Timeline for Collared boundary of a non-metrizable manifold
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 27, 2019 at 14:12 | comment | added | Mathieu Baillif | @shuhalo Yes, that's what I tried to explain in my edit: the images of the local collars must be start finite in order for the argument to work. Time allowing, I'll add some details to my answer to clarify a bit more. | |
| Dec 26, 2019 at 17:08 | comment | added | shuhalo | @kaba and Mathieu Baillif: that is interesting to read about Gauld's book. Can you comment on the following regarding his Theorem 3.10: do we need that the images of the local collars are star-finite? I believe his mapping $g_\alpha$ is well-defined only with some additional condition. For example, it is not well-defined if the images of the collars have non-empty common intersection. | |
| Oct 27, 2019 at 19:46 | vote | accept | kaba | ||
| Oct 27, 2019 at 19:45 | comment | added | kaba | Many thanks for the update! For some reason I did not get a notification of it; perhaps my MO-settings are off. Great work picking up that detail on strong paracompactness. I also wondered how a condition on the boundary could "leak" onto the interior. With the updated requirement it makes sense. | |
| Oct 21, 2019 at 10:24 | history | edited | Mathieu Baillif | CC BY-SA 4.0 |
Added details on the statement of Gauld's Thm 3.10
|
| Oct 19, 2019 at 21:10 | comment | added | Mathieu Baillif | Sorry to flood you with comments, the one just above is of course silly: a space whose components are strongly paracompact is itself strongly paracompact, so indeed strongly paracompact and metrizable are equivalent even when not connected. But then there is something strange because I am really under the impression that the Prüfer surface has a non-collared boundary, but this boundary is itself strongly paracompact. I have to think about it a bit more. | |
| Oct 19, 2019 at 20:58 | comment | added | Mathieu Baillif | Actually, a manifold with countably many connected components is metrisable if and only it is strongly paracompact. (See Gauld's diagram on p. 31 of his book for an astonishing amount of information about these things.) Hence the proof of Cor. 3.11 shows that a manifold with a boundary made of countably many components that are metrisable has a collared boundary. This is a small generalization. I did not check but it is possible that it appears in the original paper of R. Connely. | |
| Oct 19, 2019 at 20:56 | comment | added | kaba | According to this paper, also by Gould, a "paracompact locally compact space is strongly paracompact". If that holds, then I think that paracompactness, strong paracompactness, and metrizablity are equivalent even for non-connected manifolds, right? | |
| Oct 19, 2019 at 20:04 | comment | added | Mathieu Baillif | The trick is that the boundary of the Prüfer surface is a discrete collection of continuum many real lines, and hence as a topological space it is metrisable (points in different components may be thought to be at fixed distance $1$). So I believe that Gauld tries to avoid this difficulty by talking only about components. What is needed is the property that the boundary is strongly paracompact (see Theorem 3.10), which is equivalent to metrisability for connected manifolds but not in general. | |
| Oct 19, 2019 at 19:56 | comment | added | Mathieu Baillif | (cont) but if you look at the definition of the Prüfer surface, you should see what I mean (I hope). | |
| Oct 19, 2019 at 19:54 | comment | added | Mathieu Baillif | I think that he mentions components because the answer may depend on which ones or how many you take. For instance a manifold may have metrisable and non-metrisable boundary components. In the example of the Prüfer surface they are all metrisable, but if you take only the interior and countably many boundary components then the manifold is metrisable and the boundary (as a whole) is collared, and if you take all of them (that is, continuum many), the boundary as a whole is not collared though each boundary component, individually, is collared. I might not be very clear (cont) | |
| Oct 19, 2019 at 19:19 | comment | added | kaba | Nice! I've been trying to see whether connectedness is used at all in the proofs, and it seems to me that the only reason the theorem mentions components is because Gauld defines a manifold as connected. So perhaps one could obtain a collar for the whole boundary (assuming metrizable boundary). But perhaps I'm wrong; I'll take a closer look. | |
| Oct 19, 2019 at 18:06 | history | answered | Mathieu Baillif | CC BY-SA 4.0 |