Do we have $delta(ab)=delta(a)delta(b)$ implies $Delta(cd)=Delta(c)Delta(d)$?
Assume that $B$ is an algebra which is also a coalgebras (we do not assume that $B$ is a bialgebra: we do not assume $Delta(cd)=Delta(c)Delta(d)$). Assume that $A$ is $B$-comodule algebra. Then we have a map $delta: A to B otimes A$ which satisfies $(1 otimes delta)delta = (Delta otimes 1)delta$, where $Delta: B to B otimes B$ is the comultiplication on $B$. Suppose that for all $a,b in A$, we have $delta(ab)=delta(a)delta(b)$. Do we have $Delta(cd)=Delta(c)Delta(d)$ for all $c, d in B$?
Let $a,b in A$ and $delta(a)=a_{(1)} otimes a_{(2)}$, $delta(b)=b_{(1)} otimes b_{(2)}$. Then
begin{align}
(1 otimes delta)delta(ab) & = (1 otimes delta)(delta(a)delta(b)) \
& = (1 otimes delta)(a_{(1)}b_{(1)} otimes a_{(2)} b_{(2)}) \
& = a_{(1)}b_{(1)} otimes delta(a_{(2)}b_{(2)}) \
& = a_{(1)}b_{(1)} otimes delta(a_{(2)})delta(b_{(2)}) \
& = (a_{(1)} otimes delta(b_{(1)}))(a_{(2)} otimes delta(b_{(2)})) \
& = ((1 otimes delta)delta(a))((1 otimes delta)delta(b)) \
& = ((Delta otimes 1)delta(a))((Delta otimes 1)delta(b)) \
& = (Delta(a_{(1)}) otimes a_{(2)})(Delta(b_{(1)}) otimes b_{(2)}) \
& = ( Delta(a_{(1)}) Delta(b_{(1)}) ) otimes a_{(2)}b_{(2)}. quad (1)
end{align}
On the other hand,
begin{align}
(1 otimes delta)delta(ab) & = (Delta otimes 1)delta(ab) \
& = (Delta otimes 1)(delta(a)delta(b)) \
& = (Delta otimes 1)(a_{(1)}b_{(1)} otimes a_{(2)} b_{(2)}) \
& = Delta(a_{(1)}b_{(1)}) otimes a_{(2)} b_{(2)}. quad (2)
end{align}
By $(1)$ and $(2)$, do we have $Delta(cd)=Delta(c)Delta(d)$ for all $c,d in B$? Thank you very much.
Edit: assume that $B$ is an algebra which is also a coalgebras (but not necessarily bialgebra). Assume that $A$ is $B$-comodule algebra.
abstract-algebra representation-theory hopf-algebras
asked Aug 12 '15 at 8:37
LJR
6,55641749
|
show 4 more comments
Assume that $B$ is an algebra which is also a coalgebras (we do not assume that $B$ is a bialgebra: we do not assume $Delta(cd)=Delta(c)Delta(d)$). Assume that $A$ is $B$-comodule algebra. Then we have a map $delta: A to B otimes A$ which satisfies $(1 otimes delta)delta = (Delta otimes 1)delta$, where $Delta: B to B otimes B$ is the comultiplication on $B$. Suppose that for all $a,b in A$, we have $delta(ab)=delta(a)delta(b)$. Do we have $Delta(cd)=Delta(c)Delta(d)$ for all $c, d in B$?
Let $a,b in A$ and $delta(a)=a_{(1)} otimes a_{(2)}$, $delta(b)=b_{(1)} otimes b_{(2)}$. Then
begin{align}
(1 otimes delta)delta(ab) & = (1 otimes delta)(delta(a)delta(b)) \
& = (1 otimes delta)(a_{(1)}b_{(1)} otimes a_{(2)} b_{(2)}) \
& = a_{(1)}b_{(1)} otimes delta(a_{(2)}b_{(2)}) \
& = a_{(1)}b_{(1)} otimes delta(a_{(2)})delta(b_{(2)}) \
& = (a_{(1)} otimes delta(b_{(1)}))(a_{(2)} otimes delta(b_{(2)})) \
& = ((1 otimes delta)delta(a))((1 otimes delta)delta(b)) \
& = ((Delta otimes 1)delta(a))((Delta otimes 1)delta(b)) \
& = (Delta(a_{(1)}) otimes a_{(2)})(Delta(b_{(1)}) otimes b_{(2)}) \
& = ( Delta(a_{(1)}) Delta(b_{(1)}) ) otimes a_{(2)}b_{(2)}. quad (1)
end{align}
On the other hand,
begin{align}
(1 otimes delta)delta(ab) & = (Delta otimes 1)delta(ab) \
& = (Delta otimes 1)(delta(a)delta(b)) \
& = (Delta otimes 1)(a_{(1)}b_{(1)} otimes a_{(2)} b_{(2)}) \
& = Delta(a_{(1)}b_{(1)}) otimes a_{(2)} b_{(2)}. quad (2)
end{align}
By $(1)$ and $(2)$, do we have $Delta(cd)=Delta(c)Delta(d)$ for all $c,d in B$? Thank you very much.
Edit: assume that $B$ is an algebra which is also a coalgebras (but not necessarily bialgebra). Assume that $A$ is $B$-comodule algebra.
abstract-algebra representation-theory hopf-algebras
asked Aug 12 '15 at 8:37
LJR
6,55641749
What does "$ab$" mean? Are $A$ and $B$ actually bialgebras? Anyway I don't see why that would be true, at least not with some sort of unitality condition: you could just take $delta = 0$ and get a $B$-coalgebra no matter what $Delta$ is.
– Najib Idrissi
Aug 12 '15 at 8:55
I second the concern that you are assuming the existence of a multiplication, something that doesn't exist if $A$ and $B$ are just co-algebras. But I'm also curious about your definition of $B$-coalgeba. Nowhere do you make use of $A$'s comultiplication, you just have that $A$ is a co-module of $B$. I feel like you want to use $delta$ as a comultiplication, but it isn't quite. It seems to me like your definition should probably be that you have a map of coalgebras $Ato B$ (dual to the definition of $Rto S$ making $S$ into an $R$-algebra).
– Aaron
Aug 12 '15 at 9:15
@NajibIdrissi, thank you very much. I added that assume that $A, B$ are also algebras in the last line of the post.
– LJR
Aug 12 '15 at 9:31
@Aaron, thank you very much. Yes, we only need the condition that $A$ is a comodule algebra.
– LJR
Aug 12 '15 at 9:32
1
OK. But $A$ can be the zero ring. It then becomes a $B$-comodule algebra in an obvious way, but clearly this cannot say anything about $B$. So the literal answer to your question should be a "No". On the other hand, it may be worth looking for weak assumptions under which the answer can be "Yes", such as the existence of an algebra homomorphism from $A$ to $B$ which interacts with the structures nicely (whatever this means).
– darij grinberg
Aug 12 '15 at 15:17
|
show 4 more comments
Assume that $B$ is an algebra which is also a coalgebras (we do not assume that $B$ is a bialgebra: we do not assume $Delta(cd)=Delta(c)Delta(d)$). Assume that $A$ is $B$-comodule algebra. Then we have a map $delta: A to B otimes A$ which satisfies $(1 otimes delta)delta = (Delta otimes 1)delta$, where $Delta: B to B otimes B$ is the comultiplication on $B$. Suppose that for all $a,b in A$, we have $delta(ab)=delta(a)delta(b)$. Do we have $Delta(cd)=Delta(c)Delta(d)$ for all $c, d in B$?
Let $a,b in A$ and $delta(a)=a_{(1)} otimes a_{(2)}$, $delta(b)=b_{(1)} otimes b_{(2)}$. Then
begin{align}
(1 otimes delta)delta(ab) & = (1 otimes delta)(delta(a)delta(b)) \
& = (1 otimes delta)(a_{(1)}b_{(1)} otimes a_{(2)} b_{(2)}) \
& = a_{(1)}b_{(1)} otimes delta(a_{(2)}b_{(2)}) \
& = a_{(1)}b_{(1)} otimes delta(a_{(2)})delta(b_{(2)}) \
& = (a_{(1)} otimes delta(b_{(1)}))(a_{(2)} otimes delta(b_{(2)})) \
& = ((1 otimes delta)delta(a))((1 otimes delta)delta(b)) \
& = ((Delta otimes 1)delta(a))((Delta otimes 1)delta(b)) \
& = (Delta(a_{(1)}) otimes a_{(2)})(Delta(b_{(1)}) otimes b_{(2)}) \
& = ( Delta(a_{(1)}) Delta(b_{(1)}) ) otimes a_{(2)}b_{(2)}. quad (1)
end{align}
On the other hand,
begin{align}
(1 otimes delta)delta(ab) & = (Delta otimes 1)delta(ab) \
& = (Delta otimes 1)(delta(a)delta(b)) \
& = (Delta otimes 1)(a_{(1)}b_{(1)} otimes a_{(2)} b_{(2)}) \
& = Delta(a_{(1)}b_{(1)}) otimes a_{(2)} b_{(2)}. quad (2)
end{align}
By $(1)$ and $(2)$, do we have $Delta(cd)=Delta(c)Delta(d)$ for all $c,d in B$? Thank you very much.
Edit: assume that $B$ is an algebra which is also a coalgebras (but not necessarily bialgebra). Assume that $A$ is $B$-comodule algebra.
abstract-algebra representation-theory hopf-algebras
asked Aug 12 '15 at 8:37
LJR
6,55641749
Assume that $B$ is an algebra which is also a coalgebras (we do not assume that $B$ is a bialgebra: we do not assume $Delta(cd)=Delta(c)Delta(d)$). Assume that $A$ is $B$-comodule algebra. Then we have a map $delta: A to B otimes A$ which satisfies $(1 otimes delta)delta = (Delta otimes 1)delta$, where $Delta: B to B otimes B$ is the comultiplication on $B$. Suppose that for all $a,b in A$, we have $delta(ab)=delta(a)delta(b)$. Do we have $Delta(cd)=Delta(c)Delta(d)$ for all $c, d in B$?
Let $a,b in A$ and $delta(a)=a_{(1)} otimes a_{(2)}$, $delta(b)=b_{(1)} otimes b_{(2)}$. Then
begin{align}
(1 otimes delta)delta(ab) & = (1 otimes delta)(delta(a)delta(b)) \
& = (1 otimes delta)(a_{(1)}b_{(1)} otimes a_{(2)} b_{(2)}) \
& = a_{(1)}b_{(1)} otimes delta(a_{(2)}b_{(2)}) \
& = a_{(1)}b_{(1)} otimes delta(a_{(2)})delta(b_{(2)}) \
& = (a_{(1)} otimes delta(b_{(1)}))(a_{(2)} otimes delta(b_{(2)})) \
& = ((1 otimes delta)delta(a))((1 otimes delta)delta(b)) \
& = ((Delta otimes 1)delta(a))((Delta otimes 1)delta(b)) \
& = (Delta(a_{(1)}) otimes a_{(2)})(Delta(b_{(1)}) otimes b_{(2)}) \
& = ( Delta(a_{(1)}) Delta(b_{(1)}) ) otimes a_{(2)}b_{(2)}. quad (1)
end{align}
On the other hand,
begin{align}
(1 otimes delta)delta(ab) & = (Delta otimes 1)delta(ab) \
& = (Delta otimes 1)(delta(a)delta(b)) \
& = (Delta otimes 1)(a_{(1)}b_{(1)} otimes a_{(2)} b_{(2)}) \
& = Delta(a_{(1)}b_{(1)}) otimes a_{(2)} b_{(2)}. quad (2)
end{align}
By $(1)$ and $(2)$, do we have $Delta(cd)=Delta(c)Delta(d)$ for all $c,d in B$? Thank you very much.
Edit: assume that $B$ is an algebra which is also a coalgebras (but not necessarily bialgebra). Assume that $A$ is $B$-comodule algebra.
abstract-algebra representation-theory hopf-algebras
abstract-algebra representation-theory hopf-algebras
asked Aug 12 '15 at 8:37
LJR
6,55641749
asked Aug 12 '15 at 8:37
LJR
6,55641749
edited Aug 12 '15 at 13:47
asked Aug 12 '15 at 8:37
LJR
6,55641749
asked Aug 12 '15 at 8:37
LJR
6,55641749
asked Aug 12 '15 at 8:37
LJR
6,55641749
6,55641749
What does "$ab$" mean? Are $A$ and $B$ actually bialgebras? Anyway I don't see why that would be true, at least not with some sort of unitality condition: you could just take $delta = 0$ and get a $B$-coalgebra no matter what $Delta$ is.
– Najib Idrissi
Aug 12 '15 at 8:55
I second the concern that you are assuming the existence of a multiplication, something that doesn't exist if $A$ and $B$ are just co-algebras. But I'm also curious about your definition of $B$-coalgeba. Nowhere do you make use of $A$'s comultiplication, you just have that $A$ is a co-module of $B$. I feel like you want to use $delta$ as a comultiplication, but it isn't quite. It seems to me like your definition should probably be that you have a map of coalgebras $Ato B$ (dual to the definition of $Rto S$ making $S$ into an $R$-algebra).
– Aaron
Aug 12 '15 at 9:15
@NajibIdrissi, thank you very much. I added that assume that $A, B$ are also algebras in the last line of the post.
– LJR
Aug 12 '15 at 9:31
@Aaron, thank you very much. Yes, we only need the condition that $A$ is a comodule algebra.
– LJR
Aug 12 '15 at 9:32
1
OK. But $A$ can be the zero ring. It then becomes a $B$-comodule algebra in an obvious way, but clearly this cannot say anything about $B$. So the literal answer to your question should be a "No". On the other hand, it may be worth looking for weak assumptions under which the answer can be "Yes", such as the existence of an algebra homomorphism from $A$ to $B$ which interacts with the structures nicely (whatever this means).
– darij grinberg
Aug 12 '15 at 15:17
|
show 4 more comments
What does "$ab$" mean? Are $A$ and $B$ actually bialgebras? Anyway I don't see why that would be true, at least not with some sort of unitality condition: you could just take $delta = 0$ and get a $B$-coalgebra no matter what $Delta$ is.
– Najib Idrissi
Aug 12 '15 at 8:55
I second the concern that you are assuming the existence of a multiplication, something that doesn't exist if $A$ and $B$ are just co-algebras. But I'm also curious about your definition of $B$-coalgeba. Nowhere do you make use of $A$'s comultiplication, you just have that $A$ is a co-module of $B$. I feel like you want to use $delta$ as a comultiplication, but it isn't quite. It seems to me like your definition should probably be that you have a map of coalgebras $Ato B$ (dual to the definition of $Rto S$ making $S$ into an $R$-algebra).
– Aaron
Aug 12 '15 at 9:15
@NajibIdrissi, thank you very much. I added that assume that $A, B$ are also algebras in the last line of the post.
– LJR
Aug 12 '15 at 9:31
@Aaron, thank you very much. Yes, we only need the condition that $A$ is a comodule algebra.
– LJR
Aug 12 '15 at 9:32
1
OK. But $A$ can be the zero ring. It then becomes a $B$-comodule algebra in an obvious way, but clearly this cannot say anything about $B$. So the literal answer to your question should be a "No". On the other hand, it may be worth looking for weak assumptions under which the answer can be "Yes", such as the existence of an algebra homomorphism from $A$ to $B$ which interacts with the structures nicely (whatever this means).
– darij grinberg
Aug 12 '15 at 15:17
What does "$ab$" mean? Are $A$ and $B$ actually bialgebras? Anyway I don't see why that would be true, at least not with some sort of unitality condition: you could just take $delta = 0$ and get a $B$-coalgebra no matter what $Delta$ is.
– Najib Idrissi
Aug 12 '15 at 8:55
What does "$ab$" mean? Are $A$ and $B$ actually bialgebras? Anyway I don't see why that would be true, at least not with some sort of unitality condition: you could just take $delta = 0$ and get a $B$-coalgebra no matter what $Delta$ is.
– Najib Idrissi
Aug 12 '15 at 8:55
I second the concern that you are assuming the existence of a multiplication, something that doesn't exist if $A$ and $B$ are just co-algebras. But I'm also curious about your definition of $B$-coalgeba. Nowhere do you make use of $A$'s comultiplication, you just have that $A$ is a co-module of $B$. I feel like you want to use $delta$ as a comultiplication, but it isn't quite. It seems to me like your definition should probably be that you have a map of coalgebras $Ato B$ (dual to the definition of $Rto S$ making $S$ into an $R$-algebra).
– Aaron
Aug 12 '15 at 9:15
I second the concern that you are assuming the existence of a multiplication, something that doesn't exist if $A$ and $B$ are just co-algebras. But I'm also curious about your definition of $B$-coalgeba. Nowhere do you make use of $A$'s comultiplication, you just have that $A$ is a co-module of $B$. I feel like you want to use $delta$ as a comultiplication, but it isn't quite. It seems to me like your definition should probably be that you have a map of coalgebras $Ato B$ (dual to the definition of $Rto S$ making $S$ into an $R$-algebra).
– Aaron
Aug 12 '15 at 9:15
@NajibIdrissi, thank you very much. I added that assume that $A, B$ are also algebras in the last line of the post.
– LJR
Aug 12 '15 at 9:31
@NajibIdrissi, thank you very much. I added that assume that $A, B$ are also algebras in the last line of the post.
– LJR
Aug 12 '15 at 9:31
@Aaron, thank you very much. Yes, we only need the condition that $A$ is a comodule algebra.
– LJR
Aug 12 '15 at 9:32
@Aaron, thank you very much. Yes, we only need the condition that $A$ is a comodule algebra.
– LJR
Aug 12 '15 at 9:32
1
1
OK. But $A$ can be the zero ring. It then becomes a $B$-comodule algebra in an obvious way, but clearly this cannot say anything about $B$. So the literal answer to your question should be a "No". On the other hand, it may be worth looking for weak assumptions under which the answer can be "Yes", such as the existence of an algebra homomorphism from $A$ to $B$ which interacts with the structures nicely (whatever this means).
– darij grinberg
Aug 12 '15 at 15:17
OK. But $A$ can be the zero ring. It then becomes a $B$-comodule algebra in an obvious way, but clearly this cannot say anything about $B$. So the literal answer to your question should be a "No". On the other hand, it may be worth looking for weak assumptions under which the answer can be "Yes", such as the existence of an algebra homomorphism from $A$ to $B$ which interacts with the structures nicely (whatever this means).
– darij grinberg
Aug 12 '15 at 15:17
|
show 4 more comments
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What does "$ab$" mean? Are $A$ and $B$ actually bialgebras? Anyway I don't see why that would be true, at least not with some sort of unitality condition: you could just take $delta = 0$ and get a $B$-coalgebra no matter what $Delta$ is.
– Najib Idrissi
Aug 12 '15 at 8:55
I second the concern that you are assuming the existence of a multiplication, something that doesn't exist if $A$ and $B$ are just co-algebras. But I'm also curious about your definition of $B$-coalgeba. Nowhere do you make use of $A$'s comultiplication, you just have that $A$ is a co-module of $B$. I feel like you want to use $delta$ as a comultiplication, but it isn't quite. It seems to me like your definition should probably be that you have a map of coalgebras $Ato B$ (dual to the definition of $Rto S$ making $S$ into an $R$-algebra).
– Aaron
Aug 12 '15 at 9:15
@NajibIdrissi, thank you very much. I added that assume that $A, B$ are also algebras in the last line of the post.
– LJR
Aug 12 '15 at 9:31
@Aaron, thank you very much. Yes, we only need the condition that $A$ is a comodule algebra.
– LJR
Aug 12 '15 at 9:32
1
OK. But $A$ can be the zero ring. It then becomes a $B$-comodule algebra in an obvious way, but clearly this cannot say anything about $B$. So the literal answer to your question should be a "No". On the other hand, it may be worth looking for weak assumptions under which the answer can be "Yes", such as the existence of an algebra homomorphism from $A$ to $B$ which interacts with the structures nicely (whatever this means).
– darij grinberg
Aug 12 '15 at 15:17