I was interested to know about coalgebraic version of "Idempotents". So I seached the web and I found the following interesting post : http://math.stackexchange.com/questions/689322/co-idempotents-algebraic-dual-of-an-idempotent-element.
I 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:
$\;$ $m\circ (T\otimes Id)= T \circ m$
$\;$ $T^{2}=T$
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:
$\;$ $(T \otimes Id) \circ \Delta= \Delta \circ T$
$\;$ $T^{2}=T$
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:
- $\;$ Is there a nontrivial coidempotent for the standard comatrix coalgebra or $\mathbb{C}[x]$, with their natural coalgebra structures, respectively ?
- $\;$ We know that the structure 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(Motivating by the algebraic process)?
Your answers or comments are very appreciated.