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Let $A$ be a non-degenerate algebra with multiplier algebra $M(A)$. Let $S: A \to M(A)$ be an antimultiplicative linear map, i.e.

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

Consider the mapping $$\iota \otimes S: A \otimes A \to A \otimes M(A) \subseteq M(A \otimes A).$$

Does it extend to a map $\iota \otimes S: M(A \otimes A) \to M(A \otimes A)$?

The results that I know for extending linear maps to the multiplier algebra require (anti)multiplicativity to make sure the extension is well-defined. Here however, $\iota \otimes S$ is neither multiplicative nor antimultiplicative, but I guess there should be some way around this problem.


To give some context, in Van Daele's paper "An algebraic framework on group duality", in proposition 3.11, he considers an expression $(\iota \otimes S)(\Delta(q))$ where $S: A \to A$ is the antipode on the regular multiplier Hopf algebra $(A, \Delta)$. Since $\Delta(q) \in M(A \otimes A)$, we need to be able to make sense of $\iota \otimes S$ on the multiplier algebra $M(A \otimes A)$. Here is a screenshot:

enter image description here

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  • $\begingroup$ Where exactly in Proposition 3.11 does van Daele consider such an expression? I cannot see it... $\endgroup$ May 24, 2021 at 7:52
  • $\begingroup$ @MatthewDaws I added a screenshot. Maybe I'm misinterpreting something? $\endgroup$
    – Andromeda
    May 24, 2021 at 8:15
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    $\begingroup$ No, no, right you are! I missed that... $\endgroup$ May 24, 2021 at 8:29
  • $\begingroup$ No worries! Things like that are happening in the entire paper. For example in proposition 4.6, the functional $\omega_1 \otimes \omega_2 \otimes \omega_3$ is applied to $(\Delta(x)\otimes 1)\Delta^{(2)}(y)\in M(A\otimes A \otimes A)$, which also doesn't make sense to me. I know in the setting of $C^*$-algebras we can extend functionals to the multiplier algebra, but I am not sure it is possible here. I believe I must have a fundamental misunderstanding somewhere. $\endgroup$
    – Andromeda
    May 24, 2021 at 8:37

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Firstly, in An Algebraic Framework for Group Duality van Daele only considers regular multiplier Hopf algebras, so $S(A)\subseteq A$, and $S$ is a bijection, see Proposition 2.7.

Then, given such $S$, we can extend it to a map $M(A)\rightarrow M(A)$ by defining, for example, $S(x) a = S( S^{-1}(a)x )$ and $a S(x) = S( x S^{-1}(a) )$, for $a\in A, x\in M(A)$. See the discussion by van Daele on pages 333-334.

We consider Proposition 3.11 (for example, as it is in the OP). Here all elements are in $A$, and we have a formula like $$ (x\otimes p)(\iota\otimes S)\Delta(q). $$ This, however, is equal to $$ (1\otimes p)(\iota\otimes S)\big( (x\otimes 1)\Delta(q) \big), $$ and this makes perfect sense, as $(x\otimes 1)\Delta(q) \in A\otimes A$. In fact, examining van Daele's proof, it's fairly clear that this must be the interpretation being used.

Notice that if we set $(x\otimes 1)\Delta(q) = \sum_i a_i \otimes b_i$ then $$ (1\otimes p)(\iota\otimes S)\big( (x\otimes 1)\Delta(q) \big) = \sum_i a_i \otimes pS(b_i) = \sum_i a_i \otimes S(b_i S^{-1}(p)) = (\iota\otimes S)\big( (x\otimes 1)\Delta(q)(1\otimes S^{-1}(p) \big) = (x\otimes 1)(\iota\otimes S)\big( \Delta(q)(1\otimes S^{-1}(p) \big) , $$ and so we find that $$ (1\otimes p)(\iota\otimes S)\Delta(q) = (\iota\otimes S)\big( \Delta(q)(1\otimes S^{-1}(p) \big). $$

I think it is correct to believe that $(\iota\otimes S)\Delta(q)$ does not make (complete) sense in isolation, but we can always argue in ways similar to the above.

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  • $\begingroup$ Later in that same proof, he defines $a= (\iota \otimes \varphi)((1 \otimes p)(\iota \otimes S)(\Delta(q))$ and the logic seems to break here! $\endgroup$
    – Andromeda
    May 25, 2021 at 8:17
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    $\begingroup$ I added some more to the answer which I hope takes care of this. $\endgroup$ May 25, 2021 at 9:52
  • $\begingroup$ Thanks for the help! $\endgroup$
    – Andromeda
    May 25, 2021 at 9:54
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    $\begingroup$ Also notice that van Daele also applies the Haar functional $\varphi$ or $\psi$, and I think an argument using the modular properties of these (i.e. that they are "somewhat like traces") would also work to make sense of the formulae. $\endgroup$ May 25, 2021 at 9:57
  • $\begingroup$ Coming back to this old question (especially with regard to your last sentence), it might be interesting to note that $(\iota \otimes S)(\Delta(q))$ actually makes complete sense as a multiplier in $M(A \otimes A)$. Indeed, just define the left multiplier by $(\iota \otimes S)(\Delta(q))(a\otimes b) = (\iota \otimes S)(\Delta(q)(a \otimes 1))(1 \otimes b)$ and define the right multiplier similarly. Of course, this agrees with what you wrote in the first part of the answer, but it also allows us to make sense of what Van Daele writes. I am of course assuming regularity of the MHA. $\endgroup$
    – Andromeda
    Jan 25, 2022 at 21:01

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