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Suppose that $H$ is a Hopf algebra with normalised invariant integral (appropriate side) $\int:H\to \mathbb{C}$. The $H$ right comodule algebra $P$ is a Hopf Galois extension, so the canonical map $P\otimes_A P\to P\otimes H$ is a 1-1 correspondence, where $A$ is the $H$ invariant part of $P$.

There is an averaging map $E:P\to A$ given by $E(p)=p_{(0)} \ \int(p_{(1)})$ which gives a projection to the subalgebra $A$ and is an $A$ bimodule map.

Now let us also suppose that $P$ is a $C^*$ algebra, and that $H$ is a Hopf star algebra with whatever other properties we need, and that the integral preserves positivity. Just when is $E$ a conditional expectation for $C^*$ algebras, i.e. when does it preserve positivity? Any nice sufficient conditions would be interesting.

Apologies if I have missed something obvious in the literature...

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  • $\begingroup$ Let $B\subset A$ be C*-algebras. Then a bounded, linear idempotent $P\colon A\to B$ is a conditional expectation if and only if it has norm 1. $\endgroup$
    – Tomasz Kania
    Commented Sep 12, 2015 at 14:03

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First off, as you can see, the definition of $E$ doesn't require the Hopf-Galois condition. The positivity is a consequence of that for slicing by states. If $\phi$ is a state on a C$^*$-algebra $C$, the map $\iota\otimes\phi\colon B \otimes_{\rm min} C \to B$ characterized by $b \otimes c \mapsto \phi(c) b$ is (completely) positive, as the GNS representation for $\phi$ provides a Stinespring factorization of $\iota\otimes\phi$. Now, with $\phi = \int$ being the Haar state of a compact quantum group, $E$ can be presented as the composition of $(\iota \otimes \phi)$ and the coaction $*$-homomorphism $P \to P \otimes_{\rm min} H$ which is (completely) positive. So $E$ is (completely) positive as expected for a conditional expectation.

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  • $\begingroup$ That looks right! Is there a proof of the slice map result in the literature which is really explicit? Alternatively I guess that it is constructing the inner product space for the slice map in the KSGNS theorem, and checking that it really is positive... $\endgroup$
    – Edwin Beggs
    Commented Sep 16, 2015 at 10:29
  • $\begingroup$ It can be found for example here, but I'm afraid it's not much verbose than what I wrote already. $\endgroup$
    – Makoto Yamashita
    Commented Sep 16, 2015 at 15:09

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