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12/15/2017 - 12:53 PM US EDST Please note: as a result of the decision by
mercio, Math_QED, user8795, HK Lee, kjetil, and b halvorsen to put the question on hold because it is unclear to them, I have added a further clarification which should address their concern. I hope it does, and I want to thank them for their help in making sure that this question is as clear as possible.

12/14/2017 - 11:54 US EDST - Please note: as a result of previous discussion with @Joppy in comments, I have completely reworded this question - I think the question is much clearer this way.


In standard group-theory, the inverse operation is always defined in terms of just one bijection, so far as I know (and I am the first to admit that I don't know very much!).

Is there a "probabilistic" flavor of group-theory in which the inverse operation is defined in terms of an EQUIVALENCE CLASS of bijections from which one bijection must be chosen at random?

Here is an alternative way to ask the same question, which may be clearer.

I'm asking if there exists anything like a non-standard probabilistic version of standard group-theory in which:

1) the bijection which establishes a group's inverse can be selected from a set of equally acceptable alternative bijections which are part of the definition of the group;

2) the definition of the group also includes an instruction for how to choose the operative bijection when instantiating a copy of the group.

With respect to (2), there might be two different kinds of instructions - one kind for a single "one-off" construction of the group, and one kind for repeated constructions of the group such as might occur during a Monte Carlo run or other type of experiment.

For "one-off" constructions of the group, the instruction would simply be:

2)a. "randomly choose one of the alternative bijections provided in the definition of the group".

For repeated constructions of the group, the instruction would be something like the following for a case in which the definition of the group specified four different alternative bijections $B_1$, $B_2$, $B_3$, $B_4$.

2b). "for repeated constructions of the group 8k times, use the four alternative bijections $B_1$, $B_2$, $B_3$, $B_4$ in the ratios 1:3:3:1 respectively."

I hope the above makes my question clearer.

Edited 12/14/2017 07:19 US EDST to add:

Actually, in the case I'm looking at, there would be an "(super)class of equivalence classes", because there would be different equivalence classes for different subsets of the entire set. But the (super)class would not itself be an equivalence class, except indirectly by virtue of the equivalence classes which it contains.

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closed as unclear what you're asking by mercio, user370967, user8795, HK Lee, kjetil b halvorsen Dec 15 '17 at 15:21

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Is the big set that both these subsets drawn from a group? It's unclear in what sense you mean "inverse" to each other. $\endgroup$ – Joppy Dec 14 '17 at 9:46
  • $\begingroup$ No, because what we're trying to figure out is if a group can be defined on the big set, and therefore need to figure out if we do or don't have a group-theoretic inverse. Imagine the case of two opposed hypercubes in a larger hypercube, but instead of considering just the vertices of these two hypercubes, consider two "clouds" with the same number of points within each of the two hypercubes. We certainly have good reason to pair-off vertex to vertex. but suppose we can't find any good reason to pair off points within the clouds in some particular way. $\endgroup$ – David Halitsky Dec 14 '17 at 9:55
  • $\begingroup$ So, could we phrase the question more precisely? Here is my best guess at your question: Suppose we have subsets $A, B \subseteq X$, and a bijection $i: A \to B$. Under what conditions is it possible to put a multiplication on $X$ (or perhaps a subset of $X$ containing $A, B$) such that $X$ becomes a group, and $i$ is the inversion operation on $A$? $\endgroup$ – Joppy Dec 14 '17 at 11:32
  • $\begingroup$ @Joppy - I think that's correct, but not sure. When you say "X becomes a group",, what are the elements of this group? Are they subsets A and B, or are they the elements of A and B. If the elements of the group are the elements of A and B, then yes, your formulation is correct. Thank you very much for taking the time to teach me here - I am very grateful. $\endgroup$ – David Halitsky Dec 14 '17 at 12:37
  • $\begingroup$ I'm not sure - you are the one formulating the question, so you should say whether you want elements of the group to be elements of $X$, or subsets of $X$! $\endgroup$ – Joppy Dec 14 '17 at 12:39
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In a comment above, @ChristianBlatter wrote:

"There is "homotopy equivalence" which comes from a weakening of bijectivity. E.g., a full 3D donut is homotopy equivalent to $S^1$,

Since his response provides a possible KNOWN context which might be a good framework for me investigate the formal situation with which I am presently confronted, I am unilaterally upgrading his comment to an official answer. I hope that he doesn't mind!

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