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Probably an easy question, but here goes:

I'm reading the paper 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?

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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!

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  • $\begingroup$ Thanks. This completely solves the issue! $\endgroup$
    – user167952
    Commented Apr 8, 2021 at 20:55
  • $\begingroup$ If you have time and are interested, you might be able to help me with mathoverflow.net/questions/389873/… and mathoverflow.net/questions/389870/… . Both questions have a bounty on them. Thanks in advance! $\endgroup$
    – user167952
    Commented Apr 17, 2021 at 19:59

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