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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?

მამუკა ჯიბლაძე
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    I disagree strongly with the claim that, in any context "one has to speak about pairs of such forms". You don't have to speak about anything - you can speak about more-or-less anything you want. You just have to be careful that, if you speak about something that depends on a choice, you pay attention to how that thing changes if you ever change your choice. – Will Sawin Apr 22 '21 at 21:08
  • @WillSawin Sorry, what I actually wanted to say there was that while it is a canonical way to decompose any square matrix $A$ into the sum of the symmetric matrix $(A+A^\intercal)/2$ and the skew-symmetric matrix $(A-A^\intercal)/2$, I don't see a canonical way to do such thing to a nondegenerate bilinear form without choice of a basis. – მამუკა ჯიბლაძე Apr 22 '21 at 21:12
  • Doesn't any pair of $n$-dimensional vector spaces over the same field have this property? I'm trying to understand the heart of the question. – Simon Wadsley Apr 22 '21 at 21:28
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    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. – Bertram Arnold Apr 22 '21 at 21:45
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    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, $\langle v, w\rangle_\epsilon = \frac1 2(\langle v, w\rangle + \epsilon\langle w, v\rangle)$. – LSpice Apr 22 '21 at 22:20
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    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. – Benjamin Steinberg Apr 22 '21 at 22:58
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    @SimonWadsley Well, that would count as an example I guess. The corresponding structure is a trivialization of a $GL(V)$-$GL(W)$-bitorsor. Further generalizing to a pair of vector bundles of equal rank, obstruction to trivializing this torsor should give most likely the top Chern class of $V^*\otimes W$. – მამუკა ჯიბლაძე Apr 23 '21 at 03:52
  • @BertramArnold Won't you need something like group completion for that? I mean, passing from $\coprod_nBGL(n)$ to $\mathbb Z\times\varinjlim_nBGL(n)$? – მამუკა ჯიბლაძე Apr 23 '21 at 03:55
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    @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?? – მამუკა ჯიბლაძე Apr 23 '21 at 03:59
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    @LSpice Of course! I was stupid enough to forget the canonical symmetric group actions on tensor powers of objects in any symmetric monoidal category! – მამუკა ჯიბლაძე Apr 23 '21 at 04:03
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    @მამუკაჯიბლაძე : re-your comment about the dual: restricting to the core groupoid of a category, you can see a contravariant functor as a covariant functor by plugging in $f^{-1}$ instead of $f$. This gives a reasonable definition of "canonical isomorphism" (in that it means exactly "it does not depend on the choice of a basis") – Maxime Ramzi Apr 23 '21 at 11:38
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    There's no canonical isomorphism between two proofs of the same theorem. Right now that's a bit of a joke, but perhaps someone can make it into a serious answer? – Zach Teitler Apr 24 '21 at 05:45
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    Related: Example of an unnatural isomorphism – sdcvvc Apr 25 '21 at 18:43
  • Expanding on Benjamin Steinberg's comment, an isomorphism between a finite dimensional vector space and its dual could be based on an bijection between chosen bases for each of them. This in turn is could be based on a proof that these have the same cardinality. Such a proof need not use dual bases and so will not produce a canonical bijection. As Mees de Vries answer points out, there is no canonical isomorphism between two sets of the same cardinality. So the non-canonical nature of the original isomorphism is a special case of Mees de Vries answer! – Kapil Apr 26 '21 at 03:31
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    @sdcvvc Oh thanks for finding that! This most likely renders mine as a duplicate... – მამუკა ჯიბლაძე Apr 26 '21 at 06:22
  • Mees de Vries's answer referenced by @Kapil. – LSpice Apr 26 '21 at 15:19
  • @ZachTeitler Sorry for reacting that late, but only now I realized that the question you link to has an answer involving homotopy type theory, which is (I believe) certainly no joke :D – მამუკა ჯიბლაძე Nov 19 '22 at 13:16

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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.

LSpice
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Mark Wildon
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    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? – მამუკა ჯიბლაძე Apr 23 '21 at 11:48
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    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 $\omega + 1$). – Bertram Arnold Apr 23 '21 at 13:50
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    @მამუკაჯიბლაძე: Indeed; to say it another way, there is a natural bijection $\mathrm{Perm}(X) \times \mathrm{Ord}(X) \cong \mathrm{Ord}(X) \times \mathrm{Ord}(X)$. – Peter LeFanu Lumsdaine Apr 23 '21 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).

Maxime Ramzi
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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.

Igor Khavkine
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    Even easier: every fibre $F_x$ is isomorphic to $F$, but not canoincally. – Konrad Waldorf Apr 23 '21 at 08:46
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    Similarly, any two $G$-torsors (sets on which $G$ acts freely and transitively) are isomorphic, but not canonically. – Konrad Waldorf Apr 23 '21 at 08:47
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    @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. – Igor Khavkine Apr 23 '21 at 09:28
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    Ok! But $B$ has to be simply connected in order to get a specific isomorphism from a flat connection. – Konrad Waldorf Apr 23 '21 at 11:57
  • @KonradWaldorf, re, what is $F$ in this notation except a name for some $F_y$, where the choice of which one doesn't matter? (Not a rhetorical question; I thought that's what it meant, but maybe I don't understand.) That is, what is the difference between saying "$F_x$ and $F_y$ are isomorphic, but not canonically" and "$F_x$ is isomorphic to $F$, but not canonically"? – LSpice Apr 23 '21 at 14:41
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    @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 https://en.wikipedia.org/wiki/Fiber_bundle. I thought this is what Igor meant, and I was referring to this. – Konrad Waldorf Apr 23 '21 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.)

Greg Martin
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    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 $n$. – Oscar Cunningham Apr 25 '21 at 09:23
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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.

Alexander Betts
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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.

Steve Costenoble
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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.

Emil Jeřábek
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  • What does $\widetilde{\mathbb Q}_p$ mean? – LSpice Apr 26 '21 at 15:18
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    $\mathbb Q_p$ 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. – Emil Jeřábek Apr 26 '21 at 15:38
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    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.) – LSpice Apr 26 '21 at 15:39
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    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 $\mathbb Q$). – Emil Jeřábek Apr 26 '21 at 17:00
  • Ha, right you are. I'm not sure what I meant. – LSpice Apr 26 '21 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 $\operatorname{Perm}(X)$.

Mees de Vries
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    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 $\mathbb N$. – LSpice Apr 26 '21 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.

Bertram Arnold
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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.

Zach Teitler
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    What is your definition of $\mathbb{C}$? –  Apr 24 '21 at 10:17
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    @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 ${\alpha_1,\alpha_2} = {\pm i}$. – François Brunault Apr 24 '21 at 10:53
  • @FrançoisBrunault, what would it even mean to have a canonical isomorphism to a structure defined only up to isomorphism? – LSpice Apr 26 '21 at 15:18
  • @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 $\mathbb{R}[x]/(x^2+1)$. – François Brunault Apr 26 '21 at 16:07
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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).

user21820
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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.

Daniel Robert-Nicoud
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