I don't recall if we ever discussed the 11-cell, one of only two amazing 4-dimensional "abstract regular polytopes" (ARP) built from projective planes. See < http://en.wikipedia.org/wiki/11-cell > and < http://en.wikipedia.org/wiki/Abstract_polytope > and < http://en.wikipedia.org/wiki/Abstract_polytope#Regular_abstract_polytopes > for the precise definitions of ARP and 11-cell. But roughly, an ARP is a combinatorial generalization of a usual regular polytope that may seem too abstract at first but can grow on you. Anyhow, the 11-cell was discovered (apparently) independently by both Branko Grunbaum in 1977 and H.S.M. Coxeter in 1984. It consists of 11 projective planes P^2, represented as hemi-icosahedra -- the quotient of a regular icosahedron by its antipodal map -- so each P^2 is 10 equilateral triangles. (In fact, as a combinatorial object, metric properties are irrelevant to the 11-cell, but I like to make everything as regular as possible.) Each triangular face of each P^2 is identified with another such face of another P^2, so the 11-cell has 55 triangular faces in all. When this is done right, the 11-cell has a combinatorial automorphism group that is transitive on flags. A flag is a quadruple (F, T, E, V) where F is any of the 11 P^2's (so-called "3-dimensional faces") of the 11-cell, T is any triangle of F, E is any edge of T, V is any vertex of E. So there are 11x10x3x2 = 660 automorphisms in all. Also, this bizarrely symmetrical object is self-dual. I mean, who would've thought there could be such a thing with *11* faces. (The only other such thing has 55 faces -- it's 55 hemi-dodecahedra.) I'd like to know more about the 11-cell as a topological object: What is its universal covering space, its homology and its homotopy? --Dan Sometimes the brain has a mind of its own.
On 5/24/11, Dan Asimov <dasimov@earthlink.net> wrote:
I don't recall if we ever discussed the 11-cell, one of only two amazing 4-dimensional "abstract regular polytopes" (ARP) built from projective planes.
Dan mentioned this in April 2007 --- but our discussion rapidly became rather technical, and took place off-list.
Also, this bizarrely symmetrical object is self-dual. I mean, who would've thought there could be such a thing with *11* faces. (The only other such thing has 55 faces -- it's 55 hemi-dodecahedra.)
Coxeter's other polytope has 57 cells.
I'd like to know more about the 11-cell as a topological object: What is its universal covering space, its homology and its homotopy?
Perhaps it's time to give more air to (some of) what was discussed then? Below, lightly edited, are notes I made at the time, sketching in particular a proof that the fundamental group of the 11-cell is trivial: I hope they're not too incomprehensibly terse! Fred Lunnon (*Def 1*) Combinatorial "road-map" representation of 11-cell --- facets indexed by "height" h = 0,...,10; addition h + [vector] increments every component modulo 11. Vertices: h + [0]; Edges: h + {[0,j], j = 1,...,5}; Faces: h + {[0,1,4], [0,1,7], [0,2,5], [0,2,7], [1,2,4]}; or h + {[1,2,5], [0,4,5], [2,4,7], [1,5,7], [4,5,7]}; Solids: S_h = h + [0,1,2,4,5,7]; The second set of faces is identical to the first: together the two lines specify the 10 faces contained in a solid. Vertices and edges contained in a face or solid are given by all possible 1- and 2-subsets. [Other incidences may be inferred from the vertex lists above; unnecessary for the task at hand, these are omitted for the present.] (*Def 2*) Symmetry group PSL(2, 11) representation by 11-permutations --- generators as products of cycles: a = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10), b = (0)(10)(2, 3, 6)(1, 8, 5)(4, 7, 9). Order 660, number of conjugacy classes 8. Automorphism group PGL(2, 11) has order 1320, no 11-permutation representation. [The complete subgroup structure, relevant to thorough study of the polytope, is omitted.] Alternative "string C-group" presentation by reflections e,f,g,h: e^2 = f^2 = g^2 = h^2 = (e*f)^5 = (f*g)^3 = (g*h)^5 = (e*g)^2 = (e*h)^2 = (g*h)^2 = (e*f*g)^5 = (f*g*h)^5 = 1. [see sect. 4 of D.~Leemans, E.~Schulte "Groups of type L_2(q) acting on polytopes" (2006) http://arxiv.org/pdf/math/0606660]. (*Def 3*) Polarity (duality) between edges and faces: [0,1] -> [3,4,7], [0,2] -> [5,7,10], [0,3] -> [1,9,10], [0,4] -> [1,5,6], [0,5] -> [2,4,9]. The order-12 symmetry subgroup fixing a face fixes also its polar edge. For this polytope there is no analogous polarity between vertices and solids. (*Lem 4*) The "road map" satisfies the conditions for an abstract polytope, with incidence statistics tabulated below --- number of n-facets (across) per m-facet (down). m\n -1 0 1 2 3 4 -1 1 11 55 55 11 1 0 1 1 10 15 6 1 1 1 2 1 3 3 1 2 1 3 3 1 2 1 3 1 6 15 10 1 1 4 1 11 55 55 11 1 [Note table symmetric under half-turn, indicating self-dual polytope.] There doesn't seem any way to verify this without a great deal of inspection; frustratingly since it was originally constructed manually in a single day, entirely from the assumption that there existed a polytope of rank 4 with the given table! (*Lem 5*) The symmetry group of this polytope is PSL(2, 11). At present verification of this employs (and relies upon) the group-theoretic symbolic algebra package Magma. However it should be possible to demonstrate a set of generators corresponding to the more intuitive "string C-group" presentation reflections. Notice that the presence of 11-cycle generator "a" ensures that a road-map property valid for some height h is valid for all h modulo 11: for instance, the polarity relation extends by addition of h to every face and edge. (*Lem 6*) Any loop of edges is homotopic to a sequence of tripods, each containing the basepoint vertex. Given a sequence of vertices defining a basepointed loop L of finite length k, say L = [a, b, c, d, ..., a]. To simplify the reasoning, a vertex may be repeated consecutively; also short cases with b, c, d, ... absent are implicitly included. There are now several possibilities: Case (i): L = [a] is a singleton, incapable of further reduction. Case (ii): a = b: let L' = [a, c, ...] of length k-1. Case (iii): a = c: let L' = [a, d, ...] of length k-2. Case (iv): {a, b, c} is a face (order irrelevant); let L' = [a, c, d, ...] of length k-1. Case (v): {a, b, c} is a tripod; let L' = [a, b, c, a]*[a, c, d, ..., a] the tail of length k-2. In each case for k > 1, L' is homotopic to L and comprises a shorter tail loop L", possibly prefixed by a tripod L' = T * L" as in case (v). Now substitute L" for L and iterate. Finally k = 1, and the original loop L is reduced to a homotopically equivalent product of tripods L = T * T' * T" * ... where each factor contains the basepoint. (*Thm 7*) The fundamental group of the 11-cell is trivial. tripod [1,2,6] = [0,1,4] - [0,1,7] + [0,4,5] - [0,7,8] + [0,5,9] - [0,8,9] + [1,2,8] - [1,4,10] + [1,5,6] - [1,5,7] + [1,8,9] + [1,9,10] + [2,6,8] - [4,5,8] - [4,8,10] - [5,6,8] + [5,7,10] - [5,9,10] + [7,8,10] = 0, since every RHS term is a face; Checks out over |Z; no computer involved. QED.
Doesn't the fact that the fundamental group is trivial imply that the 11-cell is homeomorphic to the 3-sphere? (This is implied by the Poincare Conjecture, now, I guess, to be called Perelman's Theorem.) Then it must be orientable. That would be sort of weird, because the surfaces of it's 3-cells are *not*orientable. (I think they are topologically spheres with one crosscap.) But they are all nicely embedded in an orientable 3-manifold. How is this even possible? What am I missing? On Tue, May 24, 2011 at 7:43 PM, Fred lunnon <fred.lunnon@gmail.com> wrote:
On 5/24/11, Dan Asimov <dasimov@earthlink.net> wrote:
I don't recall if we ever discussed the 11-cell, one of only two amazing 4-dimensional "abstract regular polytopes" (ARP) built from projective planes.
Dan mentioned this in April 2007 --- but our discussion rapidly became rather technical, and took place off-list.
Also, this bizarrely symmetrical object is self-dual. I mean, who would've thought there could be such a thing with *11* faces. (The only other such thing has 55 faces -- it's 55 hemi-dodecahedra.)
Coxeter's other polytope has 57 cells.
I'd like to know more about the 11-cell as a topological object: What is its universal covering space, its homology and its homotopy?
Perhaps it's time to give more air to (some of) what was discussed then? Below, lightly edited, are notes I made at the time, sketching in particular a proof that the fundamental group of the 11-cell is trivial: I hope they're not too incomprehensibly terse! Fred Lunnon
(*Def 1*) Combinatorial "road-map" representation of 11-cell --- facets indexed by "height" h = 0,...,10; addition h + [vector] increments every component modulo 11. Vertices: h + [0]; Edges: h + {[0,j], j = 1,...,5}; Faces: h + {[0,1,4], [0,1,7], [0,2,5], [0,2,7], [1,2,4]}; or h + {[1,2,5], [0,4,5], [2,4,7], [1,5,7], [4,5,7]}; Solids: S_h = h + [0,1,2,4,5,7]; The second set of faces is identical to the first: together the two lines specify the 10 faces contained in a solid. Vertices and edges contained in a face or solid are given by all possible 1- and 2-subsets. [Other incidences may be inferred from the vertex lists above; unnecessary for the task at hand, these are omitted for the present.]
(*Def 2*) Symmetry group PSL(2, 11) representation by 11-permutations --- generators as products of cycles: a = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10), b = (0)(10)(2, 3, 6)(1, 8, 5)(4, 7, 9). Order 660, number of conjugacy classes 8. Automorphism group PGL(2, 11) has order 1320, no 11-permutation representation. [The complete subgroup structure, relevant to thorough study of the polytope, is omitted.] Alternative "string C-group" presentation by reflections e,f,g,h: e^2 = f^2 = g^2 = h^2 = (e*f)^5 = (f*g)^3 = (g*h)^5 = (e*g)^2 = (e*h)^2 = (g*h)^2 = (e*f*g)^5 = (f*g*h)^5 = 1. [see sect. 4 of D.~Leemans, E.~Schulte "Groups of type L_2(q) acting on polytopes" (2006) http://arxiv.org/pdf/math/0606660].
(*Def 3*) Polarity (duality) between edges and faces: [0,1] -> [3,4,7], [0,2] -> [5,7,10], [0,3] -> [1,9,10], [0,4] -> [1,5,6], [0,5] -> [2,4,9]. The order-12 symmetry subgroup fixing a face fixes also its polar edge. For this polytope there is no analogous polarity between vertices and solids.
(*Lem 4*) The "road map" satisfies the conditions for an abstract polytope, with incidence statistics tabulated below --- number of n-facets (across) per m-facet (down). m\n -1 0 1 2 3 4 -1 1 11 55 55 11 1 0 1 1 10 15 6 1 1 1 2 1 3 3 1 2 1 3 3 1 2 1 3 1 6 15 10 1 1 4 1 11 55 55 11 1 [Note table symmetric under half-turn, indicating self-dual polytope.]
There doesn't seem any way to verify this without a great deal of inspection; frustratingly since it was originally constructed manually in a single day, entirely from the assumption that there existed a polytope of rank 4 with the given table!
(*Lem 5*) The symmetry group of this polytope is PSL(2, 11).
At present verification of this employs (and relies upon) the group-theoretic symbolic algebra package Magma. However it should be possible to demonstrate a set of generators corresponding to the more intuitive "string C-group" presentation reflections. Notice that the presence of 11-cycle generator "a" ensures that a road-map property valid for some height h is valid for all h modulo 11: for instance, the polarity relation extends by addition of h to every face and edge.
(*Lem 6*) Any loop of edges is homotopic to a sequence of tripods, each containing the basepoint vertex.
Given a sequence of vertices defining a basepointed loop L of finite length k, say L = [a, b, c, d, ..., a]. To simplify the reasoning, a vertex may be repeated consecutively; also short cases with b, c, d, ... absent are implicitly included. There are now several possibilities: Case (i): L = [a] is a singleton, incapable of further reduction. Case (ii): a = b: let L' = [a, c, ...] of length k-1. Case (iii): a = c: let L' = [a, d, ...] of length k-2. Case (iv): {a, b, c} is a face (order irrelevant); let L' = [a, c, d, ...] of length k-1. Case (v): {a, b, c} is a tripod; let L' = [a, b, c, a]*[a, c, d, ..., a] the tail of length k-2. In each case for k > 1, L' is homotopic to L and comprises a shorter tail loop L", possibly prefixed by a tripod L' = T * L" as in case (v). Now substitute L" for L and iterate. Finally k = 1, and the original loop L is reduced to a homotopically equivalent product of tripods L = T * T' * T" * ... where each factor contains the basepoint.
(*Thm 7*) The fundamental group of the 11-cell is trivial. tripod [1,2,6] = [0,1,4] - [0,1,7] + [0,4,5] - [0,7,8] + [0,5,9] - [0,8,9] + [1,2,8] - [1,4,10] + [1,5,6] - [1,5,7] + [1,8,9] + [1,9,10] + [2,6,8] - [4,5,8] - [4,8,10] - [5,6,8] + [5,7,10] - [5,9,10] + [7,8,10] = 0, since every RHS term is a face; Checks out over |Z; no computer involved. QED.
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On Saturday 28 May 2011 02:21:20 Allan Wechsler wrote:
Doesn't the fact that the fundamental group is trivial imply that the 11-cell is homeomorphic to the 3-sphere? (This is implied by the Poincare Conjecture, now, I guess, to be called Perelman's Theorem.) Then it must be orientable.
That would be sort of weird, because the surfaces of it's 3-cells are *not*orientable. (I think they are topologically spheres with one crosscap.) But they are all nicely embedded in an orientable 3-manifold. How is this even possible? What am I missing?
My interpretation of what Fred and others wrote was: - The 11-cell is a *4-dimensional* thing. - Its fundamental group is trivial. - Its other homotopy groups may not be trivial. - In particular, what's been said so far plus the 4-dimensional version of Poincare doesn't imply that the 11-cell itself is homeomorphic to the 4-sphere. - The boundary of the 11-cell is 3-dimensional. - Its fundamental group may not be trivial. - In particular, what's been said so far plus Poincare doesn't imply that the boundary of the 11-cell is homeomorphic to the 3-sphere. ("x may not be trivial" means only "nothing I've seen said so far in this discussion obviously implies that x is trivial".) But I've forgotten most of the topology I ever knew, and never knew much about polytopes, so the above should be treated skeptically. -- g
That's a perfectly plausible explanation of how I misread Fred. I have a reflex, especially in topological contexts, to assume that when we are talking about a polyhedral complex of any sort, we are in fact talking about the surface. If it is the interior of the 11-cell that is simply-connected, I suppose that does not imply that the boundary is also (though I confess I am still a little boggled, even at that). On Fri, May 27, 2011 at 9:45 PM, Gareth McCaughan < gareth.mccaughan@pobox.com> wrote:
On Saturday 28 May 2011 02:21:20 Allan Wechsler wrote:
Doesn't the fact that the fundamental group is trivial imply that the 11-cell is homeomorphic to the 3-sphere? (This is implied by the Poincare Conjecture, now, I guess, to be called Perelman's Theorem.) Then it must be orientable.
That would be sort of weird, because the surfaces of it's 3-cells are *not*orientable. (I think they are topologically spheres with one crosscap.) But they are all nicely embedded in an orientable 3-manifold. How is this even possible? What am I missing?
My interpretation of what Fred and others wrote was:
- The 11-cell is a *4-dimensional* thing. - Its fundamental group is trivial. - Its other homotopy groups may not be trivial. - In particular, what's been said so far plus the 4-dimensional version of Poincare doesn't imply that the 11-cell itself is homeomorphic to the 4-sphere.
- The boundary of the 11-cell is 3-dimensional. - Its fundamental group may not be trivial. - In particular, what's been said so far plus Poincare doesn't imply that the boundary of the 11-cell is homeomorphic to the 3-sphere.
("x may not be trivial" means only "nothing I've seen said so far in this discussion obviously implies that x is trivial".)
But I've forgotten most of the topology I ever knew, and never knew much about polytopes, so the above should be treated skeptically.
-- g
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An utterly irrelevant anecdote --- As an undergraduate, I attended a course entitled "Homology Theory" given by J. F. Adams --- a figure so august that he eventually had his own journal (in which lesser spirits like M. F. Atiyah might occasionally be granted an appendix to themselves). Halfway through the term, I felt obliged to draft a letter, to the effect that the Homology Theory wished to inform him that he had left it behind. It was signed by the entire final year class, and greeted him on the podium at the start of his next lecture. He very graciously started again from the beginning (and afterwards remarked that he intended to have the letter framed). But all in vain: for ever since, I have regarded anything to do with homology theory (or topology generally) with superstitious dread. Anyhow, the point of this tale is simply that nobody else can possibly be more hopelessly ignorant than I about homotopy theory; and any pronouncement I might be incautious enough to bring forth should treated with the gravest suspicion. Having got that out of the way, these abstract polytopes (or "polychora") are defined initially by a lattice of sets of abstract points, with an associated symmetry group [several introductory papers by Egon Schulte & co are available online.] I found it initially difficult to accustom myself to the fact that the lattice structure may be only partially realisable geometrically. In the case of Coxeter's 11-cell, although it has rank 4 --- and therefore might be expected to have associated a 3-dimensional geometrical "surface" --- the component "solid" cells would be the interiors of homeomorphs of the projective plane, which is one-sided and so doesn't have any! The way I visualise this situation is to embed the thing as a subset of a simplex in 10-space: vertices, edges and faces present no problem, but "solids" have no bounded triangulation by tetrahedra. Pretty pictures (damned if I can make sense of 'em, mind) at Carlo H. Séquin, Jaron Lanier "Hyperseeing the Regular Hendecachoron" http://www.cs.berkeley.edu/~sequin/PAPERS/2007_ISAMA_11Cell.pdf Carlo H. Séquin, James F. Hamlin "The Regular 4-Dimensional 57-Cell" http://www.cs.berkeley.edu/~sequin/PAPERS/2007_SIGGRAPH_57Cell.pdf Fred Lunnon On 5/28/11, Allan Wechsler <acwacw@gmail.com> wrote:
That's a perfectly plausible explanation of how I misread Fred. I have a reflex, especially in topological contexts, to assume that when we are talking about a polyhedral complex of any sort, we are in fact talking about the surface. If it is the interior of the 11-cell that is simply-connected, I suppose that does not imply that the boundary is also (though I confess I am still a little boggled, even at that).
On Fri, May 27, 2011 at 9:45 PM, Gareth McCaughan < gareth.mccaughan@pobox.com> wrote:
On Saturday 28 May 2011 02:21:20 Allan Wechsler wrote:
Doesn't the fact that the fundamental group is trivial imply that the 11-cell is homeomorphic to the 3-sphere? (This is implied by the Poincare Conjecture, now, I guess, to be called Perelman's Theorem.) Then it must be orientable.
That would be sort of weird, because the surfaces of it's 3-cells are *not*orientable. (I think they are topologically spheres with one crosscap.) But they are all nicely embedded in an orientable 3-manifold. How is this even possible? What am I missing?
My interpretation of what Fred and others wrote was:
- The 11-cell is a *4-dimensional* thing. - Its fundamental group is trivial. - Its other homotopy groups may not be trivial. - In particular, what's been said so far plus the 4-dimensional version of Poincare doesn't imply that the 11-cell itself is homeomorphic to the 4-sphere.
- The boundary of the 11-cell is 3-dimensional. - Its fundamental group may not be trivial. - In particular, what's been said so far plus Poincare doesn't imply that the boundary of the 11-cell is homeomorphic to the 3-sphere.
("x may not be trivial" means only "nothing I've seen said so far in this discussion obviously implies that x is trivial".)
But I've forgotten most of the topology I ever knew, and never knew much about polytopes, so the above should be treated skeptically.
-- g
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participants (4)
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Allan Wechsler -
Dan Asimov -
Fred lunnon -
Gareth McCaughan