# 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:

(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

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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