This basketball is combinatorially equivalent to?
Recent Comments

Recent Posts
 AviFest, AviStories and Amazing Cash Prizes.
 Polymath 10 post 6: The ErdosRado sunflower conjecture, and the Turan (4,3) problem: homological approaches.
 Polymath 10 Emergency Post 5: The ErdosSzemeredi Sunflower Conjecture is Now Proven.
 Mind Boggling: Following the work of Croot, Lev, and Pach, Jordan Ellenberg settled the cap set problem!
 More Math from Facebook
 The Erdős Szekeres polygon problem – Solved asymptotically by Andrew Suk.
 The Quantum Computer Puzzle @ Notices of the AMS
 Three Conferences: Joel Spencer, April 2930, Courant; Joel Hass May 2022, Berkeley, Jean Bourgain May 2124, IAS, Princeton
 Math and Physics Activities at HUJI
Top Posts & Pages
 Believing that the Earth is Round When it Matters
 Answer: Lord Kelvin, The Age of the Earth, and the Age of the Sun
 Polymath 10 Emergency Post 5: The ErdosSzemeredi Sunflower Conjecture is Now Proven.
 Seven Problems Around Tverberg's Theorem
 AviFest, AviStories and Amazing Cash Prizes.
 Amazing: Peter Keevash Constructed General Steiner Systems and Designs
 The KadisonSinger Conjecture has beed Proved by Adam Marcus, Dan Spielman, and Nikhil Srivastava
 A Breakthrough by Maryna Viazovska Leading to the Long Awaited Solutions for the Densest Packing Problem in Dimensions 8 and 24
 About Conjectures: Shmuel Weinberger
RSS
Two cellular decompositions A and B are combinatorially equivalent if there is an order preserving bijection between the faces (of all dimensions) of A and the faces of B.
Intuitively this is surprising to me because the faces of the basketball and those of the octahedron have different numbers incident edges. What does “order preserving” mean in this context. I don’t think I understand the definition. What is the order which must be preserved?
Dear Louigi, every face of the basketbal is incident to three (curved) edges.
Another way to describe combinatorial equivalence is that there is a bijection between the vertices of A to the vertices of B that induces a map between the edges of A to the edges of B and the faces of A to the faces of B. It is easier to see if you hold a basketball and follow the edges and not just look at the picture.
To see from the picture that every face has three edges requires some imagining of the hidden part; but you see easily that every vertex belongs to 4 edges and that there are 6 vertices. Therefore you conclude that there are 12 edges. (8 streight and 4 curved.) It takes a bit of imagining to realize that every pair of vertices are the end points of at most one edge. In the picture you see (partially) eight faces and it is not hard to be convinced that these are all and they are all distinct (instrad you can use Euler formula). This is already enough information.
Of course, every professional basketball player is fully aware of this isomorphism.
Thanks, I was imagining the wrong hidden part. It’s easy to extend the above picture to get a “basketball” that has four faces that are only incident to two edges, and four faces that are incident to four edges. But that’s not what a real basketball is like.
Although I noticed it a long time ago, without a real basketball at hand it is very confusing now. I think all 8 faces of the basketball are isometric but there are 2 types of vertices and 3 types of edges. So an advanced excercise is: what is the symmetry of the basketball? Is there a convex polyhedron which is combinatorially equivalent and has precisely the same symmetry?
Gil,
the symmetry group is of order 8 (knowing where one curved edge goes fixes the isometry). Is not having all the faces isometric the key (for production purposes)? That also explains why they are not the (spherical) triangles (that is why basketballs are not so obviously octahedra): one wants to keep the width of the pieces as small as possible, to ease the material stress caused by the mismatch of the curvatures.
Dear Yuliy, that’s interesting.
Here is another link I found about it
http://www.isama.org/hyperseeing/06/0611.pdf
(I suppose I do not know the asnwer to the question I asked about finding a combinatorially isomorphic polytope with the same symmetry group.)
actually the black lines B drawn on the basketball are homotopical to 3 circles in generic position (this is just a restatement but may sound like another counterintuitive sentence XD)
and they are also equal to 2 “equators” E1,E2 and one “tennis ball drawing” T.
Actually T has not many symmetries: the points with maximal curvature are 4, and fixing
where one of them goes
if we preserve or reverse the orientation on T
fixes the selfmap of B on T. Also where E_1, E_2 go will be fixed.This shows that a tennis ball has the same 8 symmetries as the basket ball.
I think you can deform the octahedron to obtain the same symmetries: consider the convex hull of the six points
2,0,0
1,0,1
1,0,1
1,1,0
1,1,0
2,0,0
..this should work, right? recall that the regular octahedron is given by
1,0,0
0,0,1
0,0,1
0,1,0
0,1,0
1,0,0
so we just “shifted triples of edges” to kill symmetries.
Dear Gil,
Although I noticed it a long time ago, without a real basketball at hand it is very confusing now. I think all 8 faces of the basketball are isometric but there are 2 types of vertices and 3 types of edges. So an advanced excercise is: what is the symmetry of the basketball? Is there a convex polyhedron which is combinatorially equivalent and has precisely the same symmetry?
As regards the second question, since one has a planar 3connected graph here why can’t one just apply Peter Mani’s Theorem to realize the automorphism group of this graph as the regular octahedron? However, for subgroups of the automorphism group one may not be able to realize these with 3polytopes having that exact symmetry and no more.
Cheers,
Joe
Dear Mircea, it looks promising, but I cannot see it right a way but will have to think about it.
Dear Joseph, I had a similar line of thought, first I thought that such a polytope can be realized because of Peter Mani’s theorem that asserts that every 3connected planar graph can be realized by a polyhedran so that the group of automorphisms of the graph is the group of summetries of the polytopes. (BTW, there is a simple argument for this theorem based on KoebeAndreevThurston circle packing theorems, and which allow to look at even larger group wher polarity are added.)
Then I realized that we want to realize only a specified subgroup as the group of symmetries and this does not follow from Mani’s theorem. I remember seeing once a paper about a systematic study of non regular cubes and what kind of symmetries they may have, but I forgot where.
Dear Gil,
actually afterwards I realized that the construction is equivalent to fixing the crossing points, on the basketball, and using them as vertices of our polyhedron…does it look better this way?
The symmetries are the same as those of a streched tetrahedron (where you take opposite sides and move them away): you don’t really need all six vertices to realize the symmetries.. 2 of them are there just to give combinatorial equivalence.
Dear Mircae, yes, yes this looks right.
Dear Gil,
Yes, I noted Mani’s result only applies to the full automorphism group.
The graph of the 4cycle has a group of order 8 which can be realized as a metric square with a symmetry group of order 8, However, the rotation subgroup of order 4 can not be realized by a metrical 4 sided polygon whose symmetry group only has order 4.
The same issue comes up in 3dimensions.
Best,
Joe
Dear Joseph, that’s a good example to know, thanks! And this was the longest discussions thread over my blog so far, and a very interesting one.
If you dont mind me asking I was wondering, do you mind telling me where you aquired theme from?
Great Blog, Dude! I am constantly on the watch for new and interesting sports sites and postings… which is what led me here. I certainly plan on visiting again! Adios
a cool post there mate respect to post author cool you made some clear points there