& f(x+\lambda )=f(x)+\lambda Df+\frac{(\lambda D{)}^2}{2}f+...=e^{\lambda D}f\end{array}$$

Study the group action of $\mathbb{R}$$\mathbb R$ on real analytic function space is equivalence to study the group of $e^{\lambda D}$$e^{\lambda D}$

For perodic function, we have

$$\begin{array}{}\text{(13)}& f(x)=f(x+T)=e^{TD}f(x)\end{array}$$