I am trying to find out whether there are mathematically important or useful properties (of some object(s)) that are nevertheless not invariant under some usual choice of isomorphism?

Are there any such "natural" examples of objects having some property (expressed in some language) which are non-invariant?

What I mean by "natural" is roughly that I want a non ad-hoc example, i.e. something actually used/known by some mathematician(s) that they find important for some purpose or other.

The background to my questioning is that I would like to probe whether some of the claims made by structuralist philosophers of mathematics, specifically that "structural" properties are the only mathematically relevant properties, hold any water. By structural properties I mean properties invariant under isomorphisms in some category (in whatever relevant sense).

So far, I have not been able to find a convincing counter-example (one that isn't ad-hoc) to the above structuralist claim, hence my question.

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    $\begingroup$ Perhaps properties such as "the first component of a vector $v \in \mathbb{R}^3$"? Automorphisms of $\mathbb{R}^3$ do not preserve the components of vectors. $\endgroup$ – Carl Mummert May 31 '17 at 18:21
  • $\begingroup$ The problem with the structuralist claim (as you describe it - and I have no reason to object to your description) is that it is useless: given any class of properties of some class of objects, you can define a category in which there is an arrow from object $A$ to object $B$ iff $A$ satisfies every property satisfied by $B$, so that the properties are trivially invariant under isomorphisms. So the really important part is the qualification "in whatever relevant sense": then the answers below indicate that the notion of "relevance" is nuanced and interesting. $\endgroup$ – Rob Arthan Jun 1 '17 at 23:09
  • $\begingroup$ @RobArthan Exactly! I was trying to see whether categorical structuralism ,(I believe it is called) which is sometimes taken to be a more refined version of structuralism (I think ?), holds up to scrutiny. I somewhat think it does hold up, but I was curious as to the claims of "isomorphism implies identity" questions surrounding structuralism in general. It seems that the category theoretic approach is a little more helpful since here one always specifies which morphisms are considered in what categories. $\endgroup$ – Bartuc Jun 2 '17 at 9:56
  • $\begingroup$ However, I still find it somewhat unsatisfactory as a philosophical position since it isn't always evident why such-and-such morphisms are considered and not others, but I guess that isn't the main point of structuralist positions like that anyhow. For example, I don't find it illuminating to be told that the Gelfand transform is a natural transformation of functors, but it does help to see it in action when applied to give a continuous calculus for normal elements in $C^\ast$-algebras. $\endgroup$ – Bartuc Jun 2 '17 at 10:00

In setting up the categories of Banach spaces (or Hilbert spaces or inner product spaces or normed spaces), we choose to take continuous linear mappings as the morphisms rather than isometric linear mappings. The metric structure is not invariant under this choice for the morphisms, but the resulting categories are far more interesting and useful.

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    $\begingroup$ Which, with regards to the OPs comments about "structuralism", points out an issue. An object can be an object of multiple categories. Two objects that are isomorphic in one category may not be in another. This doesn't really undermine the position of structuralism, but it makes what that position actually is more nuanced. $\endgroup$ – Derek Elkins Jun 1 '17 at 2:54
  • $\begingroup$ @DerekElkins could one say that the "identity" (read isomorphism class) of an object $X$ depends on which category $\mathcal{C}$ it is considered an object in? Which again depends on what the morphisms in $\mathcal{C}$ are (or chosen to be)? $\endgroup$ – Bartuc Jun 1 '17 at 8:26

An elliptic curve is said to be in Weierstrass normal form if it is defined by an equation of the form $$ y^2 = x^3 + ax + b $$ for constants $a,b$. This property is not preserved under automorphisms of elliptic curves: in fact every elliptic curve over a field of characteristic not 2 or 3 is isomorphic to an elliptic curve in Weierstrass form. Being in Weierstrass normal form is therefore not an intrinsic property of the elliptic curve itself, but of the presentation of the elliptic curve. The reason that Weierstrass normal forms are useful, despite not being preserved under isomorphism, is that it simplifies some calculations.

Something similar happens for matrices over $\mathbb C$. Every square matrix over $\mathbb C$ is similar to a square matrix in Jordan normal form. Clearly, the property of `being in Jordan normal form' is not preserved under similarities, which is the natural notion of isomorpism for square matrices in most situations. Again, the reason that we are studying the Jordan normal form (and similar decompositions) anyway, is that it helps when doing explicit calculations.

  • $\begingroup$ Thank you Marc. That is a very good elementary example I think! $\endgroup$ – Bartuc Jun 1 '17 at 8:21

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