Comments on: Equivalence relations

By: Jon Awbrey

Jon Awbrey — Sun, 20 May 2012 17:54:25 +0000

See “Minimal Negation Operator” for a concise way of describing partitions in propositional calculus.

By: Walter Sinclair

Walter Sinclair — Mon, 17 May 2010 06:21:21 +0000

A structure’s isomorphism class may be asymmetric — it may lack any non-trivial automorphism group — however, it may have a completion within a more blessed superstructure. The connectivity equivalence classes, for instance, may seem to have no more group to them than the passage of one point to another, until the key is supplied.

By: Peter Cameron

Peter Cameron — Thu, 06 May 2010 08:19:52 +0000

It depends how you look at it, I think.

On the one hand, for any kind of structure at all, there are so many structures (even in a single isomorphism class) that they don’t form a set, and I get a bit nervous about this.

On the other hand, if you don’t mind that, then the cartesian product of the symmetric groups on the isomorphism classes is a group whose orbits are these classes. (The elements are just those mappings which act on each object as an isomorphism, so it isn’t really cheating – or is it?)

By: Qiaochu Yuan

Qiaochu Yuan — Thu, 06 May 2010 03:12:01 +0000

I might be mistaken about this, but in most categories that I can think of, the equivalence relation coming from isomorphism doesn’t come from a group action in any natural way. For example, the category of topological spaces.

By: Peter Cameron

Peter Cameron — Thu, 01 Apr 2010 09:00:17 +0000

Douglas: I agree that you could argue that the transitive closure of any symmetric relation (or, indeed, the symmetric and transitive closure of any relation) is an equivalence relation. Not all of these are orbit relations. However, some are. If the moves (like your cycle switches) are invertible, then they generate a group.

A relation may be messy and still be an orbit relation. Groups can be messy things!

Matrices with given determinant over a field form an orbit of the special linear group acting by right multiplication.

In my previous comment, second paragraph, I meant “preserve the equivalence classes” not “preserve the equivalence relation”.

By: Douglas Stones

Douglas Stones — Wed, 31 Mar 2010 23:42:20 +0000

Some examples spring to mind. You could define an equivalence relation on the set of Latin squares by: L and L’ are equivalent if L’ can be formed from L by a sequence of cycle switches.

I’ve recently been studying the set of isotopisms theta for which there exists a Latin square L and theta(L)=L (i.e. theta is an autotopism). The equivalence “is (or is not) an autotopism of some Latin square” amongst isotopisms is quite messy.

Also, is there some group orbit that describes the equivalence class of matrices with determinant X?

By: Peter Cameron

Peter Cameron — Wed, 31 Mar 2010 16:42:15 +0000

I accept that the first paragraph is a possible counterexample. If I knew more about foliations I would no doubt simply agree.

The second paragraph doesn’t meet the bill since every equivalence relation is a group orbit relation – groupoids not required. Just take all permutations of the set which preserve the equivalence relation. I don’t regard this as properly answering my own question, of course.

By: javier

javier — Wed, 31 Mar 2010 15:30:43 +0000

Concerning the “All eq. relations are orbit actions”, I cannot quite agree with that, unless the statement is relaxed a bit. An example that comes to mind is the space of leaves of a foliation. Whilst in many cases foliations on manifolds come from some group action, that doesn’t need to be the case. Something similar could be said for orbifolds.

Of course, one might argue that these geometrical situations equivalence classes still “look like orbits”. And they are in a way, if we relax “orbits under a group action” to “orbits under a groupoid action”. To each equivalence relation R (seen as a subset of XxX) in a set X one can associate a groupoid (the “equivalence groupoid”) with objects X and arrows indexed by elements of R. Composition is given by (y,z)*(x,y) = (x,z), well defined by transitivity of the equivalence relation, identity in x is (x,x) (in R by reflexivity) and inverse of (x,y) is (y,x) (in R by symmetry). The equivalence classes determined by R coincide of course with the orbits by the action of the equivalence groupoid, so in that sense your statement holds true.