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I have two binary functions $f_i: D \times D \rightarrow \mathbb{Q}$ for $i \in \{1,2\}$ with the following properties:

  • $f_i(b,a) = f_i(a,b) \cdot g_i(a,b)$
  • $f_i(a,c) = g_{i,1}(a,b) \cdot f_i(a,b) + g_{i,2}(b,c) \cdot f_i(b,c)$

for a certain functions $g_i(a,b)$ which does not involve $f_j(b,a)$ for any $j$, and certain functions $g_{i,1}(a,b)$, $g_{i,2}(b,c)$ which do not involve $f_j(a,c)$ for any $j$.

The whole point of this is that if $f_1(a,b)$ and $f_2(a,b)$ are known, one can compute $f_1(b,a)$ and $f_2(b,a)$ (since $g_1(a,b)$ doesn't mention $f_1(b,a)$ nor $f_2(b,a)$). Similarly, the relation of $a$, $c$ can be computed if the relations of $a$, $b$ and $b$, $c$ are known respectively.

The best naming which comes to my mind is symmetry and transitivity, respectively. Both terms seem to be wrong in this context as they refer to relations. For the first property, commutativity also seems to be similar, but this would mean $f(b,a) = f(a,b)$.

So, are there mathematical terms for these properties or similar properties? If not, which terms would you use?

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What is $E$???? – Ittay Weiss Apr 18 '13 at 21:08
A function satisfying $f(x,y)=f(y,x)$ would definitely be called symmetric Since your definition suggests that it is something less than symmetric, you might try skew-symmetric or twisted. (People might object to skew-symmetric. That has a rather special meaning in linear algebra.) – rschwieb Apr 18 '13 at 21:10
Pardon my obtuseness, but doesn't almost every function $f$ have this property? One can take $g(a,b) = f(b,a)\div f(a,b)$ and that provides a $g$ for any function for which $f(a,b)=0$ implies $f(b,a) = 0$. – MJD Apr 18 '13 at 21:15
You're right, in my case the point is that $g$ is pretty simple, particularly it doesn't contain $f(b,a)$. – chs Apr 18 '13 at 21:20
up vote 1 down vote accepted

If $f$ satisfies $f(b,a)=f(a,b)\cdot g(a,b)$ you might call symmetric up-to $g:D\times D\to \mathbb Q$. But, it's not a very interesting concept without further conditions on $g$ and/or $f$ since you can take $g(a,b)=\frac{f(b,a)}{f(a,b)}$ (if $f(a,b)\ne 0$) and $f$ will be symmetric up-to $g$.

As for the second condition, it looks much like the triangle inequality for $f$ up-to $g_1,g_2$. Again, without further conditions this notion seems too broad to me (not as broad as symmetry up-to $g$ though).

So, you might call $(D,f,g,g_1,g_2)$ a partial (i.e., $d(x,x)=0$ is not demanded) metric space (valued in $\mathbb Q$) up-to a $(g,g_1,g_2)$ deformation. But without any further properties of these functions I doubt such spaces will have a nice theory.

Added after the comment above that your $g$ does not contain $f(a,b)$: then the structure functions $g,g_1,g_2$ are very special and far from arbitrary functions. Then there is hope. You might want to add that information (i.e., how special these functions are) in the question in order to get better results.

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Thanks! I tried to expand on $g$ and $g_1, g_2$. – chs Apr 18 '13 at 21:34

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