Timeline for Collared boundary of a non-metrizable manifold
Current License: CC BY-SA 4.0
38 events
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| Dec 5, 2019 at 19:44 | history | edited | kaba | CC BY-SA 4.0 |
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| Dec 3, 2019 at 9:00 | comment | added | Mathieu Baillif | My email is mathieu.baillif at hesge.ch, it will be better to continue there than to flood this thread with comments, I think. | |
| Dec 2, 2019 at 12:52 | history | edited | kaba | CC BY-SA 4.0 |
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| Dec 2, 2019 at 10:50 | comment | added | kaba | Do you mean that $\partial X'_i$ is not closed when $i > 0$? I'm not sure which points you are referring to. At the latest bullet point I added a theorem which I think to have proved, which provides a more general relation between local collars and a collar. I think it applies to my triangle-construction. Each triangle provides a local Connelly-collar, and the collection of those triangles is $X$-locally finite. The constructed function is also injective. By this theorem $\partial X'_i$ should be Connelly-collared for $i > 0$ too. Probably I'm missing something. | |
| Dec 2, 2019 at 10:10 | comment | added | kaba | I'd email you, but my google-fu wasn't strong enough at finding your email address. | |
| Dec 2, 2019 at 9:58 | history | edited | kaba | CC BY-SA 4.0 |
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| Dec 2, 2019 at 9:21 | comment | added | Mathieu Baillif | Your example for $X_0$ is right, but you cannot do the same thing with $X_1$, as the collar is not closed. In your construction, there are points in the closure of the collar in the line $\{\omega\}\times\{0\}\times[0,1)$. I have more or less finished a first (draft) version of my notes on the problem (I decided to turn them into something that might be published one day, hence with all the features of an official paper, so to say, but it is still not polished at all). Just email me if you are interested. | |
| Dec 2, 2019 at 8:57 | history | edited | kaba | CC BY-SA 4.0 |
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| Dec 1, 2019 at 23:17 | history | edited | kaba | CC BY-SA 4.0 |
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| Dec 1, 2019 at 23:11 | history | edited | kaba | CC BY-SA 4.0 |
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| Dec 1, 2019 at 23:03 | history | edited | kaba | CC BY-SA 4.0 |
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| Dec 1, 2019 at 22:56 | history | edited | kaba | CC BY-SA 4.0 |
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| Nov 28, 2019 at 8:46 | comment | added | kaba | Made some typos. $h((x, 0), t) = (x, (1/2 - |(x \mod 1) - 1/2|) * t)$ and $h((0, y), t) = ((1/4 - |y - 1/2|/2)*t, y)$. | |
| Nov 28, 2019 at 8:38 | comment | added | kaba | ... Since paracompactness does not hold when $i > 0$, I'll try to provide an explicit construction for a boundary-collar instead. We have that $\partial X' = \{0\} \times (0, 1) \cup \bigcup \{(n, n + 1) \times \{0\} : n \in \mathbb{N}\}$. Let $h : \partial X' \times [0, 1]$ be such that $h((x, 0), t) = (x, 1/2 - |(x \mod 1) - 1/2| * t)$ and $h((0, y), t) = (1/4 - |y - 1/2| * t / 2, y)$. I think I can show that this is a Connelly-collar of $\partial X'$. But if so, I think I can extend this construction to show that $X'_i$ is also Connelly-collared. Where do I go wrong? | |
| Nov 28, 2019 at 8:38 | comment | added | kaba | Let $X_i = \omega_i \times [0, 1) \times [0, 1)$, and $X'_i = X_i \setminus (\omega_i \times \{0\} \times \{0\})$. Then your example is $X'_1$. To gain intuition, I considered $X'_0$. Let $X = [0, \infty) \times [0, 1) \approx X_0$, and $X' = [0, \infty) \times [0, 1) \setminus (\mathbb{N} \times \{0\}) \approx X'_0$. Since $X$ and $X'$ are paracompact manifolds-with-boundaries, they both have a Connelly-collared boundary... | |
| Nov 27, 2019 at 21:20 | comment | added | Mathieu Baillif | Sorry, there is some confusion in the notation, but it is my fault. $\mathbb{L}_{\ge 0}$ is indeed $\omega_1\times[0,1)$ with lexicographic order topology, but what I had in mind is $\mathbb{L}_{\ge 0}\times [0,1)$ (with product topology) with $\omega_1\times\{0\}\times\{0\}$ removed. It is usual to identify $\omega_1$ with $\omega_1\times\{0\}\subset\mathbb{L}_{\ge 0}$, hence the confusion. | |
| Nov 27, 2019 at 10:17 | history | edited | kaba | CC BY-SA 4.0 |
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| Nov 27, 2019 at 10:10 | history | edited | kaba | CC BY-SA 4.0 |
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| Nov 27, 2019 at 8:36 | comment | added | kaba | The above is Connelly-collared as a disjoint sum of manifolds-with-boundary homeomorphic to $(0, 1) \times [0, 1)$, so perhaps something else... | |
| Nov 27, 2019 at 7:45 | comment | added | kaba | @MathieuBaillif Thanks for the example; I could only now get back to this. Can you spell out the product in more detail? The sets don't seem to match up. Say $\mathbb{L}_{\geq 0} = \omega_1 \times [0, 1)$, and $A = \omega_1 \times \{0\} \subset \mathbb{L}_{\geq 0}$. Is it $(\mathbb{L}_{\geq 0} \setminus A) \times [0, 1)$ or something else? | |
| Nov 20, 2019 at 10:58 | history | edited | kaba | CC BY-SA 4.0 |
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| Nov 20, 2019 at 8:48 | comment | added | Mathieu Baillif | Yes, that's exactly what happens: you can have an embedding of $\partial M\times [0,1)$ which gives a neighborhood of the boundary, but none that is a closed embedding. Such an example is given by the product of the longray with $[0,1)$ with $\omega_1\times\{0\}$ removed. If you know what these spaces are, it is not difficult to check that it is the case. If not, then maybe I'll try to finish writing up these notes and make them available. Or you can check Gauld's book where the longray (and longline) are discussed in detail. | |
| Nov 19, 2019 at 23:07 | comment | added | kaba | Do you mean that there is a Brown-collar (collar being the embedding function, rather than its image) which cannot be extended to a Connelly-collar? (By definition, if I've understood correctly, the image of a Brown-collar is open, and the image of a Connelly-collar is closed.) What would be an example of this? Do link the notes if you end up writing them. | |
| Nov 19, 2019 at 19:28 | comment | added | Mathieu Baillif | So, thanks for asking the question in the first place. | |
| Nov 19, 2019 at 19:28 | comment | added | Mathieu Baillif | I'll add that your question got me thinking because I noticed as well that there are cases where we cannot ensure that the collar is a closed subset of the manifold in the non-metrisable case. Connelly's and Brown's definitions of collars differ on this point, as the former includes closedness while the latter does not. I found these subtle points interesting enough to prepare a small note (for myself mainly) on the subject summarizing what I know. I might publish it when I am done writing it if some people are interested. (Nothing in the note is really new, just compiled in one place.) | |
| Nov 19, 2019 at 6:37 | history | edited | kaba | CC BY-SA 4.0 |
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| Nov 15, 2019 at 21:44 | history | edited | kaba | CC BY-SA 4.0 |
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| Nov 15, 2019 at 21:32 | history | answered | kaba | CC BY-SA 4.0 |