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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 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?

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  • $\begingroup$ Let $(C,\Delta_{C})$ be a coalgebra and $\Delta_{n}$ be the standard coproduct structure on $M_{n}(\mathbb{C}$. Let $\delta_{ij}$ be the element of $M{n}(\mathbb{C})$ with $1$ in i-j place and zero other wise. for $c\in C$ let $\delta_{ij}(c)$ be the matric in $M_{n}(C)$ with only one non zero element $c$ in i-j place. Define a special product as follows; the product of $x\otimes y \in C\otimes C$ and $\delta_{ij}\otimes \delta_{jk} \in M_{n}(\mathbb{C}) \otimes M_{n}(\mathbb{C}$ is $\delta_{ij}(x) \otimes \delta_{jk}(y)$, as an element in $M_{n}(C) \otimes M_{n}(C)$. $\endgroup$
    – Ali Taghavi
    Commented Apr 15, 2014 at 21:35
  • $\begingroup$ Now define $\tilde{\Delta_{n}}$ with $\tilde {\Delta_{n}}(\delta{ij}(c))$= the above product of $\Delta_{C}(c)$ by $\Delta_{n}(\delta_{ij})$. Is this $\tilde{\Delta_{n}}$ a coproduct on $M_{n}(C)$? $\endgroup$
    – Ali Taghavi
    Commented Apr 15, 2014 at 21:43
  • $\begingroup$ The álgebra of matrices with entries in an algebra is just the tensor product of that álgebra with $M_n(\mathbb C)$; it is natural to define the comatrix coalgebra with entries in a coalgebra C as the tensor product of C and the comatrix coalgebra (which is just the coalgebra dual to the álgebra $M_n(\mathbb C)$) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Apr 16, 2014 at 5:11
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    $\begingroup$ Your definition of idempotent in an algebra is simply that of an idempotent in the endomorphism álgebra of its right regular module; your definition of a coidempotent is, similarly, that of an idempotent in the álgebra of endomorphisms of the regular right comodule. In particular, coidempotents correspond to direct summands of the regular right comodule. In the case of comatrix coalgebras, the right regular comodule is a direct sum of n simple comodules —just as in the álgebra case); $\mathbb C[x]$ with the coalgebra structure which makes x primitive, is indecomposable as a comodule $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Apr 16, 2014 at 5:13
  • $\begingroup$ over itself, so there are no non-trivial coidempotents. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Apr 16, 2014 at 5:20

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