Wikipedia:Reference desk/Archives/Mathematics/2021 October 28

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October 28

9-coloured genus-3 complete map

8-coloured genus-2 complete map

For the Heawood conjecture article, I had previously drawn this diagram based on "Tietze's Berühmte Mathematische Probleme, originally published in 1911. It appears in (and on the cover of) the translation, Famous Problems of Mathematics, Graylock Press, New York; 2nd edition (1965)."

I would like to draw a similar one for genus 3. http://www.ams.org/publicoutreach/feature-column/fc-2018-05 gives a hint of adjacent regions as a table: http://www.ams.org/images/fcarc-may2018-Heffter-genus-3.jpg and proposes a challenge:

We encourage the reader to try to construct a more geometrically intelligible complete map with nine regions on the genus-three surface, and perhaps to render it in crochet.

Would anyone have a method to translate that into a legible diagram?

Thanks,
cmɢʟee⎆τaʟκ 21:52, 28 October 2021 (UTC)

• That looks like it's been inspired by a Mayan/Aztec headdress. -- Jack of Oz ^{[pleasantries]} 21:36, 29 October 2021 (UTC)
  • Perhaps 😅 Would you have an answer to the question, though? cmɢʟee⎆τaʟκ 00:20, 31 October 2021 (UTC)

• • • Seems like you could layout the regions on a genus 3 surface by brute force search? 9 regions isn't all that many. Then you'd want to munge the result around til it looked nice. I don't fully understand the notation in that table: what do the little arcs connecting some of the numbers mean? 2601:648:8202:350:0:0:0:D4A (talk) 00:48, 31 October 2021 (UTC)

      It is slightly confusing that the terminology in the journal article from which the table is copied is dual to that of the web page; the web page's regions are vertices in Hefter's article, and his faces are the web page's vertices. I'll stick to the terminology of the web page; then there are 9 regions, numbered 1 through 9. Then, for example ,

      $6)~7~1~5_{\mathbf {\!\,\smile } }\!2~4~9~8~3$

      means that region $6$ borders the eight regions $7$, $1$, $5$, $2$, $4$, $9$, $8$ and $3$ in that order along its perimeter (presumably in the same orientation for all lines of the table), that the three regions $6$, $7$ and $1$ border each other and so meet in a tri-valent vertex, and that $6$ borders $5$ and $2$, but that $5$ and $2$ do not border each other; the three meet in the six-valent vertex, so around that vertex there are some more regions wedged in between $5$ and $2$. (The last line implies that two different corners of region $2$ contribute to the six-valence.)  --Lambiam 12:10, 31 October 2021 (UTC)

@Cmglee: — Perhaps the following is somewhat helpful. I imagine the edges forming the perimeter of a region being traversed anti-clockwise, viewed from the interior of the region. Let

${\overset {a}{\underset {b}{\longrightarrow }}}$

represent an edge of the perimeter of region $a$, bordering $b$, traversed anti-clockwise. If the next edge along the perimeter is

${\overset {a}{\underset {c}{\longrightarrow }}},$

and the inbetween vertex is trivalent, let us denote it by

$({}_{b}{}^{\!a}{}_{\!c}).$

Putting the edges together, showing the shared vertex, we obtain the picture

${\overset {a}{\underset {b}{\longrightarrow }}}({}_{b}{}^{\!a}{}_{\!c}){\overset {a}{\underset {c}{\longrightarrow }}}.$

(Think of a clock face on region $a$; then this vertex is at six o'clock.) Note that $({}_{b}{}^{\!a}{}_{\!c}),$ $({}_{c}{}^{\!b}{}_{\!a})$ and $({}_{a}{}^{\!c}{}_{\!b})$ are the same vertex viewed from different regions. To obtain a view-independent, simple notation for the 22 trivalent vertices, we'll use instead the first 22 capital Greek letters:

$~~1:(\mathrm {A} )=({}_{3}{}^{\!1}{}_{\!4})~~~~~~~~~~12:(~\!\!\mathrm {M} ~\!\!)=({}_{9}{}^{\!2}{}_{\!7})$

$~~2:(\mathrm {B} )=({}_{4}{}^{\!1}{}_{\!5})~~~~~~~~~~13:(\mathrm {N} )=({}_{5}{}^{\!3}{}_{\!9})$

$~~3:(\Gamma )=({}_{5}{}^{\!1}{}_{\!6})~~~~~~~~~~14:(\Xi )=({}_{6}{}^{\!3}{}_{\!8})$

$~~4:(~\!\!\Delta ~\!\!)=({}_{6}{}^{\!1}{}_{\!7})~~~~~~~~~~15:(\mathrm {O} )=({}_{7}{}^{\!3}{}_{\!6})$

$~~5:(\mathrm {E} )=({}_{7}{}^{\!1}{}_{\!8})~~~~~~~~~~16:(\Pi )=({}_{8}{}^{\!3}{}_{\!5})$

$~~6:(\mathrm {Z} )=({}_{8}{}^{\!1}{}_{\!9})~~~~~~~~~~17:(\mathrm {P} )=({}_{9}{}^{\!3}{}_{\!4})$

$~~7:(\mathrm {H} )=({}_{9}{}^{\!1}{}_{\!2})~~~~~~~~~~18:(\Sigma )=({}_{7}{}^{\!4}{}_{\!5})$

$~~8:(\Theta )=({}_{4}{}^{\!2}{}_{\!6})~~~~~~~~~~19:(\mathrm {T} )=({}_{8}{}^{\!4}{}_{\!7})$

$~~9:(\,\mathrm {I} \,)=({}_{5}{}^{\!2}{}_{\!8})~~~~~~~~~~20:(\Upsilon )=({}_{9}{}^{\!4}{}_{\!6})$

$10:(\mathrm {K} )=({}_{7}{}^{\!2}{}_{\!3})~~~~~~~~~~21:(\Phi )=({}_{7}{}^{\!5}{}_{\!9})$

$11:(\Lambda )=({}_{8}{}^{\!2}{}_{\!4})~~~~~~~~~~22:(\mathrm {X} )=({}_{9}{}^{\!6}{}_{\!8})$

The notation $(\Omega )$ is reserved for the vertex of valence six.

Using this notation, the edge walks along the perimeters of each of the nine octagonal regions are then given by:

$1:~(\mathrm {H} ){\overset {1}{\underset {2}{\longrightarrow }}}(\Omega ){\overset {1}{\underset {3}{\longrightarrow }}}(\mathrm {A} ){\overset {1}{\underset {4}{\longrightarrow }}}(\mathrm {B} ){\overset {1}{\underset {5}{\longrightarrow }}}(\Gamma ){\overset {1}{\underset {6}{\longrightarrow }}}(~\!\!\Delta ~\!\!){\overset {1}{\underset {7}{\longrightarrow }}}(\mathrm {E} ){\overset {1}{\underset {8}{\longrightarrow }}}(\mathrm {Z} ){\overset {1}{\underset {9}{\longrightarrow }}}(\mathrm {H} )$

$2:~(\Theta ){\overset {2}{\underset {6}{\longrightarrow }}}(\Omega ){\overset {2}{\underset {1}{\longrightarrow }}}(\mathrm {H} ){\overset {2}{\underset {9}{\longrightarrow }}}(~\!\!\mathrm {M} ~\!\!){\overset {2}{\underset {7}{\longrightarrow }}}(\mathrm {K} ){\overset {2}{\underset {3}{\longrightarrow }}}(\Omega ){\overset {2}{\underset {5}{\longrightarrow }}}(\,\mathrm {I} \,){\overset {2}{\underset {8}{\longrightarrow }}}(\Lambda ){\overset {2}{\underset {4}{\longrightarrow }}}(\Theta )$

$3:~(\mathrm {P} ){\overset {3}{\underset {4}{\longrightarrow }}}(\mathrm {A} ){\overset {3}{\underset {1}{\longrightarrow }}}(\Omega ){\overset {3}{\underset {2}{\longrightarrow }}}(\mathrm {K} ){\overset {3}{\underset {7}{\longrightarrow }}}(\mathrm {O} ){\overset {3}{\underset {6}{\longrightarrow }}}(\Xi ){\overset {3}{\underset {8}{\longrightarrow }}}(\Pi ){\overset {3}{\underset {5}{\longrightarrow }}}(\mathrm {N} ){\overset {3}{\underset {9}{\longrightarrow }}}(\mathrm {P} )$

$4:~(\Sigma ){\overset {4}{\underset {5}{\longrightarrow }}}(\mathrm {B} ){\overset {4}{\underset {1}{\longrightarrow }}}(\mathrm {A} ){\overset {4}{\underset {3}{\longrightarrow }}}(\mathrm {P} ){\overset {4}{\underset {9}{\longrightarrow }}}(\Upsilon ){\overset {4}{\underset {6}{\longrightarrow }}}(\Theta ){\overset {4}{\underset {2}{\longrightarrow }}}(\Lambda ){\overset {4}{\underset {8}{\longrightarrow }}}(\mathrm {T} ){\overset {4}{\underset {7}{\longrightarrow }}}(\Sigma )$

$5:~(\Omega ){\overset {5}{\underset {6}{\longrightarrow }}}(\Gamma ){\overset {5}{\underset {1}{\longrightarrow }}}(\mathrm {B} ){\overset {5}{\underset {4}{\longrightarrow }}}(\Sigma ){\overset {5}{\underset {7}{\longrightarrow }}}(\Phi ){\overset {5}{\underset {9}{\longrightarrow }}}(\mathrm {N} ){\overset {5}{\underset {3}{\longrightarrow }}}(\Pi ){\overset {5}{\underset {8}{\longrightarrow }}}(\,\mathrm {I} \,){\overset {5}{\underset {2}{\longrightarrow }}}(\Omega )$

$6:~(\mathrm {O} ){\overset {6}{\underset {7}{\longrightarrow }}}(~\!\!\Delta ~\!\!){\overset {6}{\underset {1}{\longrightarrow }}}(\Gamma ){\overset {6}{\underset {5}{\longrightarrow }}}(\Omega ){\overset {6}{\underset {2}{\longrightarrow }}}(\Theta ){\overset {6}{\underset {4}{\longrightarrow }}}(\Upsilon ){\overset {6}{\underset {9}{\longrightarrow }}}(\mathrm {X} ){\overset {6}{\underset {8}{\longrightarrow }}}(\Xi ){\overset {6}{\underset {3}{\longrightarrow }}}(\mathrm {O} )$

$7:~(\mathrm {T} ){\overset {7}{\underset {8}{\longrightarrow }}}(\mathrm {E} ){\overset {7}{\underset {1}{\longrightarrow }}}(~\!\!\Delta ~\!\!){\overset {7}{\underset {6}{\longrightarrow }}}(\mathrm {O} ){\overset {7}{\underset {3}{\longrightarrow }}}(\mathrm {K} ){\overset {7}{\underset {2}{\longrightarrow }}}(~\!\!\mathrm {M} ~\!\!){\overset {7}{\underset {9}{\longrightarrow }}}(\Phi ){\overset {7}{\underset {5}{\longrightarrow }}}(\Sigma ){\overset {7}{\underset {4}{\longrightarrow }}}(\mathrm {T} )$

$8:~(\mathrm {X} ){\overset {8}{\underset {9}{\longrightarrow }}}(\mathrm {Z} ){\overset {8}{\underset {1}{\longrightarrow }}}(\mathrm {E} ){\overset {8}{\underset {7}{\longrightarrow }}}(\mathrm {T} ){\overset {8}{\underset {4}{\longrightarrow }}}(\Lambda ){\overset {8}{\underset {2}{\longrightarrow }}}(\,\mathrm {I} \,){\overset {8}{\underset {5}{\longrightarrow }}}(\Pi ){\overset {8}{\underset {3}{\longrightarrow }}}(\Xi ){\overset {8}{\underset {6}{\longrightarrow }}}(\mathrm {X} )$

$9:~(~\!\!\mathrm {M} ~\!\!){\overset {9}{\underset {2}{\longrightarrow }}}(\mathrm {H} ){\overset {9}{\underset {1}{\longrightarrow }}}(\mathrm {Z} ){\overset {9}{\underset {8}{\longrightarrow }}}(\mathrm {X} ){\overset {9}{\underset {6}{\longrightarrow }}}(\Upsilon ){\overset {9}{\underset {4}{\longrightarrow }}}(\mathrm {P} ){\overset {9}{\underset {3}{\longrightarrow }}}(\mathrm {N} ){\overset {9}{\underset {5}{\longrightarrow }}}(\Phi ){\overset {9}{\underset {7}{\longrightarrow }}}(~\!\!\mathrm {M} ~\!\!)$

This fixes the topology. Embedding it in 3D is a different matter. It may be best to start with stitching the regions $4,~7,~8$ and $9$ together; they all have only trivalent corner vertices. Region $2$ is an obvious candidate for being stitched on last; it meets itself, having two of its corners on $(\Omega )$. (Or perhaps it is better to go the other way and start with $2$ and the other four regions meeting in $(\Omega ),$ regions $1,~3,~5$ and $6.$)  --Lambiam 14:00, 2 November 2021 (UTC)

Note that regions $1,~5$ and $6$ also meet in $(\Gamma ),$ which is adjacent to $(\Omega ).$  --Lambiam 19:29, 2 November 2021 (UTC)

@Lambiam: Thank you so much for your thorough analysis and beautiful presentation. I'm still unsure how to proceed after sketching layouts for regions 4, 7, 8, 9 and 2. There are many repeated regions, in particular, 3, and I've no idea how to join them together. I also can't see how there can be a vertex with more than 3 regions (trivalent): doesn't that mean regions diagonally opposite just touch at the vertex and are not considered adjacent?

Instead, I tried adapting the genus 2 diagram and am short of just one region to connect. If I add a region where red, pink and yellow regions meet (between the second and third bubbles in the top left) and another region between the dark blue and green regions (the gap between the third and fourth bubbles in the top right), and join them together with a bridge (giving genus 3), I connect all 8 regions except the orange one, the largest region covering most of the back of the object. Can this be improved or is it a dead end?

Thanks,
cmɢʟee⎆τaʟκ 00:15, 4 November 2021 (UTC)

Mathematicians realized early on that the four-color theorem could be expressed in the language of graph theory, avoiding all kinds of potential ambiguity or weirdness surrounding the notions of "region" and "adjacency of regions". The concept of chromatic number is defined in purely graph-theoretical terms. To apply this to the four-color theorem, you need the notion of a planar graph. It has an obvious generalization to arbitrary two-dimensional manifolds, such as the torus, the Klein bottle, and surfaces of any genus, whether orientable or non-orientable. When a graph is embedded in a surface such that edges intersect only at their endpoints, it divides the surface into regions, usually (the interiors of topological) polygons. There is no ambiguity if some of these polygons have more than three edges; adjacency of vertices is explicitly defined in the original graph. It is only when switching back to the dual viewpoint, in which vertices become regions and regions become vertices, that a potential ambiguity appears. The original formulations of the four-colour conjecture defined adjacency as having a common boundary line, so an $N$-point with $N{\,>\,}3$ does not introduce additional adjacencies. For crocheting, the graph view is not very interesting, but for construction on a surface of genus 3, it may be the simpler approach – I just don't know. Perhaps another reader here can help you with your last question.  --Lambiam 09:22, 4 November 2021 (UTC)

Thanks again, Lambiam. I'm aware of the region-vertex equivalence. Just wondering that more than 3 edges from a vertex means that some regions are adjacent only at a point. Anyway, I appreciate your effort. Cheers, cmɢʟee⎆τaʟκ 19:35, 4 November 2021 (UTC)

P.S. Not exactly what i wanted, but close:

• http://youtube.com/watch?v=8yGarv8GXL8
• http://icerm.brown.edu/topical_workshops/tw-20-glu/posters/Stewart-poster.pdf

@Cmglee:


Parcly Taxel 12:42, 8 October 2024 (UTC)

Thanks a lot, @Parcly Taxel:. I've redrawn it with tunnels highlighted and without pseudoquadripoints. Is that correct?

A redrawing of Parcly Taxel's drawing of a complete map of 8 regions on a triple torus (genus-three surface) with each region touching every other. The black blobs are ends of their respective tunnels.

Cheers, cmɢʟee⎆τaʟκ 20:26, 8 October 2024 (UTC)

I was going to try and render my map in Blender, but yes, your redrawing is correct.

Note that this is obviously not Heffter's embedding since it has three 4-faces instead of one 6-face. I found both embeddings using trimbed.py here; the other files in that gist together form a program for automatically constructing minimal-genus complete graph embeddings at any number of vertices, thus proving the Heawood conjecture for orientable surfaces. Parcly Taxel 01:37, 9 October 2024 (UTC)

Thanks for explaining and reviewing it. Good to know. I look forward to seeing your Blender render. Cheers, cmɢʟee⎆τaʟκ 08:30, 9 October 2024 (UTC)