# Antipode on a multiplier Hopf-algebra

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?

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!

• Thanks. This completely solves the issue!
– user167952
Apr 8, 2021 at 20:55
• 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!
– user167952
Apr 17, 2021 at 19:59