Timeline for The connection between Lie algebroids and foliations
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 19, 2017 at 22:58 | comment | added | Joe Pollard | Yes. If I'm seeing this right, in the first case the leaves are circles on the torus, and in the second case it's like the circles have been 'infinitely wrapped around' so they're dense. I'll have to think about it a bit more but this does help, thanks. | |
| Nov 19, 2017 at 22:32 | comment | added | Qfwfq | I don't think the leaves look the same in general. Consider the distribution generated by $\partial_x$ in the torus $\mathbb{R}^2/\mathbb{Z}^2$, and the foliation $\mathcal{F}_t$ generated by $\partial_{x} + t\cdot \partial_y$ where $t$ is a constant. I haven't thought carefully, but I think the distributions for $\mathcal{F}_0$ and for $\mathcal{F}_{\sqrt 2}$ are isotopic, but the first one has compact leaves while the second has dense leaves. | |
| Nov 19, 2017 at 21:03 | comment | added | Joe Pollard | I interpret it like this: if $\mathcal{F}$ is a foliation we construct the corresponding Lie algebroid to be the bundle $T\mathcal{F} \to TM$, which anchor map $\rho = \text{id}$, the identity (or the inclusion). If we keep the bundle the same but replace the map with some injective map $\tilde \rho$, then the image $\tilde \rho(T \mathcal{F})$ defines a new foliation, say $\mathcal{G}$, and if we restrict to the leaves of this foliation then the map $\tilde \rho$ is surjective. My first question was really whether the leaves of this look the same but are embedded in $M$ differently. | |
| Nov 19, 2017 at 20:59 | comment | added | Joe Pollard | I have missed a requirement. The Lie algebroid $A \to TM$ is said to be regular if the map $\rho$ (called the anchor) is of locally constant rank. The exact comment in the book is then "If $A$ is regular then the image of the anchor defines a foliation, the characteristic foliation of $A$, and over each leaf of the foliation, the Lie algebroid is transitive." There's no further explanation so I suppose it's meant to be 'obvious', but it's not entirely obvious to me. | |
| Nov 18, 2017 at 23:32 | comment | added | Qfwfq | I don't know, but maybe it's meant that $T\mathcal{F}|_{\mathcal{L}}\to T\mathcal{F}|_{\mathcal{L}}$ is surjective in this case (of course it is), not the analogous map to $TM$ which is certainly not surjective (for a foliation of positive codimension). Am I missing something? | |
| Nov 17, 2017 at 10:09 | history | edited | Joe Pollard | CC BY-SA 3.0 |
Clarified a point.
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| Nov 17, 2017 at 10:08 | comment | added | Joe Pollard | @Qfwfq A Lie algebroid is called transitive if $\rho$ is fibrewise surjective. The textbook I am using (Mackenzie, General Theory of Lie Groupoids and Lie Algebroids) says that if the image of $\rho$ gives a foliation, then when we restrict the algebroid to a leaf $L$ of the foliation, it is transitive. | |
| Nov 17, 2017 at 0:08 | comment | added | Qfwfq | What do you mean by "the map $\rho$ is surjective on the leaves"? If $\mathcal{L}\subseteq M$ is any leaf, then $\rho:T\mathcal{F}|_{\mathcal{L}}\to TM|_{\mathcal{L}}$ is a map of vector bundles which is still injective on its fibers. | |
| Nov 16, 2017 at 17:05 | review | First posts | |||
| Nov 16, 2017 at 17:43 | |||||
| Nov 16, 2017 at 17:01 | history | asked | Joe Pollard | CC BY-SA 3.0 |