Do you know if the following set of rules has an algebra more general than the usual complex numbers?

Maybe someone can also help me state the rules in some mathematically rigorous (fancy :) ) way so that I have a solid statement.

The elements are objects indexed with two real numbers. The multiplication is associative and satisfies

$X_{a,b}\cdot X_{c,d}=X_{ac,b+d}$

Moreover the second index is periodic


[EDIT: from further research it seems I actually define a 'field' and require the two above rules]

Now I always require that some operation called addition satisfying all axioms of common addition (associative, commutative, zero, negation) yields an element within the set of these elements

[EDIT: I think I forgot the rule $X_{a,0}+X_{b,0}=X_{a+b,0}$ or is it not needed?]

$\exists e,f:X_{a,b}+X_{c,d}=X_{e,f}$

The multiplication from above is distributive over this addition.

Do these rules already define how $e$ and $f$ should be? I have a shaky proof that the have to be complex number rules, but I'm not sure if there is something more general. One shaky part is that I require square roots.

(Extra: what if I remove the condition about periodicity in the second index?)

[EDIT: Motivation of this algebra: I imagined the first parameter being some sort of growth/scaling parameter and the second parameter as some sort of aging parameter. The aging parameter is periodic. The rules mentioned are supposed to logically represent growth and aging. Addition relates to growth. Now I wondered whether addition is uniquely determined if I assume that for some reason this a sum should always give back a single growth and a single aging.] ]

  • $\begingroup$ It's clear that one can define a map $f:X_{a,b}\mapsto a e^{i2\pi b}$ that preserves the multiplicative and periodic structure of your space. But it's not obvious that your addition should be related to addition of complex numbers without adding some more constraints. $\endgroup$ – Raskolnikov Mar 1 '12 at 10:49
  • $\begingroup$ Sure, but the question is whether this is the only solution. $\endgroup$ – Gerenuk Mar 1 '12 at 10:51
  • $\begingroup$ I think this will depend on the interplay between your "addition" and the other operations. $\endgroup$ – Raskolnikov Mar 1 '12 at 10:54
  • $\begingroup$ As I added, all I know is that it is a field and I know the rules of multiplication. Will that restrict the only solution to be the complex number field? I can try something like $f(t)=X_{0.5,t}+X_{0.5, -t}$. With $f(t)^2=X_{0.5,0}(f(2t)+X_{1,0})$ I can sort of show that it corresponds to complex numbers. I'm just not sure whether this is rigorous. $\endgroup$ – Gerenuk Mar 1 '12 at 11:09
  • $\begingroup$ It might be possible that there is a magma satisfying your properties, but if you want it to be an algebra you'd have to specify over which ring $R$ it is one, how it's an $R$-module and how the multiplication is $R$-bilinear. $\endgroup$ – kahen Mar 1 '12 at 11:45

This is not really an answer (yet), just some thinking out loud that got too long for a comment.

So, as I understand it, you have a field $F$ (with the operations $+$ and $\cdot$) equipped with a surjection $f: \mathbb R^2 \to F$ and the following additional axioms (which hold for all $a, b, c, d \in \mathbb R$):

$$\tag{1} f(a,b+1) = f(a,b)$$ $$\tag{2} f(a,b) \cdot f(c,d) = f(ac, b+d)$$ $$\tag{3} f(a,0) + f(b,0) = f(a + b, 0)$$

Let's start by finding the additive and multiplicative identities $\mathbf 0$ and $\mathbf 1$ of $F$. This turns out to be pretty simple:

  • By axiom $(2)$, $f(a,b) \cdot f(1,0) = f(a,b)$, so $f(1,0) = \mathbf 1$.
  • By axiom $(3)$, $f(0,0) + f(0,0) = f(0,0)$. By subtracting $f(0,0)$ from both sides, we have $f(0,0) = \mathbf 0$.

From axiom $(2)$, it follows that $f(a,b) \cdot f(0,0) = f(0,b)$, but from the field axioms, we know that $x \cdot \mathbf 0 = \mathbf 0$ for any $x$. The only way to reconcile these results is to have, for all $b \in \mathbb R$:

$$f(0,b) = f(0,0)$$

This suggests one trivial model of these axioms: $F = \mathbb R$ and $f(a,b) = a$. You can easily check that this model satisfies the axioms $(1)$ to $(3)$ and makes $F$ a field and $f$ a surjection.

There is also at least one non-trivial model, as suggested by Raskolnikov in the comments: $F = \mathbb C$ and $f(a,b) = a e^{i 2 \pi b}$. In fact, for any positive integer $k$, $F = \mathbb C$ and $f(a,b) = a e^{i 2 k \pi b}$ yield another model of the axioms.

But of course, I assume you've already figured out everything up to this point, and now want to find out if there are any non-trivial models where $F \ne \mathbb C$. I'll have to think about that a bit more, but in the mean time, perhaps these notes may at least give you a new perspective on the problem.

OK, some more thoughts:

  • Axiom $(2)$ gives us a very useful decomposition: $f(a,b) = f(a,0) \cdot f(1,b)$. Using this and distributivity, we can generalize axiom $(3)$ to: $$\tag{3*} f(a,c) + f(b,c) = f(a+b,c)$$ This hopefully gets us a step closer to a general addition rule.

  • Let's call an element $x \in F$ "real" if there exists $a \in \mathbb R$ such that $f(a,0) = x$. The subset $F_R$ of real elements of $F$ contains both identities and is closed under addition, multiplication and taking of inverses, so it's a subfield of $F$. In fact, $a \mapsto f(a,0)$ is a field isomorphism from $\mathbb R$ to $F_R$.

    In all the models described above, $f(a,\frac12)$ is also real for all $a$. In fact, we have two cases: either $f(a,\frac12) = f(a,0)$ or $f(a,\frac12) = -f(a,0)$. Can we prove that these are all the possible cases? In particular, can we prove that $f(1,\frac12)$ is either $\mathbf 1$ or $-\mathbf 1$?

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  • $\begingroup$ My method to proceed was defining $f(t)$ as in the comments above. Basically this way I defined something like $sin$ and $cos$ which in the end possibly enables me to deduce all numbers from rule (3) $\endgroup$ – Gerenuk Mar 1 '12 at 15:48
  • $\begingroup$ F is an R-algebra, since f:Rx0→F is a ring homomorphism (so an injective field homomorphism). f(1,1/2) is a root of the polynomial X^2 - 1. If f(1,1/2) = 1, then f(1,1/4) is a root of the polynomial X^2 - 1. If we continue in this way, then either f(a,m/2^n) = a for all integers m,n or there is some f(1,1/2^n) = -1. In the first case, it'd be nice to use a continuity assumption (should be compatible with "age") that says if the function is identically 1 on a dense set, then it is identically 1 on the closure. In the second case, i = f(1,1/4) should give us F=C. $\endgroup$ – Jack Schmidt Mar 1 '12 at 16:58
  • $\begingroup$ For your last question: in a field, X^2-1 has only two roots, 1 and -1. $\endgroup$ – Jack Schmidt Mar 1 '12 at 17:31
  • $\begingroup$ In the second case, I am still needing some sort of continuity. b is chosen from the group R/Z, and on the group Q/Z it behaves like b→exp(2πib/n), but I think it can be sort of arbitrary on the direct complement (let $g:\mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z} \times \mathbb{R}(x)$ be an isomorphism, and I think F becomes C(x) instead of just C, but the map f is not even close to continuous or measurable or anything). $\endgroup$ – Jack Schmidt Mar 1 '12 at 17:45
  • $\begingroup$ Yeah, the multiplication is never used in the second coordinate, so we don't have the ring R as the second coordinate, we just have the abelian group that is an uncountable vector space over Q (which we mod out by one copy of Z). Without continuity, there are many possibilities, more or less any field of the right cardinality containing R. $\endgroup$ – Jack Schmidt Mar 1 '12 at 19:14

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