The proof of $S(gh)=S(h)S(g)$ of an antipode in Hopf algebras

Question

Let $H=(H, m, \Delta, \mu, \epsilon, S)$ be a Hopf algebra. Then there is a property of antipde $S$: $$S(gh)=S(h)S(g)$$ where $g,h \in H$.

I have seen in some materials the proof of this property need the following properties: $S \star1= 1 \star S= \mu \circ \epsilon$. I want to know whether we could prove $S(gh)=S(h)S(g)$ using just $S \star 1= \mu \circ \epsilon$ or just $1\star S= \mu \circ \epsilon$?

I have seen here a proof, it seems do not use both properties, but I don't know why the place marked green holds.

Answer 1

The displayed proof above uses left-inverse in the first block of equations and right-inverse in the second. The green part follows by using right-inverses, namely: $$ x_{(1)}y_{(1)}S(y_{(2)})S(x_{(2)})=x_{(1)}\epsilon(y)S(x_{(2)})=\epsilon(y)x_{(1)}S(x_{(2)})=\epsilon(y)\epsilon(x)=\epsilon(xy)\,. $$ The proof uses both properties. Whether there exists a left Hopf algebra with a left antipode which is not an algebra antiendomorphism is an interesting question. Example 21 in "Left Hopf algebras, Green, Nichols, Taft, 1979" seems not to qualify. But I still doubt that it is true.

Further reading: https://mathoverflow.net/questions/332085/a-concrete-example-of-a-one-sided-hopf-algebra

Answer 2

This is not trivial, allow me to share a complete proof, we need first to establish a convolution in $Hom(H\otimes H, H)$.

For any coalgebra $C$ and algebra $A$, one can endow the $k$-module $Hom(C, A)$, the set of all $k$-linear maps from $C$ to $A$, with an associative algebra structure called the convolution algebra: we define the product $f*g$ of two maps $f, g \in Hom(C, A)$ by $(f*g)(c)= \sum f(c_1)g(c_2)$, using the Sweedler notation $\Delta(c)= \sum c_1 \otimes c_2$ . \ Equivalently, $f*g$ is the composition of maps $C \xrightarrow{\Delta} C\otimes C \xrightarrow{f\otimes g} A\otimes A \xrightarrow{m} A$
The associativity of this $*$ operation is easy to check, in fact
$(f*g)*h\ (c)= \sum f*g(c_1)h(c_2)= \sum f(c_{1})g(c_{3}))h(c_2)= \sum f(c_{1})g(c_{2}))h(c_3)$ and
$f*(g*h)\ (c)= \sum f(c_1)h*g(c_2)= \sum f(c_1) g(c_{2})h(c_{3})$ .

The map $u \circ\epsilon $ is a two-sided identity element for $*$, meaning that every $f\in Hom(C, A)$ satisfies $(f*u \circ\epsilon)(c)= \sum f(c_1)\epsilon(c_2)= f(c)= \sum \epsilon(c_1)f(c_2)= (u \circ\epsilon*f)(c)\ ,\ \forall c\in C$ .

In particular, when $H$ is a Hopf algebra, the convolution product $*$ gives a unital associative algebra structure on $End(H) := Hom(H, H)$ .

We consider now $H\otimes H$ as a coalgebra (being a tensor product of two coalgebras) and $H$ as an algebra. Then $Hom(H\otimes H, H) $ is an associative algebra with a convolution product $\star$ , with the element $u_H\epsilon_{H\otimes H}$ being a two-sided identity element, and to be distinguished from the convolution $*$ on $End(H)$ that we established earlier.
We define three elements $f, g, h $ of $Hom(H \otimes H, H) $ by $$f(a \otimes b)= ab\ ,\quad g(a \otimes b)= S(b)S(a)\ ,\quad h(a \otimes b)= S(ab)$$ We seek to show that $h= g$ , we will show that these three elements satisfy $h\star f = u_H\epsilon_{H\otimes H}= f\star g$, for any $a \otimes b\in H\otimes H$, if we denote $\Delta(a)= \sum_{(a)} a_1 \otimes a_2$ and $\Delta(b)= \sum_{(b)} b_1 \otimes b_2$, as $\Delta$ is an algebra morphism we have $\Delta(ab)= \sum_{(a),(b)} a_1b_1 \otimes a_2b_2$ , then \begin{align*} (u_H\epsilon_{H\otimes H})(a \otimes b)&= u_H(\epsilon_{H}(a) \epsilon_{H}(b))= u_H(\epsilon_{H}(ab)) \\ (h\star f)(a \otimes b)&= \sum_{(a),(b)}h(a_1\otimes b_1)f(a_2\otimes b_2) \\ &= \sum_{(a),(b)}S(a_1b_1)a_2b_2 \\ &= (S*Id_H)(ab)= u_H(\epsilon_{H}(ab)) \\ (f\star g)(a \otimes b)&= \sum_{(a),(b)} f(a_1\otimes b_1)g(a_2\otimes b_2) \\ &= \sum_{(a),(b)} a_1b_1S(b_2)S(a_2)= \sum_{(a)} a_1. (Id_H*S)(b). S(a_2) \\ &= u_H(\epsilon_{H}(b)) \sum_{(a)} a_1S(a_2)= u_H(\epsilon_{H}(b))u_H(\epsilon_{H}(a))= u_H(\epsilon_{H}(ab)) \end{align*} So in fact $h*f = u_H\epsilon_{H\otimes H}= f*g$ , then by associativity
$h= h\star(u_H\epsilon_{H\otimes H})= h\star(f\star g)= (h\star f)\star g = (u_H\epsilon_{H\otimes H})\star g= g$
We conclude that $h= g$ as desired.