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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?

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    $\begingroup$ There is a $C_2$-action on the category of finite-dimensional vector spaces and isomorphisms which sends a vector space to its dual and a linear isomorphism to the inverse of its transpose (this is a baby case of the cobordism hypothesis). A vector space with a symmetric non-degenerate bilinear form is a homotopy fixed point of this action, and antisymmetric forms have a similar description. In general, given an object of your category, a lift to homotopy fixed points need not exist, and if it does, need not be unique; that seems to encompass most of your examples. $\endgroup$ – Bertram Arnold Apr 22 at 21:45
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    $\begingroup$ The involution needed to decompose a bilinear form $\langle\cdot, \cdot\rangle$ into its symmetric and antisymmetric $\langle\cdot, \cdot\rangle_\pm$ parts is the "swap the arguments" involution. (I think this might be related to what @BertramArnold is saying, but I am not enough of a geometer really to be sure.) Namely, $\langle v, w\rangle_\epsilon = \frac1 2(\langle v, w\rangle + \epsilon\langle w, v\rangle)$. $\endgroup$ – LSpice Apr 22 at 22:20
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    $\begingroup$ If the question is just about specific objects, then the isomorphism of a finite dimensional vector space with its dual is no different than the isomorphism between any two vector spaces of the same dimension. What's interesting there is that the dual is functorial but the isomorphism is not natural. $\endgroup$ – Benjamin Steinberg Apr 22 at 22:58
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    $\begingroup$ @BenjaminSteinberg I keep oscillating between understanding and misunderstanding your comment. Certainly "no natural isomorphism" is more precise and focussed than "no canonical isomorphism". But on the other hand the functor you mention is contravariant, and there can be no natural transformations between identity and this functor. Or does it still make sense to talk about isomorphisms between covariant and contravariant self-equivalences?? $\endgroup$ – მამუკა ჯიბლაძე Apr 23 at 3:59
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    $\begingroup$ Related: Example of an unnatural isomorphism $\endgroup$ – sdcvvc Apr 25 at 18:43

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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.

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    $\begingroup$ A nice example! I believe $\operatorname{Ord}(X)$ has a canonical $\operatorname{Perm}(X)$-torsor structure, no? In fact, $\operatorname{Ord}(X)$ is the set of bijections between the cardinal of $X$ and $X$. So in fact there is a subtlety when $X$ is infinite. Shall one talk about well-orderings in this case? $\endgroup$ – მამუკა ჯიბლაძე Apr 23 at 11:48
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    $\begingroup$ For $X$ infinite, $\operatorname{Perm}(X)$ does not act transitively on the set of well-orderings of $X$ (consider $X = \mathbb N$ with the two well-orderings corresponding to the ordinals $\omega$ and $\omega + 1$). $\endgroup$ – Bertram Arnold Apr 23 at 13:50
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    $\begingroup$ @მამუკაჯიბლაძე: Indeed; to say it another way, there is a natural bijection $\mathrm{Perm}(X) \times \mathrm{Ord}(X) \cong \mathrm{Ord}(X) \times \mathrm{Ord}(X)$. $\endgroup$ – Peter LeFanu Lumsdaine Apr 23 at 17:22
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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).

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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.

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    $\begingroup$ Even easier: every fibre $F_x$ is isomorphic to $F$, but not canoincally. $\endgroup$ – Konrad Waldorf Apr 23 at 8:46
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    $\begingroup$ Similarly, any two $G$-torsors (sets on which $G$ acts freely and transitively) are isomorphic, but not canonically. $\endgroup$ – Konrad Waldorf Apr 23 at 8:47
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    $\begingroup$ @KonradWaldorf Of course, that is the core "non-canonicity". What I like about my version is that the extra data needed to provide an isomorphism itself has a nice geometric interpretation: a connection. I think that's partly what the OP wanted to see. $\endgroup$ – Igor Khavkine Apr 23 at 9:28
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    $\begingroup$ Ok! But $B$ has to be simply connected in order to get a specific isomorphism from a flat connection. $\endgroup$ – Konrad Waldorf Apr 23 at 11:57
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    $\begingroup$ @LSpice: there is a standard definition of "fibre bundle with typical fibre $F$"; in which $F$ is a fixed topological space that is part of the structure; for example, see en.wikipedia.org/wiki/Fiber_bundle. I thought this is what Igor meant, and I was referring to this. $\endgroup$ – Konrad Waldorf Apr 23 at 19:04
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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.)

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    $\begingroup$ Even more elementary: any two sets of cardinality $n$ are isomorphic, but (when $n\geq 2$) there is no preferred isomorphism between any two given sets of cardinality $n$. $\endgroup$ – Oscar Cunningham Apr 25 at 9:23
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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.

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    $\begingroup$ If you squint, this is a special case of Igor's example, but the geometric intuition is different. $\endgroup$ – Alexander Betts Apr 23 at 19:22
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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.

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    $\begingroup$ Well one can make it come from torsors: $H^n(-;-)$ is an $\operatorname{Ext}(H_{n-1}(-;\mathbb Z),-)$-torsor over $\operatorname{Hom}(H_n(-;\mathbb Z),-)$ $\endgroup$ – მამუკა ჯიბლაძე Apr 23 at 21:25
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    $\begingroup$ Fair point. Though that's perhaps the fanciest way I've seen of saying that the Ext group is the kernel of the surjective map from $H^n(-;-)$ to $\text{Hom}(H_n(-;\mathbb Z),-)$... $\endgroup$ – Steve Costenoble Apr 23 at 22:36
  • $\begingroup$ Well in fact if you want to express the fact that there is no natural splitting, it is more or less optimal to say that this torsor is nontrivial (in the slice of an appropriate category (of certain functors)). $\endgroup$ – მამუკა ჯიბლაძე Apr 27 at 4:30
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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.

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  • $\begingroup$ What does $\widetilde{\mathbb Q}_p$ mean? $\endgroup$ – LSpice Apr 26 at 15:18
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    $\begingroup$ $\mathbb Q_p$ is the field of $p$-adic numbers, $\widetilde{\mathbb Q}_p$ is the algebraic closure of $\mathbb Q_p$, and $\mathbb C_p$ is the completion of $\widetilde{\mathbb Q}_p$ with respect to the $p$-adic valuation. $\endgroup$ – Emil Jeřábek Apr 26 at 15:38
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    $\begingroup$ Ah, I'm used to a bar rather than a tilde for algebraic closure, but I understand that when completion and algebraic closure interact only one can get the bar. Thanks! (I guess I should have figured it couldn't be the completion, since that would give a non-algebraically-closed, hence not isomorphic to $\mathbb C$, field.) $\endgroup$ – LSpice Apr 26 at 15:39
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    $\begingroup$ Right. Also note that there is not much point in completing $\mathbb Q_p$, as it is already complete (or rather, it itself arises as a completion of $\mathbb Q$). $\endgroup$ – Emil Jeřábek Apr 26 at 17:00
  • $\begingroup$ Ha, right you are. I'm not sure what I meant. $\endgroup$ – LSpice Apr 26 at 17:01
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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 $\operatorname{Perm}(X)$.

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    $\begingroup$ It'd probably be easier to agree on a preferred bijection $\mathbb N \to \mathbb Z$ than to get everyone to agree on the same subset of $\mathbb Z$ to call $\mathbb N$. 😁 $\endgroup$ – LSpice Apr 26 at 15:20
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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.

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

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$\mathbb{R}[x]/(x^2+1)$ 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.

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    $\begingroup$ What is your definition of $\mathbb{C}$? $\endgroup$ – Oniqa Apr 24 at 10:17
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    $\begingroup$ @Oniqa One can define $\mathbb{C}$ as an algebraic closure of $\mathbb{R}$. It is unique only up to isomorphism. Explicitly one can take $\mathbb{R}[\alpha_1,\alpha_2]/(\alpha_1+\alpha_2,\alpha_1 \alpha_2 -1)$ and then $\{\alpha_1,\alpha_2\} = \{\pm i\}$. $\endgroup$ – François Brunault Apr 24 at 10:53
  • $\begingroup$ @FrançoisBrunault, what would it even mean to have a canonical isomorphism to a structure defined only up to isomorphism? $\endgroup$ – LSpice Apr 26 at 15:18
  • $\begingroup$ @LSpice Indeed, this quotient of $\mathbb{R}[\alpha_1,\alpha_2]$ is just one algebraic closure. There is no canonical isomorphism with the algebraic closure someone else may come up with, like $\mathbb{R}[x]/(x^2+1)$. $\endgroup$ – François Brunault Apr 26 at 16:07
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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.

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    $\begingroup$ Is there any nice interpretation for the non-canonical isomorphism once it is selected? $\endgroup$ – Igor Khavkine Apr 24 at 21:04

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