TITLE: Convolution in Hopf algebras
QUESTION [8 upvotes]: For each Hopf algebra $H$ its space ${\mathcal L}(H)$ of operators $A:H\to H$ is usually endowed with the operation of convolution by the identity
$$
A*B = \mu \circ (A\otimes B)\circ \varDelta
$$
where $\mu$ is the multiplication, $\varDelta$ the comultiplication in $H$, and $A,B\in{\mathcal L}(H)$.
Can anybody advise me a text where this notion is described in detail? I wonder, in particular, which identities connect $A*B$ with the usual operation $A\circ B$ of composition of operators.

REPLY [10 votes]: You can try "Hopf algebras. An introduction" by Dascalescu, Nastasescu, Raianu. They have a detailed account with full proofs and further examples and exercises. 
Kassel's book "Quantum groups" and Sweedler's book on "Hopf algebras" (see esp. p. 69-79, 314-315) might also prove useful. 
Finally, you might  find some interest at 
https://math.stackexchange.com/a/1980032/195021. 
Edit: Motivated by your question on the relation between convolution and the composition, i could not find some direct reference, however i recalled some old notes of mine and found the following relations (in what follows, $\mu$, $\Delta$, $\eta$, $\varepsilon$, $S$, stand for the corresponding multiplication, comultiplication, unity, counity and antipode maps respectively): 

the composition distributes over convolution: i.e. if $f,g\in Hom(F,G)$, $p\in Hom(H,F)$ where $H,F,G$ are Hopf algebras and $Hom(.,.)$ stands for the corresponding Hopf algebra maps, then:
$$
\boxed{ (f\star g)\circ p=(f\circ p)\star(g\circ p) }
$$
Proof: by definition: 
$$
(f\star g)\circ p=\mu_G\circ(f\otimes g)\circ\Delta_F \circ p
$$ 
and 
$$
(f\circ p)\star(g\circ p)=\mu_G\circ\big((f\circ p)\otimes (g\circ p)\big)\circ\Delta_H = \mu_G\circ (f\otimes g)\circ (p\otimes p)\circ\Delta_H =\\=\mu_G\circ(f\otimes g)\circ\Delta_F\circ p
$$
where in the last line we have made use of the fact that since $f,g,p$ are assumed to be Hopf algebra morphisms they are also coalgebra morphisms and thus $\Delta_F \circ p=(p\otimes p)\circ\Delta_H$. 
$f\circ S_F=S_G\circ f$, is the $\star$-inverse of $f$, i.e.:
$$
\boxed{ (S_G\circ f)\star f=f\star(S_G\circ f)=\eta_G\circ\varepsilon_F}   
$$
Proof:  Applying the definition of the convolution we get: 
$$
\big((S_G\circ f)\star f\big)=\mu_G\circ\big((S_G\circ f)\otimes f\big)\circ\Delta_F=  
\mu_G\circ(S_G\otimes Id)\circ(f\otimes f)\circ\Delta_F=\\=\mu_G\circ(S_G\otimes Id)\circ\Delta_G\circ f=\eta_G\circ\varepsilon_G\circ f=\eta_G\circ\varepsilon_F
$$
where we have used:
$\bullet \ (f\otimes f)\circ\Delta_F=\Delta_G\circ f$, because $f\in Hom(F,G)$,
$\bullet \ \mu_G\circ(S_G\otimes Id)\circ\Delta_G=\eta_G\circ\varepsilon_G$, by the definition of the antipode,
$\bullet \ \varepsilon_G\circ f=\varepsilon_F$, because $f\in Hom(F,G)$.  
A more general property, which has a categorical "flavor" and is related to the preceding properties, is the following one: 

Let $\mathcal{H}$ be the Category of commutative, cocommutative, finite dimensional Hopf algebras. Let the corresponding Hom-sets $\mathcal{H}om(F,G)$ be equipped with sum (convolution) and product (composition). Then, $\mathcal{H}$ becomes an abelian Category.
  (see Sweedler's book, sect. 16.2, p. 314-315, where this is result is cited).   

What is essentialy happening here, is that $\mathcal{H}om(F,G)$ becomes an abelian group -under convolution- with neutral element: 
$\eta_G\circ\varepsilon_F$.
Furthermore, it can be shown that there is an equivalence of abelian Categories, between the Category of commutative, cocommutative, finite dimensional Hopf algebras $\mathcal{H}$ and the category $\mathcal{Ab}_{fin}$ of the finite, abelian groups. It is possible to construct a fully faithful and essentially faithful functor between $\mathcal{H}$ and $\mathcal{Ab}_{fin}$. See more details on this point at: 
Reference on correspondence between commutative Hopf Algebras and Groups