[MSE-SOLVED] Subfunction is functorial!!!! MphLee Long Time Fellow    Posts: 374 Threads: 30 Joined: May 2013 06/02/2021, 12:14 PM (This post was last modified: 06/02/2021, 04:50 PM by MphLee.) Idk if someone is interested but this night I had an eureka moment. Since years I'm trying to find the missing ingredient in the functorial approach to hyperoperations. I found a new way to define a functor out of the subfunction operator, i.e. the one that takes a function and "de-iterate" it or, if you want, that decrease the rank by one. It is the inverse of the superfunction operator (I won't annoy you with the fine details). The problem is that I always found some obstruction in this specific point. Tonight I had a nice idea and finally I was albe to give half of the definition... the missing half is not certain so I'm asking on MSE and if ignored on MO. Maybe nobody can help but at least you can get a glimpse of how my work on hyperoperations looks. https://math.stackexchange.com/questions...left-given Regards. UPDATE I solved it! Subfucntion operator is functorial! That's great news. I wasn't able to see it caus I worked on it all the night so my brain was super tired but it was trivial. What this means? We have a very general result that holds for all the groups but I can spell to you its corollary in the particular case of real and complex functions. Take an invertible function $f:{\mathbb C}\to{\mathbb C}$ in this case we fix our base function to be $g(z)=S(z)=z+1$. The operation $\Sigma[f](z):=f(1+f^{-1}(z))$ is functorial!!!! To be precise be  ${\mathfrak B}({\mathbb C})$ the group of bijective functions over the complex numbers then we have a functor $\Sigma: S\setminus{\mathfrak B}({\mathbb C})^{\mathbb N}\to S\setminus{\mathfrak B}({\mathbb C})^{\mathbb N}$ It takes discrete dynamicals system over C and gives the dynamics of the subfunction. This means that subtracting one rank is functorial!!!! 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: 1,214 Threads: 126 Joined: Dec 2010 06/03/2021, 03:10 AM Does this imply the "rank up" operation is a functorial by inversion? MphLee Long Time Fellow    Posts: 374 Threads: 30 Joined: May 2013 06/03/2021, 01:28 PM (06/03/2021, 03:10 AM)JmsNxn Wrote: Does this imply the "rank up" operation is a functorial by inversion? That's a great question! The answer is: no, but maybe yes. The point is subtle but is relevant for the problem of chosing canonically a superfunction among all possible solutions. It is a long but interesting story, I hope. The fact that subfunction is functorial is amazing: it just means that it has some kind of "naturality" because it respect structure and is well behaved (eg. continuous functions respect the given topology). Not only that! It is functorial for every group and for every base function g (not only for g=successor) is beautiful imho. Why? If we restrict ourselves to general groups then subfunction is never surjective (the identiy can not have a superfunction) and the best case we can get is when our group is superfunction-complete. In that case the only function that cannot have a superfunction is the identity. The other obstacle for invertibility is that subfunction is never injective because subfunction of the identity is a the fixed point of subfunction $\Sigma_g[1_G]=1_Gg1_G=g=\Sigma_g[g]$ unless the group has only two element. So subfunction is never invertible in a strict sense. But it has many injective  inverses! Now we are forced to make a choice ($\Psi$). Every possible inversion is a possible choice of superfunction. We can define concretely a choice as any function $\Psi:G\to G$ such that performing the choice and then applying the subfunction brings us back where we started (mathematicians call it a section). $\Sigma_g[\Psi[f]]=f$ thus $\Psi[f]g=f\Psi[f]$ Now as long as we are dealing with mere sets, and subfunction is just a set theoretic function we are in deep shit! Because in set theory every element of a set is the same and we have no preferences over elements, no systematic way of performing a choice. We needed more structure on the preimages and on the operator itself to hope in a universal way to invert it! So this is just the beginning of the research. Brief note on why this is important. Sadly Subfunction does not respect group structure and it is not clear if it is an important operation at all. This is in fact a hint that maybe subfunction is not relevant to real mathematics...this is exaclty the feeling we get by reading Qiaochu Yuan answer here https://math.stackexchange.com/questions...c-f-circ-1 . It seems that subfunction is the wrong way to look at conjugation. And conjugation is the real important concept here. This would imply that maybe iterating it (hyperoperations) is just an artifact, there is nothing deep about it. This proof of functoriality changes this narrative! It says that subfunction respect the structure of two categories. At this point one could make an objection... well, giving to a set the hand-made ad-hoc topology you can turn every set theoretic function into a continuous one relatively to that made up topology, but we, serious people, only consider the good old classic topologies, the only topologies that are meaningful. It turn out that the category structure I had to define to make subfunction functorial is pretty natural and it comes straight from the dynamical information contained (trapped?) in the group of functions considered! So extracting that dynamical information and organizing it under the shape of a category automatically makes the subfunction operator "natural" and well behaved! This is so exciting! Back to why now we have new hopes. The functor itself is not invertible in a straight way, for the same reason set-theoretic subfunction is not. The functor maps many different objects (dynamical systems) to the same one (not injective on objects). For example take the two S-dynamical systems ${\rm X}=({\mathbb C}, {\rm Add}_b, \phi)$ and ${\rm Y}=({\mathbb C}, {\rm Add}_b, \psi)$. Here I mean that we have $\phi(z+1)={\rm Add}_b(\phi(z))$ and $\psi(z+1)={\rm Add}_b(\psi(z))$. Those two objects are elements of the category that makes the subfunction operator a functor. We haven't injectivity for the subfunction functor (I should name it.. maybe the subdynamics functor?) ${\mathbb \Sigma}{\rm X}={\mathbb \Sigma}{\rm Y}$ To check this consider that ${\mathbb \Sigma}{\rm X}=({\mathbb C}, \Sigma[{\rm Add}_b], \Sigma[\phi])=({\mathbb C}, S,{\rm Add}_b)$ But now we have a functor! And now the preimages have a richer structure. An elementary example. To make clear why this could be the case consider basic undergraduate point-set topology. A topological space is a set equipped with a system of open subsets ${\mathcal X}=(X,\tau)$. So to every top, space we can assign a set, the set of its points. ${\rm Pts}({\mathcal X})=X$ That is a functor that forgets information, it forgets the topology, the "shape of the space" ${\rm Pts}:{\bf Top}\to{\bf Set}$. Now, can we invert it? Can we restore the lost information? Can we assign a topology to a set of points? Of course we can in an infinite number of ways (how to make a choice?). All the topologies on the same set can be ordered by the the coarser/finer relation. The preimage of a set under the point functor is the collection of all the possible topologies on a given set. Just like the preimage of a function under the subfunction operator is the set of all its superfunctions. With a small but crucial difference. The collection of topologies on a given set is a lattice ordered by coarser/finer relation and has a top and a bottom, two special elements: the discrete and the trivial topology. So in the case of the functor of points we have two canonical ways to invert it: two canonical way to assign a topology to given set, i.e. its trivial (co-discrete) and it's discrete topology. Those two "sections" are not exactly inverse functors, but a weaker concept that is possible only when we replace functions with functors. They are called adjoint functors of the Pts functor. So the next question we should ask is: does the subfunction functor have adjoints?? looks very promising... but that is just the beginning. Now we can finally unleash the full power of category theory on the theory of hyperoperations! 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: 1,214 Threads: 126 Joined: Dec 2010 06/04/2021, 01:07 AM (This post was last modified: 06/04/2021, 01:11 AM by JmsNxn.) (06/03/2021, 01:28 PM)MphLee Wrote: ... So the next question we should ask is: does the subfunction functor have adjoints?? I can't remember off the top of my head; but there's a relation between adjoints in category theory and adjoints in functional analysis, and the dual space. That would be super cool. My dream was always a functional space where you can do hyper operators... I think it might be beneficial for me to dust off my hilbert space textbooks. I think looking at, $\uparrow f= \frac{d^{z-1}}{dw^{z-1}}|_{w=0} \sum_{n=0}^\infty f^{\circ n+1}(a)\frac{w^n}{n!}\\$ with an inner product may be important. Of course we'll have to be in a restricted kind of space; and only consider bounded super-functions with an attracting fixed point of real-positive multiplier. MphLee Long Time Fellow    Posts: 374 Threads: 30 Joined: May 2013 06/04/2021, 01:00 PM (This post was last modified: 06/04/2021, 08:29 PM by MphLee.) There is a link indeed. Every Hilbert space $H$ is a vector space equipped with a special bilinear form, i.e. a homomorphism antilinear in the first variable and linear in the second one, of vector spaces valued in the base field $\langle -\,,\,-\rangle_H:H\times H\to \mathbb C$, called inner product, whose induced metric space structure is complete. Every(modulo some abstract nonsense) Category $\mathcal C$ comes equipped with a special bifunctor, i.e. a homomorphism of categories, contravariant in the first variable and covariant in the second, valued in the cat. of sets ${\rm Hom}_{\mathcal C}(-\,,\,-):{\mathcal C}^{op}\times {\mathcal C}\to {\bf Set}$, called Hom functor, a category can be asked to be complete under some limits of diagrams (eg. every sequence has a categorical limit). Given two Hilbert spaces $H_1, H_2$ and two operators $H_1\array{\overset{F}{\rightarrow}\\ \underset{G}{\leftarrow}}H_2$: we say that $G$ is adjoint of $F$ if we have the identity $\langle f,G(g)\rangle_{H_1}=\langle F(f),g\rangle_{H_2}$ Given two categories $\mathcal C, \mathcal D$ and two functors ${\mathcal C}\array{\overset{F}{\rightarrow}\\ \underset{G}{\leftarrow}}{\mathcal D}$: we say that $G$ is LEFT adjoint of $F$ if we have the natural isomorphism ${\rm Hom}_{\mathcal C}( X,G(Y))\simeq {\rm Hom}_{\mathcal D}( F(X),Y)$ So adjunctions in category theory subsume and immensely extend the "classical" setting in functional analysis. Adjunctions arise everywhere and are probably the most profound, deep and meaningful concept that mathematicians has ever defined. Some example of adjunctions are: betwen algebra and geometry; between existential and universal quantification (logic), between Necessity and Possibility (modal logic) between set of point and discrete topology functor (topology), an honorable example of adjunction is the following law in arithmetic and real algebra $w^{(\lambda z)}=(w^\lambda)^z$ The curryng isomorphism $X^{T\times Y}\simeq(X^T)^Y$ I leave you not with an example but with two suggestive views $|x-c|_X<\delta(\epsilon)\Rightarrow|f(x)-\lim_c f|_Y<\epsilon$ $\sigma\bullet S(x)=F(\sigma)\bullet x$ 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: 1,214 Threads: 126 Joined: Dec 2010 06/04/2021, 10:25 PM (This post was last modified: 06/04/2021, 10:26 PM by JmsNxn.) (06/04/2021, 01:00 PM)MphLee Wrote: There is a link indeed. Every Hilbert space $H$ is a vector space equipped with a special bilinear form, i.e. a homomorphism antilinear in the first variable and linear in the second one, of vector spaces valued in the base field $\langle -\,,\,-\rangle_H:H\times H\to \mathbb C$, called inner product, whose induced metric space structure is complete. Every(modulo some abstract nonsense) Category $\mathcal C$ comes equipped with a special bifunctor, i.e. a homomorphism of categories, contravariant in the first variable and covariant in the second, valued in the cat. of sets ${\rm Hom}_{\mathcal C}(-\,,\,-):{\mathcal C}^{op}\times {\mathcal C}\to {\bf Set}$, called Hom functor, a category can be asked to be complete under some limits of diagrams (eg. every sequence has a categorical limit). Given two Hilbert spaces $H_1, H_2$ and two operators $H_1\array{\overset{F}{\rightarrow}\\ \underset{G}{\leftarrow}}H_2$: we say that $G$ is adjoint of $F$ if we have the identity $\langle f,G(g)\rangle_{H_1}=\langle F(f),g\rangle_{H_2}$ Given two categories $\mathcal C, \mathcal D$ and two functors ${\mathcal C}\array{\overset{F}{\rightarrow}\\ \underset{G}{\leftarrow}}{\mathcal D}$: we say that $G$ is LEFT adjoint of $F$ if we have the natural isomorphism ${\rm Hom}_{\mathcal C}( X,G(Y))\simeq {\rm Hom}_{\mathcal D}( F(X),Y)$ So adjunctions in category theory subsume and immensely extend the "classical" setting in functional analysis. Adjunctions arise everywhere and are probably the most profound, deep and meaningful concept that mathematicians has ever defined. Some example of adjunctions are: betwen algebra and geometry; between existential and universal quantification (logic), between Necessity and Possibility (modal logic) between set of point and discrete topology functor (topology), an honorable example of adjunction is the following law in arithmetic and real algebra $w^{(\lambda z)}=(w^\lambda)^z$ The curryng isomorphism $X^{T\times Y}\simeq(X^T)^Y$ I leave you not with an example but with two suggestive views $|x-c|_X<\delta(\epsilon)\Rightarrow|f(x)-\lim_c f|_Y<\epsilon$ $\sigma\bullet S(x)=F(\sigma)\bullet x$ Yes! That was it; I'm glad I don't have to go searching for it. I'm gonna spend some time fiddling with, $\vartheta(w) = \sum_{n=0}^\infty f(n+1)\frac{w^n}{n!}\\ H\vartheta(w) = \sum_{n=0}^\infty f^{\circ n+1}(1) \frac{w^n}{n!}\\$ Where the chain is, $\frac{d^{z-1}}{dw^{z-1}}|_{w=0} \vartheta = f\\ \frac{d^{z-1}}{dw^{z-1}}|_{w=0} H \vartheta = \uparrow f\\$ Now $H$ is not linear; but there may be a way to pull out an adjoint on the inner product, $(f,g) = \int_0^\infty f(x)\overline{g(x)}\,dx/x\\$ Where the law is, $(f,x^{\overline{z}}) = \Gamma(z) \frac{d^{-z}}{dw^{-z}}\vartheta\\ (Hf,x^{\overline{z}}) = \Gamma(z) \frac{d^{-z}}{dw^{-z}}H\vartheta\\$ And if we had something like $H^*$ which satisfies, $(Hf,g) = (f,H^*g)\\$ Then if we could iterate $H^*$ on $x^{\overline{z}}$ we'd have the fractional hyper-operator nonsense. BTW, I use the convention that the inner product is linear in the first, and antilinear in the second variable. I'm a mathematician, stop this quantum physics nonsense notation. We are assuming that $|f(z)| < M$ is bounded for $\Re(z) > 0$. We are assuming it takes $\mathbb{R}^+ \to \mathbb{R}^+$; and additionally that $f : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C}_{\Re(z) > 0}$. This is enough for these transforms to converge. MphLee Long Time Fellow    Posts: 374 Threads: 30 Joined: May 2013 06/04/2021, 10:37 PM (This post was last modified: 06/04/2021, 10:41 PM by MphLee.) (06/04/2021, 10:25 PM)JmsNxn Wrote: Yes! That was it; I'm glad I don't have to go searching for it. I'm gonna spend some time fiddling with,  I expected to blow your mind dropping those analogies... xD but you are cold as ice. I mean, those regularities are fire... something magic. Quote:$(f,x^{\overline{z}}) = \Gamma(z) \frac{d^{-z}}{dw^{-z}}\vartheta\\ (Hf,x^{\overline{z}}) = \Gamma(z) \frac{d^{-z}}{dw^{-z}}H\vartheta\\$ ... We are assuming that $|f(z)| < M$ is bounded for $\Re(z) > 0$. We are assuming it takes $\mathbb{R}^+ \to \mathbb{R}^+$; and additionally that $f : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C}_{\Re(z) > 0}$. This is enough for these transforms to converge. Mmhh but what  $x^{\overline{z}}$ is? I can't follow properly. 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: 1,214 Threads: 126 Joined: Dec 2010 06/04/2021, 11:22 PM (This post was last modified: 06/04/2021, 11:26 PM by JmsNxn.) (06/04/2021, 10:37 PM)MphLee Wrote: (06/04/2021, 10:25 PM)JmsNxn Wrote: Yes! That was it; I'm glad I don't have to go searching for it. I'm gonna spend some time fiddling with,  I expected to blow your mind dropping those analogies... xD but you are cold as ice. I mean, those regularities are fire... something magic. Quote:$(f,x^{\overline{z}}) = \Gamma(z) \frac{d^{-z}}{dw^{-z}}\vartheta\\ (Hf,x^{\overline{z}}) = \Gamma(z) \frac{d^{-z}}{dw^{-z}}H\vartheta\\$ ... We are assuming that $|f(z)| < M$ is bounded for $\Re(z) > 0$. We are assuming it takes $\mathbb{R}^+ \to \mathbb{R}^+$; and additionally that $f : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C}_{\Re(z) > 0}$. This is enough for these transforms to converge. Mmhh but what  $x^{\overline{z}}$ is? I can't follow properly. if $z = a+ib$ then $\overline{z} = a-ib$ then $x^{\overline{z}} = x ^{a-ib}$. Sorry, I should've been clearer. Then this is just a strange way of writing the differintegral, $(f,g) = \int_0^\infty f(x)\overline{g(x)}\,dx/x\\ (f,x^{\overline {z}}) = \int_0^\infty f(x)x^{z-1}\,dx\\$ Which is; as we've talked about before; the Kernel of the differintegral. And I wanna see if we can do something to the kernel to get the $\uparrow$ operator somehow. Not too sure if this is possible; as it may imply linearity if done too obviously. And we know it isn't linear. I'm just trying to think of ways we can make $(\frac{d^{z}}{dw^z}|_{w=0} H)^n$ for complex $n$; where for natural n it just makes a hyperoperation-chain. .... I apologize for not being so surprised. I had learned all of that before, I just needed a reminder of its exact shape  MphLee Long Time Fellow    Posts: 374 Threads: 30 Joined: May 2013 06/05/2021, 12:12 AM I'm a bit lost tbh. What H* can look like in your opinion? Whatever it is, it is applied to exponentiation so it should be a good news. Also operators are linear, usually, so Idk how to manage non-linearity inside an ambient, that of vector spaces/Hilbert spaces, where things should be linear. But I'm curious of what you can carve out of this. Even if you don't get result I believe that what we are doing here is usefull, brushing the dust from the hidden spots, making clear what the mechanisms are. I went full speed on categories and non-commutative groups just for that reason. Because I wanted a landscape where I couldn't care less about linearity because we don't want to start with a god-given abelian ground (i.e. linearity) by default. 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: 1,214 Threads: 126 Joined: Dec 2010 06/05/2021, 12:39 AM (This post was last modified: 06/05/2021, 01:30 AM by JmsNxn.) (06/05/2021, 12:12 AM)MphLee Wrote: I'm a bit lost tbh. What H* can look like in your opinion? Whatever it is, it is applied to exponentiation so it should be a good news. Also operators are linear, usually, so Idk how to manage non-linearity inside an ambient, that of vector spaces/Hilbert spaces, where things should be linear. But I'm curious of what you can carve out of this. Even if you don't get result I believe that what we are doing here is usefull, brushing the dust from the hidden spots, making clear what the mechanisms are. I went full speed on categories and non-commutative groups just for that reason. Because I wanted a landscape where I couldn't care less about linearity because we don't want to start with a god-given abelian ground (i.e. linearity) by default. Hf's adjoint H* depends on f; that's really about the most of it. So we'd be talking about $H^*_f x^{-z}$. We're really just looking for $g_s^z$ such that, $(f,g_s^z) = \uparrow^s f\\$ I've screwed up some of my capitals a bit; but think we're trying to solve the equation with an inner product.... Give me a bit Mphlee; I can explain what I mean by adjoint in full detail. This is pretty straightforward, I think this is really important. « Next Oldest | Next Newest »

 Possibly Related Threads… Thread Author Replies Views Last Post Rational operators (a {t} b); a,b > e solved JmsNxn 30 88,071 09/02/2016, 02:11 AM Last Post: tommy1729

Users browsing this thread: 1 Guest(s) 