## triangulations of torus, general, and Euler number. (Hopefully more interesting/relevant)

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

Hi, everyone:

I have been going over some simplicial homology recently, hoping to get
some geometric insight that I don't know how to get from the algebraic machinery alone.

I have been trying to find the homology of the torus this way, i.e., by triangulating it ( i.e., finding a carrier for the torus), but the smallest triangulation I have been able to do , has 18 triangles/faces --I checked it works; there are 8 vertices and 26 edges. Still: does anyone know of a simpler triangulation, ie., one with a smaller total number of triangles (and, of course, fewer vertices and edges resp.). ?

I had tried the long shot of solving the very simple equation:

V-E+F =0

in positive integers.

but this alone does not seem to help . Any ideas.?. Any ideas for finding minimal triangulations of surfaces, or higher-dimensional manifolds.?

Thanks.

-
My apology. I mistakenly, and carelessly, entered triangulated categories as tags. Sorry. – Herb Apr 25 2010 at 3:53
This can be found in an undergraduate textbook titled "Basic Concepts of Algebraic Topology" by Fred Croom. We really, really shouldn't be working standard questions from algebraic topology classes here. And yet, most of the algebraic topology questions are things I assign as exercises in my graduate classes. – Charlie Frohman Apr 26 2010 at 17:36
Well, the question about "finding minimal triangulations of ... higher-dimensional manifolds" is not at all trivial. – John Palmieri Apr 26 2010 at 18:08

1
The pattern is pretty clear. Low points, because they just created an account, name that can't be traced back to an individual. Homework question dressed up to look like a little more. – Charlie Frohman Apr 26 2010 at 18:44
To Charlie Frohman: This is not a homework question. I am computing the actual simplicial homology of spaces to get insights I do not know how to get by using the algebraic machinery alone (e.g., with simplicial homology) As to not stating my name, I have to admit I feel somewhat intimidated in this forum, being a first-year student at a school other than one of the top 10, specially after having read the resumes/CV's of many here. If this is against MO policy, I apologize, and I will drop out. – Herb Apr 28 2010 at 4:11
show 2 more comments

## 2 Answers

If you're just looking to glue triangles together along their edges, you can do it with two triangles, glued together to form a square, and then with opposite sides of the square glued to form a torus in the usual way. The resulting mesh has one vertex and three edges.

But if the triangles have to form a simplicial complex (meaning that the intersection of any two triangles is empty, a single vertex, or an edge) then I think the smallest mesh for a torus has 14 triangles, connected to each other in the pattern of the Heawood graph. The resulting mesh has seven vertices and 21 edges. It can be embedded into space as the Császár polyhedron.

-

3
Herb, if you want to employ David's "triangulation" for computing homology of the torus, take a look at Hatcher's algebraic topology book. He explains how to use Delta complexes in place of simplicial complexes (David's decomposition of the torus into two triangles is a Delta complex). This gives a theory in which it's easy to find (generalized) triangulations of spaces you may encounter, and also makes the resulting homology computations very clean. – Dan Ramras Apr 25 2010 at 23:22

### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

For the particular case of a simplicial complex structure for a torus, David Eppstein is right: the minimal triangulation has 7 vertices, 21 edges, and 14 triangles. For a sphere, the minimal triangulation has $(v,e,f) = (4, 6, 4)$. For a real projective plane, the minimal triangulation has $(v,e,f) = (6, 15, 10)$.

For the general situation of finding minimal triangulations of manifolds, Frank Lutz has written a nice preprint, and he also has some information and other references on The Manifold Page. There are plenty of unsolved problems in this area, it seems...

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9357728958129883, "perplexity_flag": "middle"}