What is the definition of "canonical"?

Question

I just received a referee report criticizing that I would too often use the word "canonical". I have a certain understanding of what "canonical" should stand for, but the report shows me that other people might think differently. So I am asking:

Is there a definition of "canonical"?
What are examples where the use of "canonical" is undoubtedly correct?
What are examples where the use of "canonical" is undoubtedly incorrect?

VERY LATE EDIT: I just came across this wonderful passage written by R.Lipschitz, In a correspondence with André Weil (Oeuvres, vol. 2, page 558):

I can assure you, at any rate, that [...] my results are invariant, probably canonical, perhaps even functorial.

Nice question but the math police will probably close it down. I don't think there is a canonical definition of 'canonical'. — Idoneal, Mar 28 '10 at 18:35
I think this question is fine, because: it admits answers where the respondent can bring external knowledge to bear; it is based on a particular point of interest and importance to practising mathematicians, esp. those in early stages of their careers; and there is some sense of consensus/tradition, so that something like a "right" or "wrong" answer could be attempted. — Yemon Choi, Mar 28 '10 at 18:43
I think this question is a great example of how bringing in context can make a question less likely to be closed. Without the comment about the referee report, I probably would have thought the OP was being a wag, but in that context the question makes a lot of sense. — Ben Webster, Mar 28 '10 at 19:26
I'm surprised that noone has complained about the term "canonical bundle" in algebraic geometry, meaning the top exterior power of the cotangent bundle. Let me do so here. — Allen Knutson, Mar 22 '14 at 16:57
So according to the late edit, does canonical means invariant? :) — Hosein Rahnama, Aug 22 '16 at 19:17
It's clear that there is no clear or single answer as to what canonical means — luysii, Jul 17 '18 at 17:58
@AllenKnutson the funny thing is that the canonical bundle (and its powers, positive or negative) are the only "functorially constructed" line bundles on a variety that have a chance to be ample (well, it is not a theorem, but this opinion was expressed to me by an expert in birational geometry). So in some sense there is not much to complain about. — , May 31 '19 at 18:21
Blogging by @KevinBuzzard now complements his top-voted answer. — Francois Ziegler, Jun 07 '19 at 22:33

score 70 · Answer 1 · edited Aug 23 '16 at 07:30

70

I always had the following working definition of canonical (which I think Gordon James told me and he might have said it was due to Conway? Not sure): a map $A\to B$ is canonical if you construct a candidate, and the guy in the office next to you constructs a candidate, and you end up with the same map twice.

Somehow there is something more to it than that though. For example if $A$ is an abelian group and we want a map $A\to A$ then I will choose the identity, but I know for sure that the wag in the office next door to me will choose the map sending $a$ to $-a$ because that's his sense of humour. What has happened here is that there are in fact two canonical maps $A\to A$. This issue shows up in class field theory, which is an isomorphism between two rather fancy abelian groups $X$ and $Y$, and where no-one could decide for a long time which one of the two canonical isomorphisms was "best". So you often see statements in number theory papers saying "we normalise our class field theory isomorphisms so that geometric Frobenii go to uniformizers" (the alternative being the inverse of this). It also shows up in the Weil pairing on an elliptic curve: it's canonical, but because we're in an abelian situation, its inverse is too. So you see in e.g. Katz-Mazur an explicit spelling out of which of the two canonical choices one is going to make (and hang all the non-canonical ones!).

edited Aug 23 '16 at 07:30

A_S

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answered Mar 28 '10 at 20:27

Kevin Buzzard

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I like the working definition, but I can't trust my neighbors with that kind of responsibility. To paraphrase something I hear very often from one of my colleagues: "You mean a (blah) that would occur in real mathematics, not any (blah) that I would come up with." – François G. Dorais Mar 28 '10 at 20:40
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@Francois: somehow it's worse than that with the class field theory isomorphism. It's not that you can't trust your neighbour---there really is a 50-50 chance with that one. But somehow what saves us is that it's not that there's one canonical pair of maps, there really are two canonical maps, so if you were to confer beforehand then you'd both get the same map, whereas however much conferring you did, if you were presented with two abstract 1-dimensional real vector spaces where you couldn't see anything at all but fog, you'd never both manage to choose the same basis. – Kevin Buzzard Mar 28 '10 at 20:52
@Francois: the class field theory situation is just a fancy variant of your shoe puzzle in your answer. – Kevin Buzzard Mar 28 '10 at 21:52
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@Kevin: I like your examples. I think they illustrate very well how the line between what I called 'uniform' and 'canonical' is sometimes blurry. You appear to be one who regards 'uniform' as basically the same as 'canonical'. That's fine with me but many will disagree. I would prefer to avoid 'canonical' altogether (though I often forget force of habit) and use 'uniform' instead. In Computability Theory, for example, I rarely see the word 'canonical' but 'uniform' is used all over the place. – François G. Dorais Mar 28 '10 at 22:06
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Please tell me people do not use the plural Frobenii! – Mariano Suárez-Álvarez Mar 28 '10 at 23:34
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I've heard it too. I've even heard the verb "Frobenate" for applying the Frobenius map to something. – Tyler Lawson Mar 29 '10 at 04:16
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I once saw the misspelling "Frobenious", followed by someone's suggestion that this was the adjectival form. – Mark Meckes Mar 29 '10 at 13:00
Mark, I think that "misspelling" is quite intentional. Poor Frobenius. – Todd Trimble Feb 16 '13 at 15:40
@Todd: I doubt it. The person who wrote that misspelling rarely spelled a name of 8 or more letters correctly. – Mark Meckes Feb 16 '13 at 19:13
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But Mark, I know for a fact that some people use that misspelling quite intentionally! – Todd Trimble Feb 19 '13 at 15:48
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I believe that, Todd, I just doubt it was the case in the instance I saw. – Mark Meckes Feb 21 '13 at 23:24
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Mathscinet gives 10 hits for "Anywhere=(Frobenious)" but it looks like they are all just spelling errors. – Gerry Myerson Jul 17 '18 at 23:42
@FrançoisG.Dorais can you trust yourself with that kind of responsibility? I am not implying anything, I just would not trust myself really. – May 31 '19 at 18:23
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@MarianoSuárez-Álvarez Herr Scholze himself is using Frobenii, so there is not much choice by this point (http://www.math.uni-bonn.de/people/scholze/Berkeley.pdf) – May 31 '19 at 18:49
By this definition (comparing with the other answers), is this a canonical answer? – R. van Dobben de Bruyn Dec 16 '20 at 01:12
"There are in fact two canonical maps $A \rightarrow A$" What about the zero map? – seub Aug 22 '21 at 23:46

François G. Dorais · Answer 2 · 2010-03-28T20:00:27.770

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I think there is a multi-level classification associated to "canonicalness," which explains why some clashes of definition occur.

Arbitrary — No requirements.
Uniform — There may be a few options but these options can be selected by making a few global choices.
Canonical — As in the uniform case, but there is only one natural choice of options which applies globally.

Canonical examples à la Russell:

Choose one sock from each pair in a collection of sock pairs — There is no way to make a uniform choice.
Choose one shoe from each pair in a collection of shoe pairs — There are two obvious global solutions, left shoe or right shoe, but no way to prefer one over the other.
Choose one object from each set in a collection of sets each consisting of a bowtie and possibly other items — There is only one obvious global solution.

I think the main point of contention is distinguishing uniform and canonical. Some will argue that it's not canonical if there is a choice to be made, while some will argue that a finite number of global choices is still canonical.

There is yet another use of canonical to mean something like 'universally sanctioned' (this is closer to the religious term). The second occurrence of canonical above is of this type.

edited Mar 28 '10 at 20:00

answered Mar 28 '10 at 19:36

François G. Dorais

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@François: In my example about ring homomorphisms Z[x] -> R I think there are two global solutions, but one seems clearly better than the other. Would you consider this "uniform" or "canonical"? – Reid Barton Mar 28 '10 at 20:00
@Reid: I'm not very fond of the distinction and I usually prefer the term uniform. That said, I think $x \mapsto r$ is arguably more natural than $x \mapsto -r$. – François G. Dorais Mar 28 '10 at 20:04
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@Reid: I think the distinction is that Z[x] implicitly comes equipped with a pointer to the generator x. Or I guess one can define Z[x] by the fact that it represents the forgetful functor from Ring to Set, and then I imagine that the two functors given by sending x to r and x to -r are naturally isomorphic, so the issue is moot. – Qiaochu Yuan Mar 28 '10 at 20:37
FWIW this answer above isn't how I think of things. My interpretation of the shoe question is that there are two canonical ways to do things there. It's not the finiteness that matters: there are two isomorphisms between the group of cube roots of unity in Q-bar and the group of cube roots of unity in an alg closure of Z/2Z, but neither are canonical. Similarly there are two canonical bijections R=Hom(Z[x],R), and both are canonical, but one is eminently more sensible in this particular case. But that's OK. It's not about sensibleness, it's about distinguishability. – Kevin Buzzard Mar 28 '10 at 21:41
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@Kevin: There are more than two choices for Hom(Z[x],R) = R, viz. x, -x, 1+x, 1-x,... all generate Z[x]. They are all equivalent, but I wouldn't say that they are all canonical. – François G. Dorais Mar 28 '10 at 21:59
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Yeah, I was wrong about there being only two, but the point remains. – Reid Barton Mar 28 '10 at 22:05
A particularly good example of 'canonical' which agrees with Kevin's view is the notion of 'canonical function' as in the Canonical Ramsey Theorem. – François G. Dorais Mar 28 '10 at 22:10
(Sorry, my previous comment wasn't supposed to sound that harsh.) – Reid Barton Mar 28 '10 at 22:12
(@Reid: I don't think it sounded harsh, and I still think the Hom(Z[x],R) = R example is a very good one.) – François G. Dorais Mar 28 '10 at 22:19

score 28 · Answer 3 · edited Apr 13 '17 at 12:58

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Not a definition, exactly; I would say the situation is similar to that of forgetful functor. If I say there is a canonical isomorphism between X and Y, then what I mean is that if asked, pretty much everyone would choose the same isomorphism. A canonical isomorphism is very often a natural isomorphism in the sense of category theory, but the converse need not hold. A canonical isomorphism does not need to be the unique isomorphism between X and Y, though sometimes it is when X and Y are considered as equipped with some additional structure.
"There is a canonical isomorphism between the set of elements of a ring R and the set of ring maps $\mathbb{Z}[x] \to R$." Obviously, I mean for $r \in R$ to correspond to the ring map sending $x$ to $r$, although I could just as well send $x$ to $-r$.
"There is a canonical isomorphism between a finite-dimensional vector space V and its dual." No explanation needed, I suppose.

Maybe more interesting would be an example where the word "canonical" is arguably correct or incorrect; I can't think of one off-hand.

Addendum, after reading some of the other answers: I would emphasize that for me there is a difference between "natural" in the formal category-theoretic sense and "canonical". For one thing there is a linguistic distinction: if I am considering an isomorphism F between X and Y then "Theorem: F is a natural isomorphism" is perfectly acceptable but "Theorem: F is a canonical isomorphism" is very strange to me. There should be only one canonical isomorphism between two things, though what that isomorphism is could depend on context, e.g., "the canonical isomorphism $A \otimes B \to B \otimes A$" where $A$ and $B$ are graded abelian groups might mean different things to an algebraic geometer and an algebraic topologist.

Finally, this is hardly a definition, more of a rule of thumb: there is a canonical isomorphism between X and Y if and only if you would feel comfortable writing "X = Y".

edited Apr 13 '17 at 12:58

Community

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answered Mar 28 '10 at 18:36

Reid Barton

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bothers me. Don't you want it to say "...and the dual of the dual," so the statement is correct?

AndrewLMarshall

Mar 28 '10 at 18:45

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The OP asked for an incorrect use of the word "canonical", which I took to mean a false statement which becomes true if "canonical" is removed. – Reid Barton Mar 28 '10 at 18:46

I think a correct precise definition would be a refinement of Reid's answer #1. There should be a unique isomorphism between any other candidate Y to the "canonical" X. As Reid mentions, this will require listing precisely the amount of information to yield only one isomorphism. – David Jordan Mar 28 '10 at 19:49

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+1 for the rule of thumb. I think this is a very good litmus test. – François G. Dorais Mar 28 '10 at 20:13

@Reid: Returning to the question you asked me, I don't think I'm very comfortable writing "Hom(Z[x],R) = R." – François G. Dorais Mar 28 '10 at 20:15

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but the converse need not hold.

Could you please give an example for this statement?

– Dmitri Pavlov Mar 28 '10 at 20:28

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Sure, for example, the natural isomorphism $V \to V^{**}$ given by $x \mapsto (f \mapsto 3f(x))$. – Reid Barton Mar 28 '10 at 20:34

apologies, I read it wrong. – AndrewLMarshall Apr 05 '10 at 20:10

See the answer referring to page vii of the Introduction to the 1996 edition of Sheaf Theory by Glen E. Bredon for a comment relating to your rule of thumb. – David Corwin Jun 04 '12 at 20:29

score 17 · Answer 4 · answered Mar 28 '10 at 18:52

17

For me the word “canonical” always means “functorial in some sense”, usually without using any form of the axiom of choice. For example, every finite-dimensional vector space is canonically isomorphic to its double dual, because there is an isomorphism of functors id → **, but there is no canonical isomorphism between a finite-dimensional vector space and its dual, because one cannot construct an isomorphism of functors id → * without using some form of the axiom of choice. Likewise, the construction of an algebraic closure is not canonical because there is no functor that sends a field to its algebraic closure, even though every two algebraic closures are (non-canonically) isomorphic.

I presume that one can allow using the axiom of choice and still get the same results, but in this case one needs to use the language of 2-categories. For every well-pointed elementary topos T (basically, a set theory), we can construct the category of finite-dimensional vector spaces in this topos and isomorphism of functors id → **. I think that this isomorphism depends 2-functorially on T. On the other hand, even if we use the axiom of choice to construct an isomorphism of functors id → * for every well-pointed elementary topos T, there is no way to make it depend functorially on T. I must say that I have never tried to prove any of these statements, so they might as well be totally wrong.

answered Mar 28 '10 at 18:52

Dmitri Pavlov

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Your usage of "axiom of choice" is wrong -- the fact that a finite-dimensional vector space is isomorphic to its dual is provable without any use of the axiom of choice (even in topoi if you use the correct definition of finite).
What you meant is probably "without making arbitrary choices" or "by making uniform choices." (The latter is how I usually understand the phrase "canonical choice.")
– François G. Dorais Mar 28 '10 at 19:04
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To construct an isomorphism between the identity functor and the dual functor in the category of finite-dimensional vector spaces, isn't it necessary to choose, for each vector space, an explicit isomorphism to its dual? I think that's what Dmitri means. – Qiaochu Yuan Mar 28 '10 at 19:10
@Qiaochu: You're right! I read too quickly. – François G. Dorais Mar 28 '10 at 19:16
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Qiaochu is right. I only said that “one cannot construct an isomorphism of functors id → * without using some form of the axiom of choice”, not that one cannot construct an isomorphism between a finite-dimensional vector space and its dual without using some form of the axiom of choice. – Dmitri Pavlov Mar 28 '10 at 19:17
Your answer is the kind I was looking for, although I have the impression that you're talking about "natural" instead of "canonical". Or would you say that both words have the same meaning? Then you contradict Reid's answer above. – Konrad Waldorf Mar 28 '10 at 19:39
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@Dmitri: Yes, and your usage is correct, I apologize for the confusion. I'm leaving my comment there to since there is apparently a way to misunderstand what you wrote. (Though, after rereading carefully once more, I don't think what you wrote is ambiguous in any way. My guess is that accidentally skipping over a few words changes the meaning.) – François G. Dorais Mar 28 '10 at 19:43
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@Konrad: For me natural = functorial = canonical. In particular, the natural isomorphism V → V** given by x ↦ (f ↦ -f(x)) is a canonical isomorphism for me, even though Reid Barton claims that it isn't. (I replaced 3 by -1 in his example to avoid problems with characteristic 3.) In my papers I avoid the term “canonical” and use more precise words such as “functorial” instead. – Dmitri Pavlov Mar 28 '10 at 21:06
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I m not sure, whether canonical implies functorial. For example assign to an object of a category its automorphism group. This seems canonical to me, however it is not functorial. – Jan Weidner Mar 29 '10 at 10:12
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@Jan: The automorphism group of an object is functorial with respect to the subcategory consisting of isomorphisms in the original category. Therefore it is canonical. Notice that sometimes we need to pass to a different category to achieve functoriality. – Dmitri Pavlov Mar 29 '10 at 18:03
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@Dmitri: Pass to the subcategory that has the same objects but only has identity morphisms. Then any assignment is functorial. Would you then consider any such assignment "canonical"? – Chris Schommer-Pries Apr 09 '10 at 11:35
@Chris: Well, usually we want to keep all isomorphisms in the subcategory, but your question still stands, because some categories only have identity morphisms as isomorphisms. In this case I am tempted to use the 2-functoriality condition in the second paragraph of my answer (alternatively, disallow the axiom of choice). For example, using the axiom of choice we can construct an algebraic closure functor on the category of fields with identity maps as morphisms, but I do not think there is a way to make this functor depend 2-functorially on the underlying elementary topos. – Dmitri Pavlov Apr 11 '10 at 05:05
I'm late to the party but: how can the identity functor possibly be (naturally or otherwise) isomorphic to the dualization functor?! The identity is covariant while the dualization is contravariant! – Qfwfq Jul 17 '18 at 18:18
@Qfwfq: Both functors id and * are covariant endofunctors on the category of vector spaces and isomorphisms of vector spaces: id(X)=X, id(f)=f, X=Hom(X,k), f=Hom(f,k)^{−1}. – Dmitri Pavlov Jul 18 '18 at 02:28
@Dmitri Pavlov: well, okay, that's not the usual dualization functor $f^{}:Y^ \to X^$, $f^{}(\varphi):=\varphi \circ f$. – Qfwfq Jul 18 '18 at 09:46
Maybe there is a way of defining what a natural transformation between a covariant and a contravariant functor (not just their restrictions to a subcategory) should be. Consider $F:\mathcal C \to \mathcal D$ and $G:\mathcal{C}^{op}\to \mathcal{D}$. Then you can say a natural transformation between $F$ and $G$ is a natural transformation (in the usual sense) between $F$ and $G\circ i$ where $i:\mathcal C \to \mathcal{C}^{op}$ is the "formal opposite" $i:x\mapsto x$, $i: (f:x\to y)\mapsto (f^{op}: y\to x)$. – Qfwfq Jul 18 '18 at 09:57
@Qfwfq: The functor i does not always exist: in the category {0→1} there is a morphism 0→1, but no morphism 1→0. However, i always exists in a groupoid, where we can take f↦f^{−1}. This is exactly what I did above. – Dmitri Pavlov Jul 18 '18 at 16:53
My $i:\mathcal{C}\to\mathcal{C}^{op}$ was not $f\mapsto f^{-1}$, but $f\mapsto f^{op}$, where, for $f: x\to y$ in $\mathcal C$, $f^{op}$ is the same map but thought of as $y\to x$ in $\mathcal{C}^{op}$. – Qfwfq Jul 18 '18 at 20:30
If you want to spell out "my" definition of $\alpha:F\Rightarrow G$ where $F$ is covariant and $G$ contravariant, without mentioning $i$, you can equivalently say that you give a collection $\alpha_x$ for $x\in\mathrm{ob}(\mathcal{C})=\mathrm{ob}(\mathcal{C}^{op})$ such that, for every $f:x\to y$ in $\mathcal{C}$, $G(f)\circ\alpha_{y}\circ F(f)=\alpha_x$. – Qfwfq Jul 18 '18 at 20:35
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@Qfwfq: f↦f^op does not define a functor C→C^op (but rather only a functor C^op→C^op). Your last definition is different, however, and is known under the name of a dinatural transformation: http://ncatlab.org/nlab/show/dinatural+transformation – Dmitri Pavlov Jul 18 '18 at 23:57
I realize now you're obviously right: $i$ is indeed not a functor $C\to C^{op}$. I thought there ought to be a well known structure under which my definition (the one that makes sense) was a particular case of: thanks for the info! – Qfwfq Jul 19 '18 at 10:06

score 14 · Answer 5 · answered Mar 28 '10 at 20:58

I was taught to think that there is a precise definition of "canonical" in differential topology, at least in the context of linear algebra constructions. A construction is canonical if it is a smooth functor. (There is a Wikipedia page about smooth functors but it is not very insightful). And since it is hard to invent a non-smooth functor, it practically boils down to just being a functor.

The categories involved are usually not mentioned explicitly, and they are not things like vectors spaces with linear maps. They are rather things like vector spaces with linear isomorphisms as morphisms. Or, more generally, isomorphisms of whatever structure you happen to have on them. For example, dual vector space is a canonical construction but an isomorphism between a vector space and its dual is not. On the other hand, there is a canonical one if your spaces carry Euclidean structure.

The idea is that a canonical construction can be applied fiber-wise in fiber bundles. Sometimes this feature is advertised as a poor man's definition of "canonical" but this is not quite correct: for example, every vector bundle (over a paracompact base) is isomorphic to its dual, but this is not really canonical.

The observation that "canonical constructions" can be applied fiber-wise to bundles is an excellent one. I now realize that in practice, this perspective underlies a lot of my use of "canonical". — Tom Church, Mar 28 '10 at 22:49
This seems to be the same as the use at this nlab page: http://ncatlab.org/nlab/show/canonical+morphism — Reid Barton, Mar 29 '10 at 00:42
It is not the same that what I wrote, but maybe it is more correct. Although I don't see how it covers cases like "canonical orientation on $X\times X$". — Sergei Ivanov, Mar 29 '10 at 02:58

score 12 · Answer 6 · answered Jun 04 '12 at 20:28

On page vii of the Introduction to the 1996 edition of Sheaf Theory by Glen E. Bredon, the author discusses the difference between "canonical" and "natural" and points to a historical context:

Occasionally, we use the equal sign to mean a "canonical" isomorphism, perhaps not, strictly speaking, an equality. The word "canonical" is often used for the concept for which the word "natural" was used before category theory gave that word a precise meaning. That is, "canonical" certainly means natural when the latter has meaning, but it means more: that which might be termed "God-given." We shall make no attempt to define that concept precisely. (Thanks to Dennis Sullivan for a theological discussion in 1969.)

score 9 · Answer 7 · answered Feb 16 '13 at 15:02

9

I have two competing interpretations of the word canonical.

One, apparently the one used by Bourbaki, is mathematically informal. In various contexts, some objects (maps, modules, etc.) are defined unambiguously and called canonical. For example, the canonical basis of the free module $A^{(I)}$ over a set $I$, the canonical surjection from a set $X$ to its set $X/R$ of equivalence classes with respect to some equivalence relation $R$, the canonical bilinear map from a product of two modules to their tensor product, etc.

The other interpretation is categorical. The given context defines (often implicitly) some categories and the canonical object is functorial with respect to isomorphisms. It is more or less what Kevin Buzzard says in its answer, when he defines canonical by the property that he and a colleague, when asked to define the object, would agree on the same object.

answered Feb 16 '13 at 15:02

ACL

12,762

Dear Antoine, how do you apply this categorical interpretation to prove the following fact: there is no canonical generator for the multiplicative group of the field with $p$ elements? – Matthieu Romagny Feb 16 '13 at 19:50
Well (except when $p=2$), for every generator of the multiplicative group $\mathbf F_p^\times$, there is an automorphism of this group which moves the generator.
But there is a canonical generator in the former sense: I would say that the one with smallest positive representative in $\mathbf Z$ qualifies.
– ACL Feb 16 '13 at 23:48
OK, thanks! (Note: for $p=3$ also, the generator is canonical.) – Matthieu Romagny Feb 17 '13 at 13:02
If we are considering the category of fields with $p$ elements, then the canonical generator $c$ in Bourbaki's sense works also in the functorial sense since an isomorphism between two fields with $p$ elements takes $c$ to $c$. (Of course there is only one isomorphism, and it is used to identify $c$ everywhere...) Maybe a more satisfying non-existence statement in the functorial sense of canonicity is that cyclic groups of order $n$ do not have a canonical generator if $n\ge 3$. – Matthieu Romagny Feb 17 '13 at 13:30
@Mathieu. As far as I am concern, this fact "there is no canonical generator for the multiplicative group of the field with p elements?" Is false. There is canonical generator : the one which is the smallest if I identify $F_p$ with {0,1,..p-1}. But I guess it depends on your interpretation of "canonical" – Simon Henry Nov 13 '13 at 23:47

score 8 · Answer 8 · edited Jun 02 '19 at 03:00

I would say that "canonical" ought to be used to describe when no choices have been made.

A nice example of a non-canonical identification: A principal bundle is made up of principal homogeneous spaces for the action of a Lie group. These are spaces which are homeomorphic but non-canonically isomorphic to the Lie group. For example, I might have a circle bundle. My Lie group would be a `concrete version' of the group such as $\{|z| = 1\}$, but my fibres are simply circles. I would need to choose a base point on each of the circles to make them into groups in the same way. This amounts to taking a global section and can't always be done (e.g. circle bundle on the sphere has no global section by hairy ball theorem), so the non-canonical-ness might actually be important

The labelling of identifications as Canonical and Non-canonical is common in linear algebra: Since one chooses bases so often, it is worth pointing out when such a choice is avoided...

To prove that $V^* \otimes V^*$ is isomorphic to $(V \otimes V)^*$, one ought to work with elements of the spaces directly rather than their representations in some basis. I would therefore call the resulting isomorphism `canonical'.

score 8 · Answer 9 · answered Mar 28 '10 at 21:44

8

I was always under the impression that canonical meant, precisely, that no arbitrary choices were necessary. But, that it was occasionally used less formally, in a more standard-English sort of way to mean traditional/obvious/well known. The informal meaning is usually used as a cheap way to avoid explaining something that's easier for the reader to guess anyway.

Ex 1: Two vector spaces of the same dimension are isomorphic. The isomorphism is not canonical.

Ex 2: A finite dimensional vector space is canonically isomorphic to its double dual.

Ex 3: Let $\pi: S^3 \to S^2$ be the canonical fibration.

I never really liked it when people use canonical as in example 3. It seems like using it this flexibly detracts from the useful technical interpretation of the word.

I've also heard some more complicated category theoretic interpretations of what canonical meant. But, after more scrutiny, it seems that these "definitions" are specific cases of the "no arbitrary choices" principle.

answered Mar 28 '10 at 21:44

Fabrizio Polo

671

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I would never call the Hopf fibration canonical, but maybe that's just me. – Deane Yang Mar 28 '10 at 21:51
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The third sense is using "canonical" to mean "that which is found in canon," e.g. the literature. I think "standard" or "classical" would work better. Or "Hopf." – Qiaochu Yuan Mar 28 '10 at 23:41
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I wouldn't call the Hopf fibration "canonical" either. But it used exactly that way, and it does fit the standard English definition of the word. If I was writing, I would definitely use something like "classical" instead. – Fabrizio Polo Mar 29 '10 at 06:23
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IF I had a shiny 4D brain, I'd probably call the Hopf fibration "canonical". – horse with no name Feb 16 '13 at 15:03
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Ex 3 might be an interesting counterexample to the "person in the office next door" rule of thumb. All four of the commenters so far seem to agree that $\pi$ is obviously supposed to be the Hopf fibration, so this usage of "canonical" passes the "office next door" test... but I still think it's wrong! – Vectornaut Nov 14 '13 at 02:59

Matthieu Romagny · Answer 10 · 2013-02-16T17:18:02.827

Regarding the (widely considered to be false) statement that "There is a canonical isomorphism between a finite-dimensional vector space V and its dual" in Reid Barton's answer, I think that the situation is a bit more interesting than that. It is a good illustration of the idea that an object may be defined to be "canonical" if it is constructed without making any choices, and the interesting point here is that there are various degrees of (in-)tolerance to choices. If we work with vector spaces of fixed finite dimension, then an isomorphism $i_E:E\to E^*$ between a vector space $E$ and its dual $E^*$ may be called canonical if

it does not depend on the choice of a basis for $E$ but we need a basis to define it, or
it does not depend on the choice of a basis and may be defined without choosing a basis, or
it does not depend on basis choices as above and does not even depend on $E$, in the sense that whenever $u:E\to F$ is an isomorphism between dim. vector spaces of the same finite dimension, then $u^*\circ i_F\circ u=i_E$ where $u^*$ is the transpose of $u$. This third notion of canonicity is essentially functoriality.

Concretely, given a basis $B=\{e_j\}$ of $E$ with dual basis $B^*=\{e_j^*\}$, we can construct an isomorphism $i=i_{E,B}:E\to E^*$ that maps $e_j$ to $e_j^*$. This map does not depend on the choice of basis if and only if $u^*\circ i\circ u=i$ for all $u\in \text{GL}(E)$. It is easily seen that this is equivalent to the fact that $\text{GL}_n(k)=\text{O}_n(k)$, where $k$ is the base field, $n$ is the dimension, and $\text{O}_n(k)$ is the orthogonal group of the standard (sum of squares) quadratic form. Exercise: this equality holds if and only if $n=1$ and $k$ has at most $3$ elements. Thus for $n=1$ and $\text{card}(k)\leq 3$ the map $i_{E,B}:E\to E^*$ does not depend on $B$, so we may write it simply $i_E$. When $\text{card}(k)=2$ this is not so surprising because any two one-dimensional vector spaces over the field with two elements are uniquely (hence canonically in whichever sense you like) isomorphic, but for $\text{card}(k)=3$ this is a bit more exotic. Having reached this point, we might think that we are in the funny notion 1 of canonicity (and this is what I thought some minutes ago). But in fact, still assuming that $n=1$ and $\text{card}(k)\leq 3$, we can exhibit an isomorphism $i:E\to E^*$ without any reference to a basis. Namely, define $i(0)=0$ and if $x\in E$ is nonzero, then it is a basis of $E$, and we can define $i(x)=x^*$, the only element of the dual basis. The point is that since $a^2=1$ for all nonzero scalars in $k$, this map $i$ is linear.

Conclusion: if $n=1$ and $\text{card}(k)\leq 3$ there is an isomorphism $i:E\to E^*$ that is constructed without a choice of basis, and it is functorial for isomorphisms of one-dimensional vector spaces. If $n\ge 2$ or $\text{card}(k)\ge 4$, the map $i_{E,B}:E\to E^*$ is not independent of the basis $B$.

I would guess that it is possible to find examples of phenomena like 1 above.

Here is an example I learned in Rennes, and which, I believe, originates from Bas Edixhoven. If $k$ is the field with two elements and $E$ has dimension $2$, then there is a unique alternate form on $E$, which gives rise to a canonical isomorphism between $E$ and $E^*$. — ACL, Feb 16 '13 at 14:52
You are right that the last-but-one sentence "there is no such basis-choice-free isomorphism" is not true and must be replaced by "the map $i_{E,B}:E\to E^*$ above is not independent of the basis. — Matthieu Romagny, Feb 16 '13 at 17:16

score 5 · Answer 11 · answered Nov 13 '13 at 23:17

Since this question has just popped up again to the the top of the stack, I can't resist adding one more response.

It occurs to me that constructive logic (or type theory) suggests a way to formalize the meaning of "canonical". The idea is that when you say "the canonical $x$ such that $P(x)$" (or "the canonical $x$ of type $P$"), your statement is implicitly justified by some proof that $\exists x . P(x)$ (or $\exists x : P$). If this proof is constructive, then it actually constructs a specific $x$ such that $P(x)$ or term $x$ of type $P$. When you say "the canonical $x$ ...", you "extract" that constructed $x$.

Whenever we talk about formalizing math, there must be some notion of "implicit" parts of a mathematical argument which must be reconstructed in order to formalize it. You might think that the implicit parts consist simply of implicit proofs, but the use of the word "canonical" actually allows us to leave implicit certain constructions. What the constructive logic (or type theory) does is to view all the implicit stuff, proofs and constructions alike, in the same light, since it treats a construction as a kind of proof.

It's also interesting that normally, if an argument relies on an implicit proof, then it doesn't matter what proof the reader supplies, so long as it's correct. But if your argument relies on an implicit construction witnessing an existence statement, and if you want to refer back to this construction as "the canonical $x$...", then it does matter what proof is supplied insofar as the proof has to construct the right witness. Maybe this starts to point toward homotopy aspects of the type theory involved?

Grant Olney Passmore · Answer 12 · 2010-03-29T12:54:35.537

5

Not a definition, but an example of use in logic:

In model theory, "canonical" is often used in the phrase "the canonical model" to mean "intended structure." For instance, in first-order logic, one may speak of "the canonical model of Peano Arithmetic" to mean the structure of the natural numbers, or "the canonical model of the theory of real-closed fields" to mean the field of real numbers. Intuitively, "the canonical model" of a theory is the structure one was trying to pin down when the axiomatisation of the theory was written. It's just that in first-order logic, it is hard to pin down (infinite) structures! No first-order theories admitting infinite models are categorical (they admit non-isomorphic models; indeed, they admit models of every infinite cardinality), and compactness/ultraproduct/(many other) constructions can often be used to build "non-standard" models of theories. "Non-standard" models of Peano Arithmetic or the theory of real-closed fields would in this context be called "non-canonical" (even though there are many canonically studied "non-standard" models of those theories!).

But, many commonly studied theories do not have a notion of "canonical model." For instance, one would not say "the canonical model of group theory."

edited Mar 29 '10 at 12:54

answered Mar 28 '10 at 22:03

Grant Olney Passmore

971

1

I probably just never noticed, but I don't recall seeing 'canonical' used instead of 'standard' in this context. Can you point to an example? – François G. Dorais Mar 28 '10 at 22:16
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I do think you're right that 'standard' is used more often. But, a few examples:
- Peano Arithmetic (p 18): http://arxiv.org/pdf/0905.1680
- model theory of geometry: http://people.maths.ox.ac.uk/zilber/q_tori.pdf
- modal logic (questioning just how 'canonical' a certain model is!): http://ideas.repec.org/a/spr/jogath/v28y1999i3p435-442.html
If I remember correctly, Hodges' model theory texts also use 'canonical' regularly to mean 'intended' or 'standard.'
– Grant Olney Passmore Mar 28 '10 at 22:44
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Thanks! Zilber's use of canonical is particularly interesting. – François G. Dorais Mar 28 '10 at 22:58
Hmmmmm, this seems slightly weird to me. I associate the term 'standard model' with the model that the theory was designed to describe. For example, the standard model of Peano arithmetic.
In my head, 'canonical model' refers to the model constructed in the proof of the completeness theorem. I think this is pretty standard too (see Van Dalen, for instance). I'm sure it's standard in modal logic as well, I wrote my thesis on a method of constructing these kind of canonical models (see http://en.wikipedia.org/wiki/Kripke_semantics#Canonical_models).
– Brendan Cordy Mar 29 '10 at 14:07
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@Brendan Cordy: Great point. I've seen that use (in regards to Henkin constructions) quite a bit as well. I guess the moral of the story here is that the term "canonical model" is used in a number of different - even incompatible - ways! – Grant Olney Passmore Mar 29 '10 at 19:44

score 5 · Answer 13 · answered Mar 29 '10 at 13:07

Vague definition of canonical:

Let $X$ and $Y$ be collections (often sets) for which assumptions has been made (has been given structures and/or are related somehow). A function $f\colon X \to Y$ is canonical if it is given by a rule using only the already given structure.

This explains the relation to the greek word rule (kanon). The precise meaning of the above are open for enterpretation: how much structure can the rule itself contain (maybe this can be made precise)! This "definition" somewhat contradicts many of the other answers, which for some reason is under the impression that canonical implies unique (or almost unique), which in my point of view is very wrong since different rules may define different maps. E.g. if we let $X$ be the objects in the category of abelian groups and $Y$ the morphisms then the definitions makes all the group homomorphisms $A \to A$ given by multiplication with an element in $\mathbb{Z}$ canonical, which to me is not a problem.

Usually when there is an especially simple rule it is often assumed without mentioning that this is the rule defining the function. E.g. most will understand the following:"there is a canonical endemorphism of any object in a category". This emphasizes the multiplication with 1 above as somehow speciel or "more canonical" than the rest. This is simply because the rule works in much greater generality and is shorter.

Usually if a rule is very simple the function will have nice properties. E.g. simply rules in category theory often define functors, natural transformations, e.t.c. This leeds many people to confuse the notion of canonical with "something behaving nicely".

I am somewhat puzzled by the use of the word uniform in one of the answers. The nature of the word uniform is "of the same form" and relates more to symmetries and things looking the same every where. This often leeds to canonical maps, since a choice at one point can sometimes be extended to a choice at every point. Please someone comment on this since maybe this is just a use of the word I have not seen before!

And rule is from the latin regula, which is the exact correspondent of κανών (straight piece of wood/measuring rod/ruler). Thus "canonical" is (constructed) following a rule. Moreover, the uniqueness of the canonical choice produces further regularity between objects: E.g. if $X$ is a Hilbert space and $Y$ a closed subspace of its, and $f$ maps $x$ to its closest point in $Y$, this canonical choice makes the projector $f$ linear. And I agree with your point of view that the building rule maybe was given arbitrarily (as it is in the human world). — Pietro Majer, Jun 02 '19 at 11:05
An example you mentioned: if $X$ and $Y$ are categories (say $X$ small) with a given continuous functor $g:Y\to X$, the rule to associate to any object $x\in X$ a certain limit in $Y$ produces a map $f:X\to Y$, and this is automatically a functor, according to the adjoint functor theorem of Freyd. We call this map canonical, although the whole construction may rely on some assumed arbitrary choice (possibly non canonical!) for the representative in the isomorphism class of each limit. — Pietro Majer, Jun 02 '19 at 11:05

score 5 · Answer 14 · answered Aug 23 '16 at 06:35

5

I remember this exchange from a lecture by Prakash Panangaden:

"And so every quantum algorithm can be written in this canonical form."

"What do you mean by canonical?"

"Canonical means I like it a lot."

answered Aug 23 '16 at 06:35

Robin Adams

203

score 4 · Answer 15 · answered Mar 29 '10 at 05:06

4

Hopefully I don't say something too stupid. I just wonder whether the definition of canonical might be relative.

For example, if we look at $ \mathbb Z / p$ as an additive group in fact there is no non-zero element which stands out. But if we look at $ \mathbb Z / p$ as a field $1$ stands out as a non-zero element.

Another example. From the geometrical reason alone, there is no good reason to choose a positive direction (Essentially there is no way to distinguish from left hand and right hand). But in a universe where there is electro magnetic force, we then have a canonical way to choose a positive direction.

Yet another example, there is a canonical way to choose whether you want a left shoe or a right shoe: If you are left-handed then choose the left one, if you are right handed choose the right one.

Perhaps what counts as canonical depends on where we are standing. A suggestion for a heuristic definition: canonical is definable with respect to the structure you are standing at.

answered Mar 29 '10 at 05:06

abcdxyz

2,744

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Certainly. If you take the functorial point of view, everything depends on the ambient category. – Qiaochu Yuan Mar 29 '10 at 05:10
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Tran Chieu Minh, this suggests the horrifying possibility of considering canonically canonical objects …. – LSpice Mar 29 '10 at 15:29
May be it is because I am not very familiar with the functorial point of view. But why canonical and functorial are the same is still not very clear to me. Canonical as I perceive is close in meaning to being naturally suggested. Functorial is however somesort of a commutative condition. Though in practice they often coincide but is there any conceptual gap? – abcdxyz Mar 29 '10 at 17:01
@ L Spice: Not quite sure what you mean. Are you saying that there should be no fixed meaning of being canonial? Or are you worrying about some type of self reference that might be paradoxical? – abcdxyz Mar 29 '10 at 17:04
Tran Chieu Minh, I was only joking. While I agree with the statement mathematically, the idea that canonicity is relative seems etymologically very funny. – LSpice Mar 29 '10 at 17:08
Ah, I see. I was worrying about it just now. – abcdxyz Mar 29 '10 at 17:15
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To use physics to distinguish left from right, you'll need the weak force, not electromagnetic. The apparent left-right distinction in electromagnetic theory (right-hand rules for the direction of magnetic fields induced by currents, etc.) arises solely from our conventions (particularly, our tendency to conflate polar and axial vectors); the actual physics of electromagnetism is left-right symmetric. – Andreas Blass Jun 05 '12 at 11:38

Douglas S. Stones · Answer 16 · 2010-03-29T14:02:57.163

For me, if we have a partition P of a set S, then we can define a set of representatives, one from each part of P, each of which is called canonical.

Typically, the partition P is formed by the orbits of a group G acting on S. If we choose G so that every element in S has a trivial stabiliser, then we can find |S| by instead counting the canonical representatives since |S|=|G|*|P| by the Orbit-Stabiliser Theorem.

Often, the elements that are chosen to be canonical can be quite contrived - e.g. just because your program outputs a certain element of P first, i.e. "lexicographical order".

To add some examples:

a) An orthomorphism of $\mathbb{Z}_n$ is a permutation $\sigma$ such that $i \mapsto \sigma(i)-i \pmod n$ is also a permutation. We partition the orthomorphisms of $\mathbb{Z}_n$ into equivalence classes under the transformation $E_g$ for which $E_g\[\sigma\](i)=\sigma(i)+g \pmod n$. Therefore the parts each have cardinality $n$ and we define the canonical representatives to be the orthomorphisms $\sigma$ for which $\sigma(0)=0$. Therefore the total number of orthomorphisms is $n$ times the number of canonical orthomorphisms.

b) A Latin square is an $n \times n$ matrix containing $n$ distinct symbols in which each symbol occurs exactly once in each row and each column. For instance, $$\begin{matrix} 1 & 3 & 2 \\\\ 3 & 2 & 1 \\\\ 2 & 1 & 3 \end{matrix}$$ is a $3 \times 3$ Latin square. We can put it in a canonical form (which I call normalised) by permuting the columns so that the first row is in order, i.e. $$\begin{matrix} 1 & 2 & 3 \\\\ 3 & 1 & 2 \\\\ 2 & 3 & 1 \end{matrix}$$ Here the total number of Latin squares is $n!$ times the number of normalised Latin squares. There's another canonical form (which I call reduced) which has the first row and first column in order.

Finding representatives for equivalence classes is often impossible to do canonically... I'm not really sure what sense you mean. — Qiaochu Yuan, Mar 29 '10 at 05:12
Just take the map $i \mapsto j$ where $j$ is the canonical element in the part containing $i$. However, it's not always a pretty function. — Douglas S. Stones, Mar 29 '10 at 12:39
chicken/egg: all based spaces have a canonical point, but there is no canonical way of picking a point in all (non-empty) spaces. The terminology:"picking a canonical point" should really be:"we give all the sets the structure of being based by picking a point in each - and hence forth may talk about the canonical point in each set". In short it is not a part of the structure to have a canonical point is a consequence of the structure. — Thomas Kragh, Mar 31 '10 at 08:50
Yes exactly. Here structures gives rise to canonical maps. E.g. the Normal $n$ by $n$ matrices over $\mathbb{C}$ is sufficiently structured such that there is a canonical map to $\mathbb{C}^n$ modulo the action of the symmetric group on the image, and this map classifies the matrices up similarity. So the canonical form is an equivalence class of diagonal matrices. I.e. for each matrix we get a canonical point in $\mathbb{C}^n$ modulo the symmetric group action using the inherent structure of the space of normal matrices. — Thomas Kragh, Apr 06 '10 at 11:03

Patrick I-Z · Answer 17 · 2012-06-05T15:54:19.210

It is not an answer but it must have to do with the way our brain pick up a sample between a few ones. It must be the minimum of some function which can be implemented for real in the brain. I do not believe that there is a pure logical definition of "canonical" independently of the way our brain works. Experience : give me a number? What do you answer? 0 or 1 rarely $\pi$ or even 115674. The numbers 0 and 1 are canonical in some sense. Give me a basis of ${\bf R}^3$. The same holds $((1,0,0),(0,1,0),(0,0,1))$ I minimize the number of different digits and I pick them in my "basis" of canonical numbers. Well, interesting question.

What is the canonical circle ? Ce circle in ${\bf R}^2$, centered at $(0,0)$ with radius $1$.

I know two numbers $0$ and $1$, the radius cannot be $0$ because it is not a (true) circle, so the radius is $1$, now the center could be $(0,0)$, $(0,1)$, $(1,0)$ or $(1,1)$ ? I prefer $(0,0)$, $0$ is simpler than $1$. How do you fit this example with category arguments?

BTW I have nothing against category theory, I like it. But I'm curious to see if this example fits general categorical arguments.

I have the impression that any definition of canonical should refer to a choice to be made in several analogous situations, following a rule. So I would agree with the definition of "canonical circle" in your example in that it refers to certain implicitly assumed principles of choices: "the canonical sphere of a normed space is the sphere centered at the origin with radius $1$"; "the canonical $n$-dimensional normed space is $\mathbb{R}^n$ with the Euclidean norm". — Pietro Majer, Jun 02 '19 at 10:13

Fictional · Answer 18 · 2019-10-25T22:44:09.540

For me canonical often means derived from existing structure. Let's say I have an inner product space $S$ with some nice inner product ($\langle{}x,y\rangle{}$ for $x,y \in S$ ). Now let's say I need a norm on this space. I could pick an arbitrary one. There are many possible choices. I would consider the one induced by the inner product ( i.e. $\sqrt{\langle{}x,x\rangle{}}$ ) to be the canonical choice. Even if another was more useful for the problem at hand.

score 0 · Answer 19 · answered Dec 16 '20 at 00:51

There are two kinds of meanings for terms used in mathematics:

stipulated definitions, where the meaning of a term is only what is given by the symbolic definition, like 'field' or 'normal' as in a vector. The word may connote certain things, but whenever in doubt it's -only- the stipulated definition that matters.
loose, informal terms that have only their native language meaning, like 'number' (by itself).

And some terms have multiple meanings, and so can have stipulated and loose versions.

There are a lot of uses of 'canonical' which are loose. I do know of one meaning that is mathematically defined. It comes out of proof theory, specifically rewrite systems.

In this area there are very formal, stipulated definitions of 'canonical': roughly (in non-stipulated terms) you specify a partial order on terms and then a set of rewrite rules from one term to another. Then if after a succession of rewrite rules has occurred to a term, and no more rules apply, then you are at a minimal element of the order, and that element is called 'canonical'.

Since all these loose ideas are formalized in the rewrite rule literature, the idea 'canonical' then has a formal definition.

What is the definition of "canonical"?

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