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 Qs on extension of continuous iterations from analytic functs to non-analytic JmsNxn Ultimate Fellow Posts: 902 Threads: 111 Joined: Dec 2010 06/30/2022, 12:15 AM Oh my god, I'm stupid! I get it now, I thought it was something fancy. The double norm screwed me up, lmao. If that's the case, it cannot be analytic in $$z$$ as stated. It fails the Cauchy-Riemann equations by construction. It's a perfectly weird $$C^\infty$$ construction though. Weird looking. MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 07/01/2022, 12:13 AM Quote:Weird looking. Hey, I'll show you that is very natural as a construction, and it is something very general relating extensions on iterations and the glueing of iterations. I'm preparing a post explaining all that business from zero to hero. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ Leo.W Junior Fellow Posts: 47 Threads: 5 Joined: Apr 2021 07/05/2022, 12:27 PM (06/18/2022, 02:55 PM)MphLee Wrote: ...Reaaaally fascinates me! to rectify myself, I wrote $$\mathbb{C}^0$$ for discontinuous infinitely many "slices" of parellel lines, so that f cannot have a complex derivative, but does have a directional derivative. Your claim is right about that we'd apply the study of group or monoid structures to this but I now have more a strong (belief?)sense  that even that can be still inadequate, and we may need more complex algebra system to extend the iterations of a map, what I mean is that, for instance if we really do have re-defined the complex numbers as $$z=\{Re(z),Im(z),0\}\in\mathbb{R}^3$$ and then$$f^t: \{Re(z),Im(z),0\}\to\{Re(z),Im(z)cos(\pi t),sin(\pi t)\}$$ as a common function (linear transformation by rotation matrices) which map $$\mathbb{R}^3$$ to $$\mathbb{R}^3$$, then we can have naturally an extended iteration of conjugate function...... so I feel no surprise that there's limitation on the operators like discontinuous iterations, lol I may give up on continuous iterations on conj(z) only in $$\mathbb{C}$$ I consider that any partition that has the same in "count" (2 partitions has this "count" 2) can be transfered to each other if they can be measured, by some underlying function $$g:X_i\to Y_i$$ for different partition X and Y, or at least analogous. For the conj(z) as you partitioned the plane into $$\bigcup_{r\ge0} rS^1$$, and you brought up about K_r function, I think its structure can be similar to the iterations of $$f:[0,2\pi]\to[0,2\pi],f(\theta)\equiv-\theta \pmod{2\pi}$$, or maybe more $$f:\mathbb{R}\to\mathbb{R},f(x)=-x$$, the iterations then be transferred into a question that whether there's a continuous function $$F:\mathbb{R}\times[0,2\pi]\to[0,2\pi],F(t+s,\theta)=F(t,F(s,\theta)),F(0,\theta)=\theta,F(1,\theta)=2\pi-\theta$$, closely to a continuous iteration of $$f:\mathbb{R}\to\mathbb{R},f(x)=-x$$, I don't know if such function exists, but at least it can be shown that there is no solution to $$f(f(x))=-x$$ on real axis Regards, Leo JmsNxn Ultimate Fellow Posts: 902 Threads: 111 Joined: Dec 2010 07/07/2022, 12:55 AM (This post was last modified: 07/07/2022, 01:01 AM by JmsNxn.) Your posts are fantastic, Leo. As far as I'm concerned that's a novel proof of $$f(f(x)) = -x$$ can't be real valued. We have a couple of these problems on these forums. I hope you post more. You have a great brain. To iterate $$\text{conj(z)} = z^* = \overline{z}$$ you need to do Grothendiek level shit. Iterates only exist in exotic metric spaces, that are probably better represented using p-adics, or something like that. It'll never be analytic/continuous/differentiable. Nothing good is to come from looking at it like that. But as an abstract algebraic idea, it does exist. Assuming the axiom of choice there are uncountably many automorphisms of $$\mathbb{C}\to \mathbb{C}$$ as a field. But the only continuous one is conjugation and the identity mapping. Not to rain on your parade, but this question is pretty much settled. I admire your spirit. Love your energy. Regards, James MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 07/07/2022, 09:25 PM Meanwhile I have two minutes. Conj is a field automorphism of the complexes but that doesn't mean that we have to look for it's fractional iterates inside the group $${\rm Aut}(\mathbb C)$$ unless we ask for an $$\mathbb R$$-action over $$\mathbb C$$ that acts by automorphisms. Sure since conjugation is involution, hence isomorphism, we'd like its fractional iterates to be iso, since composition preserves bijectivity... and also because we'd like fractional iterations to respect the field structure...This however seems too restrictive. We'd get something very rigid and similar to a weakened scalar multiplication... Let $$\alpha(r,z)$$ be an action over the complexes: $$\alpha(0,z)=z$$ and $$\alpha(r+s,z)=\alpha(r,\alpha(s,z))$$ s.t. $$\alpha(1,z)={\rm conj}(z)$$. First notice that if we ask it to act by automorphisms then we induce a group homomorphisms $$\alpha:\mathbb R \to \rm{ Aut}(\mathbb C)$$. The "intrinsic time" of the iteration is given by the group quotient $$\mathbb R / {\rm ker}\alpha$$ where $$2\mathbb Z\subseteq {\rm ker}\alpha =\{k\in\mathbb R\, : \,\alpha(r,z)=z \}$$, since conjugation being an involution $${\rm conj}^{2n}={\rm id}_{\mathbb C}$$. If the kernel is the even numbers then the intrinsic iterates looks like the circle $$S^1 \simeq \mathbb R/2\mathbb Z$$. In other words we are looking for $$S^1$$-actions $$\alpha:{}S^1\times \mathbb C\to \mathbb C$$ over $$\mathbb C$$. I wonder what this entails at the level of James' remarks. If we consider also wild automorphisms... maybe this wold entails some contradiction... maybe idk. I was thinking about failure of injectivity in $$z$$ of iterates $$\alpha(r,z)={\rm conj}^r(z)$$ when the time belongs to the circle $$r\in S^1$$... while all of them should be injective since are field morphisms.... this seems fascinating but got no time to study it more. The second critical point is that we would also have $$\alpha(t,0)=0$$ and $$\alpha(r,z+w)=\alpha(r,z)+\alpha(r,w)$$, and $$\alpha(t,1)=1$$ and $$\alpha(r,zw)=\alpha(r,z)\alpha(r,w)$$. This seems too rigid to have any room to play around... probably it is too rigid that ruels out every non-trivial solution.... Just a feeling... MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 07/14/2022, 07:18 PM (06/26/2022, 05:37 AM)JmsNxn Wrote: Mphlee just made me more confused. Hi James, I apologize, I went to fast assuming too much. Idk what part was confusing so I'd start from zero and I'll try to bring you up to speed. I've made a long expository post that belongs to my note dumps on iteration and actions. GO HERE @Leo, also made this post for you too. The decomposition mechanism is really general, but is just pasting together known stuff. In the post linked you can find an introduction to it. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ Leo.W Junior Fellow Posts: 47 Threads: 5 Joined: Apr 2021 Yesterday, 01:49 PM (07/14/2022, 07:18 PM)MphLee Wrote: (06/26/2022, 05:37 AM)JmsNxn Wrote: Mphlee just made me more confused. Hi James, I apologize, I went to fast assuming too much. ... @Leo, also made this post for you too. The decomposition mechanism is really general, but is just pasting together known stuff. In the post linked you can find an introduction to it. Thx Mphlee, I sensed this by your previous posts, you're reaaaaaaaaaaally good at rigorous-ify random statements XD, excellant! Regards, Leo JmsNxn Ultimate Fellow Posts: 902 Threads: 111 Joined: Dec 2010 Yesterday, 11:34 PM (Yesterday, 01:49 PM)Leo.W Wrote: Thx Mphlee, I sensed this by your previous posts, you're reaaaaaaaaaaally good at rigorous-ify random statements XD, excellant! Mphlee is a wizard, he's going to do great things! « Next Oldest | Next Newest »

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