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So I was considering the following functional equation:

Given $H :\Bbb{C}^2 \rightarrow \Bbb{C} $ find $\theta: \Bbb{C}^2 \rightarrow \Bbb{C}$ such that

$$ \theta(H(a,b), H(c,d)) = H(\theta(a,c), \theta(b,d)) $$

This resulted as a bit of a generalization on the solution to, given $t(x)$ find $H(x,y)$ such that

$$ t(H(x,y)) = H(t(x),t(y))$$

Which has an elegant series solutions using the theory of finite differences. By observing we can split up to any concrete case where:

$$ \begin{pmatrix} H(t(x),y) = q_1(H(x,y)) \\ H(x,t(y)) = q_2(H(x,y)) \\\text{such that} \ q_1(q_2) = t\end{pmatrix} $$

And then recover the 2-d plane of terms that when summed together form $H$.

Does anyone know if others have worked on the same equation? Also what subject is this? I've mostly been working in my black hole, and operator theory+ functional equations seem relevant but i'm not too sure if this is of interest to others

Some updates:

Suppose there exists as $u \in \Bbb{C}$ such that $H(x,u) = x$

Then the solution to the system:

$$ \begin{pmatrix} \theta ( H(x,w), y) = H(\theta(x,y), K_1 ) \\ \theta ( x, H(y,z)) = H(\theta(x,y), K_2) \end{pmatrix}$$

for $\theta (x,y) $ with suitable choice of $K_1$ and $K_2$ yields the $\theta(x,y)$ that satisfies

$$ \begin{pmatrix} \theta ( H(x,w), y) = H(\theta(x,y), \theta(w,u ) \\ \theta ( x, H(y,z)) = H(\theta(x,y), \theta(u,z) ) \end{pmatrix}$$

Which must be a solution to:

$$ \theta(H(x,w), H(y,z)) = H(\theta(x,y), \theta(w,z)) $$

So all that remains to be considered is if no such $u$ exists for a given $H$

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  • $\begingroup$ If you need to come at this from a functional analysis point, I can't help. If you don't mind a general algebraic approach, there is a notion of commutator for general algebras. Freese, Kiss, and McKenzie are some authors that come to mind. If you come back with a question related to this question and concerning a point in commutator theory, I might be able to provide a hint. Gerhard "Then Again, I Might Not" Paseman, 2015.09.17 $\endgroup$
    – Gerhard Paseman
    Commented Sep 17, 2015 at 20:44
  • $\begingroup$ I believe I may be abusing the term "commutator" as I used it in a casual context, to indicate that my functional equation involved commuting or moving around arguments. $\endgroup$
    – Sidharth Ghoshal
    Commented Sep 17, 2015 at 20:45
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    $\begingroup$ Quite possible, in which case my reference won't help. However, more general notions of commutation (including the one you are using) have been studied in the context of general algebra. I am not willing at present to answer general questions, but if you think something I've suggested might help and have a specific point to resolve, I can probably help. Good Luck. Gerhard "Longest Commute Was Four States" Paseman, 2015.09.17 $\endgroup$
    – Gerhard Paseman
    Commented Sep 17, 2015 at 20:48
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    $\begingroup$ Something like this occurs in what B. Pachuashvili nicknamed "grgr=ab" :D I mean, category of internal groups in the category of groups is equivalent to the category of abelian groups (more generally, monoids would do). In other words, if you have a monoid $M$ and a second monoid structure on $M$ such that operations for the second structure are homomorphisms with respect to the first structure, then these structures actually coincide and are commutative. In the argument your equation appears, but units are essential. I don't know what one gets without them. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 18, 2015 at 6:53

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