Timeline for Uniqueness of dualizing objects
Current License: CC BY-SA 4.0
15 events
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| Oct 31, 2018 at 14:17 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Nov 9, 2017 at 18:53 | comment | added | Tim Campion | I suppose the catch is that oftentimes we start with a category $C$ of interest, and no dualizing object exists there. Every object $D$ will have a maximal subcategory $C_D \subseteq C$ on which $D$ is dualizing; it's a delicate condition to ensure that $D \in C_D$. And there can be $D,D'$ with $C_D \neq C_{D'}$, even with $D \in C_D$ and $D' \in C_{D'}$ so that we get distinct star-autonomous full monoidal subcategories of $C$. For instance with $D$ the monoidal unit we get the rigid objects, which doesn't preclude the existence of another subcategory with a different dualizing object. | |
| Nov 9, 2017 at 15:29 | comment | added | Jeff Egger | This is spectacular! (I had no idea that such a thing could be true.) I just want to add that, with a small amount of care, the result can be extended to arbitrary (i.e., not necessarily symmetric) $*$-autonomous categories. | |
| Nov 1, 2017 at 16:38 | comment | added | Tim Campion | If we fix a dualizing object $\bot$, and set $A^\ast := [A,\bot]$, then we have dual tensor $\&$ defined by $A \&B = (A^\ast \otimes B^\ast)^\ast$ with unit $\bot = I^\ast$. Then the possible dualizing objects $D$ with respect to $\otimes$ are just the invertible objects with respect to $\&$. So one way to phrase all of this is that the Picard groups with respect to $\otimes$ and $\&$ are isomorphic. | |
| Oct 31, 2017 at 7:34 | comment | added | მამუკა ჯიბლაძე | Let me still type it out:$$A\to[[A,D],D]\to[[L,[A,D]],[L,D]]\to[[A,[L,D]],[L,D]]$$ | |
| Oct 31, 2017 at 4:29 | comment | added | მამუკა ჯიბლაძე | After all it also would not make much difference since $[L,D]=L^\vee\otimes D$, so this would amount to work with $L^\vee$ instead of $L$ | |
| Oct 31, 2017 at 3:22 | comment | added | Mike Shulman | Nice, thanks! In fact, this is one of those magical places where you don't need to show that the isomorphism is the canonical one! If there's any natural isomorphism $A \cong [[A,D],D]$, then the functor $[-,D]$ is an equivalence (since it has both a right and left inverse, namely itself). Hence since it is also adjoint to itself on the right, the unit and counit of that adjunction are isomorphisms; but those are both the canonical map $A\to [[A,D],D]$, so that is an isomorphism too. | |
| Oct 31, 2017 at 3:15 | vote | accept | Mike Shulman | ||
| Oct 30, 2017 at 22:07 | history | edited | Tim Campion | CC BY-SA 3.0 |
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| Oct 30, 2017 at 22:02 | history | edited | Tim Campion | CC BY-SA 3.0 |
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| Oct 30, 2017 at 21:52 | history | edited | Tim Campion | CC BY-SA 3.0 |
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| Oct 30, 2017 at 21:29 | comment | added | Tim Campion | Or maybe I'll leave it as-is for now. Probably the true measure of how simple a presentation is would be how easy it is to show that the isomorphism is the canonical one, which is eluding me right now. | |
| Oct 30, 2017 at 20:58 | comment | added | Tim Campion | Good call! I'll edit that in. | |
| Oct 30, 2017 at 20:54 | comment | added | მამუკა ჯიბლაძე | It might be slightly more straightforward to work with $[L,D]$ instead of $L\otimes D$ | |
| Oct 30, 2017 at 20:21 | history | answered | Tim Campion | CC BY-SA 3.0 |