Timeline for The Tensor product of algebra group
Current License: CC BY-SA 3.0
13 events
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| Jul 28, 2015 at 8:55 | comment | added | hosain | @Transcendental Thank you very much for your answer.But what do you mean of " The pullback of this norm to L1(G)⊙L1(G) "? – – | |
| Jul 26, 2015 at 18:28 | vote | accept | hosain | ||
| Jul 24, 2015 at 3:16 | comment | added | Yemon Choi | @Transcendental Looking in a book on tensor products of Banach spaces (Defant-Floret, or the introductory book of Ryan) should answer your questions :) | |
| Jul 24, 2015 at 0:54 | comment | added | Yemon Choi | @Corbennick why do you claim that the more appropriate object to consider is the $C^*$-algebra of $G$? Are bounded representations on non-Hilbertian spaces not worth considering, then? | |
| Jul 23, 2015 at 20:46 | answer | added | P.J | timeline score: 1 | |
| Jul 1, 2015 at 18:42 | comment | added | Transcendental | Whether this particular completed tensor product is isomorphic to the injective tensor product or the projective one remains to be seen. @Corbennick: I just saw your comment! Hence, what I’ve written above basically repeats what you’ve already said. :) | |
| Jul 1, 2015 at 18:34 | comment | added | Transcendental | Next, show that the image of this map is dense in $ {L^{1}}(G \times G) $ with respect to the $ L^{1} $-norm. The pullback of this norm to $ {L^{1}}(G) \odot {L^{1}}(G) $ is a cross-norm with respect to which the completion of $ {L^{1}}(G) \odot {L^{1}}(G) $ is automatically isomorphic to $ {L^{1}}(G \times G) $ as a Banach space. | |
| Jul 1, 2015 at 18:34 | comment | added | Transcendental | See if this works. Embed the algebraic tensor product $ {L^{1}}(G) \odot {L^{1}}(G) $ into $ {L^{1}}(G \times G) $ by mapping an elementary tensor $ [f] \odot [g] $ to $ [(x,y) \mapsto f(x) g(y)] $ — where $ [\cdot] $ denotes the taking of an equivalence class — and then extending this map by linearity to the entire algebraic tensor product. | |
| Jul 1, 2015 at 18:30 | comment | added | user1688 | The tensor product is certainly dense in $L^1(G\times G)$. This amounts to saying that the algebraic tensor product carries a norm such that the completion of this norm gives $L^1(G\times G)$. The more appropriate object, however, to look at is the $C^*$-algebra of $G$. | |
| Jul 1, 2015 at 8:33 | comment | added | Transcendental | I believe that you need to specify the type of tensor product to be used. The common ones are the injective and projective tensor products. | |
| Jul 1, 2015 at 7:39 | history | edited | David Roberts♦ | CC BY-SA 3.0 |
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| Jul 1, 2015 at 6:47 | review | First posts | |||
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| Jul 1, 2015 at 6:44 | history | asked | hosain | CC BY-SA 3.0 |