Two kinds of equivalence: conjugate vs. isomorphic objects - MathOverflow

Two kinds of equivalence: conjugate vs. isomorphic objects

2011-11-02T12:19:25Z

Conjugate vertices in a graph¹ or conjugate elements of a group² are equivalent (indistinguishable, essentially the same) in one specific structural sense.

Isomorphic objects in a category are equivalent in another specific structural sense.

In both cases we don't look inside the objects but declare them equivalent from the outside.

In both cases equivalence has to do with isomorphism: with structure preserving maps between the structure the conjugate elements live in and itself (automorphisms) resp. with iso arrows between the elements themselves.

Define two objects $A,B$ in a category to be conjugate when there is an isomorphism endofunctor $F$ with $F(A) = B$.

Question 1: Is it true that any two isomorphic objects are conjugate (since there is an isomorphism endofunctor that permutes them)?

The reverse is most certainly false: There are categories with conjugate objects that are not isomorphic. E.g. the graphs

in the category of graphs over two fixed vertices (with graph homomorphisms as morphisms) are two such objects (#9 and #6 in the diagram below). Note that there is no morphism at all between these two graphs.

Question 2: Might it be the case that whenever two objects are conjugate-but-not-isomorphic there is no morphism between them? Or is this true only in special categories and/or special cases?

Question 3: How "normal" is it that a category contains conjugate-but-not-isomorphic objects?

Most of all I'd like to know how to think about this bewildering pair of equivalences in general terms.

Appendix

Here is the complete category of graphs over two fixed vertices and an arrow whenever there is a graph homomorphism. Compositions and identities are omitted.

The numbers are derived from the adjacency matrices: 0 = 00|00, 1 = 10|00, ..., 15 = 11|11.

Footnotes

¹ $x,y$ are conjugate iff there is a $g \in \text{Aut}(G)$ with $g(x) = y $.

² $x,y$ are conjugate iff there is a $g \in G$ with $gx= yg$.

Answer by Karol Szumiło for Two kinds of equivalence: conjugate vs. isomorphic objects

2011-11-02T20:44:27Z

Here's a counterexample to Question 2. Let $\mathbb{Z}$ be the totally ordered set of integers regarded as a category. Then

Distinct objects of $\mathbb{Z}$ are non-isomorphic.
There is a morphism between every two objects of $\mathbb{Z}$.
All objects of $\mathbb{Z}$ are conjugate (just apply the shift that maps one object to the other).

I suspect that a finite example may be more difficult to construct (or even not exist at all).

I also believe that notions of isomorphism and conjugacy are conceptually quite apart. The first one emphasizes "categories as universes" point of view, where we are interested in properties of objects in a category. On the other hand the notion of conjugacy seems to concern symmetries of the category itself, which emphasizes "categories as structures" point of view.

Answer by Will Sawin for Two kinds of equivalence: conjugate vs. isomorphic objects

2011-11-03T02:35:01Z

Question 2 is false for some finite categories. There are two objects $A$ and $B$ that both have an idempotent endomorphism to themselves. There is a morphism from $A$ to $B$ and one from $B$ to $A$, their composition is the idempotent endomorphism. (The best realization I have for this is two finite simple groups, neither being a subgroup of the other, without their automorphisms or with the same automorphism group.)

However, Question 2 is true for finite categories where all endomorphisms are invertible.

Suppose there is an arrow $A\to B$ and an isomorphism $F(A)=B$. $F$ must have some finite order $n$. Consider the arrows $F(A\to B),F(A\to B), F^2(A\to B),...,F^{n-1}(A\to B)$. These arrows form a cycle, therefore they all have inverses.