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I was interested to know about coalgebraic version of "Idempotents". So I seached the web and I found the following interesting post : https://math.stackexchange.com/questions/689322/co-idempotents-algebraic-dual-of-an-idempotent-element.

In this question I try to look at to coidempotents with a different apparoach. First we note that an idempotent in a unital algebra $A$ with multiplication $m:A\otimes A \to A$, can be reinterpreted as a linear map $T:A\to A$ which satisfies the following properties:

1. $\;$ $m\circ (T\otimes Id)= T \circ m$

2. $\;$ $T^{2}=T$

The first condition says that $T$ is a multiplication operator $T(x)=ex$ for some $e\in A$. The second condition says that $e$ is an idempotent.

So we naturally reverse the direction of arrows of an algebras and replace the multiplication by comltipication to define a coidempotent on a coalgebra $(C, \Delta, \epsilon)$ as a linear map $T$ on $A$ which satisfies:

1. $\;$ $(T \otimes Id) \circ \Delta= \Delta \circ T$

2. $\;$ $T^{2}=T$

So a coidempotent is not an element of $C$ but is an operator on $C$

A non trivial coidempotent is an operator $T$ with the above properties which is neither $Id$ nor $0$.

Questions:

1. $\;$ Is there a nontrivial coidempotent for the standard comatrix coalgebra or $\mathbb{C}[x]$, with their natural coalgebra structures, respectively ?

2. $\;$ We know that the construction of the $K$- theory of an algebras $A$ is based on idempotents in $M_{\infty}(A)$. Now assume that $C$ is a coalgebra. Is there a natural coalgebra structure on $M_{n}(C)$? If the answer is yes, is there a natural embedding which send a coidempotent on $C$ to a coidempotent in $M_{2}(C)$, and then to higher dimensional matrices? For two coidempotent$S$ and $T$ on a coalgebra $C$, can we associate two coidempotents $\bar{S}$ and $\bar{T}$ in $M_{2}(C)$ with $\bar{S} \bar{T} =\bar{T} \bar{S}=0$?(Motivating by the algebraic process)?

3. Let $G$ be a free group. Is the "idempotent problem" equivalent to the "coidempotent problem" for $\mathbb{C}G$, when we consider $\mathbb{C}G$ as an algebra(coalgebra), respectively?

Your answers or comments are very appreciated.

Added: In the literature are there some researchs devited to a kind of co-k theiry of coalgebras or quantum groups?

I was interested to know about coalgebraic version of "Idempotents". So I seached the web and I found the following interesting post : https://math.stackexchange.com/questions/689322/co-idempotents-algebraic-dual-of-an-idempotent-element.

In this question I try to look at to coidempotents with a different apparoach. First we note that an idempotent in a unital algebra $A$ with multiplication $m:A\otimes A \to A$, can be reinterpreted as a linear map $T:A\to A$ which satisfies the following properties:

1. $\;$ $m\circ (T\otimes Id)= T \circ m$

2. $\;$ $T^{2}=T$

The first condition says that $T$ is a multiplication operator $T(x)=ex$ for some $e\in A$. The second condition says that $e$ is an idempotent.

So we naturally reverse the direction of arrows of an algebras and replace the multiplication by comltipication to define a coidempotent on a coalgebra $(C, \Delta, \epsilon)$ as a linear map $T$ on $A$ which satisfies:

1. $\;$ $(T \otimes Id) \circ \Delta= \Delta \circ T$

2. $\;$ $T^{2}=T$

So a coidempotent is not an element of $C$ but is an operator on $C$

A non trivial coidempotent is an operator $T$ with the above properties which is neither $Id$ nor $0$.

Questions:

1. $\;$ Is there a nontrivial coidempotent for the standard comatrix coalgebra or $\mathbb{C}[x]$, with their natural coalgebra structures, respectively ?

2. $\;$ We know that the construction of the $K$- theory of an algebras $A$ is based on idempotents in $M_{\infty}(A)$. Now assume that $C$ is a coalgebra. Is there a natural coalgebra structure on $M_{n}(C)$? If the answer is yes, is there a natural embedding which send a coidempotent on $C$ to a coidempotent in $M_{2}(C)$, and then to higher dimensional matrices? For two coidempotent$S$ and $T$ on a coalgebra $C$, can we associate two coidempotents $\bar{S}$ and $\bar{T}$ in $M_{2}(C)$ with $\bar{S} \bar{T} =\bar{T} \bar{S}=0$?(Motivating by the algebraic process)?

3. Let $G$ be a free group. Is the "idempotent problem" equivalent to the "coidempotent problem" for $\mathbb{C}G$, when we consider $\mathbb{C}G$ as an algebra(coalgebra), respectively?

Your answers or comments are very appreciated.

Added: In the literature are there some researchs devoted to a kind of co-k theiry of coalgebras or quantum groups?(Are there some approaches which consider the space of all f.g. comodules over $C$ as a quantum analogy of classical $K$ theory?

I was interested to know about coalgebraic version of "Idempotents". So I seached the web and I found the following interesting post : https://math.stackexchange.com/questions/689322/co-idempotents-algebraic-dual-of-an-idempotent-element.

In this question I try to look at to coidempotents with a different apparoach. First we note that an idempotent in a unital algebra $A$ with multiplication $m:A\otimes A \to A$, can be reinterpreted as a linear map $T:A\to A$ which satisfies the following properties:

1. $\;$ $m\circ (T\otimes Id)= T \circ m$

2. $\;$ $T^{2}=T$

The first condition says that $T$ is a multiplication operator $T(x)=ex$ for some $e\in A$. The second condition says that $e$ is an idempotent.

So we naturally reverse the direction of arrows of an algebras and replace the multiplication by comltipication to define a coidempotent on a coalgebra $(C, \Delta, \epsilon)$ as a linear map $T$ on $A$ which satisfies:

1. $\;$ $(T \otimes Id) \circ \Delta= \Delta \circ T$

2. $\;$ $T^{2}=T$

So a coidempotent is not an element of $C$ but is an operator on $C$

A non trivial coidempotent is an operator $T$ with the above properties which is neither $Id$ nor $0$.

Questions:

1. $\;$ Is there a nontrivial coidempotent for the standard comatrix coalgebra or $\mathbb{C}[x]$, with their natural coalgebra structures, respectively ?

2. $\;$ We know that the construction of the $K$- theory of an algebras $A$ is based on idempotents in $M_{\infty}(A)$. Now assume that $C$ is a coalgebra. Is there a natural coalgebra structure on $M_{n}(C)$? If the answer is yes, is there a natural embedding which send a coidempotent on $C$ to a coidempotent in $M_{2}(C)$, and then to higher dimensional matrices? For two coidempotent$S$ and $T$ on a coalgebra $C$, can we associate two coidempotents $\bar{S}$ and $\bar{T}$ in $M_{2}(C)$ with $\bar{S} \bar{T} =\bar{T} \bar{S}=0$?(Motivating by the algebraic process)?

3. Let $G$ be a free group. Is the "idempotent problem" equivalent to the "coidempotent problem" for $\mathbb{C}G$, when we consider $\mathbb{C}G$ as an algebra(coalgebra), respectively?

Your answers or comments are very appreciated.

I was interested to know about coalgebraic version of "Idempotents". So I seached the web and I found the following interesting post : https://math.stackexchange.com/questions/689322/co-idempotents-algebraic-dual-of-an-idempotent-element.

In this question I try to look at to coidempotents with a different apparoach. First we note that an idempotent in a unital algebra $A$ with multiplication $m:A\otimes A \to A$, can be reinterpreted as a linear map $T:A\to A$ which satisfies the following properties:

1. $\;$ $m\circ (T\otimes Id)= T \circ m$

2. $\;$ $T^{2}=T$

The first condition says that $T$ is a multiplication operator $T(x)=ex$ for some $e\in A$. The second condition says that $e$ is an idempotent.

So we naturally reverse the direction of arrows of an algebras and replace the multiplication by comltipication to define a coidempotent on a coalgebra $(C, \Delta, \epsilon)$ as a linear map $T$ on $A$ which satisfies:

1. $\;$ $(T \otimes Id) \circ \Delta= \Delta \circ T$

2. $\;$ $T^{2}=T$

So a coidempotent is not an element of $C$ but is an operator on $C$

A non trivial coidempotent is an operator $T$ with the above properties which is neither $Id$ nor $0$.

Questions:

1. $\;$ Is there a nontrivial coidempotent for the standard comatrix coalgebra or $\mathbb{C}[x]$, with their natural coalgebra structures, respectively ?

2. $\;$ We know that the construction of the $K$- theory of an algebras $A$ is based on idempotents in $M_{\infty}(A)$. Now assume that $C$ is a coalgebra. Is there a natural coalgebra structure on $M_{n}(C)$? If the answer is yes, is there a natural embedding which send a coidempotent on $C$ to a coidempotent in $M_{2}(C)$, and then to higher dimensional matrices? For two coidempotent$S$ and $T$ on a coalgebra $C$, can we associate two coidempotents $\bar{S}$ and $\bar{T}$ in $M_{2}(C)$ with $\bar{S} \bar{T} =\bar{T} \bar{S}=0$?(Motivating by the algebraic process)?

3. Let $G$ be a free group. Is the "idempotent problem" equivalent to the "coidempotent problem" for $\mathbb{C}G$, when we consider $\mathbb{C}G$ as an algebra(coalgebra), respectively?

Your answers or comments are very appreciated.

Added: In the literature are there some researchs devited to a kind of co-k theiry of coalgebras or quantum groups?

I was interested to know about coalgebraic version of "Idempotents". So I seached the web and I found the following interesting post : https://math.stackexchange.com/questions/689322/co-idempotents-algebraic-dual-of-an-idempotent-element.

In this question I try to look at to coidempotents with a different apparoach. First we note that an idempotent in a unital algebra $A$ with multiplication $m:A\otimes A \to A$, can be reinterpreted as a linear map $T:A\to A$ which satisfies the following properties:

1. $\;$ $m\circ (T\otimes Id)= T \circ m$

2. $\;$ $T^{2}=T$

The first condition says that $T$ is a multiplication operator $T(x)=ex$ for some $e\in A$. The second condition says that $e$ is an idempotent.

So we naturally reverse the direction of arrows of an algebras and replace the multiplication by comltipication to define a coidempotent on a coalgebra $(C, \Delta, \epsilon)$ as a linear map $T$ on $A$ which satisfies:

1. $\;$ $(T \otimes Id) \circ \Delta= \Delta \circ T$

2. $\;$ $T^{2}=T$

So a coidempotent is not an element of $C$ but is an operator on $C$

A non trivial coidempotent is an operator $T$ with the above properties which is neither $Id$ nor $0$.

Questions:

1. $\;$ Is there a nontrivial coidempotent for the standard comatrix coalgebra or $\mathbb{C}[x]$, with their natural coalgebra structures, respectively ?

2. $\;$ We know that the construction of the $K$- theory of an algebras $A$ is based on idempotents in $M{\infty}(A)$. Now assume that $C$ is a coalgebra. Is there a natural coalgebra structure on $M_{n}(C)$? If the answer is yes, is there a natural embedding which send a coidempotent on $C$ to a coidempotent in $M_{2}(C)$, and then to higher dimensional matrices? For two coidempotent$S$ and $T$ on a coalgebra $C$, can we associate two coidempotents $\bar{S}$ and $\bar{T}$ in $M_{2}(C)$ with $\bar{S} \bar{T} =\bar{T} \bar{S}=0$?(Motivating by the algebraic process)?

3. Let $G$ be a free group. Is the "idempotent problem" equivalent to the "coidempotent problem" for $\mathbb{C}G$, when we consider $\mathbb{C}G$ as an algebra(coalgebra), respectively?

Your answers or comments are very appreciated.

I was interested to know about coalgebraic version of "Idempotents". So I seached the web and I found the following interesting post : https://math.stackexchange.com/questions/689322/co-idempotents-algebraic-dual-of-an-idempotent-element.

In this question I try to look at to coidempotents with a different apparoach. First we note that an idempotent in a unital algebra $A$ with multiplication $m:A\otimes A \to A$, can be reinterpreted as a linear map $T:A\to A$ which satisfies the following properties:

1. $\;$ $m\circ (T\otimes Id)= T \circ m$

2. $\;$ $T^{2}=T$

The first condition says that $T$ is a multiplication operator $T(x)=ex$ for some $e\in A$. The second condition says that $e$ is an idempotent.

So we naturally reverse the direction of arrows of an algebras and replace the multiplication by comltipication to define a coidempotent on a coalgebra $(C, \Delta, \epsilon)$ as a linear map $T$ on $A$ which satisfies:

1. $\;$ $(T \otimes Id) \circ \Delta= \Delta \circ T$

2. $\;$ $T^{2}=T$

So a coidempotent is not an element of $C$ but is an operator on $C$

A non trivial coidempotent is an operator $T$ with the above properties which is neither $Id$ nor $0$.

Questions:

1. $\;$ Is there a nontrivial coidempotent for the standard comatrix coalgebra or $\mathbb{C}[x]$, with their natural coalgebra structures, respectively ?

2. $\;$ We know that the construction of the $K$- theory of an algebras $A$ is based on idempotents in $M_{\infty}(A)$. Now assume that $C$ is a coalgebra. Is there a natural coalgebra structure on $M_{n}(C)$? If the answer is yes, is there a natural embedding which send a coidempotent on $C$ to a coidempotent in $M_{2}(C)$, and then to higher dimensional matrices? For two coidempotent$S$ and $T$ on a coalgebra $C$, can we associate two coidempotents $\bar{S}$ and $\bar{T}$ in $M_{2}(C)$ with $\bar{S} \bar{T} =\bar{T} \bar{S}=0$?(Motivating by the algebraic process)?

3. Let $G$ be a free group. Is the "idempotent problem" equivalent to the "coidempotent problem" for $\mathbb{C}G$, when we consider $\mathbb{C}G$ as an algebra(coalgebra), respectively?

Your answers or comments are very appreciated.