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 Generalized Kneser superfunction trick (the iterated limit definition) MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 02/27/2021, 11:14 AM (This post was last modified: 02/28/2021, 08:31 PM by MphLee.) PROGRESS UPDATE (Feb 27, 2021): What you suggest seems reasonable. I'm drafting a comprehensive paper/post about the theory of those spaces from a general and historical point of view. What I'm posting in the thread Doubts on the domains of Nixon's method, (February 27, 2021) was part of that but at the end I convinced myself to cut out the questions and my doubts and post them separately. I'll use your feedback to tune the paper. Right now it is a 7-page paper with a detailed and commented bibliography. I hope it will be a valuable reference for the forum in general. ps: I'm sorry to be that slow but I got lot of work to do. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 03/19/2021, 11:10 PM (This post was last modified: 03/19/2021, 11:17 PM by MphLee.) PROGRESS UPDATE (Mar 19, 2021): Completed the draft of the introductory section on superfucntion complete spaces, gatheret some bibliography. The grammar needs alot of work and a deep revision is missing. This is just a preliminary draft. I'm currently working on discussing possible fixes for JmsNxn's idea about diffeomorphisms, bridging the gaps to superfunctions by limit tricks (principal solutions) and Nixon's trick. I've also worked out an interesting algebraic rephrasing of piecewise extension and a novel idea (with proofs) of a canonical extension of iteration from  the naturals (the cyclic subgroup) to the centralizer of the imput map. I'll add a sketch of possible abstract treatment of hyperoperations. When finished, I'll post something more tidy and expand more on JmsNxn comments.   Draft.pdf (Size: 676.85 KB / Downloads: 176) [WARNING: grammar and formulas were NOT doublechecked, read at you own risk] MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 03/20/2021, 07:18 PM Very, very interesting! I really enjoyed that, MphLee (by the way, what is your actual name? I feel odd you knowing mine and me not knowing yours, lol. I keep thinking it's Murphy Lee (but that's probably wrong)). I really liked the way you framed the problem--it was a very good broad stroked impression of what's going on! I'm excited to see what else you can add. Especially when you start adding category theory. I'm in the process of writing a quick analysis on making $\mathcal{C}^1$ conjugacy equations, as I think I know enough now to give examples of these things working. It's only a brief write up, I'm sure I'll have it done in a couple of days. It's mostly reusing proofs I already have, so it should be a quick write up. Regards, James. JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 03/20/2021, 09:55 PM (This post was last modified: 03/21/2021, 02:24 AM by JmsNxn.) Hey, I was inspired by your little notice, so I thought I'd give an example of your black-box convergence at work. It's not perfect, wrote it in about three hours, but I think it's a good example of what you're driving at. It nearly constructs a set, $ \mathcal{B} \subset \mathcal{C}^1(\mathbb{R},\mathbb{R})\\$ That satisfies the conjugacy property, $ \forall f,g \in \mathcal{B}\,\,\exists \phi \in \mathcal{B}\,\,\text{s.t}\,\,g(\phi(x)) = \phi(f(x))\\$ I couldn't achieve this, unfortunately. But I got very very close. If $\overline{\mathcal{B}}$ is the closure of $\mathcal{B}$ as a monoid, then, $ \forall f,g \in \mathcal{B}:\,\,\exists \phi \in \overline{\mathcal{B}}\,\,\text{s.t}\,\,g(\phi(x)) = \phi(f(x))\,\,\text{or}\,\,\exists \phi \in \overline{\mathcal{B}}\,\,\text{s.t}\,\,f(\phi(x)) = \phi(g(x))\\$ Which is so close to finding a set of functions satisfying the conjugacy property. The problem being when, $ \lim_{x\to\infty} \frac{f(x)}{g(x)} = 1\\$ Which kind of throws a wrench in the construction; but only a small one--we just need to let $\phi$ be in the closure of our monoid. I'm very confident we can use the closure of $\mathcal{B}$ as our desired set, but I'm a little wonky on how to construct superfunctions for every element in the closure in a nice clean manner as I did for all of $\mathcal{B}$ as a whole. In this space we'll find that the successor operation $s \in \overline{\mathcal{B}}$ but not in $\mathcal{B}$. This is great progress though, I think everything is coming together on trying to solve conjugacy equations in a differentiable manner. I'm going to take a closer look at $C^k$ solutions now. Regards, James PS I also think it's important to add, if we call $\mathcal{B}^{-1}$ the set of functional inverses of $\mathcal{B}$ then I believe that, $ \mathbb{B} = \overline{\mathcal{B}} \cup \overline{\mathcal{B}^{-1}}$ Will be our desired group with a conjugacy property $\forall f,g \in \mathbb{B}\,\,\exists \phi \in \mathbb{B}\,\,\text{s.t.}\,\,f\phi = \phi g$. But, I'm wary on how to show this precisely. I think I'm 90% there though. EDIT: I did notice a couple of typos in the PDF, please ignore them. I wrote this pretty quick. Especially when constructing the super function $F(x)$ and when I use the Lipschitz condition. Recall that, $ F_n(x) = f^{\circ -n} (\Phi(x+n)) = \Phi(x) + \tau_n(x)\\$ Which is a small expression I forgot to add, upon which, $ |F_{n+1}(x) - F_n(x)| \le \lambda^n|f^{-1}(\Phi(x+n)) - \Phi(x+n)|\\$ and not what I wrote, I wrote it a tad too fast. But this function does look like $x+n$, minus some error in growth, but it doesn't really matter because $\lambda^n$ will still do the job it's supposed to do. Also, this is still going to need a good amount of finesse, I'm trying to think of alternative convergence factor functions than $e^x$ (of which using $e^{\mu x}$ is necessary for $\mu > 0$ in the general case, which I didn't add as of yet), as I think this might produce problems with simple functions like $f(x) = \lambda\cdot(x+1)$ for $\lambda > 1$ unless we choose an appropriately small enough $\mu$ dependent on $\lambda$ ($\frac{e^{\mu}}{\lambda} < 1$). EDIT: Argh, this is becoming more and more frustrating, I'm having trouble making the construction work for $f'(x) \to A >1$ as $x\to\infty$; which in turn means that $\phi$ may NOT belong in the closure of $\mathcal{B}$. God damn this is hard. The exponential convergents seem to only work when $f'(x) \to \infty$. EDIT: Woohoo! I fixed the problem I was facing. The PDF I sent is slightly incorrect, but I believe I have the right result now. I made a bit too fast of a jump when considering $f'(x) \to A > 1$ as $x \to \infty$; we have to handle this case in a more special way, without using exponential convergents. So take the construction of the superfunction in this paper as only for the case $f'(x) \to \infty$ and everything is still pretty much the same. I'll post an update in a bit. Attached Files   Conjugacy_in_C_1(1).pdf (Size: 230.61 KB / Downloads: 139) MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 03/21/2021, 01:56 AM Thank you! I find your effort very precious. I'll stop working on the superfunction spaces paper and I'll investigate you construction in detail ASAP, step by step, merging it with the information you gave me in the other post (the one about diffeomorphisms). Having this concrete model seems important enough to me to pause the development the algebraic side. This time the analysis you're using seems not to far above my head, but I need some stretching. ps: I apologize if this looks odd. I don't value my contribute here up to now here enough to reveal my real name, secondly I prefer to do not reveal publicly my identity on the web (but pms can be another story ). Best regards MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 03/21/2021, 07:31 AM Hey, MphLee Give me at most a couple of days, and I'll make this conjugate property absolute on $\mathcal{B}$. I over stepped my bounds in the above PDF.  But I know why, and I know exactly how to solve it. It just takes a more strict proof in the construction of the super function. Wait to rewrite or re-evaluate everything until I get it out how I meant it. And all I think this as is an example of your black box. Regards,  James MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 03/21/2021, 12:22 PM (This post was last modified: 03/22/2021, 11:32 AM by MphLee.) Ok, but I'm sure more insight into a concrete model always improves the abstract one. Until now I only played with finite models (small finite sets of permutation/functions) or with monoid of functions over N. I'm not on it seriously yet, I'm busy with some work. But I have few quick questions. 1- With closure you mean: let $M$ be a monoid and $B\subseteq M$ be a subset (a simple subset not a submonoid). With $\overline{B}$ you mean the smaller submonoid of $M$ that contains $B$ right? In other words it is the submonoid generated by the set $B$. In this case $M={\mathcal C}^1({\mathbb R})$. 2- Is your second condition $\lim_{x\to+ \infty}f'(x)\in (1,+\infty]$? Intuitively can I see this like the graph of f being "convex" in a neighborhood of +infinity? Except for the +infinite case this seems a kind o "linearity at infinity". 3- the third condition seems to impose that the dynamics of our maps is very simple: it means that "all orbits come from the same source" and the source is external to our domain. Why do we need this? Is this to avoid fixed points and to avoid that superfunction gets a lower bound, thus non surjectivity? Excuse me, always trivial things. I know. Addendum 1: I guess the right contruction should be ${\mathbb B}= \overline{B\cup B^{-1}}$. Given $\psi\in {\mathbb B}$ then, by definition, $\psi={\Omega_{i=0}^n}f_i$ for a finite n and $f_i\in B\cup B^{-1}$. We have by anticommutativity of inversion $\psi^{-1}={\Omega_{i=0}^n}f_{n-i}^{-1}$. This expression has meaning because by definition each $f_i$ is invertible and $f_i^{-1}\in B\cup B^{-1}$. We conclude that $\psi^{-1}\in {\mathbb B}$ QED Addendum 2: if $B$ is a subset of ${\mathcal C}^1({\mathbb R})$ then $B^{-1}$ is a subset of ${\mathcal C}^1({\mathbb R})$ iff elements of $B$ are diffeomorphisms. I think the proof is trivial (but I should write it down). In that case ${\mathbb B}= \overline{B\cup B^{-1}}$ is a subgroup of the group ${\rm Diff}({\mathbb R})$. ps: Your pdf seems very excititng. I need to read it carefully. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 03/22/2021, 08:45 PM (This post was last modified: 03/22/2021, 10:29 PM by JmsNxn.) 1. By closure in the monoid, I mean specifically closure under Cauchy sequences using uniform convergence on all compact subsets. By this I mean, $ f \in \overline{\mathcal{B}}\,\,\text{implies}\,\,\exists f_n \in \mathcal{B}\,\,\text{s.t.} \forall [a,b] \subset \mathbb{R}\,\,||f(x) - f_n(x)||_{x \in [a,b]} \to 0\,\,\text{as}\,\,n\to\infty\\$ This should still be a monoid; and will look close enough to $\mathcal{B}$ for our purposes. 2. I'm not too sure how convexity enters the discussion. I've never used convexity really. The intuitive manner I view this is as an attracting fixed point at infinity. Where $f'(x) \to \lambda < \infty$ is geometrically attracting, and $f'(x) \to \infty$ as a super-attracting fixed point. 3. Yes, it is to avoid fixed points, keep the orbits simple, and keep things well adjusted. Which will in turn ensure that our superfunction is surjective. Which makes talking about the conjugate property much simpler. A1. Yes, that's my thinking exactly. A2. Yes, I did this construction in hopes this would be a diffeomorphic group (whatever the hell you call it), a subgroup of $\text{Diff}(\mathbb{R})$. I avoided this language largely because I'm not familiar enough with the language. I have attached a revamped version of this PDF. I realized that the exponential convergents will only work when $f(x) \ge e^{\mu x}$ for some $\mu > 0$ and everywhere else become sort of overkill and don't help us. So instead I decided to use convergents $p(x-j)$ where $p(x) \sim \frac{1}{\sqrt{f(x)/x}}$ as $x \to \infty$. This requires us taking our auxiliary function $\Phi(x)$ to satisfy $\Phi(x+1) \sim f(\Phi(x))/\sqrt{f(x)/x}$. This means that $\Phi(x)$ will now grow slower than our superfunction rather than faster. When I was initially designing this I was thinking too much about $\phi(s)$ which grows faster than tetration; and so I assumed we need our function $\Phi(x)$ to grow faster than our superfunction. This proves to fail, especially when we take simple functions like $f(x) \sim x^{1+\delta}$ for $x \to \infty$; where the exponential convergents don't work, and any convergent which makes $\Phi(x)$ grow faster than the super function also won't work. Anyway, here it is revamped a good amount. I'm still fiddling with this, I want it to work more rigorously. I think this has potential for a very interesting paper, so this will probably be the last update until I'm fully satisfied. I'm mostly just posting this here to correct my assumption about exponential convergents being the give all end all. We unfortunately have to use more complicated convergents in the general case. Regards, James EDIT: I think I'm going to alter the manner of convergence from what I've posted here, as I think I can choose a more universal form of the convergents; which works for pretty much every function; rather than brute forcing it based on each $f$. I'll see though. Give me a week or so and I'll try to have a comprehensive first draft. Attached Files   Conjugacy_in_C_1(4).pdf (Size: 246.74 KB / Downloads: 124) MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 03/22/2021, 11:01 PM (This post was last modified: 03/23/2021, 04:01 PM by MphLee.) (03/22/2021, 08:45 PM)JmsNxn Wrote: 1. By closure in the monoid, I mean specifically closure under Cauchy sequences using uniform convergence on all compact subsets. By this I mean,Ok, I was suspecting it but I was misguided by you emphasis on " closure of B AS a monoid". With closure as a monoid I think algebraically. I take a subset of the ambient monoid and perform the closure. Instead you are considering the closure as a subset of a topological space or a kind of completion in respect to the ambient topology, i.e. that of C^1( R). Quote:This should still be a monoid; and will look close enough to $\mathcal{B}$ for our purposes. If your $\mathcal B$, as you defined it, is a monoid, and I can se an informal proof at page 2, I don't see why the Cauchy-completion of the set of elements of this monoid should be a monoid too. Even if it is, the union of two monoids need not to be a monoid: we must do set-union on their sets and then generate the monoid from that union, i.e. perform the, this time monoid-theoretic, closure of the set under composition. From now on I'll denote with $\overline{\mathcal B}$ your Cauchy-completion and with $\langle \mathcal B\rangle$ the monoid-completion of the set (the monoid freely generated by that elements). My proof, I'm enough sure of it is the following corollary 2: Definition 1: Let $M$ be a monoid, let $X\subseteq M$ be a generic subset. Define the submonoid $\langle X\rangle$ generated by $X$ $\langle X\rangle:=\{\Omega_{i=0}^nf_i\in M \,| f_i\in X;n\in{\mathbb N} \}$ Definition 2: Let $M$ be a monoid, let $X\subseteq M$ be a subset of invertible elements of $M$. Define the set $X^{-1}:=\{f^{-1}\in M \,| f\in X\}$ Observation: Let $M$ be a monoid, let $X,Y\subseteq M$  be subsets or submonoids. The union $X\cup Y$   is not generally a monoid. Corollary 1: Let $M$ be a monoid, let $X\subseteq M$  be a subset. If $X$ is already a (sub)monoid then $\langle X\rangle=X$. Corollary 2: Let $M$ be a monoid, let $X\subseteq M$ be a subset of invertible elements of $M$. The submonoid $\langle X\cup X^{-1}\rangle$ is a group. (proof in the previous post). Questions: let $X\subseteq {\mathcal C}^1({\mathbb R})$ a submonoid. When $\overline{X}$ is monoid? If $X$ is made of invertible functions, does  $\overline{(X^{-1})}=(\overline{X})^{-1}$ ? Answers to those questions can inform us on the properties of objects like $\langle \overline{X}\cup \overline{X}^{-1}\rangle$, $\langle \overline{X\cup X^{-1}}\rangle$ or $\overline{\langle X\cup X^{-1}\rangle}$. Quote:A2. Yes, I did this construction in hopes this would be a diffeomorphic group (whatever the hell you call it), a subgroup of $\text{Diff}(\mathbb{R})$. I avoided this language largely because I'm not familiar enough with the language. Rememeber that I would not call a group diffeomorphic like it is a property of the group. I call it a group of diffeomorphism, like a group of apples or a group of permutations. "The diffeomorphisms group of R" just means "the group made of diffeomorphism of R to itself". I'm not aware if a grop can be diffeomorphic in any sense (and idc right now). MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 03/22/2021, 11:30 PM (This post was last modified: 03/22/2021, 11:36 PM by JmsNxn.) (03/22/2021, 11:01 PM)MphLee Wrote: (03/22/2021, 08:45 PM)JmsNxn Wrote: 1. By closure in the monoid, I mean specifically closure under Cauchy sequences using uniform convergence on all compact subsets. By this I mean,Ok, I was suspecting it but I was misguided by you emphasis on " closure of B AS a monoid". With closure as a monoid I think algebraically. I take a subset of the ambient monoid and perform the closure. Instead you are considering the closure as a subset of a topological space or a kind of completion in respect to the ambient topology, i.e. that of C^1( R). Quote:This should still be a monoid; and will look close enough to $\mathcal{B}$ for our purposes. If your $\mathcal B$, as you defined it, is a monoid, and I can se an informal proof at page 2, I don't see why the Cauchy-completion of the set of elements of this monoid should be a monoid too. Even if it is, the union of two monoids need not to be a monoid: we must do set-union on their sets and then generate the monoid from that union, i.e. perform the, this time monoid-theoretic, closure of the set under composition. From now on I'll denote with $\overline{\mathcal B}$ your Cauchy-completion and with $\langle \mathcal B\rangle$ the monoid-completion of the set (the monoid freely generated by that elements). My proof, I'm enough sure of it is the following corollary 2: Definition 1: Let $M$ be a monoid, let $X\subseteq M$ be a generic subset. Define the submonoid $\langle X\rangle$ generated by $X$ $\langle X\rangle:=\{\Omega_{i=0}^nf_i\in M \,| f_i\in X;n\in{\mathbb N} \}$ Definition 2: Let $M$ be a monoid, let $X\subseteq M$ be a subset of invertible elements of $M$. Define the set $X^{-1}:=\{f^{-1}\in M \,| f\in X\}$ Observation: Let $M$ be a monoid, let $X,Y\subseteq M$  be subsets or submonoids. The union $X\cup Y$   is not generally a monoid. Corollary 1: Let $M$ be a monoid, let $X\subseteq M$  be a subset. If $X$ is already a (sub)monoid then $\langle X\rangle=X$. Corollary 2: Let $M$ be a monoid, let $X\subseteq M$ be a subset of invertible elements of $M$. The submonoid $\langle X\cup X^{-1}\rangle$ is a group. (proof in the previous post). Questions: let $X\subseteq {\mathcal C}^1({\mathbb R})$ a submonoid. When $\overline{X}$ is monoid? If $X$ is made of invertible functions, does  $\overline{(X^{-1})}=(\overline{X})^{-1}$ ? Answers to those questions can inform us on the properties of objects like $\langle \overline{X}\cup \overline{X}^{-1}\rangle$, $\langle \overline{X\cup X^{-1}}\rangle$ or $\overline{\langle X\cup X^{-1}\rangle}$. Quote:A2. Yes, I did this construction in hopes this would be a diffeomorphic group (whatever the hell you call it), a subgroup of $\text{Diff}(\mathbb{R})$. I avoided this language largely because I'm not familiar enough with the language. Rememeber that I would not call a group diffeomorphic like it is a property of the group. I call it a group of diffeomorphism, like a group of apples or a group of permutations. "The diffeomorphisms groups of R" just means "the group made of diffeomorphism of R to itself". I'm not aware if a grop can be diffeomorphic in any sense (and idc right now). Lol, yes diffeomorphic group is nonsense, I meant group of diffeomorphisms . I'm horrible with jargon. The reason I call the closure of $\mathcal{B}$ a monoid is because it should look something like, $ f : \mathbb{R} \to \mathbb{R}\,\,\text{bijectively}\\ f'(x) \ge 0\\ \lim_{x\to\infty} f'(x) \ge 1\\ \lim_{n\to\infty} f^{-n}(x) = -\infty\\$ Which is certainly still a monoid; but pulling out inverses will become difficult to complete it as a group, because $f'(x) = 0$ is a possibility. And we may lose the last property (I'm not certain, I hope not), but at worst only a weakening of it. And we should also ensure convergence of the derivatives in the Cauchy sequence. For that reason we'll want $f'_n(x)$ to be Cauchy on compact subsets too. You're right about the union too; in the general case. But again, the union should look like, $ f : \mathbb{R} \to \mathbb{R}\,\,\text{bijectively}\\ f'(x) \ge 0\\ f(x) \to \infty\\ \lim_{x\to\infty} f'(x) \ge 0\\$ So at least near infinity these things should have inverses and form a group. It'll be hard to ensure it works on all of $\mathbb{R}$ though, we'll need $f'(x) > 0$ which will require more complicated tricks than just completing with Cauchy; so you're right there. I think the closure of this monoid will have to be done more carefully--you're right. I'm going to focus my attention on simply strengthening everything I put above; and covering all my bases. « Next Oldest | Next Newest »

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