6
$\begingroup$

If X and Y are non-isomorphic objects, then "[is / is not] isomorphic to [ X / Y ]" are invariants that distinguish X and Y. You can also do things like take an object Z that is not isomorphic to Y, and then "is isomorphic to X or isomorphic to Z" is another invariant that distinguishes X and Y. Similarly, if W is isomorphic to X, then "is isomorphic to W" works too.


Are there any objects X and Y known to be non-isomorphic, but all known distinguishing invariants

(a) use the concept or definition of isomorphism
or
(b) use the concept or definition of being isomorphic to some object Z, where Z is isomorphic to X or Y

Improve this question
$\endgroup$
1
  • 4
    $\begingroup$ If this question has a definitive answer, then the answer is surely "yes". If you are looking for examples, then shouldn't this be community wiki? $\endgroup$
    – Alex B.
    Commented Oct 7, 2010 at 4:45

4 Answers 4

Reset to default
15
$\begingroup$

Hereditarily indecomposable Banach spaces are strange objects that fail to be isomorphic to any of their proper subspaces. However, in a certain sense those subspaces are not "interestingly different" from the spaces themselves. So there is no hope of finding an invariant to distinguish between the space and a subspace. The way the proof actually works is that one proves that every operator from the space to itself can be approximated, in a certain sense, by a multiple of the identity, and hence either has no hope of being an isomorphism or is Fredholm with index zero. But an isomorphism to a proper subspace would have to be Fredholm with positive (or infinite) index.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ +1 for not mentioning Maurey's coauthor on one of the relevant papers... $\endgroup$
    – Yemon Choi
    Commented Oct 7, 2010 at 5:27
4
$\begingroup$

Let $X$ be an appropriate graph on $2^{100}+2$ vertices and $Y$ be the complement. Then "parity of the number of edges" is an invariant which would distinguish them, were it known. For example take the vertex pairs in a standard order and insert an edge in $X$ if the corresponding binary digit of $B_2$ is 1. Here $B_2$ is Brun's Constant which is not known to any great accuracy.

Improve this answer
$\endgroup$
4
$\begingroup$

Non-principal ultrafilters on the natural numbers don't have enough invariants to distinguish the non-isomorphic ones. (Two such ultrafilters are called isomorphic if one is sent to the other by a permutation of the underlying set $\mathbb N$.)

Improve this answer
$\endgroup$
2
  • $\begingroup$ Andreas: Reference? $\endgroup$
    – user6976
    Commented Oct 8, 2010 at 21:15
  • 1
    $\begingroup$ The number of isomorphism classes is $2^c$ where $c$ is the cardinal of the continuum. So even if we allowed a countable infinity of invariants, each with $c$ values, which is considerably more than the "known" invariants mentioned in the question, they wouldn't suffice (as `$c^{\aleph_0}$ is only $c$). A more technical answer is that there are (if Mahlo cardinals are consistent) models where there are lots of non-isomorphic selective ultrafilters, all generic, with respect to the same homogeneous notion of forcing, over the sets hereditarily ordinal definable from reals. $\endgroup$
    – Andreas Blass
    Commented Oct 11, 2010 at 0:33
3
$\begingroup$

In Nekrashevych, Volodymyr A minimal Cantor set in the space of 3-generated groups. Geom. Dedicata 124 (2007), 153--190, an infinite family of 3-generated groups is constructed which are locally non-distinguishable, that is for every $n$ and every finite generating set of one group one can find a finite generating set of another group such that the $n$-balls of the two cayley graphs corresponding to these two generating sets are isomorphic. That means these groups cannot be distinguished by any "standard" group theoretic invariants. They are distinguished, though, by certain topological dynamics objects associated to these groups (these groups are iterated monodromy groups in the sense of Nekrashevych).

Improve this answer
$\endgroup$

You must log in to answer this question.