Timeline for Local to global complexity of triangulations
Current License: CC BY-SA 4.0
23 events
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| S Jan 17 at 0:08 | history | bounty ended | CommunityBot | ||
| S Jan 17 at 0:08 | history | notice removed | CommunityBot | ||
| Jan 9 at 22:00 | history | edited | rab | CC BY-SA 4.0 |
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| Jan 9 at 22:00 | comment | added | rab | No no not at all, I am specifically using $\Delta$-complexes here - there's an additive constant which I didn't mention in the post because I was trying to highlight the $2\chi$ "order of magnitude" so to speak. The conjectured bound is specifically $2\chi(M)-2$ (which the 2-simplex $S^4$ attains). And yes we're assuming closed here (without boundary). | |
| Jan 9 at 8:52 | comment | added | Sam Nead | Oh, and we need “without boundary” as otherwise the four-ball is a counter-example. | |
| Jan 9 at 7:24 | comment | added | Sam Nead | I think that you are silently assuming that your triangulations are simplical. For example, you don’t allow pseudo-triangulations or $\Delta$-structures? For, consider doubling a four-simplex across its boundary. This gives a “triangulation” of the four-sphere with two four-simplices (which is smaller than $2 \chi(S^4)$). | |
| Jan 8 at 23:11 | history | edited | rab | CC BY-SA 4.0 |
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| S Jan 8 at 23:07 | history | bounty started | rab | ||
| S Jan 8 at 23:07 | history | notice added | rab | Current answers are outdated | |
| Jan 8 at 13:59 | comment | added | Sam Nead | Ah, in my third comment (just above) I should have said "$n-1$ simplices in all of the links". This is the same as the number of $n$ simplices in the stars. | |
| Jan 8 at 5:00 | history | edited | rab | CC BY-SA 4.0 |
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| Jan 8 at 1:59 | history | edited | rab | CC BY-SA 4.0 |
Rewrote the question giving all relevant details for the problem.
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| Jan 7 at 21:34 | comment | added | rab | Ah ok I just realised from your comment I probably shouldn't be comparing complexities (as I've defined it) like "$K\leq c(\mathrm{St}(v))\Rightarrow K\leq c(M)$" since as you pointed out, the complexity of the star is 1, and this results in an utterly vacuous statement --- rather I want to compare the number of $n$-simplices (not necessarily the complexity of star v) of $\mathrm{St}(v)$ in any triangulation of $M$ to the complexity of M. I'll edit the question to add some more details. | |
| Jan 7 at 18:08 | comment | added | Sam Nead | Ok, the following is true - you can count the number of $n-1$ simplices in all of the stars, add up, divide by $n+1$, and so obtain the number of $n$ simplices in the given triangulation. (So, if the complexity of the manifold is large, then any triangulation will need either lots of vertices, or some vertices with large stars.) But this is a combinatorial fact (a version of the hand-shaking lemma), not a topological one... | |
| Jan 7 at 18:01 | comment | added | Sam Nead | If I understand your definitions, the "star" of a vertex is always an $n$-ball and thus has "complexity" equal to one. So $K$ equals one, and so you are hoping to lower bound the complexity of a manifold by "one". | |
| Jan 6 at 22:08 | history | edited | rab | CC BY-SA 4.0 |
Added precise desire for estimate of M
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| Jan 6 at 21:58 | comment | added | rab | Hi all, my apologies for the initial vagueness, I've edited the question to hopefully be a bit more concrete. @HJRW - see my remarks regarding the sum over all verts approach. Also regarding your GB comment - yes this had occurred to me that maybe we could turn this into a geometry problem via (generalised) Chern-Gauss-Bonnet - but then I fear I'd be at even more of loss as to how to proceed haha 😅 | |
| Jan 6 at 21:54 | history | edited | rab | CC BY-SA 4.0 |
Elaborated on various details.
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| Jan 6 at 11:26 | comment | added | Sam Nead | Your question is too vague to have any answer. But it is attracting enough attention to suggest that this is an interesting area. You should rewrite to make the question more specific and then you will probably get useful answers. | |
| Jan 6 at 9:41 | comment | added | HJRW | By the way, the combinatorial Gauss—Bonnet theorem for triangulated surfaces is a nice example of the sort of phenomenon you seem to be talking about. Identify each face with a Euclidean triangle. Define the curvature of each vertex to be $2\pi$ minus the sum of the angles at the vertex. Then the sum of the curvatures of the vertices is $2\pi$ times the Euler characteristic! | |
| Jan 6 at 9:23 | comment | added | HJRW | As stated, I don’t see why you can’t just sum the local complexity over the vertices to get a global complexity. In its current form, the question seems too vague to answer. | |
| Jan 5 at 22:08 | comment | added | Calvin McPhail-Snyder | I'm not sure there's an answer to your question unless you're more specific about the kind of property that you're interested in. In general in topology the relationship between the complexity of a combinatorial representation (degree of vertices or number of simplices in a triangulation, crossings in a knot diagram, genus of a Heegard splitting) and other measures of complexity (knot genus, hyperbolic volume, etc.) is a difficult problem. | |
| Jan 5 at 21:46 | history | asked | rab | CC BY-SA 4.0 |