Timeline for Is a vector space naturally isomorphic to its dual?
Current License: CC BY-SA 4.0
16 events
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| Nov 7, 2019 at 18:55 | comment | added | Vincent R.B. Blazy | To comment your last comment to user1728, I guess a key to formalizing this informal notion of naturality can be to consider the Objects in question already are merely all Definable (in the currently used Language). Here, none of your Isomorphisms between V and its Dual are so; unlike with its Double Dual. | |
| Nov 3, 2019 at 19:47 | comment | added | Tom Ellis | I agree that mathematicians don't find my constructions interesting. There's an important reason for that. My constructions are not natural but the double dual construction is natural. I think we can all understand how the notion of "natural" applies here. But this notion of "natural" is purely informal. How can it be formalised? Is it even possible to formalise it? Until yesterday I thought the notion of "natural transformation" formalised it, but apparently it does not. | |
| Nov 3, 2019 at 18:13 | history | edited | user1728 | CC BY-SA 4.0 |
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| Nov 3, 2019 at 17:49 | comment | added | Todd Trimble | Thanks very much. | |
| Nov 3, 2019 at 17:42 | comment | added | user1728 | @ToddTrimble I struck out the text where I misunderstood what you had written. | |
| Nov 3, 2019 at 17:41 | history | edited | user1728 | CC BY-SA 4.0 |
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| Nov 3, 2019 at 17:27 | comment | added | Todd Trimble | I think we agree on the fundamentals: that in order to make sense of an assertion that the identity functor is or isn't isomorphic to the dual functor, one must first be clear on what exactly the dual functor is, including how it is defined on morphisms. I would still appreciate it if you would strike through where you say (my objection) isn't fair, which you now agree isn't the case, but you've still let it stand in your answer. This could be confusing to some readers. | |
| Nov 3, 2019 at 17:08 | history | edited | user1728 | CC BY-SA 4.0 |
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| Nov 3, 2019 at 17:02 | history | edited | user1728 | CC BY-SA 4.0 |
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| Nov 3, 2019 at 16:52 | history | edited | user1728 | CC BY-SA 4.0 |
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| Nov 3, 2019 at 16:40 | comment | added | Todd Trimble | Yes, I agree with that. In some situations, one may add "coherence conditions" which can serve to pin down the "right" natural isomorphism, as for example the associativity isomorphism $\alpha: (X \otimes Y) \otimes Z \to X \otimes (Y \otimes Z)$, but that's of course something extra. | |
| Nov 3, 2019 at 16:38 | history | edited | user1728 | CC BY-SA 4.0 |
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| Nov 3, 2019 at 16:37 | comment | added | user1728 | Okay, I see what you mean now about your objection; I had read your answer too quickly. My mathematical point still stands, namely there could be many natural isomorphisms between two specific functors and pure logic alone does not explain why one of them is chosen to be "standard" or "canonical" over all the others. | |
| Nov 3, 2019 at 16:25 | comment | added | Todd Trimble | It most certainly was a fair objection, and you mischaracterized what the objection was. It was that following the method of the OP, one defines two different functors from two different collections of isomorphisms. It wasn't that you couldn't get two different natural isomorphisms between two functors given in advance. In fact, if $\phi, \psi: F \to G$ are two different natural isomorphisms between $F, G$ given in advance, then one gets the same $G$ defined by conjugating $F$ by $\phi$ or by $\psi$ -- quite unlike the situation I was describing. | |
| Nov 3, 2019 at 16:19 | history | edited | user1728 | CC BY-SA 4.0 |
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| Nov 3, 2019 at 16:13 | history | answered | user1728 | CC BY-SA 4.0 |