Module category
I am currently reading about module categories and have been not very successful. In this case a module category is a category with an action of a monoidal category. (more information on nLab)
In specific my task is to find a module category $M$ over $G$-vect ($G$-graded vector spaces), s.th. the Functor category $Fun(M, M) cong G$-rep (representations of $G$), where G is a group.
Unfortunately I don't have any clue how that could look like.
It would already be helpful if someone knew any example of a module category! (So that I can get a feeling for what this category could look like)
If anyone knows more about this specific example, it would be of course even better!
Thanks in advance!
category-theory graded-modules
asked Dec 13 '18 at 10:39
PollyPolly
234
|
show 12 more comments
I am currently reading about module categories and have been not very successful. In this case a module category is a category with an action of a monoidal category. (more information on nLab)
In specific my task is to find a module category $M$ over $G$-vect ($G$-graded vector spaces), s.th. the Functor category $Fun(M, M) cong G$-rep (representations of $G$), where G is a group.
Unfortunately I don't have any clue how that could look like.
It would already be helpful if someone knew any example of a module category! (So that I can get a feeling for what this category could look like)
If anyone knows more about this specific example, it would be of course even better!
Thanks in advance!
category-theory graded-modules
asked Dec 13 '18 at 10:39
PollyPolly
234
1
Right, that is what I would call a representation of the category (at least with suitable additional structure, such as being additive, but that seems given here). I don't have an idea for the specific thing you are looking for, but a place to start with examples is the category acting on itself in the obvious way.
– Tobias Kildetoft
Dec 13 '18 at 10:55
1
Actually come to think of it, that example might be exactly what you are looking for, though I didn't work out the details.
– Tobias Kildetoft
Dec 13 '18 at 10:57
1
So first you need to figure out what a functor from this category to itself looks like (I assume that these are supposed to be functors compatible with the action of the category?)
– Tobias Kildetoft
Dec 13 '18 at 13:58
1
Yes. The idea to keep in mind is that such a functor will be determined on objects by what it does to a suitable collection of objects, and similar for morphisms. In this case, if you know what it does to the $1$-dimension vector space concentrated in some degree $gin G$ for all $gin G$ then you also know what it does to all objects by additivity. So what about morphisms?
– Tobias Kildetoft
Dec 13 '18 at 14:13
1
Actually, I may have been keeping the easy example of $mathbb{Z}$-graded spaces a bit too hard in my mind, since there you see how to get everything from what it does to just the $1$-dimensional space in degree $1$, which gives you a vector space for each functor, and the action of $mathbb{Z}$ should come from the morphisms. But I am now less sure if this all works out as nicely for arbitrary groups.
– Tobias Kildetoft
Dec 13 '18 at 14:15
|
show 12 more comments
I am currently reading about module categories and have been not very successful. In this case a module category is a category with an action of a monoidal category. (more information on nLab)
In specific my task is to find a module category $M$ over $G$-vect ($G$-graded vector spaces), s.th. the Functor category $Fun(M, M) cong G$-rep (representations of $G$), where G is a group.
Unfortunately I don't have any clue how that could look like.
It would already be helpful if someone knew any example of a module category! (So that I can get a feeling for what this category could look like)
If anyone knows more about this specific example, it would be of course even better!
Thanks in advance!
category-theory graded-modules
asked Dec 13 '18 at 10:39
PollyPolly
234
I am currently reading about module categories and have been not very successful. In this case a module category is a category with an action of a monoidal category. (more information on nLab)
In specific my task is to find a module category $M$ over $G$-vect ($G$-graded vector spaces), s.th. the Functor category $Fun(M, M) cong G$-rep (representations of $G$), where G is a group.
Unfortunately I don't have any clue how that could look like.
It would already be helpful if someone knew any example of a module category! (So that I can get a feeling for what this category could look like)
If anyone knows more about this specific example, it would be of course even better!
Thanks in advance!
category-theory graded-modules
category-theory graded-modules
asked Dec 13 '18 at 10:39
PollyPolly
234
asked Dec 13 '18 at 10:39
PollyPolly
234
edited Dec 13 '18 at 10:49
Polly
asked Dec 13 '18 at 10:39
PollyPolly
234
asked Dec 13 '18 at 10:39
PollyPolly
234
asked Dec 13 '18 at 10:39
PollyPolly
234
234
1
Right, that is what I would call a representation of the category (at least with suitable additional structure, such as being additive, but that seems given here). I don't have an idea for the specific thing you are looking for, but a place to start with examples is the category acting on itself in the obvious way.
– Tobias Kildetoft
Dec 13 '18 at 10:55
1
Actually come to think of it, that example might be exactly what you are looking for, though I didn't work out the details.
– Tobias Kildetoft
Dec 13 '18 at 10:57
1
So first you need to figure out what a functor from this category to itself looks like (I assume that these are supposed to be functors compatible with the action of the category?)
– Tobias Kildetoft
Dec 13 '18 at 13:58
1
Yes. The idea to keep in mind is that such a functor will be determined on objects by what it does to a suitable collection of objects, and similar for morphisms. In this case, if you know what it does to the $1$-dimension vector space concentrated in some degree $gin G$ for all $gin G$ then you also know what it does to all objects by additivity. So what about morphisms?
– Tobias Kildetoft
Dec 13 '18 at 14:13
1
Actually, I may have been keeping the easy example of $mathbb{Z}$-graded spaces a bit too hard in my mind, since there you see how to get everything from what it does to just the $1$-dimensional space in degree $1$, which gives you a vector space for each functor, and the action of $mathbb{Z}$ should come from the morphisms. But I am now less sure if this all works out as nicely for arbitrary groups.
– Tobias Kildetoft
Dec 13 '18 at 14:15
|
show 12 more comments
1
Right, that is what I would call a representation of the category (at least with suitable additional structure, such as being additive, but that seems given here). I don't have an idea for the specific thing you are looking for, but a place to start with examples is the category acting on itself in the obvious way.
– Tobias Kildetoft
Dec 13 '18 at 10:55
1
Actually come to think of it, that example might be exactly what you are looking for, though I didn't work out the details.
– Tobias Kildetoft
Dec 13 '18 at 10:57
1
So first you need to figure out what a functor from this category to itself looks like (I assume that these are supposed to be functors compatible with the action of the category?)
– Tobias Kildetoft
Dec 13 '18 at 13:58
1
Yes. The idea to keep in mind is that such a functor will be determined on objects by what it does to a suitable collection of objects, and similar for morphisms. In this case, if you know what it does to the $1$-dimension vector space concentrated in some degree $gin G$ for all $gin G$ then you also know what it does to all objects by additivity. So what about morphisms?
– Tobias Kildetoft
Dec 13 '18 at 14:13
1
Actually, I may have been keeping the easy example of $mathbb{Z}$-graded spaces a bit too hard in my mind, since there you see how to get everything from what it does to just the $1$-dimensional space in degree $1$, which gives you a vector space for each functor, and the action of $mathbb{Z}$ should come from the morphisms. But I am now less sure if this all works out as nicely for arbitrary groups.
– Tobias Kildetoft
Dec 13 '18 at 14:15
1
1
Right, that is what I would call a representation of the category (at least with suitable additional structure, such as being additive, but that seems given here). I don't have an idea for the specific thing you are looking for, but a place to start with examples is the category acting on itself in the obvious way.
– Tobias Kildetoft
Dec 13 '18 at 10:55
Right, that is what I would call a representation of the category (at least with suitable additional structure, such as being additive, but that seems given here). I don't have an idea for the specific thing you are looking for, but a place to start with examples is the category acting on itself in the obvious way.
– Tobias Kildetoft
Dec 13 '18 at 10:55
1
1
Actually come to think of it, that example might be exactly what you are looking for, though I didn't work out the details.
– Tobias Kildetoft
Dec 13 '18 at 10:57
Actually come to think of it, that example might be exactly what you are looking for, though I didn't work out the details.
– Tobias Kildetoft
Dec 13 '18 at 10:57
1
1
So first you need to figure out what a functor from this category to itself looks like (I assume that these are supposed to be functors compatible with the action of the category?)
– Tobias Kildetoft
Dec 13 '18 at 13:58
So first you need to figure out what a functor from this category to itself looks like (I assume that these are supposed to be functors compatible with the action of the category?)
– Tobias Kildetoft
Dec 13 '18 at 13:58
1
1
Yes. The idea to keep in mind is that such a functor will be determined on objects by what it does to a suitable collection of objects, and similar for morphisms. In this case, if you know what it does to the $1$-dimension vector space concentrated in some degree $gin G$ for all $gin G$ then you also know what it does to all objects by additivity. So what about morphisms?
– Tobias Kildetoft
Dec 13 '18 at 14:13
Yes. The idea to keep in mind is that such a functor will be determined on objects by what it does to a suitable collection of objects, and similar for morphisms. In this case, if you know what it does to the $1$-dimension vector space concentrated in some degree $gin G$ for all $gin G$ then you also know what it does to all objects by additivity. So what about morphisms?
– Tobias Kildetoft
Dec 13 '18 at 14:13
1
1
Actually, I may have been keeping the easy example of $mathbb{Z}$-graded spaces a bit too hard in my mind, since there you see how to get everything from what it does to just the $1$-dimensional space in degree $1$, which gives you a vector space for each functor, and the action of $mathbb{Z}$ should come from the morphisms. But I am now less sure if this all works out as nicely for arbitrary groups.
– Tobias Kildetoft
Dec 13 '18 at 14:15
Actually, I may have been keeping the easy example of $mathbb{Z}$-graded spaces a bit too hard in my mind, since there you see how to get everything from what it does to just the $1$-dimensional space in degree $1$, which gives you a vector space for each functor, and the action of $mathbb{Z}$ should come from the morphisms. But I am now less sure if this all works out as nicely for arbitrary groups.
– Tobias Kildetoft
Dec 13 '18 at 14:15
|
show 12 more comments
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1
Right, that is what I would call a representation of the category (at least with suitable additional structure, such as being additive, but that seems given here). I don't have an idea for the specific thing you are looking for, but a place to start with examples is the category acting on itself in the obvious way.
– Tobias Kildetoft
Dec 13 '18 at 10:55
1
Actually come to think of it, that example might be exactly what you are looking for, though I didn't work out the details.
– Tobias Kildetoft
Dec 13 '18 at 10:57
1
So first you need to figure out what a functor from this category to itself looks like (I assume that these are supposed to be functors compatible with the action of the category?)
– Tobias Kildetoft
Dec 13 '18 at 13:58
1
Yes. The idea to keep in mind is that such a functor will be determined on objects by what it does to a suitable collection of objects, and similar for morphisms. In this case, if you know what it does to the $1$-dimension vector space concentrated in some degree $gin G$ for all $gin G$ then you also know what it does to all objects by additivity. So what about morphisms?
– Tobias Kildetoft
Dec 13 '18 at 14:13
1
Actually, I may have been keeping the easy example of $mathbb{Z}$-graded spaces a bit too hard in my mind, since there you see how to get everything from what it does to just the $1$-dimensional space in degree $1$, which gives you a vector space for each functor, and the action of $mathbb{Z}$ should come from the morphisms. But I am now less sure if this all works out as nicely for arbitrary groups.
– Tobias Kildetoft
Dec 13 '18 at 14:15