Timeline for Constructing intertwiners between representations of compact quantum groups
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 24, 2020 at 16:58 | comment | added | user167952 | @MatthewDaws Yes I deleted it because I solved it. The inclusion map might be degenerate, but it is strict (= norm-continuous and strictly continuous on bounded subsets). Hence, it still gives a map on the multiplier $C^*$-algebras. Thanks for the very useful comments though! | |
| Nov 24, 2020 at 16:12 | comment | added | Matthew Daws | And then you seemingly deleted it! I think (a) the inclusion is degenerate; but (b) you can easily prove it exists by using the projection $H_1\oplus H_2 \rightarrow H_1$ and the inclusion going the other way, to pre and post multiply. | |
| Nov 23, 2020 at 21:20 | comment | added | user167952 | @Matthew Daws. Thanks! I will see if that is useful in this context. However, is the claim I make about the inclusion of multiplier algebras actually true? I.e.do we have an inclusion $M(K(H_1)\otimes A)\subseteq M(K(H_1\oplus H_2)\otimes A)$? | |
| Nov 23, 2020 at 21:10 | comment | added | Matthew Daws | Yes, I see why you might want this. Further, I'm not sure this would help you: what you really want is to know how to form $(x\otimes 1)u$ where $x\in\mathcal K(H_1,H_2)$ and $u\in M(\mathcal K(H_1)\otimes A)$. I think! One way around this is to observe that $x^*x \in \mathcal K(H_1)$ and so the polar decomposition $x = p|x|$ has $p$ a partial isometry $H_1\rightarrow H_2$ and $|x|\in\mathcal K(H_1)$. Then first form $(|x|\otimes 1)u$ in $\mathcal K(H_1)\otimes A$, and then multiply by $p\otimes 1$ to get a member of $\mathcal K(H_1, H_2)\otimes A$. | |
| Nov 23, 2020 at 18:59 | comment | added | user167952 | @Matthew Daws. I'm sorry to disturb you but I have a follow-up question. To make sense of the theorem statement, it looks like it is used that we have a canonical inclusion $M(\mathcal{K}(H_1) \otimes A) \subseteq M(\mathcal{K}(H_1 \oplus H_2) \otimes A)$. Do you know how we can obtain this inclusion? It seems to be natural to consider the natural inclusion map $i: \mathcal{K}(H_1) \otimes A \to \mathcal{K}(H_1 \oplus H_2) \otimes A \subseteq M(\mathcal{K}(H_1 \oplus H_2) \otimes A)$ but this does not seem to be non-degenerate, so I'm a bit unsure how to get this inclusion. | |
| Nov 21, 2020 at 18:25 | comment | added | user167952 | @Matthew Daws. I have to sincerely thank you for the suggestion to look at the original papers by Woronowicz. Some aspects have become very clear now! | |
| Nov 21, 2020 at 17:46 | history | edited | Matthew Daws | CC BY-SA 4.0 |
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| Nov 21, 2020 at 14:19 | comment | added | Ruy | Another definition of $\mathcal K(H_1, H_2)\otimes A$, more reminiscent of the usual spatial tensor product, is to take a faithful representation of $A$ on some Hilbert space $H$, and then look at the closed linear span of operators of the form $x\otimes a$ within $\mathcal B(H_1\otimes H, H_2\otimes H)$, for all $x\in \mathcal K(H_1, H_2)$ and $a\in A$. | |
| Nov 21, 2020 at 10:52 | vote | accept | CommunityBot | ||
| Nov 21, 2020 at 10:17 | history | answered | Matthew Daws | CC BY-SA 4.0 |