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Let $(A, \Delta)$ be a Hopf $C^*$-algebra, i.e. $A$ is a $C^*$-algebra, and $\Delta: A \to M(A\otimes A)$ is an injective non-degenerate $*$-homomorphism that is coassociative: $$(\iota \otimes \Delta)\Delta = (\Delta \otimes \iota)\Delta.$$ I would like to understand how the dual $A^*$ becomes a completely contractive Banach algebra for the multiplication $$\mu\star \nu := (\mu\otimes \nu)\circ \Delta.$$ I.e. why is the multiplication map $$\star: A^*\odot A^*\to A^*$$ completely contractive with respect to the projective tensor product norm on $A^*\odot A^*$? Clearly we have $$\|\mu\star \nu\| \le \|\mu\|\|\eta\|$$ so it is a contraction, but I do not understand how it is a complete contraction.

Thanks in advance for your help!

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    $\begingroup$ I don't have a complete answer, but the way this works for Hopf von Neumann algebras is that the vN spatial tensor product of M with itself can be identified with the dual of $M_* \hat\otimes M_*$ where $\hat\otimes$ is the projective tensor product of operator spaces. I suspect that one could try to adapt, or build on, this result by passing from A to its bidual, but I have not tried to check the details yet. $\endgroup$
    – Yemon Choi
    Commented Oct 2, 2023 at 2:25
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    $\begingroup$ @YemonChoi Thanks for your suggestion. I also thought about this: roughly, the idea would be to extend the map $\Delta: A \to M(A\otimes A)$ to a comultiplication $A^{**}\to A^{**}\bar{\otimes}A^{**}$. This dualises then to a multiplication map on $(A^{**})_* = A^{*}$. However, in general, it is not clear to me that we can find this extension to the bidual. In many cases of interest, this seems true: namely, it is true for those Hopf $C^*$-algebras that arise from a locally compact quantum group. Indeed, in that case the comultiplication is implemented by a multiplicative unitary. $\endgroup$
    – Andromeda
    Commented Oct 2, 2023 at 5:19

1 Answer 1

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There's already the idea of this argument in the comments, but let me flesh it out.

  1. For any $C^*$-algebra $A$ and any von Neumann algebra $M$ with $A\subseteq M$ non-degenerately we can identify $M(A)$ with the idealiser of $A$ in $M$, namely $\{ x\in M : ax,xa\in A \ (a\in A) \}$. Here "non-degenerate" means, I guess, that some bai of $A$ converges to $1\in M$ weak$^*$. (You can prove this result by representing $M$ on a Hilbert space; I like the treatment in Pedersen's book, section 3.12).
  2. $A\otimes A \subseteq A^{**}\bar\otimes A^{**}$ as a $C^*$-subalgebra.
  3. So $M(A\otimes A) \subseteq A^{**}\bar\otimes A^{**}$ by point 1.
  4. Hence we can consider $\Delta$ as a $*$-homomorphism $A\rightarrow A^{**}\bar\otimes A^{**}$, the right-hand-side being a von Neumann algebra.
  5. By the universal property of $A^{**}$ there is a normal $*$-homomorphism $\tilde\Delta: A^{**} \rightarrow A^{**}\bar\otimes A^{**}$ extending $\Delta:A\rightarrow A^{**}\bar\otimes A^{**}$.
  6. Identify $A^* \widehat\otimes A^*$ with the predual of $A^{**}\bar\otimes A^{**}$ so the pre-adjoint of $\tilde\Delta$ is a map $A^* \widehat\otimes A^* \rightarrow A^*$. (This is just operator space theory; see the chapter of Effros+Ruan about the OSS projective tensor product).
  7. As $\tilde\Delta$ is a complete contraction, so is $A^* \widehat\otimes A^* \rightarrow A^*$.
  8. Perform a diagram chase to check that $A^* \widehat\otimes A^* \rightarrow A^*$ is the product map we want.

The tricky bit is perhaps (8). Perhaps we "usually" embed $A^*\odot A^*$ into $M(A\otimes A)^*$ as follows. By Cohen-Hewitt factorisation, any $f\in A^*$ is of the form $f=ga$ for some $g\in A^*, a\in A$. Then, given $f_1\otimes f_2 = g_1a_1\otimes g_2a_2 \in A^*\odot A^*$, and $x\in M(A\otimes A)$, we define $$ \langle f_1\otimes f_2, x \rangle = \langle g_1\otimes g_2, (a_1\otimes a_2)x \rangle. $$ One checks that this is well-defined. But now this is the same as $\langle (a_1\otimes a_2)x, g_1\otimes g_2 \rangle$ in the $A^{**}\bar\otimes A^{**}, A^*\widehat\otimes A^*$ pairing, consider $A\otimes A \subseteq A^{**}\bar\otimes A^{**}$. This equals $\langle x, f_1\otimes f_2 \rangle$, when $x\in M(A\otimes A)$ is identified with an element of $A^{**}\bar\otimes A^{**}$. Then (8) follows.

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    $\begingroup$ Is there a reference for Step 1? I have a vague memory it might be in e.g. the book of Blecher and Le Merdy, but presumably there should be a more standard reference. $\endgroup$
    – Yemon Choi
    Commented Oct 2, 2023 at 16:57
  • $\begingroup$ Thanks for the answer! I'm also a bit concerned about Step 1. If $M= B(H)$, I think we need to require that $A$ is a non-degenerate $*$-subalgebra of $M$ (see e.g. Lance's book), so an extra condition seems to be involved to make this work. $\endgroup$
    – Andromeda
    Commented Oct 2, 2023 at 17:20
  • $\begingroup$ Nevertheless, I still believe that $M(A\otimes A)\subseteq A^{**}\bar{\otimes}A^{**}$ is true, upon representating $A^{**}$ on a Hilbert space. Indeed, an element $z\in M(A\otimes A)$ is a strict limit of a bounded net in $A\odot A$, and on bounded subsets of $A\otimes A$, the strict topology agrees with the strong$^*$-topology, so the limit lives in $A^{**}\bar{\otimes}A^{**}$ as well. $\endgroup$
    – Andromeda
    Commented Oct 2, 2023 at 17:24
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    $\begingroup$ Sorry, yes, I need a non-degeneracy condition. I'll edit the answer. $\endgroup$
    – Matthew Daws
    Commented Oct 3, 2023 at 14:49
  • $\begingroup$ @MatthewDaws Thanks. I decided to give the bounty before it went away, but I still need to check some details. $\endgroup$
    – Andromeda
    Commented Oct 3, 2023 at 15:13

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