Timeline for What is the extra property of this sheaf?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Apr 5, 2011 at 14:50 | comment | added | Tom Church | @unknowngoogle: good point; you're right. | |
| Apr 2, 2011 at 23:53 | comment | added | Qfwfq | @TomChurch: Perhaps you should take $U\mapsto L^p_{\mathrm{loc}}(U)$ instead, otherwise it's not a sheaf: $\exp | _{-\infty , x}$ are in $L^p(-\infty , x)$, and $\mathbb{R}$ is the union of the $(-\infty , x)$ but $\exp$ is not in $L^p(\mathbb{R})$. | |
| Apr 2, 2011 at 22:43 | comment | added | Tom Ellis | @Simon: yes my sheaves have this property. In fact one way of seeing these sheaves is that each $\mathcal{F}(I)$ is a probability space, and then your observation is (roughly) that the distributions are infinitely divisible (classically these are Levy processes, i.e. Brownian motions etc.). | |
| Apr 2, 2011 at 22:41 | comment | added | Simon Rose | Well, those aren't finitely generated... Point taken though. Not all such sheaves are horrible beasts. | |
| Apr 2, 2011 at 20:24 | comment | added | Tom Church | In one dimension the condition just means that removing one point doesn't change the sections. For example $U\mapsto L^p(U)$ seems to have this property, and it's hard to call $L^p(\mathbb{R})$ "badly behaved"... | |
| Apr 2, 2011 at 20:05 | history | answered | Simon Rose | CC BY-SA 2.5 |