04/23/2022, 02:13 PM
(This post was last modified: 07/15/2022, 04:10 PM by MphLee.

*Edit Reason: numbering*)
My time is no longer enough to work out properly the topics and concepts emerging in the latest posts. I also have the impression that I'm not going to polish my notes anytime soon, yet I'd like to not forget every little step I made in understanding some issues in the field of superfunctions. When I write in this forum I try to polish the exposition as much as I can and to include a complete reference to old posts and manuscripts: I can't afford to do this anymore and for this I apologize to the forum members. I also can not afford to study the basic math anymore atm, so what follows could be trivial, full of silly observations, or very wrong. I apologize again but I hope that seeding those notes here could be useful for some active forum members like James, who is always ready to harvest ideas and inspiration and whose work has always been inspirational to me.

The plan is to edit the first post adding more notes transcriptions, stand-alone theorems and proofs, observations and questions and prep-notes, so don't expect me to reply under this with long explanations since this post serves mostly as a personal memory tool.

INDEX

2022, april 22 - expression of a group(monoid) action using superfunctions

Let's try to make order in the logical structure of the expression of iterates in terms of a given superfunction. The following is an early attempt to study the additional properties of superfunctions when the domain has a group structure, eg. integers, real and complex numbers. Fix \((A,+_A)\) to be an abelian group, the following can be extended to noncommutative monoids. Consider two \(A\)-actions, i.e. two objects of \({\rm Set}^{BA}\) the category of \(A\)-actions: we take the regular left-translation \(\lambda_a(b):=a+_Ab\) action over the set \(A\) and an action \(\alpha_a(y)\) over the set \(Y\).

A morphism between the two actions, i.e. an element \(f\in {\rm Hom}_{{\rm Set}^{BA}}(A^{\circlearrowleft \lambda},Y^{\circlearrowleft \alpha})\) is a map \[f:A\to Y\] that is an \(A\)-equivariant map, something that can be tough as a kind of generalized superfunction, that satisfies \(\forall a,b\in A \)

Proposition 1: Let \((f,g)\) be any pair of functions s.t. \(f\in {\rm Hom}_{{\rm Set}^{BA}}(A^{\circlearrowleft \lambda},Y^{\circlearrowleft \alpha})\) and \(g:Y\to A\) inverts \(f\) on the right, i.e. \(fg={\rm Id}_Y\) then the action \(\alpha\) can be expressed as \[\alpha_a(y)=f(a+g(y))\]

Proof: by def. for every \(a\in A\) we have \(f\lambda_a=\alpha_af\). Apply \(g\) on both sides and obtain the expression. \(\square\)

Let's specialize this to the classical case. If \(A=\mathbb Z\) then a \(\mathbb Z\)-action over \(Y\) is just a choice of a bijection \(\alpha:Y\to Y\) and \(\alpha_n(y):=(\alpha_1)^{\circ n}(y)\)

Corollary 1 Let \(A=\mathbb Z\). Let \((f,g)\) be any pair of functions s.t. \(f\in {\rm Hom}_{{\rm Set}^{B\mathbb Z}}(\mathbb Z^{\circlearrowleft S},Y^{\circlearrowleft \alpha_1})\) and \(g:Y\to A\) inverts \(f\) on the right, i.e. \(fg={\rm Id}_Y\) then the action \(\alpha\) can be expressed as \[\alpha_n(y)=f(n+g(y))\]

Question 1 Going from the proposition to the corollary is matter of restriction, but going from the integer case to the general case is matter of extending iteration from integer time to \(A\)-time. When is it possible? I think I can prove formally that when \(f\) has not only a right inverse, but also a right inverse, i.e. is bijective, we can prove the closed form of the \(A\) action. The result is already valid in the real, complex case, and is the folklore expression of non-integer iteration using a superfunction/Shroeder function.

Question 2 This is an extension problem, that means that in some sense there could be some "obstructions". Cohomology is described sometimes as a way to measure obstruction. Is it possible to apply cohomology here? How?

The plan is to edit the first post adding more notes transcriptions, stand-alone theorems and proofs, observations and questions and prep-notes, so don't expect me to reply under this with long explanations since this post serves mostly as a personal memory tool.

INDEX

- 2015, September - The lattice of divisibility and fractional iterates

- 2021, february 26 - preliminary exploration of "intrinsic iteration"

- 2022, april 22 - expression of a group(monoid) action using superfunctions

- 2022, april 23 - actions, iterations and generalized elements (reboot of old notes)

- 2022, april 23 - actions, iterations and generalized elements PART 2 (reboot of old notes)

- 2022, april 23 - actions, iterations and generalized elements PART 3 (reboot of old notes)

- 2022, april 24 - actions, iterations and generalized elements PART 4 (reboot of old notes)

- 2022, april 28 - more on how to express iterates using superfunctions

- 2022, May 07 - bird's-eye view on algebraic iteration (road to reboot of old notes part 2)

- 2022, May 07 - on the method+UPDATE

- 2022, May 13 - The tales of the legendary category \({\rm ACT}\) of Actions (idea)

- 2022, June 26 - Decomposing actions/iterations part 1 - Extensions preserves coproducts (NEW)

- 2022, June 26 - Decomposing actions/iterations part 2 - On bundles and equivalence relations (NEW)

2022, april 22 - expression of a group(monoid) action using superfunctions

Let's try to make order in the logical structure of the expression of iterates in terms of a given superfunction. The following is an early attempt to study the additional properties of superfunctions when the domain has a group structure, eg. integers, real and complex numbers. Fix \((A,+_A)\) to be an abelian group, the following can be extended to noncommutative monoids. Consider two \(A\)-actions, i.e. two objects of \({\rm Set}^{BA}\) the category of \(A\)-actions: we take the regular left-translation \(\lambda_a(b):=a+_Ab\) action over the set \(A\) and an action \(\alpha_a(y)\) over the set \(Y\).

A morphism between the two actions, i.e. an element \(f\in {\rm Hom}_{{\rm Set}^{BA}}(A^{\circlearrowleft \lambda},Y^{\circlearrowleft \alpha})\) is a map \[f:A\to Y\] that is an \(A\)-equivariant map, something that can be tough as a kind of generalized superfunction, that satisfies \(\forall a,b\in A \)

Proposition 1: Let \((f,g)\) be any pair of functions s.t. \(f\in {\rm Hom}_{{\rm Set}^{BA}}(A^{\circlearrowleft \lambda},Y^{\circlearrowleft \alpha})\) and \(g:Y\to A\) inverts \(f\) on the right, i.e. \(fg={\rm Id}_Y\) then the action \(\alpha\) can be expressed as \[\alpha_a(y)=f(a+g(y))\]

Proof: by def. for every \(a\in A\) we have \(f\lambda_a=\alpha_af\). Apply \(g\) on both sides and obtain the expression. \(\square\)

Let's specialize this to the classical case. If \(A=\mathbb Z\) then a \(\mathbb Z\)-action over \(Y\) is just a choice of a bijection \(\alpha:Y\to Y\) and \(\alpha_n(y):=(\alpha_1)^{\circ n}(y)\)

Corollary 1 Let \(A=\mathbb Z\). Let \((f,g)\) be any pair of functions s.t. \(f\in {\rm Hom}_{{\rm Set}^{B\mathbb Z}}(\mathbb Z^{\circlearrowleft S},Y^{\circlearrowleft \alpha_1})\) and \(g:Y\to A\) inverts \(f\) on the right, i.e. \(fg={\rm Id}_Y\) then the action \(\alpha\) can be expressed as \[\alpha_n(y)=f(n+g(y))\]

Question 1 Going from the proposition to the corollary is matter of restriction, but going from the integer case to the general case is matter of extending iteration from integer time to \(A\)-time. When is it possible? I think I can prove formally that when \(f\) has not only a right inverse, but also a right inverse, i.e. is bijective, we can prove the closed form of the \(A\) action. The result is already valid in the real, complex case, and is the folklore expression of non-integer iteration using a superfunction/Shroeder function.

Question 2 This is an extension problem, that means that in some sense there could be some "obstructions". Cohomology is described sometimes as a way to measure obstruction. Is it possible to apply cohomology here? How?

MSE MphLee

Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)

S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)