I've been reviewing some group theory; particularly group actions. I've been working on the following problem.

Let $F(\mathbb{R})=\{f:\mathbb{R}\to\mathbb{R}\}$ denote the collection of all real-valued functions defined on $\mathbb{R}$. Let $G$ be the group $G=(\mathbb{R},+)$ and consider the map $$G\times F(\mathbb{R})\to F(\mathbb{R})\hspace{0.5in}\mathrm{with} \hspace{0.5in} (T,f(x))\mapsto f(x+T).$$

(a) Determine the stabilizer of the function $g(x)=\sin(\pi x)$.

(b) Determine the set of fixed points of this action.

I have already proven to myself that the map is indeed an action. Also, I have work for both (a) and (b) above; I'm just not sure if it's right.

Work for (a): If $G$ is a group acting on a set $X$, then the stabilizer of $x$ is the subgroup $G_{x}$ defined by $G_{x}=\{g\in G\::\:g\cdot x=x\}$. Hence,

\begin{equation} \mathbb{R}_{\sin(\pi x)}=\{T\in\mathbb{R}\::\:\sin(\pi(x+T))=\sin(\pi x)\}=\{T\in\mathbb{R}\::\:T=2k,\:k\in\mathbb{Z}.\} \end{equation}

Work for (b): If $G$ acts on a set $X$, then the set of fixed points is the subset $X^{G}$ of $X$ defined by $X^{G}=\{x\in X\::\:g\cdot x=x,\:\mathrm{for\:all}\:g\in G\}$. Then in our case, $$F(\mathbb{R})^{\mathbb{R}}=\{f(x)\in F(\mathbb{R})\::\: f(x+T)=f(x),\:\mathrm{for\:all}\:T\in\mathbb{R}\}.$$ In other words, $F(\mathbb{R})^{\mathbb{R}}$ is the collection of all functions $f:\mathbb{R}\to\mathbb{R}$ which are periodic about $T\in\mathbb{R}$.

Does my work make sense, or are there some errors? Thanks in advance for any help!


For A, the stabilizer will be $2\mathbb{Z}$ as $x \mapsto \sin \pi x$ is periodic about $2$, as you stated.

For B, you have the correct set description, but came to the wrong conclusion. If for some $x$, $f(x+T) = f(x)$ for all $T \in \mathbb{R}$, then $f$ must be constant. Hence, the fixed points of the action are constant functions.

  • $\begingroup$ Thanks, that makes sense! I thought my description for (b) was off. $\endgroup$ – Sir_Math_Cat Jul 13 '18 at 2:46

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