Timeline for Direct limits of compact surfaces with uniformly bounded topology
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Jan 13, 2020 at 3:02 | comment | added | Rohil Prasad | Sorry to revive this already long comment thread, but here is a stipulation on the embeddings that removes all of the pathology displayed above. We require that, for all $i$, the embedding $S_i \hookrightarrow S_{i+1}$ sends $S_i$ to a compact subset of the interior of $S_{i+1}$. | |
| Jan 7, 2020 at 19:25 | comment | added | Taras Banakh | @MikeMiller For every positive integer $n$ consider the disk $S_n=\{z\in\mathbb C:|z-n|\le n\}$ and observe that the union $S=\bigcup_{n=1}^\infty S_n$ is not locally compact at zero. | |
| Jan 6, 2020 at 17:40 | comment | added | mme | @TarasBanakh I am interested to hear this construction. | |
| Jan 6, 2020 at 17:38 | comment | added | Rohil Prasad | @Taras We should assume that these embeddings are smooth - Mike's picture can be modified suitably so I think this is not restrictive at all. | |
| Jan 6, 2020 at 17:36 | comment | added | Rohil Prasad | @Mike Yes, this picture perfectly clarifies the idea, thanks! | |
| Jan 5, 2020 at 10:22 | comment | added | Taras Banakh | What is the topology on the union? One can construct an increasing sequence $(S_k)$ of surfaces, homeomorphic to the closed unit disk $\{z\in\mathbb C:|z|\le 1\}$, such that for any reasonable topology on the union $S=\bigcup_{n=1}^\infty S_n$, the space $S$ is not locally compact at some point $x$ that belongs to the boundaries of all surfaces $S_n$. So, $S$ is not a surface. | |
| Jan 3, 2020 at 22:14 | comment | added | mme | Does this picture clarify the idea? I have drawn all of the discs $S_i$ as topological submanifolds of the unit disc, each nested in the next. You can see how the portion of the boundary not contained in $S_i$ undergoes the middle-thirds construction of the Cantor set. photos.app.goo.gl/WEmkhNCQgeepSpYX7 I would be surprised if the answer was much different if you demanded smooth embeddings, though it would make those corners a little more irritating. | |
| Jan 3, 2020 at 6:54 | comment | added | mme | @Rohil Minus a Cantor set on the boundary. I can try to draw a picture of the suggestion tomorrow. Lee has pointed out a subtlety I missed above. | |
| Jan 3, 2020 at 5:57 | comment | added | Lee Mosher | You have to be more careful about boundary components, or else what you say about deformation retracts is false. You can have an example where each odd surface $S_{2n+1}$ is a disc, each even surface $S_{2n}$ is an annulus, $S_{2n}$ is obtained from $S_{2n-1}$ by attaching a pair-of-pants (a 3-holed sphere) to the unique component of $\partial S_{2n-1}$, and $S_{2n+1}$ is obtained from $S_{2n}$ by attaching a disc to one component of $\partial S_{2n}$. | |
| Jan 3, 2020 at 5:45 | comment | added | Rohil Prasad | How would you construct the disk minus a Cantor set? For me this looks like the infinite type surface that looks like an infinite binary tree, which seems to be impossible since we're never adding any boundary components for large $k$. | |
| Jan 3, 2020 at 5:40 | history | edited | Rohil Prasad | CC BY-SA 4.0 |
fixed formatting on italics
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| Jan 3, 2020 at 0:44 | comment | added | mme | So long as the inclusion $f: S_k \hookrightarrow S_{k+1}$ has $f(\partial S_k) \cap \partial S_{k+1}$ a union of connected components of $\partial S_{k+1}$, all you do at each stage is glue on a cylinder, so you get exactly what you say. In general I imagine you can also cook up arbitrarily nasty punctures on the boundary; certainly the upper half plane is such a colimit, but I think so is the disc minus a Cantor set from the boundary. | |
| Jan 3, 2020 at 0:14 | history | asked | Rohil Prasad | CC BY-SA 4.0 |