Timeline for Model theory of Banach algebras
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Jul 28, 2018 at 19:35 | comment | added | Tomasz Kania | James, this will be published (not by me) but BV is indeed a counterexample. | |
| Jul 28, 2018 at 18:46 | comment | added | James E Hanson | I think I found an answer. A comment here says that the Banach algebra of functions of bounded variation is a counterexample. | |
| Jul 28, 2018 at 18:41 | comment | added | James E Hanson | Ah I see. Thank you for explaining that. Is there a known example of a Banach algebra with open multiplication that is not uniformly open? | |
| Jul 28, 2018 at 18:34 | comment | added | Nik Weaver | No, it's a little worse than that; you need the product of the $\epsilon$ balls to be boundedly away from the $z_n$. | |
| Jul 28, 2018 at 18:09 | comment | added | James E Hanson | Okay so for each $\delta$ we actually need a sequence $z_{\delta,n}$ witnessing failure of the condition for sufficiently large $n$ such that $\|x_n y_n - z_{\delta,n}\|$ is uniformly bounded away from $0$, and we don't know if we have that. Do I have that right? | |
| Jul 28, 2018 at 18:04 | history | edited | James E Hanson | CC BY-SA 4.0 |
added 80 characters in body
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| Jul 28, 2018 at 18:04 | comment | added | Nik Weaver | I am saying that it's not clear what you will use for the falsifying $z$'s. You need a different $z$ for each possible $\delta$, with $\|z - xy\| \leq \delta$. A single $z$ isn't going to work because it will equal $xy$, after you factor out null vectors. | |
| Jul 28, 2018 at 18:01 | history | edited | James E Hanson | CC BY-SA 4.0 |
added 80 characters in body
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| Jul 28, 2018 at 18:00 | comment | added | James E Hanson | I'm not sure if this is what you're saying but I think I see a problem which is that in the ultrapower we might also get new $u$ and $v$ that satisfy the condition for our putative counterexample $z$. Basically, $\varphi$ is a closed condition but $\neg \varphi$ is an open condition and really that's what we want to preserve, but while we can wiggle the inequalities to make them open or closed, the equality condition $z=uv$ is always closed. Is that what you're getting at or is there another problem too? | |
| Jul 28, 2018 at 17:34 | history | edited | James E Hanson | CC BY-SA 4.0 |
Counterexample reference
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| Jul 28, 2018 at 17:25 | comment | added | Nik Weaver | "the limits of $x_n$ and $y_n$ will be counterexamples" --- and what will you use for $z$? A limit of falsfying $z_n$ with $\|z_n - x_ny_n\| \to 0$? That certainly doesn't work ... | |
| Jul 28, 2018 at 17:00 | comment | added | James E Hanson | (The sequence $\{x_n,y_n\}$ needs to be all norm 1 vectors or at least uniformly bounded away from $0$ and infinity.) | |
| Jul 28, 2018 at 16:59 | comment | added | James E Hanson | With $\varphi$ being preserved in ultrapowers there's a subtlety with witnessing $u$'s and $v$'s maybe being of unbounded norm, but the form of the condition implies that non-trivial witnesses are of bounded norm since they need to be close to $x$ and $y$. | |
| Jul 28, 2018 at 16:54 | comment | added | James E Hanson | So then if I let $D(x,y,\varepsilon)=\{\delta:\varphi(x,y,\varepsilon,\delta)\}$, then if there exists a sequence $\{x_n,y_n\}$ and an $\varepsilon>0$ such that $\sup D(x_n,y_n,\varepsilon )\rightarrow 0$ as $n\rightarrow \infty$, then in any nontrivial ultrapower of the Banach algebra over a countable index set, the limits of $x_n$ and $y_n$ will be counterexamples. | |
| Jul 28, 2018 at 16:53 | comment | added | James E Hanson | What I want to say is that if I let $\varphi(x,y,\varepsilon,\delta) = (\forall z)(\exists u)(\exists v)(\|z-xy\|\geq\delta\vee[\|u-x\|\leq\varepsilon\wedge\|v-y\|\leq\varepsilon\wedge z=uv])$, then for fixed inputs this is preserved under passing to a modified ultrapower. | |
| Jul 28, 2018 at 16:53 | comment | added | James E Hanson | There's a lot of quantifiers, so it's entirely possible that it's more subtle than I think. If I'm not mistaken the original condition is equivalent to the following $(\forall x)(\forall y)(\forall\varepsilon>0)(\exists\delta>0)(\forall z)(\exists u)(\exists v)(\|z-xy\|\geq\delta\vee[\|u-x\|\leq\varepsilon\wedge\|v-y\|\leq\varepsilon\wedge z=uv])$ where now I'm restricting to quantification over norm 1 vectors (other than the quantification for z, u, and v). | |
| Jul 28, 2018 at 16:31 | comment | added | Nik Weaver | "If there is an $\epsilon > 0$ ... the condition will fail." Are you sure? What would be the falsifying $x$, $y$, and sequence of $z$'s? I think this is more subtle than you think ... | |
| Jul 28, 2018 at 16:14 | history | answered | James E Hanson | CC BY-SA 4.0 |