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Constructing intertwiners between representations of compact quantum groups

Consider the following paper by Van Daele en Maes Notes on compact quantum groups. For convenience of the reader, here is a picture of the relevant section:
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(1) How is compact operator defined in this context? For example, what is meant with $x$ is a compact operator from $H_1 to H_2$? Is this the usual definition of compactness? I.e. the image of the unit ball is precompact. Or is something else going on?

(2) In the proof, one considers the object $mathcal{B}_0(mathcal{H}_1, mathcal{H}_2)otimes A$. How is this tensor product defined in this context? Surely $mathcal{B}_0(mathcal{H}_1, mathcal{H}_2)$ is no $C^*$-algebra so this is not a tensor product of $C^*$-algebras.

MathOverflow Asked on January 15, 2021

1 Answers

One Answer

Yes, an operator is compact if it maps the unit ball into a precompact set. This is the usual definition for Banach spaces. For Hilbert spaces the same applies (see e.g. these notes). Indeed, a bounded linear $T:H_1rightarrow H_2$ is compact if and only if $T^*T$ is compact, iff $TT^*$ is compact, iff $T^*$ is compact.


To deal with the second question, I think one proceeds as follows. Firstly, observe that there is a $*$-algebra isomorphism $$ mathcal B(H_1oplus H_2) cong begin{pmatrix} mathcal B(H_1) & mathcal B(H_2, H_1) \ mathcal B(H_1, H_2) & mathcal B(H_2) end{pmatrix}. $$ To see this, think about the matrix acting on $H_1oplus H_2$ written as a column vector. Under this isomorphism, compact operators behave as you might hope, $$ mathcal K(H_1oplus H_2) cong begin{pmatrix} mathcal K(H_1) & mathcal K(H_2, H_1) \ mathcal K(H_2, H_1) & mathcal K(H_2) end{pmatrix}. $$ Thus I can speak of $mathcal K(H_1,H_2)$ as a "corner" of $mathcal K(H_1oplus H_2)$.

So, one can define $mathcal K(H_1,H_2) otimes A$ as the closure of $mathcal K(H_1,H_2) odot A$ inside $mathcal K(H_1oplus H_2) otimes A$. Things are nicer than this: let $p_i$ be the projection of $H_1oplus H_2$ onto $H_i$. Then $p_2 mathcal K(H_1oplus H_2) p_1$ is isomorphic to $mathcal K(H_1,H_2)$, and $p_iotimes 1in M(mathcal K(H_1oplus H_2)otimes A)$. One can check that $mathcal K(H_1,H_2) otimes A$ is isomorphic to $(p_2otimes 1)(mathcal K(H_1oplus H_2)otimes A)(p_1otimes 1)$.


The notes by Maes and Van Daele are nice, but I do find that there are various little inaccuracies, or points like this which are not (well) explained. The original papers by Woronowicz are terse, but I think a pleasure to read, and will improve your intuition about the subject. You could also look at the book of Timmermanns, but that takes a different approach.


In a comment, Ruy points our how to get this tensor product via representing on a Hilbert space.

An approach using more theory would be to use Hilbert $C^ast$-modules (I follow chapter 4 of Lance's book). $mathcal K(H_1, H_2)$ is a right module over $mathcal K(H_1)$ for the "inner-product" $(S|T) = S^*T$. $A$ is a module over itself. The exterior tensor product of modules gives $mathcal K(H_1, H_2) otimes A$ as a right module over $mathcal K(H_1)otimes A$. A rather tedious check shows that the norms of all three approaches are the same.

For the application, we need:

  • For a state $hin A^*$ we want to make sense of $iotaotimes h$ as a map $mathcal K(H_1,H_2)otimes A rightarrow mathcal K(H_1,H_2)$;
  • For a $*$-homomorphism $Phi:Arightarrow Aotimes A$ we need to make sense of $iotaotimesPhi$ as a homomorphism (suitable interpreted) $mathcal K(H_1,H_2)otimes A rightarrow mathcal K(H_1,H_2)otimes A otimes A$.

I think my original approach, of viewing things as a "corner" of $C^ast$-algebra, is probably the easiest way to get these properties.

Correct answer by Matthew Daws on January 15, 2021

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