Timeline for How can we make precise the notion that a finite-dimensional vector space is not canonically isomorphic to its dual via category theory?
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22 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 10 at 4:50 | vote | accept | Malcolm Langfield | ||
| Apr 11, 2021 at 7:03 | comment | added | C.F.G | @bxw: Why don't you accept the answer that answer to your question? please click the green check mark below the downvote triangle. What should I do when someone answers my question? | |
| Mar 9, 2021 at 20:39 | history | became hot network question | |||
| Mar 9, 2021 at 19:30 | answer | added | Simon Henry | timeline score: 16 | |
| Mar 9, 2021 at 18:51 | answer | added | Geva Yashfe | timeline score: 5 | |
| Mar 9, 2021 at 18:33 | comment | added | მამუკა ჯიბლაძე | @ChrisSchommer-Pries There is, I think, one subtlety here, which relates to the OP: it is not obvious what does it mean for a map of vector spaces to be compatible with pairings. Say, the condition $\mu'(f(x),f(y))=\mu(x,y)$ somehow looks too strong: for example, it forces $f$ to be injective. It is like defining the category of metric spaces to have only isometries as morphisms, or to define the category of categories as having only full and faithful functors as morphisms. | |
| Mar 9, 2021 at 18:00 | comment | added | Chris Schommer-Pries | You could consider the category "Pair" of pairs $(V, \mu)$ where $V$ is a vector space and $\mu: V \otimes V \to K$ is a non-degenerate pairing (i.e. an isom $V \cong V^*$. A morphism is a map of vector spaces compatible with pairings. There is a forgetful functor from Pair to Vect which forgets $\mu$. You can ask: Does the forgetful functor admit a section? This would be a functorial choice of isom $V \cong V^*$ for each $V$. The answer is no, there is no such section. | |
| Mar 9, 2021 at 16:01 | comment | added | Malcolm Langfield | @PaulTaylor Maybe you're right. What would you propose as a definition of a "natural isomorphism". (between objects)? I don't think I understand the point you're trying to make here by pointing out that these are different objects. Sorry, could you please elaborate? | |
| Mar 9, 2021 at 15:55 | comment | added | Paul Taylor | I don't see any future in attempting to define an "unnatural isomorphism". Consider the positive question instead: any isomorphism (indeed homomorphism) $V\to V^*$ is a structure, otherwise written $\mu:V\otimes V\to K$. Then $(V,\mu)$ and $(V,\mu')$ are different mathematical objects. | |
| Mar 9, 2021 at 15:36 | comment | added | Malcolm Langfield | Yes, the contravariant one. "Then there can't be a natural isomorphism by definition", right, so this is the "definition not applicable" sort of proof that I mentioned above. So does this mean the definition of an unnatural isomorphism is limited in this respect? If I were to be as frank as possible, I'd say something like "this definition seems bad, why do we use it instead of an alternative, and are there any good alternatives?" | |
| Mar 9, 2021 at 15:03 | comment | added | Najib Idrissi | What do you call the dual functor? The contravariant one? Then there can't be a natural isomorphism by definition, or a dinatural isomorphism by the MSE answer you linked. The inverse of the dual, restricted to the core groupoid? Then it's also done in the MSE answer you've linked. | |
| Mar 9, 2021 at 14:50 | comment | added | Malcolm Langfield | @NajibIdrissi That's a good point. In the context of this question, the functors are fixed. They are the identity functor $\mathbf{Vect} \to \mathbf{Vect}$ and the dual functor $\mathbf{Vect} \to \mathbf{Vect}$. In saying "cannot be extended..." I'm simply quoting the definition from wikipedia. I agree that there seems to be some missing constraint on the functors here. The constant functors is a good counterexample. | |
| Mar 9, 2021 at 14:20 | comment | added | Najib Idrissi | @bxw This is not an idle question, because surely any morphism is a natural transformation between constant functors. | |
| Mar 9, 2021 at 14:14 | comment | added | Najib Idrissi | @TimCampion It does come up (e.g. in the MSE question), and the result is that any dinatural transformation between the identity and the dual functors is zero. | |
| Mar 9, 2021 at 14:13 | comment | added | Tim Campion | There is a notion of dinatural transformation which can allow one to talk about natural transformations between covariant and contravariant functors. I was sure this had come up one of the times this question had been asked here before... | |
| Mar 9, 2021 at 14:12 | comment | added | Najib Idrissi | Okay, I misunderstood your question at first. But now I'm not sure I get it. You say "cannot be extended to a natural transformation". But a natural transformation is something between functors. So are you really asking if the map cannot be extended into a triple (functor, functor, natural transformation)? Do you have any requirement on the functors? For example, between what categories...? | |
| Mar 9, 2021 at 13:54 | comment | added | Malcolm Langfield | As far as the two examples of suggested solutions to this whole issue, I'm less interested in the specific merits of either of them than whether or not the usual definition of "unnatural isomorphism" or "canonical isomorphism" makes any approach problematic. | |
| Mar 9, 2021 at 13:51 | comment | added | Malcolm Langfield | I should clarify: the core groupoid idea is not mine, I'm quoting the MO answer linked above. I do think that is true (about it being a dinatural). | |
| Mar 9, 2021 at 13:50 | comment | added | Malcolm Langfield | @NajibIdrissi Sorry, I'm not sure I understand what you're addressing in your first question. Can you point out exactly where in Martin Brandenburg's answer this is dealt with? | |
| Mar 9, 2021 at 13:38 | comment | added | Najib Idrissi | It seems to me that Martin Brandenburg already deals with the identity not being isomorphic to the dual of the inverse of an isomorphism in his answer, no? Moreover, since $(f^{-1})^* = (f^*)^{-1}$, isn't a natural transformation from the identity to your functor a dinatural transformation from the identity to dualization (restricted to isomorphisms)? | |
| Mar 9, 2021 at 12:40 | review | First posts | |||
| Mar 9, 2021 at 12:43 | |||||
| Mar 9, 2021 at 12:36 | history | asked | Malcolm Langfield | CC BY-SA 4.0 |