Convolution algebra as a bialgebra?

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?

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

(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.

• In your point (2), you inverted $A$ and $C$. It's $\hom(C,A)$ which is a convolution algebra, not $\hom(A,C)$ (consider the equation $(\phi * \psi)(a) = \sum \phi(a_1) \psi(a_2)$: it makes sense if it's $A$ which is a coalgebra). Commented Jun 5, 2020 at 14:18
• @NajibIdrissi Of course, many thanks! Commented Jun 5, 2020 at 14:23