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...

  • $\begingroup$ 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. $\endgroup$ Sep 12, 2015 at 14:03

1 Answer 1


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.

  • $\begingroup$ 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... $\endgroup$ 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. $\endgroup$ Sep 16, 2015 at 15:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.