Timeline for How many non-homeomorphic surfaces arise from these graphs?
Current License: CC BY-SA 3.0
21 events
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| Jan 24, 2018 at 23:57 | history | edited | Christopher King | CC BY-SA 3.0 |
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| Jan 24, 2018 at 23:57 | comment | added | Christopher King | @LeeMosher Oh, yeah. Whoops! | |
| Jan 24, 2018 at 22:12 | comment | added | Lee Mosher | Something is off... you say "we count self-loops as two edges"... so does that mean you count "four self-loops" as eight edges? | |
| Jan 24, 2018 at 0:56 | answer | added | Igor Rivin | timeline score: 4 | |
| Jan 23, 2018 at 23:33 | comment | added | Joseph O'Rourke | This somewhat related question, "Fundamental polygons with infinite pairwise identifications," is notable because Bill Thurston posted an answer. Any insight into his reasoning is worthwhile. | |
| Jan 23, 2018 at 20:37 | comment | added | j.c. | I've edited the title since a "topology" usually means a collection of open sets satisfying certain conditions. | |
| Jan 23, 2018 at 20:35 | history | edited | j.c. | CC BY-SA 3.0 |
edit title, grammar, spelling
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| Jan 23, 2018 at 20:22 | comment | added | YCor | You seem to mean that you allow orientations on edges that are not cyclic. Then the number of possible choices for one vertex is $2^{d-1}(d-1)!$ which is still much larger. Here it means: $(d-1)!/2$ choices of cyclic ordering up to orientation, times $2^d$ choices of orientation. Actually, the huge product that comes is too big because changing the orientation of one edge and the incident edge does not change the resulting space. So one should get something such as $2^e\prod_v((\deg(v)-1)!/2)$, where $e$ is the total number of non-oriented edges. | |
| Jan 23, 2018 at 19:57 | comment | added | Christopher King | @YCor I mean choosing whether or not to twist the edge before you connect it. For example, a vertice with one self-loop can connect to form either an orientable or a non-orientable surface. So it has 2(2-1)!=2 different topologies. | |
| Jan 23, 2018 at 19:51 | comment | added | YCor | No more "2": the number of cyclic ordering up to orientation is $(d-1)!/2$. The number of cyclic orderings (oriented) is $(d-1)!$, not $2(d-1)!$. | |
| Jan 23, 2018 at 19:50 | history | edited | YCor |
edited tags
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| Jan 23, 2018 at 19:49 | comment | added | Christopher King | @YCor Does that clear things up? | |
| Jan 23, 2018 at 19:49 | history | edited | Christopher King | CC BY-SA 3.0 |
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| Jan 23, 2018 at 19:48 | comment | added | YCor | The number of cyclic orderings at a given vertex $v$ of degree $d$ is much more than $2d$. It's $d!/d=(d-1)!$. | |
| Jan 23, 2018 at 19:47 | comment | added | YCor | Ah that's the point. Please edit to clarify so that we can wipe out these comments. | |
| Jan 23, 2018 at 19:46 | comment | added | YCor | For each vertex, to define a polygon, you need some cyclic ordering of the set of edges at this vertex. So this sounds ill-defined. | |
| Jan 23, 2018 at 19:46 | comment | added | Christopher King | @YCor I mean if you connect them in different orders. For example, Imagine that $G$ is a vertex $v$ with four self loops. Then the topology could be a torus, klein bottle, projective plane, or even sphere. | |
| Jan 23, 2018 at 19:44 | history | edited | Christopher King | CC BY-SA 3.0 |
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| Jan 23, 2018 at 19:44 | comment | added | YCor | The answer is "1". Or maybe you mean, how many when $G$ varies... under which restriction? $G$ ranges over finite graphs? connected? with some restriction on the number of vertices?... | |
| Jan 23, 2018 at 19:44 | comment | added | Wojowu | (singular of vertices is vertex, just fyi) | |
| Jan 23, 2018 at 19:28 | history | asked | Christopher King | CC BY-SA 3.0 |