Timeline for Characterisation of positive elements in l¹(Z)
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Oct 11, 2010 at 15:44 | vote | accept | Rasmus | ||
| Sep 2, 2010 at 6:30 | answer | added | Andreas Thom | timeline score: 1 | |
| Sep 1, 2010 at 21:41 | answer | added | Yemon Choi | timeline score: 2 | |
| Sep 1, 2010 at 21:37 | comment | added | Yemon Choi | An injective star-homomorphism between star-Banach algebras still doesn't need to be an isometry. If I recall correctly, $\ell^1(\mathbb Z)$ and $C^∗({\mathbb Z})=C({\mathbb T})$ have *exactly the same positive linear functionals. It seems to me that all your approach will do, is provide a complicated way of checking that the norm on $\ell^1(\mathbb Z)$ does not satisfy the $C^∗$-condition... | |
| Sep 1, 2010 at 21:32 | comment | added | Rasmus | @Yemon Choi: But what I'm looking at is the Banach star -algebra $\ell^1(\mathbb Z)$. So the same data is given as for a C*-algebra. And one way to check that $\ell^1(\mathbb Z)$ is not a C*-algebra is by checking that its GNS-rep (let me call it that) is not isometric. Maybe this is a bit artificial, but I thought it might be worthwile to see how things fail in a slightly weaker context, which one otherwise trusts quite blindly. | |
| Sep 1, 2010 at 21:19 | history | edited | Rasmus | CC BY-SA 2.5 |
added 186 characters in body
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| Sep 1, 2010 at 21:14 | comment | added | Yemon Choi | I haven't thought about this too hard, but it seems to me that the failure of GNS to be isometric for $\ell^1(\mathbb Z)$ is less to do with it having too few positive functionals, but the fact that (as Matthew mentions) the GNS construction involves taking completions. The point is that an injective homomorphism between C*-algebras is an isometry, but if you replace C* by Banach there is no reason to expect this will work. | |
| Sep 1, 2010 at 21:11 | comment | added | Rasmus | @Matthew Daws: I want to apply the GNS-construction to all pure states on$\ell^1(\mathbb Z)$ , and then take the direct sum of all the irreps I ontain this way. I will find that this rep is not isometric. To do this I have to understand what positive functionals on $\ell^1(\mathbb Z)$ are. And therefore I have to understand the postive elements in $\ell^1(\mathbb Z)$. | |
| Sep 1, 2010 at 21:00 | comment | added | Matthew Daws | In what sense do you hope/expect/know that Gelfand-Naimark failes for $\ell^1(\mathbb Z)$? Certainly there is an injective, contractive, involution-respecting homomorphism from $\ell^1(\mathbb Z)$ to $B(H)$ for some Hilbert space $H$ (indeed, $H=\ell^2(\mathbb Z)$ works). This isn't an isometry, of course, but in the C-star world, that's proved by spectral methods, isn't it? | |
| Sep 1, 2010 at 20:43 | comment | added | t3suji | Another (almost obvious) relation: The $n\times n$ matrix $A$ with $a_{ij}=b(i-j)$ has to be Hermitian positive definite (or rather non-negative) for any $n$. This includes both of your conditions. | |
| Sep 1, 2010 at 20:19 | history | asked | Rasmus | CC BY-SA 2.5 |