A left inverse for the comultiplication on a Hopf von Neumann algebra

Edit: incorrect claim at end of earlier version; thanks to Matthew Daws for pointing this out in comments.


Let $(\cM,\Delta)$ be a Hopf von Neumann algebra: that is, $\cM$ is a von Neumann algebra and $\Delta: \cM\to \cM\stp\cM$ is a coassociative, injective, normal $*$-homomorphism.

Does there always exist a completely bounded, linear map $T:\cM\stp\cM\to \cM$ such that $T\Delta$ is the identity? If so, can we always choose $T$ to be normal?

This works for many several examples, for instance when $\cM$ is injective as a von Neumann algebra [the image of $\Delta$ is then complemented in $B(H\otimes H)$ by a norm-one projection], or $\cM$ is a locally compact quantum group in the sense of Kustermans-Vaes [use the fundamental unitary and then slice]. or if the predual $\cM_*$ has a bounded approximate identity for the natural product induced by $\Delta_*$.

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Erm... How do you use the fundamental unitary to slice? You have that $\Delta(x) = W^*(1\otimes x)W$ and so $W\Delta(x)W^* = 1\otimes x$. So it would be tempting to use the map $M\overline\otimes M \rightarrow B(H), z\mapsto (\omega\otimes\iota)(WzW^*)$ but I don't see why this lands in $M$?? Or did you have another argument in mind (in which case, I'd love to see it!) –  Matthew Daws Nov 14 '11 at 9:38
Argh, you're right - I think I had examples where $\mathcal M$ is injective or $(\mathcal M)_*$ is operator biflat in mind. Will edit accordingly –  Yemon Choi Nov 14 '11 at 21:59

This is far from a full answer (but maybe it will inspire other answers).

I think all the cases Yemon gives, we only get a cb map, but we don't get normality. However, in some special cases, you do get a normal map.

Firstly, if $M=L^\infty(G)$ with $\Delta(f)(s,t) = f(st)$, then define $T_*:L^1(G) \rightarrow L^1(G\times G)$ by $T_*(f)(s,t) = f(st) f_0(s),$ where $f_0\in L^1(G)$ is some fixed, positive function with $\|f_0\|_1=1$. This is bounded, as \begin{align} \|T_*(f)\|_1 &= \int_G \int_G |f(st) f_0(s)| \ dt \ ds = \int_G \int_G |f(ss^{-1}t)| \ dt \ |f_0(s)| \ ds \\ &= \int_G \|f\|_1 |f_0(s)| \ ds = \|f\|_1, \end{align} using the left-invariance of the Haar measure. Then the pre-adjoint $\Delta_*:L^1(G\times G)\rightarrow L^1(G)$ is $\Delta_*(f)(t) = \int_G f(s,s^{-1}t) \ ds,$ i.e. convolution (if you set $f=a\otimes b$). So then $\Delta_* T_*(f) (t) = \int_G T_*(f)(s,s^{-1}t) \ ds = \int_G f(ss^{-1}t) f_0(s) \ ds = f(t).$ So $\Delta_* T_*$ is the identity, as required.

More generally, and as Yemon alludes to in a comment, if $(M,\Delta)$ is a compact Kac algebra, then it's operator biprojective (see Aristov's paper "Amenability and compact type for Hopf-von Neumann algebras from the homological point of view"). So we can choose $T$ to be cb and normal (and with various $M_*$ module properties). If we can choose $T$ with these module properties (but not necessarily normal) then $M$ is said to be operator biflat. For $VN(G)$ this was investigated by Aristov, Runde and Spronk, "Operator biflatness of the Fourier algebra and approximate indicators for subgroups". It seems to be unknown if $VN(G)$ ever fails to be operator biflat. Of course, this is a much stronger condition that Yemon asks for.

It seems surprising to me that we can just write down a suitable (normal) map $T$ for $L^\infty(G)$, but that it seems that Yemon's question is open for $VN(G)$.

Edit (29 Nov): (This is technical; I hope I got all the details correct). Suppose $G$ is a locally compact group such that $VN(G)$ admits a faithful normal trace (I think is true if $G$ is a separable SIN group, for example). Then let $\varphi$ be the Plancheral weight on $VN(G)$; let $\omega$ be a normal tracial state. Then $\psi=\varphi\otimes\omega$ is a semifinite trace on $VN(G\times G)=VN(G)\overline\otimes VN(G)$, and for $x\in VN(G)_+$, we have that $$\psi(\Delta(x)) = \varphi((\iota\otimes\omega)\Delta(x)) = \omega(1) \varphi(x) = \varphi(x)$$ as $\varphi$ is invariant for $\Delta$ (I know this from the quantum group setting; see papers of Kustermans and Vaes). Let $(\sigma_t)$ be the modular automorphism group for $\varphi$; then $(\sigma_t\otimes\iota)\Delta(x) = \Delta(\sigma_t(x))$ for $x\in VN(G)$, and so as $\omega$ is a trace, we see that $\Delta(VN(G))$ is invariant for the modular automorphism group of $\psi$. By a theorem of Takesaki there is a normal conditional expectation $\epsilon:VN(G\times G)\rightarrow \Delta(VN(G))$ with $\psi(x) = \psi(\epsilon(x))$ for all $x$ in the definition ideal of $\psi$.

So in particular, this gives a positive answer (even with the "normal") condition in the separable SIN case, say.

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Of course, the formula for $T$ which I wrote down in the $L^\infty(G)$ case is just $(\omega\otimes\iota)W(\cdot)W^*$; it might just be blind luck that this works in the commutative case... –  Matthew Daws Nov 16 '11 at 14:14
If $G$ is discrete, $\Delta: VN(G) \to VN(G) \otimes VN(G)$ is a trace preserving normal homomorphism, its image is therefore always complemented by a normal conditional expectation, right? So the question is open in the non-discrete case only? (But perhaps this is contained in the setting of compact Kac algebra, please forgive my ignorance on the subject). –  Mikael de la Salle Nov 16 '11 at 14:32
@Mikael: Yes to all points! ($VN(G)$ is "compact" in the Kac algebra sense if and only if $G$ is discrete). –  Matthew Daws Nov 16 '11 at 17:58
I think that the answer to my question would also be yes if $G$ is almost connected, using the result proved by Connes (in one paragraph using a complicated paper by Dixmier that people don't cite enough, but that's another story) that for such $G$, $VN(G)$ is injective. Does that look right? Of course the operator biflatness question is much harder for such groups... –  Yemon Choi Nov 16 '11 at 21:00
Yes, sure-- but that doesn't give a normal map $T$, which was mainly what was bothering me (that, and somehow using that $VN(G)$ is injective seems like a big stick...) –  Matthew Daws Nov 16 '11 at 21:15