Something similar is doable for \(|\cdot|:\mathbb C\to\mathbb C\). This is idempotent because it decompose into \(|\cdot|:\mathbb C\to\mathbb R^+\to \mathbb C\) where the second arrow is an inclusion. So every eq. classes of the polar decomposition maps to a single point, the fixed point. It means that we cannot extend it to a \(\mathbb Z\) iteration, hence \(\mathbb R\) iterating it is impossible. Only iteration where the time \(f^t\) belongs to a monoid (has noninvertible elements) is possible.
The time we can iterate on could be the monoid of non-negative reals \(\mathbb R_0\) quotiented by the relation \(t\sim s\) iff \(t,s\in [1,\infty) \) and if \(s=t\) otherwise. This amounts to collapsing together all the reals bigger than one (cauz of idempotency of the complex norm), i.e. we are considering the closed unit interval \([0,1]\) equipped with the operation \(s{\bar {+}}t={\rm min}(s+t,1)\). Call that object \(I=\mathbb R_0/\sim\).
We can \(I\)-iterate the function \(|\cdot|:\mathbb C\to\mathbb C\). On each class of the partition \(rS^1=\{re^{i\theta}\}\) we consider the endomap \( f_r(z)=f_r(re^{i\theta})=r \).
The way i'd go is by selecting a family of continuous paths \(\gamma(-,\theta): [0,1]\to (-\pi,\pi]\) s.t \(\gamma(0,\theta)=\theta\) and \(\gamma(1,\theta)=0\) and such that \(\gamma(s{\bar {+}}t,\theta)=\gamma(s,\gamma(t,\theta))\) where we use the operation \(s{\bar {+}}t={\rm min}(s+t,1)\).
Take a whole family \(\gamma_r (t,\theta)\) and define \(f_r^t:rS^1\to rS^1\) as \[f^t_r(re^{i\theta})=|z|e^{ i\gamma_r(t,\theta) }\] where \(t\in I\) is in the quotient monoid we have defined before. Consider then \(f\) as \[f^t(z)=f^t_{{\rm arg}(z)}(z)\]
Questions
The question is thus reduced to the existence of at least one of such "homotopies" \(\gamma(-,\theta): [0,1]\to S^1\). How to do it? Do they exists?
Maybe we can consider \(\gamma(t,\theta)=(1-t)\theta\) but then \( \gamma(t, \gamma(s,\theta)=(1-t)(1-s)\theta=(1-(t+s)+ts)\theta =\gamma(t+s,\theta)+ts\theta\). Something is clearly off. Also, we lose continuity at -1, because all the points in the upper half plane get shrunk clockwise, and the one in the bottom go counterclockwise.
What about using the same rotation everywhere? We place our discontinuity at \(1\in S^1\)?
What if instead we just subtract angles? Consider \(\beta(t,\theta)={\rm max}(\theta-t2\pi, 0) \) then le'ts check the iteration property:
\(\beta(t,\beta(s,\theta))={\rm max}({\rm max}(\theta-s2\pi, 0)-t2\pi, 0)\) by distributivity we get
\(\beta(t,\beta(s,\theta))={\rm max}( {\rm max}(\theta-s2\pi-t2\pi, -t2\pi), 0)\), by some algebra
\(\beta(t,\beta(s,\theta))={\rm max}( {\rm max}(\theta-(s+t)2\pi, -t2\pi), 0)\), now there are two cases: if \(\theta-(s+t)2\pi< -t2\pi\leq 0\) then the overall result is zero because \(t\in [0,1]\), otherwise the first term wins so
\[\beta(t,\beta(s,\theta))={\rm max}( {\rm max}(\theta-(s+t)2\pi, 0)=\beta(t+s,\theta)\]
To complete the proof that we have an \(I\)-iteration just note how when \(1<s+t\) we have \(\beta(t+s,\theta)=\beta(1,\theta)\). The reason is because if \(1<s+t\) then \(\theta-(s+t)2\pi< 0\) thus \[ \beta(t\bar{+} s,\theta) = \beta(t,\beta(s,\theta)) \]
We have a monoid homomorphism \(\beta:I\to {\rm End}(S^1)\), i.e. an \(I\)-iteration or \(I\)-action over the circle.
The time we can iterate on could be the monoid of non-negative reals \(\mathbb R_0\) quotiented by the relation \(t\sim s\) iff \(t,s\in [1,\infty) \) and if \(s=t\) otherwise. This amounts to collapsing together all the reals bigger than one (cauz of idempotency of the complex norm), i.e. we are considering the closed unit interval \([0,1]\) equipped with the operation \(s{\bar {+}}t={\rm min}(s+t,1)\). Call that object \(I=\mathbb R_0/\sim\).
We can \(I\)-iterate the function \(|\cdot|:\mathbb C\to\mathbb C\). On each class of the partition \(rS^1=\{re^{i\theta}\}\) we consider the endomap \( f_r(z)=f_r(re^{i\theta})=r \).
The way i'd go is by selecting a family of continuous paths \(\gamma(-,\theta): [0,1]\to (-\pi,\pi]\) s.t \(\gamma(0,\theta)=\theta\) and \(\gamma(1,\theta)=0\) and such that \(\gamma(s{\bar {+}}t,\theta)=\gamma(s,\gamma(t,\theta))\) where we use the operation \(s{\bar {+}}t={\rm min}(s+t,1)\).
Take a whole family \(\gamma_r (t,\theta)\) and define \(f_r^t:rS^1\to rS^1\) as \[f^t_r(re^{i\theta})=|z|e^{ i\gamma_r(t,\theta) }\] where \(t\in I\) is in the quotient monoid we have defined before. Consider then \(f\) as \[f^t(z)=f^t_{{\rm arg}(z)}(z)\]
Questions
The question is thus reduced to the existence of at least one of such "homotopies" \(\gamma(-,\theta): [0,1]\to S^1\). How to do it? Do they exists?
Maybe we can consider \(\gamma(t,\theta)=(1-t)\theta\) but then \( \gamma(t, \gamma(s,\theta)=(1-t)(1-s)\theta=(1-(t+s)+ts)\theta =\gamma(t+s,\theta)+ts\theta\). Something is clearly off. Also, we lose continuity at -1, because all the points in the upper half plane get shrunk clockwise, and the one in the bottom go counterclockwise.
What about using the same rotation everywhere? We place our discontinuity at \(1\in S^1\)?
What if instead we just subtract angles? Consider \(\beta(t,\theta)={\rm max}(\theta-t2\pi, 0) \) then le'ts check the iteration property:
\(\beta(t,\beta(s,\theta))={\rm max}({\rm max}(\theta-s2\pi, 0)-t2\pi, 0)\) by distributivity we get
\(\beta(t,\beta(s,\theta))={\rm max}( {\rm max}(\theta-s2\pi-t2\pi, -t2\pi), 0)\), by some algebra
\(\beta(t,\beta(s,\theta))={\rm max}( {\rm max}(\theta-(s+t)2\pi, -t2\pi), 0)\), now there are two cases: if \(\theta-(s+t)2\pi< -t2\pi\leq 0\) then the overall result is zero because \(t\in [0,1]\), otherwise the first term wins so
\[\beta(t,\beta(s,\theta))={\rm max}( {\rm max}(\theta-(s+t)2\pi, 0)=\beta(t+s,\theta)\]
To complete the proof that we have an \(I\)-iteration just note how when \(1<s+t\) we have \(\beta(t+s,\theta)=\beta(1,\theta)\). The reason is because if \(1<s+t\) then \(\theta-(s+t)2\pi< 0\) thus \[ \beta(t\bar{+} s,\theta) = \beta(t,\beta(s,\theta)) \]
We have a monoid homomorphism \(\beta:I\to {\rm End}(S^1)\), i.e. an \(I\)-iteration or \(I\)-action over the circle.
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)
