Probably an easy question, but here goes:


I'm reading the paper [Multiplier Hopf algebras](https://www.researchgate.net/publication/273376321_Multiplier_Hopf_algebras) by Van Daele.


Let $(A, \Delta)$ be a multiplier Hopf algebra. Let $L(A), R(A), M(A)$ be the left, right and multiplier algebras associated to $A$.


One constructs an antimultiplicative mapping $S\_1: A \to L(A)$ and a mapping $S\_2: A \to R(A)$ (in the notation of the paper, $S\_1 = S$ and $S\_2 = S'$). Then, it is shown that actually $S\_1 = S\_2$ and let's denote the common value by $T(a):= S\_1(a) = S\_2(a)$. I guess (but it is not stated explicitely) that we define the antipode
$$S: A \to M(A)$$
by $$S(a) := (T(a), T(a)) \in M(A)$$
and it is claimed that this an antimultiplicative map:


$$S(ab) = S(b)S(a).$$


However, why is this the case?
We have
$$S(b)S(a) = (T(b)T(a), T(a)T(b))= (T(ab), T(ba)) \stackrel{?}=S(ab)$$


where the multiplication $(L\_1,R\_2)(L\_2,R\_2) = (L\_1L\_2, R\_2R\_1)$ in the multiplier algebra was used. What am I missing?



Antipode on a multiplier Hopf-algebra
So I think we need to be carefully about what *multipliers* are. For an algebra $A$, a left multiplier is a linear map $L:A\rightarrow A$ with $L(ab) = L(a)b$, and a right multiplier is a linear map $R:A\rightarrow A$ with $R(ab) = aR(b)$. A multiplier (or double multiplier) is a pair $(L,R)$ of left, right, multipliers with $aL(b) = R(a)b$. Indeed, the product on $M(A)$ is then $(L,R)(L',R') = (LL', R'R)$. If $A$ has a nondegenerate product then the natural map $A\rightarrow M(A)$ is injective, so we identify $A$ as an ideal in $M(A)$.


It is rather tedious to write things as maps, so motivated by identifying $A$ inside $M(A)$, we write elements of $M(A)$ as just vectors $x$ say, and the action by multiplication. So if $x=(L,R)$ we write $L(a) = xa$ and $R(a) = ax$. The double multiplier condition becomes $a(xb) = (ax)b$.


In van Daele's paper, Definition 4.1 defines $S\_1(a)b$ to be some element of $A$. So really here a left multiplier is defined. Then in the proof of Lemma 4.5 we define $a S\_2(b)$, so a right multiplier. Towards the end of the proof, the following formula is proved:
$$ aS\_1(b)c = aS\_2(b)c \qquad (a,b,c\in A). $$
Examining the proof closely, what is really proven is:
$$ a \big( S\_1(b)c \big) = \big( a S\_2(b) \big) c \qquad (a,b,c\in A). $$
Notice that this is exactly the double multiplier condition! Thus, we have defined $S(a)\in M(A)$ by (a slight abuse of notation) $S(a) = (S\_1(a), S\_2(a))$.


Being more careful, fix $a$, and define $L(b) = S\_1(a)b$ and $R(b) = bS\_2(a)$, for $b\in A$. Then $(L,R)$ is a multiplier, because
$$ bL(c) = b (S\_1(a) c) = (bS\_2(a))c = R(b)c. $$
Also let $a'\in A$ and define $L',R'$ analogously. Lemma 4.4 shows that $S\_1(a) S\_1(a') = S\_1(a'a)$. So if we define $L'',R''$ for $a'a$, then
$$ S(a) S(a') = (L,R)(L',R') = (LL', R'R) = (L'', R'R). $$
For $b,c\in A$ we know that $b L''(c) = R''(b) c$ but also $(L'',R'R)$ is a double multiplier, so $b L''(c) = R'(R(b))c$. By non-degeneracy, $R''(b)c=R'(R(b))c$ implies $R''(b) = R'(R(b))$. That is, $S(a) S(a') = S(a'a)$ as we hoped.


So what does van Daele mean by "$S(a)$ is also a right multiplier" and/or "$S(b)=S'(b)$"?? I think this *is* a bit misleading. I read this to mean, respectively, "Given $a$ there is a right multiplier $S'(a)$ making $(S(a), S'(a))$ into a double multiplier" and "For $b$, we have that $(S(b), S'(b))$ is a double multiplier". The point is that if $L$ is a left multiplier, then by non-degeneracy, there is *at most one* right multiplier $R$ with $(L,R)$ a double multiplier. Thus it makes sense to talk about when $L$ is "also a right multiplier" even though the right multiplier is not *literally the same map*!


