No canonical isomorphism [duplicate]

Question

I thought that it would be interesting to collect into a big list various instances of isomorphic structures with no preferred isomorphism between them. I expect the examples to be interesting since it seems that in such situations choice of a particular isomorphism is frequently an important kind of structure.

For some reason all examples that I could come up with are related to some sort of self-duality, although there must be others not related to any duality, and I am especially curious about the latter.

So let me do this: I will ask some questions about these self-dualities. If most of the answers are about these, I will not add the big list tag. If there are many examples of some other kind, then I will.

The simplest and most ubiquitous one, you have already guessed it: isomorphism between a finite-dimensional vector space and its dual. The choice of isomorphism amounts to a non-degenerate bilinear form. While we are at that, let me ask this: at a first glance, the fact that such forms can be (at least in characteristic 0) decomposed into the sum of a symmetric and a skew-symmetric form is just the consequence of the fact that eigenvalues of an involution are $\pm1$. But initially, given just a nondegenerate form, there is no involution present, unless we have another such form. So how to explain that such a decomposition still exists? Or does it in fact not, and one has to speak about pairs of such forms??

My subsequent examples will be just generalizations of the first.

A finite abelian group and its Pontryagin dual. If it is an elementary $p$-group, this is a particular case of the above (vector spaces over prime fields). What about the general case? Is a choice of isomorphism, i.e. a nondegenerate pairing $A\otimes A\to\mathbb Q/\mathbb Z$ a structure that is actually used somewhere? I've heard about Weil pairings but know too little about them to figure out whether they are an instance of such a thing.

Conjugacy classes of a finite group and its irreducible representations. Again, does choice of a bijection between these two sets come up somewhere in mathematics?

What comes next are examples when an isomorphism need not exist.

An isomorphism between an abelian variety and its dual. Is this used somewhere?

A diffeomorphism/PL-isomorphism/homeomorphism between a manifold and its dual. Here I don't even know what I am asking. Does this make sense at all?

Here I know what I am asking: a homotopy equivalence between a finite CW-complex and its Spanier-Whitehead dual. Do such equivalences have a name?

Related questions:

Are there situations when regarding isomorphic objects as identical leads to mistakes?

Equality vs. isomorphism vs. specific isomorphism

Later

Excuses to those who contributed extremely interesting answers and comments, but as it has been pointed out there actually exists a very similar question (with equally interesting answers). Besides, although closed, all of the answers will be accessible to everybody, right?

Answer 1

Let $X$ be a set. Permutations of $X$ are in bijection with total orderings on $X$, but (unless $\lvert X\rvert \le 1$) there is no canonical bijection.

In terms of Joyal's theory of species, the species of total orders and of permutations are not isomorphic (i.e. the functors are not naturally isomorphic). But for any particular set $X$, there is a bijection between $\operatorname{Ord}(X)$ and $\operatorname{Perm}(X)$.

This example is mentioned in the blog post A visual telling of Joyal’s proof of Cayley’s formula of Leinster, giving a version of Joyal's proof of Cayley's formula that there are $n^{n-2}$ labelled trees on an $n$-set. Leinster has a nice paper The probability that an operator is nilpotent on the arXiv that uses similar ideas to find the proportion of nilpotent $n \times n$ matrices with entries in a given finite field.

Answer 2

As was mentioned in the comments, the example of a vector space and its dual can be seen as being about to "two vector spaces of the same dimension".

Even in dimension one, the fact that two one-dimensional vector spaces aren't canonically isomorphic is what allows the existence of line bundles (in other words, the fact that $k^\times$ is usually nontrivial - when $k=\mathbb F_2$, it is, and thus we for instance that every compact manifold is $\mathbb F_2$-orientable).

Answer 3

In a fiber bundle $E \to B$ with typical fiber $F$, any two fibers $F_x$, $F_y$ over points $x,y \in B$ of the base are isomorphic (homeomorphic or diffeomorphic, depending on whether you are doing topology or differential topology), but not in a canonical way. A way to pick out a particular isomorphism is to introduce a connection on $E\to B$ and a path connecting $x$ and $y$, then using parallel transport. If the connection happens to be flat, than only the homotopy class of the path between $x$ and $y$ matters.

Answer 4

For a more elementary example: any two cyclic groups of order $n$ are isomorphic, but (when $n\ge3$) there is no preferred isomorphism between any two given cyclic groups of order $n$. (This is essentially the same as the fact that a cyclic group of order $n\ge3$ does not have a canonical generator.)

Answer 5

For $X$ a path-connected topological space, and two points $x,y\in X$, the fundamental group of $X$ based at $x$ is isomorphic to the fundamental group based at $y$, but not canonically. A choice of a path from $x$ to $y$ gives an isomorphism between these two fundamental groups (conjugate by the path), but there is no canonical choice in general.

Answer 6

For examples that don't come from duality or torsors, there are cases where we have short exact sequences that split, but not naturally. For example, the universal coefficients theorem for cohomology implies that we have a non-natural isomorphism $$H^n(X;A) \cong \text{Hom}(H_n(X;\mathbb Z),A) \oplus \text{Ext}(H_{n-1}(X;\mathbb Z),A)$$ for $X$ a space and $A$ an abelian group.

Answer 7

Being algebraically closed fields of characteristic $0$ and transcendence degree $2^\omega$, the fields $\mathbb C$, $\widetilde{\mathbb Q}_p$, and $\mathbb C_p$ are isomorphic for any prime $p$, but there is no preferred isomorphism between them.

Answer 8

A deliberately extreme example: an isomorphism of sets is a bijection, and two sets are isomorphic when they have the same cardinality. There is generally no preferred bijection between sets of the same cardinality. For example, there is no canonical choice of bijection between commonly used sets like $\mathbb N, \mathbb Z, \mathbb Q, \mathbb Z^2, \bar{\mathbb{Q}}$.

This ties in to Mark Wildon's example, too: if any two sets of the same cardinality had a canonical bijection between them, each finite set $X$ would have a canonical bijection to a set of the form $\{0, \ldots, |X| - 1\}$. This would give a preferred total ordering on $X$ (the one corresponding to $\leq$), which in turn would give a canonical bijection between $\operatorname{Ord}(X)$ and $\operatorname{Perm}(X)$.

Answer 9

For an example that does not (as far as I can tell) come from duality, a Drinfeld associator is an isomorphism between two operads in pro-groupoids (parenthesized braids and parenthesized chords, respectively), compare Dror Bar-Natan. On associators and the Grothendieck-Teichmuller group I. Selecta Math. (N.S.) 4:2 (1998), 183–212. It is a nontrivial result that such isomorphisms exist; once this is known, they are of course a torsor for the automorphism group (the Grothendieck-Teichmüller group) of either one of these objects.

Answer 10

$\mathbb{R}[x]/(x^2+1)$ is isomorphic to $\mathbb{C}$, but there’s not a canonical isomorphism as $x$ can map to $i$ or $-i$. I suppose it’s just a special case of $\{\pm i\}$ as a $\mathbb{Z}/2\mathbb{Z}$ torsor.

Answer 11

Algebraic closures of any given field are isomorphic, but there is no preferred isomorphism (unless the given field is already algebraically closed).

Answer 12

Minimal models

(Sullivan) minimal models of rational spaces are unique up to non-canonical isomorphism.

Minimal models of operads are unique up to non-canonical isomorphisms.