I need to find a totally discontinuous $D_4$ action on $S^\infty\subset L^2([0,1])$.
Background: Here $L^2([0,1])=\{\text{measurable }f: [0,1]\to \mathbb{C} : \int |f|^2<\infty\}$, $S^\infty:=\{f\in L^2([0,1]) : ||f||_{L^2}=1\}$ and $D_4$ is the dihedral group with 8 elements. To say $D_4$ acts "totally disconinuously" on $S^\infty$ is to say that for all $x\in S^\infty$, there exists open $U\ni x$ such that $gU\cap U\neq \emptyset \Rightarrow g=e$. Basically, this just means that every non-identity element moves every element in $S^\infty$.
My attempt: So $D_4$ is generated by $r$ and $s$ where $r^2=s^4=(rs)^2=e$. The natural choice for the action of $r$ on $S^\infty$ is $rf\to -f$. This map is (lipschitz) continuous and clearly moves every element. Now I was trying to find transformation of order 4 that moves every element of $S^\infty$ which also satisfies $(rs)^2=e$. Here are some things I have tried: Some possible actions of order 4 are $sf=if$ , $s f=g$ where $g(t)=f(1-t)$, $sf=\overline{f}$. But all of these commute with $r$. I then tried to do one thing to $f$ on half of the interval and something else to $f$ on the other half, but nothing that I could find works. I am wondering now If I chose the wrong action for $r$ but $rf=-f$ feels to me like the only natural choice. I would appreciate any sort of hint!
Edit: the maps $sf=g$ and $sf=\overline{f}$ specified above are not order 4...I meant something more like $sf=ig$ and conjugating then switching the real and imaginary parts.
Second edit: It is possible that this cannot be done... I have my doubts
Here is a partial answer: Take $rf=- \bar{f}$ and $sf=if$, then they are order $2$ and $4$ respectively and do not commute. In fact: $$ s^{-1}r(f) = s^{-1}(-\bar{f}) = i\bar{f} = r(if) = rs(f) $$ So they generate a copy of the dihedral group. However, the action is not totally discontinuous since $r$ has the imaginary functions as fixed points. I haven't found a way to improve this and am not sure this would be of interest to you.