For an algebraically closed field $k$, let $C$ be a $k$-coalgebra. Given a minimal injective cogenerator $E$, there is a so-called basic coalgebra $B_C=coend^C(E)$, s.t. the comodule categories $Mod^C$ and $Mod^{B_C}$ are equivalent. I would like to understand this object $B_C$ better, but I didn't find any elaborated examples.

I am particularly interested in the case when $C$ is a coquasitriangular Hopf algebra. In this case, is $B_C$ still a (coquasitriangular) Hopf algebra, so that we have a (braided) monoidal equivalence of categories?

A good starting point for me would be the polynomial ring $k[x_{ij}]$ with coproduct $$\Delta(x_{ij})=\sum_{k=0}^n\,x_{ik}\otimes x_{kj},$$ but I am happy with all kinds of examples/references.


In general this coend is not a Hopf algebra. To convince yourself, think of a finite dimensional example, let $C=H^*$, so that you look for a minimal projective generator $H$-module $P$ and then look at the basic algebra $End_H(P)$. Take $H=k[G]$ with $G$ finite and k algebraically closed and ch(k)=0. Then by Weddebunn $k[G]$ is isomorphic to a product of (say r copies of) Matrix algebras (of eventually different sizes). $k[G]=M_{n-1}\times\cdots M_{n_r}$. The basic algebra associated is $k\times \cdots\times k$ r-times. (by the way, $r$=number of conjugacy classes of $G$). This is a commutative algebra, if it is Hopf, it must be of the form $k^{\tilde G}$ for some group $\tilde G$. There is no obvious candidate of natural group of order $r$ associated to $G$. Isn't it?

About your example $k[x_{ij}]$, this is not a Hopf algebra, you should localize by the determinant. But assume you do that. Then $O(GL(n(k))=k[x_{ij}][det^{-1}]$ is a reductive group, you can use "Peter-Weyl" to write $H=O(GL(n,k))=\oplus_{\mu}V_\mu\otimes V_{\mu}^*$, it is an infinite sum of comatrix coalgebras. The basic coalgebra associated to it is an infinite direct sum of copies of $k$, one for each $\mu$. It is not a very explicit construction.. it is more or less equivalent to know the (co)representation theory of $H$.


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