Talk:Convex polytope

New article

I have written a significant expansion of this page, but after several hours mozilla died on me, and have no strength to start it over again right now, so I just leave a split-and-paste version for now. The text has a number of inaccuracies, to be fixed. Twri (talk) 18:13, 3 October 2008 (UTC)[reply]

Sorry to hear that. It must have been frustrating. Something similar happened to me before, too: I was in the middle of a massive rewrite of an article, when I hit the wrong key and closed the browser tab. Several hours of work vanished into the bit bucket. Nowadays I try not to make too big an edit before saving. Anyway, thanks for splitting up this article from polytope; I think polytope works better this way.—Tetracube (talk) 20:12, 3 October 2008 (UTC)[reply]

Hmmmm... I can't really add anything here, but curious on a definition, or specifically whether a polytope is considered a topological object, or as representing a geometric interior. This article talks only about the geometric interior properties.

... Okay, I added a short section for topology! Tom Ruen (talk) 22:28, 3 October 2008 (UTC)[reply]

Bounding planes

Another curiosity, I guess the definition by bounding planes still works for lower dimensional polytopes, like a 2d face in 3-space? Like Image:Permutohedron order 3.svg is constrained by 8 hyperplanes. The constraining hyperplane inequalities can be doubly defined to reduce two opposite half-spaces as the common plane. Similarly an edge in 3-space can be defined as 4 bounding hyperplanes, two for a common line, and two for the endpoints. Well, no idea what this is worth. Of course it's always better to define a lower dimensional element by a parametric subspace anyway. Tom Ruen (talk) 01:04, 4 October 2008 (UTC)[reply]

This is equivalent, of course, to setting some rows of the linear system to equalities instead of inequalities. Komei Fukuda's cddlib allows you to specify some rows of the matrix to be equalities rather than inequalities, thereby handling the case when the polytope is confined to an affine subspace. You could also think of such polytopes as being facets of a larger polytope which is full-dimensioned; after all, a polytope's face is defined by the set of points satisfying equality for one or more of rows in the linear system.—Tetracube (talk) 03:16, 4 October 2008 (UTC)[reply]

Projective space

Does anyone have good references for treatment of convex polytopes as an objects in the projective space? This treatment would remove the distinction of bounded vs. unbounded polytopes: since an infinite ridge connects to a point in the projective space, so any convex polytope may be defined in terms of convex combinations of points in the projective space (instead of the Finite Basis Theorem), if I am not mistaken. Twri (talk) 16:55, 6 October 2008 (UTC)[reply]

Vertex representation

The link on ext points to the exterior of a set, where in this context it means the extreme points. The extreme points of a set X is the smallest set S such that the convex hull of S equals X. —Preceding unsigned comment added by 72.85.2.217 (talk) 23:44, 22 October 2009 (UTC)[reply]

Error in the article?

"All bounded convex polytopes in Rⁿ, being topological (n − 1)-spheres, have an Euler characteristic of 0 for odd n and 2 for even n."

I was thinking a bounded convex polytope in R³ would be an ordinary sphere (2-sphere) topologically. The number 3 is odd, but the Euler characteristic of a sphere (2-sphere) is 2 rather than 0, right? Or maybe I am misunderstanding something. --Keith111 (talk) 17:06, 9 November 2010 (UTC)[reply]

Yes, this seems to be backwards. —David Eppstein (talk) 18:26, 9 November 2010 (UTC)[reply]

The error is worse than this. Any convex polytope is convex, hence contractible, hence has Euler characteristic 1. What the article means to say, I think, is that the boundary of any convex polytope in Rⁿ is a topological (n - 1)-sphere and hence has Euler characteristic 1 - (-1)ⁿ. This has also been discussed at Talk:Euler characteristic. Does this subject have its own notion of Euler characteristic, separate from the usual topological one? Even if so, this section must be clarified. Mgnbar (talk) 00:32, 23 February 2011 (UTC)[reply]

One has to reflect, what is a polytope? Is it a solid or a surface? Plato treated it as a solid, with the tacit assumption that the solid includes its bounding surface. By the Twentieth Century many authors (e.g. Lines, Cromwell) understood it as a surface, while Grünbaum's "Convex polytopes" defined the convex variety (at least) as a solid excluding its boundary. One might say that a convex polytopes has two Euler characteristics - χ_s for its surface and χ_b for its body or interior (though in general some non-convex polytopes, such as projective polytopes, do not have a definite χ_b). We just need to be clear about which χ we are talking about. — Cheers, Steelpillow (Talk) 10:46, 23 February 2011 (UTC)[reply]

Hi, Steelpillow. I agree. But let me point out that the article as it stands unambiguously defines the polytope to be the "solid", not the "surface". So the article is self-inconsistent. I'm hoping that David Eppstein will weigh in; he seems to have expertise in exactly this area. Mgnbar (talk) 14:18, 23 February 2011 (UTC)[reply]

The boundary of a convex polytope in Rⁿ is an (n − 1)-dimensional manifold that is topologically a sphere, that is, an (n − 1)-sphere, and has the Euler characteristic given. The polytope itself is an n-ball, and has Euler characteristic 1. Does this clarify the issue? I'll go fix the article to say something like that. —David Eppstein (talk) 17:02, 23 February 2011 (UTC)[reply]

Just what I wanted. Thanks. Mgnbar (talk) 17:18, 23 February 2011 (UTC)[reply]

So far so good. The next sentence describes the polytope a tiling of the sphere, not as a ball - that unconscious slip back to the surface again. I'll see if I can fix that. — Cheers, Steelpillow (Talk) 20:58, 23 February 2011 (UTC)[reply]

Definite some confusing issues here, and I'm not in a position to assert much. I don't think [The polytope itself, a topological ball, has Euler characteristic 1 in any dimension.] makes sense - what's a topological ball? It would seem to me that topology implies it as a surface, not a solid. Tom Ruen (talk) 21:44, 23 February 2011 (UTC)[reply]

There is no problem interpreting the whole polytope, interior and all (defined as a convex hull of fintely many many points, say) as a topological space. As for "what's a topological ball": see the already-wikilinked article, ball (mathematics)#Topological balls. —David Eppstein (talk) 23:19, 23 February 2011 (UTC)[reply]

Okay. Is that like a square (geometry) with opposite edges glued together, as topologically identical to a torus? So a square can be seen as having one vertex, 2 edges, one face, χ=F-E+V=1-2+1=0. Tom Ruen (talk) 08:19, 24 February 2011 (UTC)[reply]

"Topological ball" just means "topological space that is topologically equal (homemorphic) to a ball". That is, a space X such that there exists a one-to-one correspondence, continuous in both directions, between X and a ball in Rⁿ. Depending on context, the ball could be open (|x| < 1), closed (|x| ≤ 1), or neither; that detail doesn't matter to this problem, as all balls have Euler characteristic 1.

Just in case this is another point of confusion, the term topology applies to solids as well as surfaces. Any subset of Rⁿ can be considered a topological space, via the subspace topology. Mgnbar (talk) 14:07, 24 February 2011 (UTC)[reply]

Oh, and a ball has Euler characteristic 1 because the ball can be contracted to a point, which has Euler characteristic 1. Mgnbar (talk) 14:11, 24 February 2011 (UTC)[reply]

HI Tom, you are probably more familiar than others here of the way I bang on about the interior or body of a polytope. No, it's not like a torus as a joined-up square. A square is seen as a bounded topological surface - in this case a disc - with V=4, E+4, F=1. In one dimension higher, a polyhedral solid is a bounded topological space - in this case a ball - in its own right, with the spherical surface being its boundary. Thinking of the solid ball as a cell C, you can calculate an Euler characteristic for it: V-E+F-C. A solid torus would have a more complicated treatment of the body to derive the appropriate Euler value. Does that help? — Cheers, Steelpillow (Talk) 19:25, 24 February 2011 (UTC)[reply]

Face lattice

In the section "Face lattice", the terminology "face" and "facet" are not actually defined, and I think it would add to the article if the definitions were added. Lavaka (talk) 20:53, 18 November 2010 (UTC)[reply]

The definition of face (i.e., A face of a convex polytope is any intersection of the polytope with a halfspace such that none of the interior points of the polytope lie on the boundary of the halfspace) is not quite right. A face of the polytope is either the polytope itself, or the intersection of the polytope with a set of its supporting hyperplanes (previously defined in the article.) — Preceding unsigned comment added by Danpmoore (talk • contribs) 16:44, 18 August 2017 (UTC)[reply]

topological properties

The homeomorphism with a topological ball does not seem to be discussed in any easily accessible literature on polytopes that I know of. Can anybody provide any such reference? (This is for my own interest, I am assuming the idea is well enough sourced in the more advanced literature). — Cheers, Steelpillow (Talk) 12:41, 30 May 2011 (UTC)[reply]

See Bredon, Topology and Geometry, p. 56. For a different but similar result, see Massey, A Basic Course in Algebraic Topology, p. 46. I'll add the Bredon reference. Meanwhile, here's a hint: Let p be any interior point of the polytope. The map

x\mapsto (x-p)/|x-p|

is a homeomorphism from the boundary of the polytope to the boundary of the unit ball (i.e. the unit sphere). Use this to get the rest. Mgnbar (talk) 13:10, 30 May 2011 (UTC)[reply]

Many thanks. — Cheers, Steelpillow (Talk) 15:56, 31 May 2011 (UTC)[reply]

Another error in the article?

The article currently states

"The finite basis theorem[2] is an extension of the notion of V-description to include infinite polytopes. The theorem states that a convex polyhedron is the convex sum of its vertices plus the conical sum of the direction vectors of its infinite edges."

What about the convex polyhedron

 x+y+z <= 1
 2x-y+3z <=2

This has no vertices but it is not a conic sum of direction vectors.

????

98.155.30.229 (talk) 17:11, 25 July 2015 (UTC)[reply]

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