How are F-bialgebras defined?
$begingroup$
First, does such a notion exist?
If so, would it be as trivial as the triple $(S,alpha, beta)$, where $(S,alpha)$ is an F-algebra and $(S,beta)$ is an F-coalgebra?
I'm assuming there would be more that needs to occur, such as some interaction between $alpha$ and $beta$, but I haven't found much information out there that I can understand.
category-theory definition universal-algebra
edited Dec 31 '18 at 14:58
Shaun
9,789113684
asked Dec 31 '18 at 14:49
m1cky22m1cky22
415213
$endgroup$
add a comment |
$begingroup$
First, does such a notion exist?
If so, would it be as trivial as the triple $(S,alpha, beta)$, where $(S,alpha)$ is an F-algebra and $(S,beta)$ is an F-coalgebra?
I'm assuming there would be more that needs to occur, such as some interaction between $alpha$ and $beta$, but I haven't found much information out there that I can understand.
category-theory definition universal-algebra
edited Dec 31 '18 at 14:58
Shaun
9,789113684
asked Dec 31 '18 at 14:49
m1cky22m1cky22
415213
$endgroup$
$begingroup$
en.wikipedia.org/wiki/Bialgebra
$endgroup$
– egreg
Dec 31 '18 at 15:28
$begingroup$
There's a similar but different notion called a dialgebra.
$endgroup$
– Derek Elkins
Dec 31 '18 at 18:30
add a comment |
$begingroup$
First, does such a notion exist?
If so, would it be as trivial as the triple $(S,alpha, beta)$, where $(S,alpha)$ is an F-algebra and $(S,beta)$ is an F-coalgebra?
I'm assuming there would be more that needs to occur, such as some interaction between $alpha$ and $beta$, but I haven't found much information out there that I can understand.
category-theory definition universal-algebra
edited Dec 31 '18 at 14:58
Shaun
9,789113684
asked Dec 31 '18 at 14:49
m1cky22m1cky22
415213
$endgroup$
First, does such a notion exist?
If so, would it be as trivial as the triple $(S,alpha, beta)$, where $(S,alpha)$ is an F-algebra and $(S,beta)$ is an F-coalgebra?
I'm assuming there would be more that needs to occur, such as some interaction between $alpha$ and $beta$, but I haven't found much information out there that I can understand.
category-theory definition universal-algebra
category-theory definition universal-algebra
edited Dec 31 '18 at 14:58
Shaun
9,789113684
asked Dec 31 '18 at 14:49
m1cky22m1cky22
415213
edited Dec 31 '18 at 14:58
Shaun
9,789113684
asked Dec 31 '18 at 14:49
m1cky22m1cky22
415213
edited Dec 31 '18 at 14:58
Shaun
9,789113684
edited Dec 31 '18 at 14:58
Shaun
9,789113684
edited Dec 31 '18 at 14:58
Shaun
9,789113684
9,789113684
asked Dec 31 '18 at 14:49
m1cky22m1cky22
415213
asked Dec 31 '18 at 14:49
m1cky22m1cky22
415213
asked Dec 31 '18 at 14:49
m1cky22m1cky22
415213
415213
$begingroup$
en.wikipedia.org/wiki/Bialgebra
$endgroup$
– egreg
Dec 31 '18 at 15:28
$begingroup$
There's a similar but different notion called a dialgebra.
$endgroup$
– Derek Elkins
Dec 31 '18 at 18:30
add a comment |
$begingroup$
en.wikipedia.org/wiki/Bialgebra
$endgroup$
– egreg
Dec 31 '18 at 15:28
$begingroup$
There's a similar but different notion called a dialgebra.
$endgroup$
– Derek Elkins
Dec 31 '18 at 18:30
$begingroup$
en.wikipedia.org/wiki/Bialgebra
$endgroup$
– egreg
Dec 31 '18 at 15:28
$begingroup$
en.wikipedia.org/wiki/Bialgebra
$endgroup$
– egreg
Dec 31 '18 at 15:28
$begingroup$
There's a similar but different notion called a dialgebra.
$endgroup$
– Derek Elkins
Dec 31 '18 at 18:30
$begingroup$
There's a similar but different notion called a dialgebra.
$endgroup$
– Derek Elkins
Dec 31 '18 at 18:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
More generally, given two endofunctors $T$ and $F$ on a category $C$, we can define a $(T,F)$-bialgebra to be an object of $C$ equipped with the structure of a $T$-algebra and an $F$-coalgebra, i.e. just a triple $(S,alpha,beta)$, where $alphacolon TSto S$ and $betacolon Sto FS$.
But this is a very broad notion. As you expect, we usually want some relationship between the $T$-algebra structure and the $F$-coalgebra structure. The usual way of doing this is the notion of $lambda$-bialgebra, where $lambda$ is a distributive law (of $T$ over $F$): a natural transformation $lambdacolon TF to FT$.
A $lambda$-bialgebra is a $(T,F)$-bialgebra $(S,alpha,beta)$ such that $beta circ alpha = Falphacirc lambda_S circ Tbeta$ (I recommend drawing the diagram for yourself).
answered Dec 31 '18 at 16:51
Alex KruckmanAlex Kruckman
28.2k32658
$endgroup$
$begingroup$
Awesome! That's exactly the answer I wanted! Time to go find some papers 😊
$endgroup$
– m1cky22
Dec 31 '18 at 23:58
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
More generally, given two endofunctors $T$ and $F$ on a category $C$, we can define a $(T,F)$-bialgebra to be an object of $C$ equipped with the structure of a $T$-algebra and an $F$-coalgebra, i.e. just a triple $(S,alpha,beta)$, where $alphacolon TSto S$ and $betacolon Sto FS$.
But this is a very broad notion. As you expect, we usually want some relationship between the $T$-algebra structure and the $F$-coalgebra structure. The usual way of doing this is the notion of $lambda$-bialgebra, where $lambda$ is a distributive law (of $T$ over $F$): a natural transformation $lambdacolon TF to FT$.
A $lambda$-bialgebra is a $(T,F)$-bialgebra $(S,alpha,beta)$ such that $beta circ alpha = Falphacirc lambda_S circ Tbeta$ (I recommend drawing the diagram for yourself).
answered Dec 31 '18 at 16:51
Alex KruckmanAlex Kruckman
28.2k32658
$endgroup$
$begingroup$
Awesome! That's exactly the answer I wanted! Time to go find some papers 😊
$endgroup$
– m1cky22
Dec 31 '18 at 23:58
add a comment |
$begingroup$
More generally, given two endofunctors $T$ and $F$ on a category $C$, we can define a $(T,F)$-bialgebra to be an object of $C$ equipped with the structure of a $T$-algebra and an $F$-coalgebra, i.e. just a triple $(S,alpha,beta)$, where $alphacolon TSto S$ and $betacolon Sto FS$.
But this is a very broad notion. As you expect, we usually want some relationship between the $T$-algebra structure and the $F$-coalgebra structure. The usual way of doing this is the notion of $lambda$-bialgebra, where $lambda$ is a distributive law (of $T$ over $F$): a natural transformation $lambdacolon TF to FT$.
A $lambda$-bialgebra is a $(T,F)$-bialgebra $(S,alpha,beta)$ such that $beta circ alpha = Falphacirc lambda_S circ Tbeta$ (I recommend drawing the diagram for yourself).
answered Dec 31 '18 at 16:51
Alex KruckmanAlex Kruckman
28.2k32658
$endgroup$
$begingroup$
Awesome! That's exactly the answer I wanted! Time to go find some papers 😊
$endgroup$
– m1cky22
Dec 31 '18 at 23:58
add a comment |
$begingroup$
More generally, given two endofunctors $T$ and $F$ on a category $C$, we can define a $(T,F)$-bialgebra to be an object of $C$ equipped with the structure of a $T$-algebra and an $F$-coalgebra, i.e. just a triple $(S,alpha,beta)$, where $alphacolon TSto S$ and $betacolon Sto FS$.
But this is a very broad notion. As you expect, we usually want some relationship between the $T$-algebra structure and the $F$-coalgebra structure. The usual way of doing this is the notion of $lambda$-bialgebra, where $lambda$ is a distributive law (of $T$ over $F$): a natural transformation $lambdacolon TF to FT$.
A $lambda$-bialgebra is a $(T,F)$-bialgebra $(S,alpha,beta)$ such that $beta circ alpha = Falphacirc lambda_S circ Tbeta$ (I recommend drawing the diagram for yourself).
answered Dec 31 '18 at 16:51
Alex KruckmanAlex Kruckman
28.2k32658
$endgroup$
More generally, given two endofunctors $T$ and $F$ on a category $C$, we can define a $(T,F)$-bialgebra to be an object of $C$ equipped with the structure of a $T$-algebra and an $F$-coalgebra, i.e. just a triple $(S,alpha,beta)$, where $alphacolon TSto S$ and $betacolon Sto FS$.
But this is a very broad notion. As you expect, we usually want some relationship between the $T$-algebra structure and the $F$-coalgebra structure. The usual way of doing this is the notion of $lambda$-bialgebra, where $lambda$ is a distributive law (of $T$ over $F$): a natural transformation $lambdacolon TF to FT$.
A $lambda$-bialgebra is a $(T,F)$-bialgebra $(S,alpha,beta)$ such that $beta circ alpha = Falphacirc lambda_S circ Tbeta$ (I recommend drawing the diagram for yourself).
answered Dec 31 '18 at 16:51
Alex KruckmanAlex Kruckman
28.2k32658
answered Dec 31 '18 at 16:51
Alex KruckmanAlex Kruckman
28.2k32658
answered Dec 31 '18 at 16:51
Alex KruckmanAlex Kruckman
28.2k32658
answered Dec 31 '18 at 16:51
Alex KruckmanAlex Kruckman
28.2k32658
28.2k32658
$begingroup$
Awesome! That's exactly the answer I wanted! Time to go find some papers 😊
$endgroup$
– m1cky22
Dec 31 '18 at 23:58
add a comment |
$begingroup$
Awesome! That's exactly the answer I wanted! Time to go find some papers 😊
$endgroup$
– m1cky22
Dec 31 '18 at 23:58
$begingroup$
Awesome! That's exactly the answer I wanted! Time to go find some papers 😊
$endgroup$
– m1cky22
Dec 31 '18 at 23:58
$begingroup$
Awesome! That's exactly the answer I wanted! Time to go find some papers 😊
$endgroup$
– m1cky22
Dec 31 '18 at 23:58
add a comment |
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$begingroup$
en.wikipedia.org/wiki/Bialgebra
$endgroup$
– egreg
Dec 31 '18 at 15:28
$begingroup$
There's a similar but different notion called a dialgebra.
$endgroup$
– Derek Elkins
Dec 31 '18 at 18:30