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 [question] Local to global and superfunctions MphLee Long Time Fellow    Posts: 321 Threads: 25 Joined: May 2013 07/10/2022, 01:41 PM (This post was last modified: 07/12/2022, 08:33 AM by MphLee.) This is meant to be a quick question to begin to rectify my pov, before I begin to study seriously what I need to study (Milnor and diff. geometry in general). Recently some remarks about non uniqueness were made. I had not the time to understand it properly but I have an image in my mind... something about the direction I should take once I make clear to myself the algebraic approach to iteration/superfunctions. Algebraic approach. this is very minimal and well behaved. An iteration is an homomoprhism $$f_{t+s}=f_s\circ f_t$$, while a generalized superfunction (an $$A$$-equivariant map) is a solution of the equation $$g_s\circ \chi=\chi \circ f_s$$. The problem about this approach is that often such solutions do not exists globally. For this reason I see how we can introduce something "new" like with atlases of manifolds. The question is: is this new? Is this exactly what you are all doing here? Adopting this pov can bring progress? Definition: let $$f:X\to X$$ be a function. Let $$S:A \to A$$ the topological domain of time. Define a superfunction atlas of $$f$$ to be an open cover $$U_i\subseteq A$$ of $$A$$ such each $$S(U_i)\subseteq U_i$$ and a family of superfunctions $$\Psi_i:U_i\to X$$ s.t. $$\Psi_i(S|_{U_i}(a))=f(\Psi_i(a))$$ Note that here it is possible that $$\Psi_i:U_i\to X$$ and $$\Psi_j:U_j\to X$$ may disagree on $$U_{ij}=U_i\cap U_j$$. In that case the atlas produces different iterations depending on the element of the cover chosen... but globally we are covering all of $$A$$. We may also ask that some coherence conditions like in atlases of coordinate charts. Example: We may ask that the restrictions $$\Psi_j |_{U_ij}:U_{ij} \to X$$ and $$\Psi_i |_{U_ij}:U_{ij} \to X$$ have nice properties, like glueing maps. Tbh we could also turn this definition upside down asking instead for a cover of $$(X,f)$$... but I'm not sure how to go because It is not clear to me what goals we would like to fulfill and what are the obstacles in the classical approach to locally/globally iterating stuff. ps: I feel that untangling this is the first step before bringing all the algebraic theory to a new level merging topological and analytical phenomena into a new abstract framework. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ JmsNxn Ultimate Fellow     Posts: 977 Threads: 114 Joined: Dec 2010 07/11/2022, 08:56 PM I'd say that's a very good minimalist description of what we do, lmao. It's also very important to add, that more times than not the different super functions do not agree. And if they do agree, it's probably a large global solution. This isn't really a mathematically quantifiable statement. But a lot of the time the super function either works for a tiny domain, or it works almost everywhere. Very rarely is there a middle ground. The trouble tends to be the global solutions are ugly and useless, the local solutions are awesome but so restricted they can become useless. Very rarely do we have some breakthrough case like Kneser which has the best of both worlds. I highly suggest a complex analysis book too, if you are going into milnor. It definitely helps to read complex analysis before hand, and at least have a moderate grasp of how it works. My vote, for a somewhat beginner text, is John B. Conway Functions Of One Complex Variable. Super clean and well written book (And as a fun piece of trivia it's a different John Conway that made game of life, this is John B Conway, lol) It's not my favourite book, but it'd be an easier read then Reinhold Remmert's two part series on complex analysis (which is a fucking beautiful tome). But it's very very German style math. Conway's book is lighter and more concise. MphLee Long Time Fellow    Posts: 321 Threads: 25 Joined: May 2013 07/12/2022, 09:01 AM I have that Conway... tbh honest I have a shit-ton of books... just in case... you know... xD Anyways... let's be sure I understand. I guess that we are not on the same page... What do you exactly mean when you say Quote:super function either works for a tiny domain, or it works almost everywhere Question: Assume $$S:A\to A$$,$$f:X\to X$$.... the two case you are describing can be formalized as..."works for a tiny domain"=the f.eq $$\chi S=f \chi$$ has a solutions only on some tiny sets $$U\subseteq A$$. I.e. we can only find functions $$\chi:U\to Y$$ satisfying $$\chi S=f \chi$$ only for "some small sets" $$U\subseteq A$$... "or it works almost eveywhere"=the f.eq $$\chi S=f \chi$$ has an ugly solution on some set $$A'=A\setminus D$$, where $$D$$ is very small. I.e. we can only find functions $$\chi:A'\to Y$$ satisfying $$\chi S=f \chi$$ except on a small set of points $$D$$, where $$A'=A\setminus D$$... I don't think that it is what you mean... or it is...? Maybe It is what you mean because when you say Quote:different super functions do not agree I read it as follows: Assume $$S:A\to A$$,$$f:X\to X$$ and $$U_0$$ and $$U_1$$ subsets of $$A$$, we know how to compute two functions $$\chi_0:U_0\to X$$ and $$\chi_1:U_1\to X$$ s.t. $$\chi_0 S|_{U_0}=f \chi_0\quad and\quad \chi_1 S|_{U_1}=f \chi_1$$ saying that they do not agree, to me can be expressed as... assume $$U_0\cap U_1$$ is non empty, the restrictions $$(\chi_0)|_{U_0\cap U_1}$$ is not the same as $$(\chi_1)|_{U_0\cap U_1}$$. In other words, exists an $$a\in U_0\cap U_1$$ s.t. $$\chi_0(a)\neq \chi_1(a)$$ This seems good but in my mind it is not consistent with your use of the phrase "superfunction about something (e.g a fixed point)". . In my mind this read as the superfunction $$\chi:A\to X$$ is about a neighborhood of a fixed point $$p\in X$$... so locality isn't about a small set in the domain of the superfunction but is about a small set in the codomain, i.e. $$X$$, where the fixed points of $$f:X\to X$$ live. Pleas help me make this clear. My mind is breaking around this terminology. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ JmsNxn Ultimate Fellow     Posts: 977 Threads: 114 Joined: Dec 2010 07/13/2022, 06:47 PM Lol, I apologize. I am rather loose with my language, so I'll try to be clearer. The answer to your first question is no. I did mean in the codomain. So when I said we get a small domain, we usually get a super function $$F : \mathbb{C}_{\Re(z) > 0} \to \mathcal{U}$$ where $$\mathcal{U}$$ is small. I apologize, that's what I meant. I meant we can get a local solution, so that it's in a tiny domain. For example take the Schroder iteration about $$|\lambda| < 1$$ for a fixed point $$p$$, and assume $$\lambda$$ is complex valued. There exists a half plane $$\mathcal{H}$$ and a neighborhood $$U$$, such that we can easily construct $$F: \mathcal{H} \to U$$. But as we try to pull back, extend $$\mathcal{H}$$, we encounter loads of problems. So this is still what I'd refer to as a local solution, largely because: $$F(z) = p + \lambda^z \xi_0 + \mathcal{O}(\lambda^{2z})\\$$ Is a very accurate approximation. So when I mean local, I'm referring to the codomain primarily, but remember that $$F(z)$$ is also a function of $$\xi_0$$, and you can move $$\xi_0$$ around, and it too usually lives in the codomain. So, when I refer to local solutions I mean: $$F(z,\xi) : \mathcal{H} \times U \to U$$, and $$U$$ is the "local" in a local iteration. Then similarly, when I mean a global iteration, I mean $$F(z,\xi) : \mathcal{H} \times A \to A$$ where $$A$$ is a "large" set let's say.  Now this will usually involve growing the domain $$\mathcal{H}$$ to be larger than a half plane. For example with the Schroder iteration, just as above, if $$A$$ is the immediate basin of $$p$$, then $$F$$ can always exist as above. It can actually be extended even further for $$\mathbb{C}/B$$ for some branch cuts $$B$$ (area measure zero). This is what I would call a global solution. Another example of global would be kneser's tetration, because it's holomorphic almost everywhere on $$\mathbb{C}$$, yet also having a huge codomain. For transcendental entire functions $$f$$, these ideas are fairly synonymous because any super function $$F$$ defined for $$\mathbb{C}/B$$ for some measure zero set $$B$$, will send to a rather large domain. This is intrinsic to how the dynamics of transcendental entire functions behave. If $$f$$ has a geometrically attracting fixed point at $$p$$, then $$f^{-1}$$ has a geometric repelling fixed point at $$p$$, and is in the julia set. So as we take $$F(z-n)$$, we are iterating $$f^{-1}$$ about the Julia set, which is always crazy chaotic, and approaches infinity. So the two ideas are fairly close "large domain in z" and "non local codomain". I apologize for the confusion, I can get ahead of myself sometimes. I hope that clears things up? MphLee Long Time Fellow    Posts: 321 Threads: 25 Joined: May 2013 07/15/2022, 04:17 PM (This post was last modified: 07/15/2022, 04:18 PM by MphLee.) Thanks James I suspected that, I'm starting to think that local in state space is linked to local in the time space... I just need to get this straight since this is crucial... your previous post about how not every superfunction is an iteration... this is all linked... I'm collecting items in order to paint a complete map of the landscape. This is like passing from affine varieties to manifolds... from curves to Riemann manifolds. This is fundamental, I'll get back to this... everything said will stay here waiting to click something in my brain. btw... can you pls expand  a bit on this Quote:So the two ideas are fairly close "large domain in z" and "non local codomain". MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ JmsNxn Ultimate Fellow     Posts: 977 Threads: 114 Joined: Dec 2010 07/15/2022, 06:07 PM (This post was last modified: 07/15/2022, 10:46 PM by JmsNxn.) (07/15/2022, 04:17 PM)MphLee Wrote: Quote:So the two ideas are fairly close "large domain in z" and "non local codomain". Just note that I mean this statement fairly imprecisely. That is to mean it's a loose description of what things look like. I like to think using half planes, so let's use half planes. Take an arbitrary local iteration: $$f^{\circ s}(z)$$, and let's let $$\Re(s) > K$$ for $$K$$ large. Let's call the fixed point $$p$$, and the multiplier $$\lambda$$--which for expositional purposes we'll assume is between $$0 < \lambda < 1$$. Then: $$|f^{\circ s}(z) - p| \le r^K\\$$ For some $$0 < |\lambda| < r < 1$$. So the codomain of this function looks a lot like a disk about $$p$$. We haven't mentioned $$z$$ yet, but since we've defined our codomain, let $$z$$ exist there too; inside this tiny disk. Call it $$E$$. Now this is still a very large domain that $$s$$ lives in, it is a half plane. But! It's a rather restrictive half plane. What happens when we limit $$K \to -\infty$$? We'll we are effectively taking $$f^{-1}$$ repeatedly of this little disk. This little disk behaves oddly underneath $$f^{-1}$$ though. First of all, since $$f$$ has an attracting fixed point at $$p$$, we must have $$f^{-1}$$ has a repelling fixed point. Any repelling fixed point is in the Julia set of its function, so call $$J$$ the julia set of $$f^{-1}$$, then $$J \cap E \neq \emptyset$$.  Here is where things get interesting. Pick a neighborhood $$\mathcal{N}$$ of a point $$p$$ in the julia set $$J$$; then the orbits $$\bigcup_{n \ge 0} f^{-n}(\mathcal{N})$$ is dense in $$J$$. Now what is very common, is that $$J$$ includes the point at infinity, this is almost always, though there are exceptions. So this means, as we take repeated $$f^{-1}$$'s we are getting closer and closer to eventually hitting infinity (on the Riemann sphere I mean)--or we hit wild essential singularity chaos. All this is happening as we are letting $$K \to -\infty$$, as we are growing the half plane (and by consequence growing the codomain from a tiny disk to a wild chaotic fractal that will eventually include infinity, or at least a singularity). I guess what I'm trying to say, is there is a correlation to just how local we are, and the size of the domain of $$s$$. The more it looks like a disk, the "smaller" the half plane of the iteration, the more local it is. The more it looks like chaos and fractally, the "larger" the half plane of the iteration, the less local it is. You can see this in almost all of our iterations too, they happen even with Kneser which is pretty far from a fixed point iteration. But for $$\Re(s) < -K$$ and $$\Im(s) > 0$$, then more and more tetration just looks like $$L + e^{Ls}$$. So if you were to do a local tetration in this quarter plane with very large $$K$$, the codomain would just look like a disk around $$L$$.  But as you grow the domain of $$s$$, we let $$K \to -\infty$$, the more chaotic we get, and more and more we get closer to the codomain being exactly $$\mathbb{C}$$. MphLee Long Time Fellow    Posts: 321 Threads: 25 Joined: May 2013 07/16/2022, 04:49 PM K, I have to slow down. finally in Devaney I'm starting the chapter on complex dynamics... Finally I'll get what those Julia set are... I hope... but before I'll have to read again two chapters of Devaney that shocked me... the ones about attractors and the one about Sarkovskii's theorem... I'm at chapter 13 but need to read again most of it... since I was too fast to digest everything. 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 07/16/2022, 09:03 PM (This post was last modified: 07/16/2022, 09:49 PM by MphLee.) OMG... I was able to parse your first two paragraphs... that inequality is incredible... that means that the superfunction shrinks that right halfplane into a small disk around the fixed point...... so... more we extend to the left, bigger smaller $$\Re s$$, more the disk we are sending the half-plane to should grow...in the other direction, more is on the right our halfplane, the bigger is the real part and more we are approaching the attracting fixed point....more the halfplane is sent near the fixed point because the $$f^{\Re s}$$ in $$f^s(z)=f^{\Re s}f^{\Im s}(z)$$ dominates the behavior... This all makes very sense... I can visualize it... I think I have to wait until I discover what Julia sets are... and study again what means for orbit to be dense... (Devaney uses this conditions to characterize chaos)... Pls give me time... I need to absorb all of this. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ JmsNxn Ultimate Fellow     Posts: 977 Threads: 114 Joined: Dec 2010 07/17/2022, 06:46 AM Yes, Mphlee Also remember I chose $$0 < \lambda < 1$$ in this example. If $$0 < | \lambda| < 1$$, then the half plane $$H$$ is rotated such that $$|\lambda^z| < 1$$. So it no longer becomes the real part dominates it, you have to do a slight coordinate change first. Exact same principles across the board though. You can still bound based on the real argument, you just have to be more careful, because the imaginary argument can grow unbounded. « Next Oldest | Next Newest »

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