Convolution algebra as a bialgebra?

Question

1. Context
   Let $(A, \mu, \eta, \Delta, \epsilon)$ be a bialgebra over a field $k$. Consider the vector space $\mathrm{End}(A)$ over $k$.
   Define the convolution product $$*: \mathrm{End}(A)\otimes \mathrm{End}(A) \rightarrow \mathrm{End}(A); \qquad f \otimes g \mapsto \mu \circ (f \otimes g)\circ \Delta.$$
   Define the unit map $$\overline \eta: k \rightarrow \mathrm{End}(A); \qquad 1 \mapsto \eta \circ \epsilon.$$ Then $(\mathrm{End}(A), *, \overline \eta)$ becomes an associative, unital algebra.

2. Questions
   

   • Can $(\mathrm{End}(A), *, \overline \eta)$ be made into a bialgebra?
   • Does it become a Hopf algebra that way?
   • Is there a canonical way?

Answer 1

This is not a complete answer, but it is too long to be a comment.

Let me answer your questions in the finite-dimensional case.

(1) Assume that $A$ is finite-dimensional as a vector space. Then, as vector spaces, $$ \begin{array}{ccc} \mathrm{End}(A) & \cong & A \otimes A^* \\ f & \to & \sum_if(e_i) \otimes e_i^* \\ \left[b\mapsto a\varphi(b)\right] & \leftarrow & a \otimes \varphi \end{array}$$ where $\{e_i\}$ is a basis of $A$ and $\{e_i^*\}$ is the corresponding dual basis of $A^*.$

(2) The construction of the algebra structure can be replicated for every $\mathrm{Hom}(C,A)$ where $A$ is an algebra and $C$ a coalgebra. In particular, $A^*$ always admits an algebra structure: $(\varphi*\psi)(a) = \sum \varphi(a_1)\psi(a_2)$.

(3) Being $A$ finite-dimensional, $A^*$ admits in fact a bialgebra structure, where $\Delta_*(\varphi) = \sum \varphi_1 \otimes \varphi_2$ is uniquely determined by the rule $$\sum \varphi_1(a)\varphi_2(b) = \varphi(ab)$$ for all $a,b \in A$ and $\varepsilon_*(\varphi) = \varphi(1)$.

(4) Since both $A$ and $A^*$ are bialgebras, $A \otimes A^*$ is a bialgebra as well with $$ (a \otimes \varphi)(b \otimes \psi) = ab \otimes \varphi * \psi, \\ \Delta_{A \otimes A^*}(a \otimes \varphi) = \sum \left(a_1 \otimes \varphi_1\right) \otimes \left(a_2 \otimes \varphi_2\right), \\ u_{A \otimes A^*} = u_A \otimes u_{A^*},\\ \varepsilon_{A \otimes A^*} = \varepsilon_A \otimes \varepsilon_{A^*}. $$

(5) If you consider the algebra structure on $\mathrm{End}(A)$ you gave above and the foregoing algebra structure on $A \otimes A^*$, you will realize that $\mathrm{End}(A) \cong A \otimes A^*$ as algebras. In particular, if you transfer the coalgebra structure on $\mathrm{End}(A)$ then you obtain a bialgebra structure. Therefore, in this case the answer to your first and third question is yes.

(6) The answer to the second question instead is: with the above construction, in general no, unless $A$ is already a Hopf algebra. Assume that you manage to endow $\mathrm{End}(A)$ with an antipode $S_E$. Consider the composition $$S:= \left(A^* \xrightarrow{1\otimes A^*} A \otimes A^* \xrightarrow{S_E} A \otimes A^* \xrightarrow{\varepsilon \otimes A^*} A^*\right).$$ It satisfies $$S(\varphi_1)*\varphi_2 = (\varepsilon \otimes A^*)\left(S_E(1 \otimes \varphi_1)(1 \otimes \varphi_2)\right) = (\varepsilon\otimes A^*)(1_A \otimes \varepsilon_*(\varphi)1_{A^*}) = \varepsilon_*(\varphi)1_{A^*}$$ and analogously on the other side (and you may perform the same construction for $A$). Thus you have an antipode on $A^*$ and on $A$.

For the infinite dimensional case, I would say that the answer is no (at least, not "canonically"), but I don't have any counter example to exhibit presently.