Lifting ring homomorphism of Grothendieck rings to functor of semisimple categoriesAbout the closed structure on the modules of a monoidal closed symmetrical category What is the free monoidal category generated by a monoid?Coherence in pseudo.monoidsA question about braiding represented as pseudofunctorsKan extensions in the $2$-category of monoidal categoriesRectifying the definition of a closed categoryWhat's the (monoidal) image of a monoidal functor?Quotient-free monoidal categoriesWhat is the name of this construction on monoidal categories?monoidality of $ Aotimes (-) $ with $ A $ monoid belonging to the center
Lifting ring homomorphism of Grothendieck rings to functor of semisimple categories
About the closed structure on the modules of a monoidal closed symmetrical category What is the free monoidal category generated by a monoid?Coherence in pseudo.monoidsA question about braiding represented as pseudofunctorsKan extensions in the $2$-category of monoidal categoriesRectifying the definition of a closed categoryWhat's the (monoidal) image of a monoidal functor?Quotient-free monoidal categoriesWhat is the name of this construction on monoidal categories?monoidality of $ Aotimes (-) $ with $ A $ monoid belonging to the center
$begingroup$
I have two $mathbbC$-linear semisimple tensor categories $C$ and $D$. Let $K(C)$ and $K(D)$ be their Grothendieck groups. I have a specific ring homomorphism $f colon K(C) to K(D)$ that I would like to realize as coming from a monoidal functor $F colon C to D$. What I know is that for every simple object $X in C$, $f([X])$ is the genuine class of an object in $D$ (not just a virtual object).
The tensor products on both $C$ and $D$ have a symmetry, so $K(C)$ and $K(D)$ are commutative rings. I do know that there is no monoidal functor that preserves the symmetry, but I don't care about preserving the symmetry.
Can this always be done or is there some kind of obstruction I need to consider?
Edit: Since there are many adjectives you can place in front of monoidal functor, I would like the version where $F(X otimes Y) cong F(X) otimes F(Y)$ (satisfying the appropriate axioms).
Edit 2: In light of the Simon Henry's comment, let me add that both categories are linear over the complex numbers and that the endomorphisms of a simple object are just scalar multiples of the identity.
ct.category-theory monoidal-categories
asked Aug 5 at 0:51
Steven SamSteven Sam
7,4271 gold badge31 silver badges64 bronze badges
$endgroup$
add a comment |
$begingroup$
I have two $mathbbC$-linear semisimple tensor categories $C$ and $D$. Let $K(C)$ and $K(D)$ be their Grothendieck groups. I have a specific ring homomorphism $f colon K(C) to K(D)$ that I would like to realize as coming from a monoidal functor $F colon C to D$. What I know is that for every simple object $X in C$, $f([X])$ is the genuine class of an object in $D$ (not just a virtual object).
The tensor products on both $C$ and $D$ have a symmetry, so $K(C)$ and $K(D)$ are commutative rings. I do know that there is no monoidal functor that preserves the symmetry, but I don't care about preserving the symmetry.
Can this always be done or is there some kind of obstruction I need to consider?
Edit: Since there are many adjectives you can place in front of monoidal functor, I would like the version where $F(X otimes Y) cong F(X) otimes F(Y)$ (satisfying the appropriate axioms).
Edit 2: In light of the Simon Henry's comment, let me add that both categories are linear over the complex numbers and that the endomorphisms of a simple object are just scalar multiples of the identity.
ct.category-theory monoidal-categories
asked Aug 5 at 0:51
Steven SamSteven Sam
7,4271 gold badge31 silver badges64 bronze badges
$endgroup$
4
$begingroup$
I would say there are absolutely no chances of doing something like this as the Grothendieck ring essentially remembers nothing of the morphisms. If you are not convince look at when $C$ and $D$ are the category of vectors spaces over two different fields. They have both $mathbbZ$ as Grothendieck ring, but a monoidal functors would corresponds to a fields homomorphism.
$endgroup$
– Simon Henry
Aug 5 at 1:17
$begingroup$
Fair point. I will update my question.
$endgroup$
– Steven Sam
Aug 5 at 1:55
add a comment |
$begingroup$
I have two $mathbbC$-linear semisimple tensor categories $C$ and $D$. Let $K(C)$ and $K(D)$ be their Grothendieck groups. I have a specific ring homomorphism $f colon K(C) to K(D)$ that I would like to realize as coming from a monoidal functor $F colon C to D$. What I know is that for every simple object $X in C$, $f([X])$ is the genuine class of an object in $D$ (not just a virtual object).
The tensor products on both $C$ and $D$ have a symmetry, so $K(C)$ and $K(D)$ are commutative rings. I do know that there is no monoidal functor that preserves the symmetry, but I don't care about preserving the symmetry.
Can this always be done or is there some kind of obstruction I need to consider?
Edit: Since there are many adjectives you can place in front of monoidal functor, I would like the version where $F(X otimes Y) cong F(X) otimes F(Y)$ (satisfying the appropriate axioms).
Edit 2: In light of the Simon Henry's comment, let me add that both categories are linear over the complex numbers and that the endomorphisms of a simple object are just scalar multiples of the identity.
ct.category-theory monoidal-categories
asked Aug 5 at 0:51
Steven SamSteven Sam
7,4271 gold badge31 silver badges64 bronze badges
$endgroup$
I have two $mathbbC$-linear semisimple tensor categories $C$ and $D$. Let $K(C)$ and $K(D)$ be their Grothendieck groups. I have a specific ring homomorphism $f colon K(C) to K(D)$ that I would like to realize as coming from a monoidal functor $F colon C to D$. What I know is that for every simple object $X in C$, $f([X])$ is the genuine class of an object in $D$ (not just a virtual object).
The tensor products on both $C$ and $D$ have a symmetry, so $K(C)$ and $K(D)$ are commutative rings. I do know that there is no monoidal functor that preserves the symmetry, but I don't care about preserving the symmetry.
Can this always be done or is there some kind of obstruction I need to consider?
Edit: Since there are many adjectives you can place in front of monoidal functor, I would like the version where $F(X otimes Y) cong F(X) otimes F(Y)$ (satisfying the appropriate axioms).
Edit 2: In light of the Simon Henry's comment, let me add that both categories are linear over the complex numbers and that the endomorphisms of a simple object are just scalar multiples of the identity.
ct.category-theory monoidal-categories
ct.category-theory monoidal-categories
asked Aug 5 at 0:51
Steven SamSteven Sam
7,4271 gold badge31 silver badges64 bronze badges
asked Aug 5 at 0:51
Steven SamSteven Sam
7,4271 gold badge31 silver badges64 bronze badges
edited Aug 5 at 1:56
Steven Sam
asked Aug 5 at 0:51
Steven SamSteven Sam
7,4271 gold badge31 silver badges64 bronze badges
asked Aug 5 at 0:51
Steven SamSteven Sam
7,4271 gold badge31 silver badges64 bronze badges
asked Aug 5 at 0:51
Steven SamSteven Sam
7,4271 gold badge31 silver badges64 bronze badges
7,4271 gold badge31 silver badges64 bronze badges
4
$begingroup$
I would say there are absolutely no chances of doing something like this as the Grothendieck ring essentially remembers nothing of the morphisms. If you are not convince look at when $C$ and $D$ are the category of vectors spaces over two different fields. They have both $mathbbZ$ as Grothendieck ring, but a monoidal functors would corresponds to a fields homomorphism.
$endgroup$
– Simon Henry
Aug 5 at 1:17
$begingroup$
Fair point. I will update my question.
$endgroup$
– Steven Sam
Aug 5 at 1:55
add a comment |
4
$begingroup$
I would say there are absolutely no chances of doing something like this as the Grothendieck ring essentially remembers nothing of the morphisms. If you are not convince look at when $C$ and $D$ are the category of vectors spaces over two different fields. They have both $mathbbZ$ as Grothendieck ring, but a monoidal functors would corresponds to a fields homomorphism.
$endgroup$
– Simon Henry
Aug 5 at 1:17
$begingroup$
Fair point. I will update my question.
$endgroup$
– Steven Sam
Aug 5 at 1:55
4
4
$begingroup$
I would say there are absolutely no chances of doing something like this as the Grothendieck ring essentially remembers nothing of the morphisms. If you are not convince look at when $C$ and $D$ are the category of vectors spaces over two different fields. They have both $mathbbZ$ as Grothendieck ring, but a monoidal functors would corresponds to a fields homomorphism.
$endgroup$
– Simon Henry
Aug 5 at 1:17
$begingroup$
I would say there are absolutely no chances of doing something like this as the Grothendieck ring essentially remembers nothing of the morphisms. If you are not convince look at when $C$ and $D$ are the category of vectors spaces over two different fields. They have both $mathbbZ$ as Grothendieck ring, but a monoidal functors would corresponds to a fields homomorphism.
$endgroup$
– Simon Henry
Aug 5 at 1:17
$begingroup$
Fair point. I will update my question.
$endgroup$
– Steven Sam
Aug 5 at 1:55
$begingroup$
Fair point. I will update my question.
$endgroup$
– Steven Sam
Aug 5 at 1:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Without the symmetry it’s easy to produce oodles of examples where a map of fusion rings doesn’t lift to the category level. The simplest example is probably $mathrmVec(G,w)$ the category of G-graded vector spaces with associator given by a 3-cocycle $w$. The fusion ring has an automorphism for any automorphism of G, but this only lifts to the category when it preserves $w$ in cohomology.
Giving a symmetric example is harder, but I think the following works. Let $D_8$ be the $8$-element dihedral group, and let $mathrmRep(D_8)$ be its category of representations. The fusion ring has an $S_3$ of automorphisms which permute the three non-trivial invertible objects. (Side note: $mathrmRep(Q_8)$ has the same fusion ring, and there this $S_3$ comes from an action on $Q_8$ and so lifts to the category level.) However, only one of the non-trivial elements of this $S_3$ lifts to the category level. Morally the reason is simple, the 1-dimensional representations each have a kernel, two of them have kernel isomorphic to the Klein 4-group while one of them has kernel isomorphic to $mathbbZ/4mathbbZ$ and the only the permutation swapping the two Klein 4-group ones will lift to the category level. Unfortunately this isn't quite a proof because it's not clear how to describe the kernel of the representation using only the tensor category structure (and not the symmetric tensor category structure).
Nonetheless you can see that the other permutations don't lift by following the classification of Tambara-Yamagami categories in Section 9.2 of Etingof-Nikshych-Ostrik, but it's a little technical. Namely, each 1-dimensional rep generates a $mathrmRep(mathbbZ/2mathbbZ)$ category, and the 2-dimensional rep is fixed by tensoring on either side by this category, so this exhibits $mathrmVec$ as a bimodule category over $mathrmRep(mathbbZ/2mathbbZ)$, and for the Klein 4-group kernel reps this bimodule is invertible while for the $mathbbZ/4mathbbZ$ kernel reps this bimodule is not invertible. Since this was all stated in the language of tensor categories, there's no tensor autoequivalence which mixes the two kind of 1-dimensional objects.
answered Aug 5 at 3:21
Noah SnyderNoah Snyder
19.9k2 gold badges68 silver badges137 bronze badges
$endgroup$
$begingroup$
I'm probably being too naive: but my original thinking was to first replace the monoidal categories by their strict monoidal versions. So in your example when C=D, cycling the 1-dimensional irreps doesn't seem like it should create any problems. But my main problem is I don't have a good intuition for tihnking about coherence issues so it's hard to know where to look for inconsistencies.
$endgroup$
– Steven Sam
Aug 5 at 4:25
2
$begingroup$
You can’t make that example both strict and skeletal at the same time. If you make it skeletal then it’s easy to see what the associator is, it’s just a 3-cocycle. If you make it strict, then you have to make the category much bigger and it’s harder to see what’s going on.
$endgroup$
– Noah Snyder
Aug 5 at 4:43
$begingroup$
Edited to upgrade my deleted comment on a symmetric example to the main text since I think I have an argument that shows it works now.
$endgroup$
– Noah Snyder
Aug 5 at 14:56
1
$begingroup$
Everyone when they first learn about strictification imagines that it's making a smaller category, but actually it's making a much bigger one. You're essentially adding new objects which are formal tensor products of your old ones. So if $1_g$ denotes the 1-dimensional vector space in degree g, in the skeletal version $1_g otimes 1_h = 1_gh$ while in the strict version there's an extra new object $1_g 1_h$ which is isomorphic but not equal to $1_gh$. You can still see the associator by comparing the two isomorphisms $1_x 1_y 1_z rightarrow 1_xyz$.
$endgroup$
– Noah Snyder
Aug 5 at 16:06
1
$begingroup$
Here is a related (and perhaps easier) example: the categories $textRep(D_8)$ and $textRep(Q_8)$ have isomorphic Grothendieck rings (with an isomorphism sending classes of simple objects to classes of simple objects), both categories are symmetric and not monoidally equivalent (which was proved by Tambara-Yamagami).
$endgroup$
– Victor Ostrik
Aug 6 at 16:08
|
show 1 more comment
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$begingroup$
Without the symmetry it’s easy to produce oodles of examples where a map of fusion rings doesn’t lift to the category level. The simplest example is probably $mathrmVec(G,w)$ the category of G-graded vector spaces with associator given by a 3-cocycle $w$. The fusion ring has an automorphism for any automorphism of G, but this only lifts to the category when it preserves $w$ in cohomology.
Giving a symmetric example is harder, but I think the following works. Let $D_8$ be the $8$-element dihedral group, and let $mathrmRep(D_8)$ be its category of representations. The fusion ring has an $S_3$ of automorphisms which permute the three non-trivial invertible objects. (Side note: $mathrmRep(Q_8)$ has the same fusion ring, and there this $S_3$ comes from an action on $Q_8$ and so lifts to the category level.) However, only one of the non-trivial elements of this $S_3$ lifts to the category level. Morally the reason is simple, the 1-dimensional representations each have a kernel, two of them have kernel isomorphic to the Klein 4-group while one of them has kernel isomorphic to $mathbbZ/4mathbbZ$ and the only the permutation swapping the two Klein 4-group ones will lift to the category level. Unfortunately this isn't quite a proof because it's not clear how to describe the kernel of the representation using only the tensor category structure (and not the symmetric tensor category structure).
Nonetheless you can see that the other permutations don't lift by following the classification of Tambara-Yamagami categories in Section 9.2 of Etingof-Nikshych-Ostrik, but it's a little technical. Namely, each 1-dimensional rep generates a $mathrmRep(mathbbZ/2mathbbZ)$ category, and the 2-dimensional rep is fixed by tensoring on either side by this category, so this exhibits $mathrmVec$ as a bimodule category over $mathrmRep(mathbbZ/2mathbbZ)$, and for the Klein 4-group kernel reps this bimodule is invertible while for the $mathbbZ/4mathbbZ$ kernel reps this bimodule is not invertible. Since this was all stated in the language of tensor categories, there's no tensor autoequivalence which mixes the two kind of 1-dimensional objects.
answered Aug 5 at 3:21
Noah SnyderNoah Snyder
19.9k2 gold badges68 silver badges137 bronze badges
$endgroup$
$begingroup$
I'm probably being too naive: but my original thinking was to first replace the monoidal categories by their strict monoidal versions. So in your example when C=D, cycling the 1-dimensional irreps doesn't seem like it should create any problems. But my main problem is I don't have a good intuition for tihnking about coherence issues so it's hard to know where to look for inconsistencies.
$endgroup$
– Steven Sam
Aug 5 at 4:25
2
$begingroup$
You can’t make that example both strict and skeletal at the same time. If you make it skeletal then it’s easy to see what the associator is, it’s just a 3-cocycle. If you make it strict, then you have to make the category much bigger and it’s harder to see what’s going on.
$endgroup$
– Noah Snyder
Aug 5 at 4:43
$begingroup$
Edited to upgrade my deleted comment on a symmetric example to the main text since I think I have an argument that shows it works now.
$endgroup$
– Noah Snyder
Aug 5 at 14:56
1
$begingroup$
Everyone when they first learn about strictification imagines that it's making a smaller category, but actually it's making a much bigger one. You're essentially adding new objects which are formal tensor products of your old ones. So if $1_g$ denotes the 1-dimensional vector space in degree g, in the skeletal version $1_g otimes 1_h = 1_gh$ while in the strict version there's an extra new object $1_g 1_h$ which is isomorphic but not equal to $1_gh$. You can still see the associator by comparing the two isomorphisms $1_x 1_y 1_z rightarrow 1_xyz$.
$endgroup$
– Noah Snyder
Aug 5 at 16:06
1
$begingroup$
Here is a related (and perhaps easier) example: the categories $textRep(D_8)$ and $textRep(Q_8)$ have isomorphic Grothendieck rings (with an isomorphism sending classes of simple objects to classes of simple objects), both categories are symmetric and not monoidally equivalent (which was proved by Tambara-Yamagami).
$endgroup$
– Victor Ostrik
Aug 6 at 16:08
|
show 1 more comment
$begingroup$
Without the symmetry it’s easy to produce oodles of examples where a map of fusion rings doesn’t lift to the category level. The simplest example is probably $mathrmVec(G,w)$ the category of G-graded vector spaces with associator given by a 3-cocycle $w$. The fusion ring has an automorphism for any automorphism of G, but this only lifts to the category when it preserves $w$ in cohomology.
Giving a symmetric example is harder, but I think the following works. Let $D_8$ be the $8$-element dihedral group, and let $mathrmRep(D_8)$ be its category of representations. The fusion ring has an $S_3$ of automorphisms which permute the three non-trivial invertible objects. (Side note: $mathrmRep(Q_8)$ has the same fusion ring, and there this $S_3$ comes from an action on $Q_8$ and so lifts to the category level.) However, only one of the non-trivial elements of this $S_3$ lifts to the category level. Morally the reason is simple, the 1-dimensional representations each have a kernel, two of them have kernel isomorphic to the Klein 4-group while one of them has kernel isomorphic to $mathbbZ/4mathbbZ$ and the only the permutation swapping the two Klein 4-group ones will lift to the category level. Unfortunately this isn't quite a proof because it's not clear how to describe the kernel of the representation using only the tensor category structure (and not the symmetric tensor category structure).
Nonetheless you can see that the other permutations don't lift by following the classification of Tambara-Yamagami categories in Section 9.2 of Etingof-Nikshych-Ostrik, but it's a little technical. Namely, each 1-dimensional rep generates a $mathrmRep(mathbbZ/2mathbbZ)$ category, and the 2-dimensional rep is fixed by tensoring on either side by this category, so this exhibits $mathrmVec$ as a bimodule category over $mathrmRep(mathbbZ/2mathbbZ)$, and for the Klein 4-group kernel reps this bimodule is invertible while for the $mathbbZ/4mathbbZ$ kernel reps this bimodule is not invertible. Since this was all stated in the language of tensor categories, there's no tensor autoequivalence which mixes the two kind of 1-dimensional objects.
answered Aug 5 at 3:21
Noah SnyderNoah Snyder
19.9k2 gold badges68 silver badges137 bronze badges
$endgroup$
$begingroup$
I'm probably being too naive: but my original thinking was to first replace the monoidal categories by their strict monoidal versions. So in your example when C=D, cycling the 1-dimensional irreps doesn't seem like it should create any problems. But my main problem is I don't have a good intuition for tihnking about coherence issues so it's hard to know where to look for inconsistencies.
$endgroup$
– Steven Sam
Aug 5 at 4:25
2
$begingroup$
You can’t make that example both strict and skeletal at the same time. If you make it skeletal then it’s easy to see what the associator is, it’s just a 3-cocycle. If you make it strict, then you have to make the category much bigger and it’s harder to see what’s going on.
$endgroup$
– Noah Snyder
Aug 5 at 4:43
$begingroup$
Edited to upgrade my deleted comment on a symmetric example to the main text since I think I have an argument that shows it works now.
$endgroup$
– Noah Snyder
Aug 5 at 14:56
1
$begingroup$
Everyone when they first learn about strictification imagines that it's making a smaller category, but actually it's making a much bigger one. You're essentially adding new objects which are formal tensor products of your old ones. So if $1_g$ denotes the 1-dimensional vector space in degree g, in the skeletal version $1_g otimes 1_h = 1_gh$ while in the strict version there's an extra new object $1_g 1_h$ which is isomorphic but not equal to $1_gh$. You can still see the associator by comparing the two isomorphisms $1_x 1_y 1_z rightarrow 1_xyz$.
$endgroup$
– Noah Snyder
Aug 5 at 16:06
1
$begingroup$
Here is a related (and perhaps easier) example: the categories $textRep(D_8)$ and $textRep(Q_8)$ have isomorphic Grothendieck rings (with an isomorphism sending classes of simple objects to classes of simple objects), both categories are symmetric and not monoidally equivalent (which was proved by Tambara-Yamagami).
$endgroup$
– Victor Ostrik
Aug 6 at 16:08
|
show 1 more comment
$begingroup$
Without the symmetry it’s easy to produce oodles of examples where a map of fusion rings doesn’t lift to the category level. The simplest example is probably $mathrmVec(G,w)$ the category of G-graded vector spaces with associator given by a 3-cocycle $w$. The fusion ring has an automorphism for any automorphism of G, but this only lifts to the category when it preserves $w$ in cohomology.
Giving a symmetric example is harder, but I think the following works. Let $D_8$ be the $8$-element dihedral group, and let $mathrmRep(D_8)$ be its category of representations. The fusion ring has an $S_3$ of automorphisms which permute the three non-trivial invertible objects. (Side note: $mathrmRep(Q_8)$ has the same fusion ring, and there this $S_3$ comes from an action on $Q_8$ and so lifts to the category level.) However, only one of the non-trivial elements of this $S_3$ lifts to the category level. Morally the reason is simple, the 1-dimensional representations each have a kernel, two of them have kernel isomorphic to the Klein 4-group while one of them has kernel isomorphic to $mathbbZ/4mathbbZ$ and the only the permutation swapping the two Klein 4-group ones will lift to the category level. Unfortunately this isn't quite a proof because it's not clear how to describe the kernel of the representation using only the tensor category structure (and not the symmetric tensor category structure).
Nonetheless you can see that the other permutations don't lift by following the classification of Tambara-Yamagami categories in Section 9.2 of Etingof-Nikshych-Ostrik, but it's a little technical. Namely, each 1-dimensional rep generates a $mathrmRep(mathbbZ/2mathbbZ)$ category, and the 2-dimensional rep is fixed by tensoring on either side by this category, so this exhibits $mathrmVec$ as a bimodule category over $mathrmRep(mathbbZ/2mathbbZ)$, and for the Klein 4-group kernel reps this bimodule is invertible while for the $mathbbZ/4mathbbZ$ kernel reps this bimodule is not invertible. Since this was all stated in the language of tensor categories, there's no tensor autoequivalence which mixes the two kind of 1-dimensional objects.
answered Aug 5 at 3:21
Noah SnyderNoah Snyder
19.9k2 gold badges68 silver badges137 bronze badges
$endgroup$
Without the symmetry it’s easy to produce oodles of examples where a map of fusion rings doesn’t lift to the category level. The simplest example is probably $mathrmVec(G,w)$ the category of G-graded vector spaces with associator given by a 3-cocycle $w$. The fusion ring has an automorphism for any automorphism of G, but this only lifts to the category when it preserves $w$ in cohomology.
Giving a symmetric example is harder, but I think the following works. Let $D_8$ be the $8$-element dihedral group, and let $mathrmRep(D_8)$ be its category of representations. The fusion ring has an $S_3$ of automorphisms which permute the three non-trivial invertible objects. (Side note: $mathrmRep(Q_8)$ has the same fusion ring, and there this $S_3$ comes from an action on $Q_8$ and so lifts to the category level.) However, only one of the non-trivial elements of this $S_3$ lifts to the category level. Morally the reason is simple, the 1-dimensional representations each have a kernel, two of them have kernel isomorphic to the Klein 4-group while one of them has kernel isomorphic to $mathbbZ/4mathbbZ$ and the only the permutation swapping the two Klein 4-group ones will lift to the category level. Unfortunately this isn't quite a proof because it's not clear how to describe the kernel of the representation using only the tensor category structure (and not the symmetric tensor category structure).
Nonetheless you can see that the other permutations don't lift by following the classification of Tambara-Yamagami categories in Section 9.2 of Etingof-Nikshych-Ostrik, but it's a little technical. Namely, each 1-dimensional rep generates a $mathrmRep(mathbbZ/2mathbbZ)$ category, and the 2-dimensional rep is fixed by tensoring on either side by this category, so this exhibits $mathrmVec$ as a bimodule category over $mathrmRep(mathbbZ/2mathbbZ)$, and for the Klein 4-group kernel reps this bimodule is invertible while for the $mathbbZ/4mathbbZ$ kernel reps this bimodule is not invertible. Since this was all stated in the language of tensor categories, there's no tensor autoequivalence which mixes the two kind of 1-dimensional objects.
answered Aug 5 at 3:21
Noah SnyderNoah Snyder
19.9k2 gold badges68 silver badges137 bronze badges
edited Aug 5 at 14:55
answered Aug 5 at 3:21
Noah SnyderNoah Snyder
19.9k2 gold badges68 silver badges137 bronze badges
answered Aug 5 at 3:21
Noah SnyderNoah Snyder
19.9k2 gold badges68 silver badges137 bronze badges
answered Aug 5 at 3:21
Noah SnyderNoah Snyder
19.9k2 gold badges68 silver badges137 bronze badges
19.9k2 gold badges68 silver badges137 bronze badges
$begingroup$
I'm probably being too naive: but my original thinking was to first replace the monoidal categories by their strict monoidal versions. So in your example when C=D, cycling the 1-dimensional irreps doesn't seem like it should create any problems. But my main problem is I don't have a good intuition for tihnking about coherence issues so it's hard to know where to look for inconsistencies.
$endgroup$
– Steven Sam
Aug 5 at 4:25
2
$begingroup$
You can’t make that example both strict and skeletal at the same time. If you make it skeletal then it’s easy to see what the associator is, it’s just a 3-cocycle. If you make it strict, then you have to make the category much bigger and it’s harder to see what’s going on.
$endgroup$
– Noah Snyder
Aug 5 at 4:43
$begingroup$
Edited to upgrade my deleted comment on a symmetric example to the main text since I think I have an argument that shows it works now.
$endgroup$
– Noah Snyder
Aug 5 at 14:56
1
$begingroup$
Everyone when they first learn about strictification imagines that it's making a smaller category, but actually it's making a much bigger one. You're essentially adding new objects which are formal tensor products of your old ones. So if $1_g$ denotes the 1-dimensional vector space in degree g, in the skeletal version $1_g otimes 1_h = 1_gh$ while in the strict version there's an extra new object $1_g 1_h$ which is isomorphic but not equal to $1_gh$. You can still see the associator by comparing the two isomorphisms $1_x 1_y 1_z rightarrow 1_xyz$.
$endgroup$
– Noah Snyder
Aug 5 at 16:06
1
$begingroup$
Here is a related (and perhaps easier) example: the categories $textRep(D_8)$ and $textRep(Q_8)$ have isomorphic Grothendieck rings (with an isomorphism sending classes of simple objects to classes of simple objects), both categories are symmetric and not monoidally equivalent (which was proved by Tambara-Yamagami).
$endgroup$
– Victor Ostrik
Aug 6 at 16:08
|
show 1 more comment
$begingroup$
I'm probably being too naive: but my original thinking was to first replace the monoidal categories by their strict monoidal versions. So in your example when C=D, cycling the 1-dimensional irreps doesn't seem like it should create any problems. But my main problem is I don't have a good intuition for tihnking about coherence issues so it's hard to know where to look for inconsistencies.
$endgroup$
– Steven Sam
Aug 5 at 4:25
2
$begingroup$
You can’t make that example both strict and skeletal at the same time. If you make it skeletal then it’s easy to see what the associator is, it’s just a 3-cocycle. If you make it strict, then you have to make the category much bigger and it’s harder to see what’s going on.
$endgroup$
– Noah Snyder
Aug 5 at 4:43
$begingroup$
Edited to upgrade my deleted comment on a symmetric example to the main text since I think I have an argument that shows it works now.
$endgroup$
– Noah Snyder
Aug 5 at 14:56
1
$begingroup$
Everyone when they first learn about strictification imagines that it's making a smaller category, but actually it's making a much bigger one. You're essentially adding new objects which are formal tensor products of your old ones. So if $1_g$ denotes the 1-dimensional vector space in degree g, in the skeletal version $1_g otimes 1_h = 1_gh$ while in the strict version there's an extra new object $1_g 1_h$ which is isomorphic but not equal to $1_gh$. You can still see the associator by comparing the two isomorphisms $1_x 1_y 1_z rightarrow 1_xyz$.
$endgroup$
– Noah Snyder
Aug 5 at 16:06
1
$begingroup$
Here is a related (and perhaps easier) example: the categories $textRep(D_8)$ and $textRep(Q_8)$ have isomorphic Grothendieck rings (with an isomorphism sending classes of simple objects to classes of simple objects), both categories are symmetric and not monoidally equivalent (which was proved by Tambara-Yamagami).
$endgroup$
– Victor Ostrik
Aug 6 at 16:08
$begingroup$
I'm probably being too naive: but my original thinking was to first replace the monoidal categories by their strict monoidal versions. So in your example when C=D, cycling the 1-dimensional irreps doesn't seem like it should create any problems. But my main problem is I don't have a good intuition for tihnking about coherence issues so it's hard to know where to look for inconsistencies.
$endgroup$
– Steven Sam
Aug 5 at 4:25
$begingroup$
I'm probably being too naive: but my original thinking was to first replace the monoidal categories by their strict monoidal versions. So in your example when C=D, cycling the 1-dimensional irreps doesn't seem like it should create any problems. But my main problem is I don't have a good intuition for tihnking about coherence issues so it's hard to know where to look for inconsistencies.
$endgroup$
– Steven Sam
Aug 5 at 4:25
2
2
$begingroup$
You can’t make that example both strict and skeletal at the same time. If you make it skeletal then it’s easy to see what the associator is, it’s just a 3-cocycle. If you make it strict, then you have to make the category much bigger and it’s harder to see what’s going on.
$endgroup$
– Noah Snyder
Aug 5 at 4:43
$begingroup$
You can’t make that example both strict and skeletal at the same time. If you make it skeletal then it’s easy to see what the associator is, it’s just a 3-cocycle. If you make it strict, then you have to make the category much bigger and it’s harder to see what’s going on.
$endgroup$
– Noah Snyder
Aug 5 at 4:43
$begingroup$
Edited to upgrade my deleted comment on a symmetric example to the main text since I think I have an argument that shows it works now.
$endgroup$
– Noah Snyder
Aug 5 at 14:56
$begingroup$
Edited to upgrade my deleted comment on a symmetric example to the main text since I think I have an argument that shows it works now.
$endgroup$
– Noah Snyder
Aug 5 at 14:56
1
1
$begingroup$
Everyone when they first learn about strictification imagines that it's making a smaller category, but actually it's making a much bigger one. You're essentially adding new objects which are formal tensor products of your old ones. So if $1_g$ denotes the 1-dimensional vector space in degree g, in the skeletal version $1_g otimes 1_h = 1_gh$ while in the strict version there's an extra new object $1_g 1_h$ which is isomorphic but not equal to $1_gh$. You can still see the associator by comparing the two isomorphisms $1_x 1_y 1_z rightarrow 1_xyz$.
$endgroup$
– Noah Snyder
Aug 5 at 16:06
$begingroup$
Everyone when they first learn about strictification imagines that it's making a smaller category, but actually it's making a much bigger one. You're essentially adding new objects which are formal tensor products of your old ones. So if $1_g$ denotes the 1-dimensional vector space in degree g, in the skeletal version $1_g otimes 1_h = 1_gh$ while in the strict version there's an extra new object $1_g 1_h$ which is isomorphic but not equal to $1_gh$. You can still see the associator by comparing the two isomorphisms $1_x 1_y 1_z rightarrow 1_xyz$.
$endgroup$
– Noah Snyder
Aug 5 at 16:06
1
1
$begingroup$
Here is a related (and perhaps easier) example: the categories $textRep(D_8)$ and $textRep(Q_8)$ have isomorphic Grothendieck rings (with an isomorphism sending classes of simple objects to classes of simple objects), both categories are symmetric and not monoidally equivalent (which was proved by Tambara-Yamagami).
$endgroup$
– Victor Ostrik
Aug 6 at 16:08
$begingroup$
Here is a related (and perhaps easier) example: the categories $textRep(D_8)$ and $textRep(Q_8)$ have isomorphic Grothendieck rings (with an isomorphism sending classes of simple objects to classes of simple objects), both categories are symmetric and not monoidally equivalent (which was proved by Tambara-Yamagami).
$endgroup$
– Victor Ostrik
Aug 6 at 16:08
|
show 1 more comment
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$begingroup$
I would say there are absolutely no chances of doing something like this as the Grothendieck ring essentially remembers nothing of the morphisms. If you are not convince look at when $C$ and $D$ are the category of vectors spaces over two different fields. They have both $mathbbZ$ as Grothendieck ring, but a monoidal functors would corresponds to a fields homomorphism.
$endgroup$
– Simon Henry
Aug 5 at 1:17
$begingroup$
Fair point. I will update my question.
$endgroup$
– Steven Sam
Aug 5 at 1:55