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Hi, I know this is not directly related to tetration. It is related to iteration theory and solving functional equations.

https://math.stackexchange.com/questions...alpha-beta
The question is the following \( f(g,\beta\alpha,x+y)=f(g^\alpha,\beta,x)f(g,\alpha,y)

\)

Mind that here x,y can be thought to be numbers. But the crucial part is that \( \alpha \), \( \beta \) and \( g \) are intended to be matrices, formal powerseries or functions (non commutative objects in general) under composition.

I ask also here because with JmsNxn deep familiarity with iterated composition (in the question i rendered the Omega notation as a product to not scare away normies) I could get some knowledge.

MSE MphLee

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

S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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06/11/2021, 02:10 AM
(This post was last modified: 06/11/2021, 03:50 AM by JmsNxn.)
Is this some kind of semi-direct product? Jesus; that would be cool...

Let me think on this... I'll see if I can think of anything.

Is \( g^\alpha \) really supposed to be an element of \( \text{Aut} \); rather than necessarily a conjugation?

........

Update:

So I'm not certain I can answer this; but, did you understand when I introduced the congruent integral (if you read the paper)? This is a different form of the compositional integral that is in a "modded out space". I can reduce (I think); your question into the abelian case; but I need to know how much of the congruent integral I should explain.

Also; what the solutions \( f \) are referred to as is homormorphisms of the semi-direct products between two groups.

In this case you are taking an inner automorphism \( g^{\alpha}\in \text{InnAut}(G) \subset \text{Aut}(G) \). Then you are constructing what is typically written,

\(

G \propto_{g^\alpha} N\\

\)

And you are looking for

\(

f : G \propto_{g^\alpha} N \to G

\)

There's a word for this; I can't remember it exactly.

Also, this forum doesn't have the best latex implementation; so \( \propto \) should actually be the symbol \( \rtimes \); which is more angular. It should look more like this

.....

I'll explain this better tomorrow. Long night; but it's at least SOMETHING like this. Nonetheless; your answer lies in semi-direct products.

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06/13/2021, 08:38 AM
(This post was last modified: 06/13/2021, 09:42 AM by JmsNxn.)
I finally can answer your question. I was a little confused about the details before; but it's definitely a semi-direct product.

Let's first of all, ignore the function \( f \). Let's write a product,

\(

(xy,\beta \alpha) = (\varphi(\alpha)(x),\beta)(y, \alpha)\\

x,y \in N\,\,\alpha,\beta \in G\\

\)

And call this group,

\(

N \propto_\varphi G\\

\)

Where \( \varphi(\alpha) : G \to \text{Aut}(N) \). Where \( \text{Aut}(N) \) is the group of isomorphisms of \( N \).

Now; there are many choices of \( \varphi(\alpha) \) (something to do with Euler's phi function will be involved in the estimate of how many). Now when you write \( g^{\alpha} = \alpha g \alpha^{-1} \in \text{Aut}(G) \), you are choosing an isomorphism of \( \text{Aut}(G) \to \text{Aut}(N) \); let's call this \( \psi : \text{Aut}(G) \to \text{Aut}(N) \). We can do this because we're only going to care about \( f \) in the final result. And this is just considering \( f(g,\alpha,x) \) at an implicit level in the preimage and considering it equivalent.

Now, here is where I wasn't making any sense before. You wrote your equation backwards from the usual semi-direct product. The right way I should've written; which I apologize for saying. Is that it's,

\(

N \propto_{\psi(g)} G = N \propto_{\varphi} G\\

\)

You are now looking for projections,

\(

f(N \propto_{\psi(g)} G) \to G\\

\)

I fucking knew it was semi-direct products! Took me a while to think about it though...

Regards, James. I hope this helps.

Long story short; you have a lot of group theory at your disposal, Mphlee. May I recommend dummit & foote.
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Yes!

That's very promising. I'll have to work hard on this. Thank you so much! I believe that it is the solution: I just need to rephrase it, play with it a bit and prove it with my hands.

This problem popped out at the intersection between the abstract Jabotinky foundation and superfunction-spaces. Can't wait to find time to work on the details.

Thank you again.... nobody there on MSE even had the kindness to drop a single keyword. For group theorist this should be the abc... for sure my question is not well written, my English su*x, but I don't feel is a bad question.

MSE MphLee

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

S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
Posts: 1,214

Threads: 126

Joined: Dec 2010

I think your question was written very well. It was probably more suited to Mathoverflow though; stack exchange tends to be mostly for undergraduate homework problems. But even then, the community can be a tad abrasive; and they always poke holes in technicalities while ignoring the larger picture. I deleted my stack exchange accounts a long time ago; bored by how little they actually help. Sometimes, you have to be left to your own wits

.

I suspect though, no one saw that it was a semi-direct product... just a weird one. Only reason I saw it was because I was thinking about bullet notation. And I was initially trying to give an example of your function.

Regards, James