Very interesting stuff, Mphlee!
It'll take me awhile to reread and absorb what you've written; but is it possible to do something similar in the non-abelian case?
By this I mean, when we write,
\(
y' = \text{ilog}(y)\\
y(0) = z\\
\)
We are inducing an abelian group. But if we change into,
\(
y' = f(s,y)\\
\)
Then writing this,
\(
\int_b^c f(s,z)\,ds\bullet \int_a^b f(s,z)\,ds \bullet z = \int_a^c f(s,z)\,ds\bullet z\\
\)
Forms a non-abelian group across composition. Would we be able to talk about \( \{T,0_T,+_T\} \) (or something like that), but with a non abelian \( +_T \)? I think you should get something similar because,
\(
\lim_{h\to 0} \frac{\int_a^{x+h} f(s,z)\,ds \bullet z - \int_a^x f(s,z)\,ds \bullet z}{h} = \lim_{h\to 0} \frac{\int_x^{x+h} f(s,z)\,ds \bullet z - z}{h} \bullet \int_a^x f(s,z)\,ds \bullet z\\
\)
And the term,
\(
\lim_{h\to 0} \frac{\int_x^{x+h} f(s,z)\,ds \bullet z - z}{h} = f(x,z)\\
\)
Which is just the fundamental property of the compositional integral.
EDIT:
This is also very similar to what I was trying to say with closed contours, Mphlee. When you take an integral about a singularity; it looks abelian (upto conjugation). And you can decompose it in such a manner that everything looks abelian...upto conjugation. Which is how I described my property:
\(
\int_\tau f(s,z)\,ds\bullet \int_\gamma f(s,z)\,ds\bullet z = \int_\varphi f(s,z)\,ds\bullet \int_\tau f(s,z)\,ds\bullet z\\
\)
Where \( \gamma,\varphi \) are Jordan curves (just think they're a closed contour), and \( \tau \) is just an arc. So create an equivalence class,
\(
\int_\gamma f(s,z) \, ds\bullet z \simeq \int_\varphi f(s,z)\,ds\bullet z\\
\)
If they are conjugate similar. It turns out, each conjugate class depends ONLY ON WHAT SINGULARITIES ARE WITHIN THE CONTOUR. And then we mod out by this equivalence class; we get an abelian group and we're back to a discussion of \( y' = \text{ilog}(y) \) (at least in spirit, I tried to draw this out as best I could). And even better; we can decompose,
\(
\int_\gamma f(s,z)\,ds\bullet z = \Omega_j \text{Rsd}(f,\zeta_j;z)\bullet z\\
\)
For the list of singularities \( \zeta_j \) within the contour \( \gamma \). Which means, for SOME sequence of closed contours \( \gamma_j \) which only encircle \( \zeta_j \);
\(
\int_\gamma f(s,z) \,ds\bullet z = \Omega_j \int_{\gamma_j} f(s,z)\,ds\bullet z\\
\)
Where, \( \int_{\gamma_j} f(s,z)\,ds\bullet z \in \text{Rsd}(f,\zeta_j) \). But! we can also rearrange this, because once you mod out, it's effectively abelian. If I reindex the singularities \( \zeta_i = \zeta_{\sigma(i)} \); there exists another sequence of closed contours \( \varphi_i \) which only encircle \( \zeta_i \) in which:
\(
\int_\gamma f(s,z) \,ds\bullet z = \Omega_i \int_{\varphi_i} f(s,z)\,ds\bullet z\\
\)
To such an extent, I can apply \( \int_{\varphi_i^{-1}} f(s,z)\,ds\bullet z \) and I get:
\(
\int_\gamma f(s,z) \,ds\bullet \int_{\varphi_i^{-1}} f(s,z)\,ds\bullet z = \Omega_{j\neq i} \text{Rsd}(f,\zeta_j;z)\bullet z\\
\)
This was essentially the thesis of that paper; because once you do that you can do a lot of stuff you do in traditional analysis, but with compositions and compositional integrals. Basically; everything you get from Cauchy's residue theorem; you get a slightly more difficult to handle equivalent.
If I'm understanding you correctly; from what you've written here; this is directly the second chapter of The Compositional Integral: The Narrow and The Complex Looking Glass and the fifth section An additive to composition homomorphism. Except, you have entirely rephrased what was just a couple pages into a very different flavour using your fancy arrows, lol. The rest of the book essentially just tries to make this homomorphism exist in more extravagant scenarios by "modding out." But this looks really pretty, how you're writing it. And it definitely breaks it down into its atoms much more. If anything it's a sigh of relief that someone else is beginning to see what I see, lol!
The difficulty I've always had though; is that it's not very obvious how, let's say, tetration is related to it's i-logarithm. Or how they behave. I have existence; but I don't really know any algorithm to relate the two in a nice readable manner. Or, any way, say, where we can hunt for a logarithm and then just plug and play to get tetration.
I know there is SOME function such that,
\(
\exp^{\circ s}(\xi) = \int_0^s h(\xi)\,dw\bullet \xi\\
\)
But, there's no obvious way to find that (and trust me, I've tried). The best I could do is attempt to related these things to contour integrals. But I couldn't figure out any kind of decision process, where we construct a specific \( h \) associated to what you're calling \( \overline{f} \). I was only able to sort of classify what these constructions WOULD look like; and their behaviour in complicated scenarios. But nothing like a decision process.
Anyway; you've given me much to think about Mphlee. I'll try and digest your graphs better; but I think I'm getting what you're saying much more clearly; especially as I reread.
It'll take me awhile to reread and absorb what you've written; but is it possible to do something similar in the non-abelian case?
By this I mean, when we write,
\(
y' = \text{ilog}(y)\\
y(0) = z\\
\)
We are inducing an abelian group. But if we change into,
\(
y' = f(s,y)\\
\)
Then writing this,
\(
\int_b^c f(s,z)\,ds\bullet \int_a^b f(s,z)\,ds \bullet z = \int_a^c f(s,z)\,ds\bullet z\\
\)
Forms a non-abelian group across composition. Would we be able to talk about \( \{T,0_T,+_T\} \) (or something like that), but with a non abelian \( +_T \)? I think you should get something similar because,
\(
\lim_{h\to 0} \frac{\int_a^{x+h} f(s,z)\,ds \bullet z - \int_a^x f(s,z)\,ds \bullet z}{h} = \lim_{h\to 0} \frac{\int_x^{x+h} f(s,z)\,ds \bullet z - z}{h} \bullet \int_a^x f(s,z)\,ds \bullet z\\
\)
And the term,
\(
\lim_{h\to 0} \frac{\int_x^{x+h} f(s,z)\,ds \bullet z - z}{h} = f(x,z)\\
\)
Which is just the fundamental property of the compositional integral.
EDIT:
This is also very similar to what I was trying to say with closed contours, Mphlee. When you take an integral about a singularity; it looks abelian (upto conjugation). And you can decompose it in such a manner that everything looks abelian...upto conjugation. Which is how I described my property:
\(
\int_\tau f(s,z)\,ds\bullet \int_\gamma f(s,z)\,ds\bullet z = \int_\varphi f(s,z)\,ds\bullet \int_\tau f(s,z)\,ds\bullet z\\
\)
Where \( \gamma,\varphi \) are Jordan curves (just think they're a closed contour), and \( \tau \) is just an arc. So create an equivalence class,
\(
\int_\gamma f(s,z) \, ds\bullet z \simeq \int_\varphi f(s,z)\,ds\bullet z\\
\)
If they are conjugate similar. It turns out, each conjugate class depends ONLY ON WHAT SINGULARITIES ARE WITHIN THE CONTOUR. And then we mod out by this equivalence class; we get an abelian group and we're back to a discussion of \( y' = \text{ilog}(y) \) (at least in spirit, I tried to draw this out as best I could). And even better; we can decompose,
\(
\int_\gamma f(s,z)\,ds\bullet z = \Omega_j \text{Rsd}(f,\zeta_j;z)\bullet z\\
\)
For the list of singularities \( \zeta_j \) within the contour \( \gamma \). Which means, for SOME sequence of closed contours \( \gamma_j \) which only encircle \( \zeta_j \);
\(
\int_\gamma f(s,z) \,ds\bullet z = \Omega_j \int_{\gamma_j} f(s,z)\,ds\bullet z\\
\)
Where, \( \int_{\gamma_j} f(s,z)\,ds\bullet z \in \text{Rsd}(f,\zeta_j) \). But! we can also rearrange this, because once you mod out, it's effectively abelian. If I reindex the singularities \( \zeta_i = \zeta_{\sigma(i)} \); there exists another sequence of closed contours \( \varphi_i \) which only encircle \( \zeta_i \) in which:
\(
\int_\gamma f(s,z) \,ds\bullet z = \Omega_i \int_{\varphi_i} f(s,z)\,ds\bullet z\\
\)
To such an extent, I can apply \( \int_{\varphi_i^{-1}} f(s,z)\,ds\bullet z \) and I get:
\(
\int_\gamma f(s,z) \,ds\bullet \int_{\varphi_i^{-1}} f(s,z)\,ds\bullet z = \Omega_{j\neq i} \text{Rsd}(f,\zeta_j;z)\bullet z\\
\)
This was essentially the thesis of that paper; because once you do that you can do a lot of stuff you do in traditional analysis, but with compositions and compositional integrals. Basically; everything you get from Cauchy's residue theorem; you get a slightly more difficult to handle equivalent.
If I'm understanding you correctly; from what you've written here; this is directly the second chapter of The Compositional Integral: The Narrow and The Complex Looking Glass and the fifth section An additive to composition homomorphism. Except, you have entirely rephrased what was just a couple pages into a very different flavour using your fancy arrows, lol. The rest of the book essentially just tries to make this homomorphism exist in more extravagant scenarios by "modding out." But this looks really pretty, how you're writing it. And it definitely breaks it down into its atoms much more. If anything it's a sigh of relief that someone else is beginning to see what I see, lol!
The difficulty I've always had though; is that it's not very obvious how, let's say, tetration is related to it's i-logarithm. Or how they behave. I have existence; but I don't really know any algorithm to relate the two in a nice readable manner. Or, any way, say, where we can hunt for a logarithm and then just plug and play to get tetration.
I know there is SOME function such that,
\(
\exp^{\circ s}(\xi) = \int_0^s h(\xi)\,dw\bullet \xi\\
\)
But, there's no obvious way to find that (and trust me, I've tried). The best I could do is attempt to related these things to contour integrals. But I couldn't figure out any kind of decision process, where we construct a specific \( h \) associated to what you're calling \( \overline{f} \). I was only able to sort of classify what these constructions WOULD look like; and their behaviour in complicated scenarios. But nothing like a decision process.
Anyway; you've given me much to think about Mphlee. I'll try and digest your graphs better; but I think I'm getting what you're saying much more clearly; especially as I reread.