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I have a web page at wordpress.com (URL:http://polyhedron100.wordpress.com) with a variety of Nolidean Polyhedra including some nice Crown Polyhedra in wood. I am hoping I might have a link to them here at the polyhedron page as there is no mention of these types of polyhedra elsewhere at wikipedia. Please advise if this would be acceptable? Thanks and Take Care User:Bertimusminimus 10:15, 29 November 2014 — Preceding unsigned comment added by Bertusminimus (talk • contribs)

Hi again. These are really beautiful, they certainly deserve a wider audience. I will post a link on my facebook page for sure. But they are not normally regarded as polyhedra, because the surfaces are bounded. If they are regarded as toroidal nolids, even then they are not usually understood as polyhedra: for example two (coincident) edges may share the same two vertices, which is not allowed in conventional polyhedron theory, but only if one chooses to extend the theory specially. As a footnote, a "crown polyhedron", sometimes called a stephanoid, is a particular kind of axially-symmetric (pyramid/prism symmetries) polyhedron. In the end, your creations are at heart beautiful symmetrical mathematical sculptures, and best appreciated as such. Not perhaps what you wanted to hear, but I hope it helps clarify things for you. — Cheers, Steelpillow (Talk) 16:47, 29 November 2014 (UTC)[reply]

Thank you for responding and your kind words. It's a little disappointing to be sure, but I will of course honor your decision. That said, I do realize that they are not polyhedra in the traditional sense but I thought that they still would fall into that category if one allows for a relaxation of the terms of their definition such as having gaps or holes between polygons. In any event thanks again and Take Care, User:Bertimusminimus 18:15, 29 November 2014 (UTC) — Preceding unsigned comment added by 75.120.178.107 (talk) [reply]

This section currently defines only topological polytopes. Are we to take it that "polyhedron" and "polytope" are synonyms in this context, or that a topological polyhedron is a topological 3-polytope? — Cheers, Steelpillow (Talk) 18:58, 21 December 2014 (UTC)[reply]

I don't even know, and it's unsourced. What I would normally do in such situations is try quick Google scholar and Google books searches to determine whether there is in fact a standard meaning for this term; if so, add the sources and clarify the meaning, and if not just delete the section. —David Eppstein (talk) 19:42, 21 December 2014 (UTC)[reply]

I found several definitions, each built on more impenetrable buzzwords than the last, so I have no idea even whether they are equivalent or not. Then there is this possibly related remark from Grünbaum & Shephard, 1969:

"A topological polytope P' is the image of a convex polytope P under a homeomorphism Φ : P -> Eⁿ. The faces of P' are the images of the faces of P under Φ, and the dimension of P' is defined to be the dimension of P. Sometimes we shall use the term geometric polytope for a convex polytope when we wish to emphasise the difference from a topological polytope."[1]

To my poor understanding, Eⁿ is Euclidean n-space and anything injected into it is perforce a real geometric polytope. But here it is a topological polytope and is being distinguished from the geometric variety. This kind of apparent non sequitur, supported invariably by the most impenetrable of jargon salads, always utterly baffles me. Hence my appeal to someone schooled in such ways. — Cheers, Steelpillow (Talk) 21:33, 21 December 2014 (UTC)[reply]

This passage seems clear enough to me, at least. It's describing what you get from a polyhedron when you warp space with a continuous but 1-1 transformation. E.g. you could make a topological cube with six Coons patches whose boundary curves and corners have the same combinatorial structure as the edges and vertices of a geometric cube. —David Eppstein (talk) 22:15, 21 December 2014 (UTC)[reply]

Thank you, yes, that makes sense now - as far as polytopes go. The relation to "topological polyhedra" is still undefined. I also begin to wonder as to what extent things like CW complexes are related. — Cheers, Steelpillow (Talk) 23:22, 21 December 2014 (UTC)[reply]

In the Polyhedron#Euler_characteristic section we see the folowing statement.

For a convex polyhedron or more generally for any simply connected polyhedron whose faces are also simply connected, χ = 2.

This statement isn't true for a solid made by two tetrahedron joining in one of their vertices. Since this is a simply connected solid (bounded by planes) with simply connected faces and χ = 3, either this object isn't a polyhedron, or the above statement is false. I don't see anything in this article that would exclude this wedge-sum object from among polyhedra, so it would be helpful to improve the either the defintion of polyhedrion, or the statement above. 89.135.19.75 (talk) 07:36, 18 May 2015 (UTC)[reply]

"A solid made by two tetrahedron joining in one of their vertices" is not a polyhedron as defined in this article. Nor is it a simply-connected manifold as linked to in the article. Now I know some mathematicians squawk in horror at that because they have any of several different specialised definitions of a polyhedron in mind from their favourite standard text on their chosen field. But this is a Wikipedia article introducing polyhedra, and it uses the classical definition unless some variation is expressly stated. I have tidied the article to help clarify one or two aspects. — Cheers, Steelpillow (Talk) 12:04, 18 May 2015 (UTC)[reply]

Sorry, but I don't find this statement in Richeson's book. Where is it exactly?

And the other thing: exactly which sentence excludes this object from among the polyhedra in the article? The link points to siply connected space, and this object is also a simply connected space, so the link doesn't. 89.135.19.75 (talk) 17:12, 18 May 2015 (UTC)[reply]

p.255 states "We know from the classification of surfaces that the sphere is the only simply connected closed surface", p.182 gives χ for the sphere = 2. I have added a parenthetical comment to separate out the two logical steps for you. Any better? — Cheers, Steelpillow (Talk) 18:25, 18 May 2015 (UTC)[reply]

1. I see nowhere in this article, that every polyhedron should be topologically a connected closed surface.
2. It isn't true, that if a polyhedron is topologically a sphere, then its Euler characteristic is 2, see for example a small cube on the middle of the top of a bigger cube (the interior of the contacting face parts are removed). This is topologically a sphere, but its Euler characteristic is 3. The simply connectedness of the faces is an additional necessary condition, so it should not have been removed from the text (see also here) 89.135.19.75 (talk) 20:03, 18 May 2015 (UTC)[reply]

[Edit conflict] First, thank you for pointing out the omission re. vertex-connected objects. It is implicit in standard definitions, such as that of an abstract polytope, so I don't think it need be made explicit in the discussion on definition. But I have now added a commentary in the section on the surface characteristics. Richeson illustrates your figure and discusses the problem, so too does Cromwell at a more basic level. Do you think it needs a citation?

The requirement to have a closed surface is expressed in simpler language as a requirement that the surface must not end abruptly, perhaps why you missed it. The cube-on-a-cube is of course excluded by the need for every face to be a polygon: a square with a hole in is not a valid face and the figure is not in fact a polyhedron. So the definition of a face needs clarifying too. I can think of at least one other possible omission. I don't have time to think the changes through now, I'll try and remember to take a proper look tomorrow (assuming nobody beats me to it). — Cheers, Steelpillow (Talk) 20:34, 18 May 2015 (UTC)[reply]

So,this concave heptahedron Isn't a polyhedron at all? 89.135.19.75 (talk) 23:04, 18 May 2015 (UTC)[reply]

It isn't according to the definition Steelpillow is using. One can find sources that use definitions that would allow it but that may be a minority view of the subject. —David Eppstein (talk) 23:07, 18 May 2015 (UTC)[reply]

That is correct. The article takes as its backbone the modern formulation of the definition used by well-known authors such as Euclid, Coxeter, Cromwell and (at an elementary level) Grünbaum and found in every school geometry textbook. From this it classifies "polyhedra" found in the wider literature into families (i.e. sub-species), generalisaztions (broadly compatible) and alternatives (incompatible in some profound way). One may hope that the heptahedron shown comes under one of these alternatives, although its discussion in the linked article is uncited, is not supported by the article sources given, and including it as a "polyhedron" under any sensible definition could just be a lapse of rigour. — Cheers, Steelpillow (Talk) 10:31, 19 May 2015 (UTC)[reply]

We shouldn't pretend that there is a single universally-accepted definition, but I don't think our article does that. —David Eppstein (talk) 16:14, 18 May 2015 (UTC)[reply]

Thanks! 89.135.19.75 (talk) 20:04, 20 May 2015 (UTC)[reply]

The text of this article allows us to regard a polyhedron as a 2-dimensional CW-complex as well as a 3-dimensional one. However the definition of the Euler-characteristic implicitly assumes that it is taken as a 2-dimensional CW-complex (i.e. the definition of the Euler characteristic given here is true for polyhedral surfaces, but it isn't true, if the interior volume is considered also to be part of the polyhedron. Perhaps it would be useful to tell this in the article (see Euler_characteristic#Topological_definition). 89.135.8.194 (talk) 22:40, 25 August 2015 (UTC)[reply]

What the text allows and what it should say are very different things. There is no need to discuss the niceties of CW-complexes in an article which does not mention them. In particular, the Euler characteristic as defined for a polyhedron references only vertices, edges and faces: contrary to what you say, whether or not there is an interior has no relevance. A figure decomposed into say tetrahedral cells is no longer just a polyhedron but a more general topological object - and of no relevance to an article on polyhedra. — Cheers, Steelpillow (Talk) 10:05, 26 August 2015 (UTC)[reply]

I mean this:

Different approaches - and definitions - may require different realisations. Sometimes the interior volume is considered to be part of the polyhedron, sometimes only the surface is considered, and occasionally only the skeleton of edges or even just the set of vertices.[1]

and this:

For example a convex or indeed any simply connected polyhedron is a topological sphere or ball (depending on whether its body is taken into account).

The problem is, that if I regard a simply connected polyhedron as being topologically a ball, then it is contractible, hence in this case, its Euler-characteristic should be 1 instead of 2. Euler characteristic should be homotopically invariant. 86.101.236.13 (talk) 10:55, 26 August 2015 (UTC)[reply]

And that shows clearly what I mean. The conventional definition of the Euler characteristic for a polyhedron is given in the article. It does NOT invoke the body and therefore does NOT address the polyhedron as a ball. You are using a different definition applicable in different and more advanced circumstances. — Cheers, Steelpillow (Talk) 11:14, 26 August 2015 (UTC)[reply]

And this is exactly what I mean: "Euler characteristic for a polyhedron is given in the article does NOT invoke the body" I miss this sentence from the article. 86.101.236.13 (talk) 11:18, 26 August 2015 (UTC)[reply]

It is there to be read. The section refers to "any simply connected polyhedron (i.e. a topological sphere)". — Cheers, Steelpillow (Talk) 11:33, 26 August 2015 (UTC)[reply]

I think the confusion was caused more by the introduction of a ball in this context. Although more complete it brings complexities seldom addressed at this level and best left out. Accordingly, I have edited the general remarks about topological characteristics to confine the discussion to the surface. Any better? — Cheers, Steelpillow (Talk) 11:42, 26 August 2015 (UTC)[reply]

This is better already, but still isn't explicit enough in my taste. What about something like this:

From this perspective, a polyhedron is regarded as its surface. Any polyhedral surface may be classed as certain kind of topological manifold. For example a convex or indeed any simply connected polyhedron is a topological sphere.

instread of the current

From this perspective, any polyhedral surface may be classed as certain kind of topological manifold. For example a convex or indeed any simply connected polyhedron is a topological sphere.

86.101.236.13 (talk) 13:08, 26 August 2015 (UTC)[reply]

I don't think that would be accurate. One can perfectly well be considering the surface of a solid polyhedron. The phrase "polyhedral surface" deliberately covers both possibilities. — Cheers, Steelpillow (Talk) 14:10, 26 August 2015 (UTC)[reply]

Yes, but (according the current text) the Euler characteristic is assigned to the "polyhedron", not to the "polyhedral surface". If you don't like my proposal,we should say at least, that we mean the "Euler characterisic of a polyhedron" the "Euler characterisic of its (polyhedral) surface". Should't we? 86.101.236.13 (talk) 14:24, 26 August 2015 (UTC)[reply]

You mean, as in "The topological class of a polyhedron is defined by its Euler characteristic and orientability"? The problem we face is that most if not all mainstream sources associate the Euler characteristic in this way, whether or not they note the stricture about its surface. What I have tried to do with my recent edits is to lead the reader from this widely-stated but not wholly rigorous picture to something closer to the truth. But on Wikipedia one cannot say "reliable sources are not rigorous" unless there is a reliable source telling us this explicitly. We are stuck with some measure of woolliness and I am not sure how the current text can be improved on. — Cheers, Steelpillow (Talk) 15:28, 26 August 2015 (UTC)[reply]

I've made an attempt.89.135.8.194 (talk) 06:15, 27 August 2015 (UTC)[reply]

Shouldn't we define what does "polygonal face" mean? Shouldn't we require that the polygon must be simple (non self-intersecting)? — Preceding unsigned comment added by 89.135.8.194 (talk) 06:51, 29 September 2015 (UTC)[reply]

The faces of star polyhedra are not simple. Precise definitions differ widely and a full treatment would not be useful in the present introductory article. See for example Lakatos, Proofs and Refutations.— Cheers, Steelpillow (Talk) 07:58, 29 September 2015 (UTC)[reply]

Even nonplanar or skew polygon faces are not excluded in Polyhedron#Abstract_polyhedra! Simple would seem to be a 2D-space concept?! p.s. I forgot how passionately this article hates pictures. Tom Ruen (talk) 09:55, 29 September 2015 (UTC)[reply]

OK, but then what excludes the not simply connected faces? What is a polygonal face at all? For example, what kind of face is defined by the fourth polygon here?

89.135.8.194 (talk) 21:44, 29 September 2015 (UTC)[reply]

Could you clarify your question? By "simply connected" do you mean a synonym for "simple" or something more general? The polygon you ask about is not simple, but nor is it excluded. Would you regard it as simply connected? — Cheers, Steelpillow (Talk) 07:51, 30 September 2015 (UTC)[reply]

A simply-connected polygon faces is a good requirement, excluding disconnected sets (compound forms), and excluding coinciding vertices, edges or faces (degenerate forms). Like this isogonal decagon, left, is approaching a degenerate case if vertices come together. Still, simply connected allows edges to intersect where the interior of polygon is ambiguous. Tom Ruen (talk) 08:18, 30 September 2015 (UTC)[reply]

The case of skew polygon faces is the least referenced generalization of a face, like the Petrial cube. You can see the hexagons on the right as red, orange, blue and green edge-paths around a cube. Every edge has 2 colors (2 skew polygon faces). Tom Ruen (talk) 08:24, 30 September 2015 (UTC)[reply]

@Steelpillow: Consider this part of the article:

The Euler characteristic χ relates the number of vertices V, edges E, and faces F of a polyhedron:

${\displaystyle \chi =V-E+F.\ }$

This is equal to the topological Euler characteristic of its surface. For a convex polyhedron or more generally any simply connected polyhedron (i.e. with surface a topological sphere), χ = 2.

If nothing rules out simply connected faces, then we should explicitly include this condition here as

generally any simply connected polyhedron with simply connected faces (i.e. with surface a topological sphere), χ = 2.

Bul last time you told that a"polygonal face" is always simply connected, so we don't need to specify here the simply connectedness of the faces explicitly.89.135.8.194 (talk) 08:23, 30 September 2015 (UTC)[reply]

Yes, the boundary of a real polygon is by definition simply connected (This is not true of complex polygons but that is an obscure side issue, as they are defined very differently and inhabit the complex plane). Polyhedra of course need not be simply connected. The condition arises in abstract polytope theory as a consequence of the rules for the partial ordering of the set, which are explained in that article. In more traditional topology it arises because a piece cut from a contiguous, smooth surface such as a real plane must always have a simply connected boundary. Any polygon may be used as a face of a polyhedron. For example the polygon you ask about can form an end face of a self-intersecting hexagonal prism. — Cheers, Steelpillow (Talk) 08:52, 30 September 2015 (UTC)[reply]

Guy, going up a dimension, can a "real" 4-polytope exist with toroidal polyhedron cells, not topological spheres? I'd have to work a bit to find a full example.... perhaps like a tesseract's 8 cubes could be merged into 2 sets of 4 cubes by removing 8 square faces, and leaving 2 toroid cells?! Tom Ruen (talk) 09:31, 30 September 2015 (UTC)[reply]

Yes. They are called locally toroidal polytopes. Locally projective polytopes also exist, for example having one or more hemicubes as cells. I don't know for sure whether such things necessarily can or can't be faithfully realised in real n-space, but a few minutes' thought suggests to me that some can and some can't. They can make topological analysis difficult as they don't obey the usual Euler formula and its generalisations, for example they cannot be subdivided into simplexes without changing the values of such topological "invariants". — Cheers, Steelpillow (Talk) 10:10, 30 September 2015 (UTC)[reply]

This isn't really different from the phenomenon that three-dimensional polyhedra with annular faces also don't obey the Euler formula (even when they have spherical topology), I don't think. —David Eppstein (talk) 06:32, 1 October 2015 (UTC)[reply]

The key difference is that a toroidal face does not have a continuous boundary It is therefore not a valid polygon and cannot be used to construct higher polytopes. A three-dimensional toroid is a valid polyhedron and so it can be used. The disruption to topological analysis is the same though. — Cheers, Steelpillow (Talk) 08:49, 1 October 2015 (UTC) [Updated 09:39, 1 October 2015 (UTC)][reply]

Cool! Oh, I see my tesseract reconstruction would have two flat tori cells (two sides of a common flat tori surface), as Coxeter's {4,4|4} regular skew polyhedra, a 4D folding of a 4×4 grid from a square tiling! Tom Ruen (talk) 12:11, 30 September 2015 (UTC)[reply]

Why not the regular small stellated 120-cell {5/2,5,3}, which has genus-4 small stellated dodecahedron cells? Double sharp (talk) 12:53, 30 September 2015 (UTC)[reply]

Yes indeed. The "reduced tessaract" is a toroidal 4D equivalent of a dihedron. {5/2,5,3} is a really nice example. — Cheers, Steelpillow (Talk) 13:28, 30 September 2015 (UTC)[reply]

@Steelpillow : I see a general misunderstanding between us. I talk abot this sentence:

for any simply connected polyhedron (i.e. with surface a topological sphere), χ = 2

I thought originally, that simply connectedness refers here to the body of the polyhedron, i.e. to the the 3-dimensional domain bounded by its surface. This belief was supported by the fact that this sentence was originally (up to 12:00, 18 May 2015)

For a convex polyhedron or more generally for any simply connected polyhedron whose faces are also simply connected, χ = 2

, and that this is in accordance with Lakatos's book:

For a simple polyhedron, with all its faces simply-connected, V-E+F = 2

As far as I see, the referred book of Richeson also doesn't say else.

But now, since you talk about self-intersecting polyhedra, I have doubts about what do you mean. Has at all sense to talk about the body of a self-intersecting polyhedron? What does this sentence mean in your opinion? 89.135.8.194 (talk) 06:23, 1 October 2015 (UTC)[reply]

The following applies whether we consider the surface or the interior of a polygon or polyhedron, for if one is simple or simply-connected then the other must be as well: wherever space is locally flat, a topological sphere will always surround a topological ball.

There is a distinction between structural incidence or connectedness on the one hand and geometric coincidence on the other. "Simple" is a geometric property, "simply-connected" is a structural or topological property. In topology, whether a particular geometrical form is simple or self-intersecting has no significance. For example a cross-quadrilateral (a butterfly or hourglass shape) has four vertices each connected to, i.e. incident with, two sides. The crossing point in the centre has no such connection and the two sides are merely coincident at that point. If the quadrilateral is unwound and made convex then it is easy to comment on the fact that it is now a simple polygon, and the fact that it is simply connected is easy to see. When twisted up it is no longer geometrically simple but structurally it is still simply-connected. If we make it the end of a four-sided prism, the prism remains simply connected no matter how we squash or morph it around. Another simply connected polyhedron is the great stellated dodecahedon {5/2, 3}, because (as Cayley noted) it is just such a morph of the convex regular dodecahedron {5, 3}. But of course it is not simple like {5, 3} because it self-intersects. On the other hand the small stellated dodecahedron {5/2, 5} is a toroid of genus around 4 (I can't remember exactly) and is neither simple nor simply-connected.

It is relatively easy to find out whether a certain figure is simple, especially if it is convex. It is much harder to see, by looking or by analysis of things like half-spaces or vertex connectivity, whether a self-intersecting polyhedron is simply-connected or not. The sure way to find out is by discovering its Euler characteristic. This is the analysis which the article tries to explain by first introducing the Euler characteristic of the structural sphere, whether it be geometrically simple or self-intersecting. Does this clarify the situation for you? — Cheers, Steelpillow (Talk) 09:27, 1 October 2015 (UTC)[reply]

(Yes, the genus of {5/2, 5} is indeed 4, as its Euler characteristic is −6 and it is orientable.) Double sharp (talk) 08:37, 2 October 2015 (UTC)[reply]

Not exactly, but trying to comprehend your words. Of course, we can talk about abstract graphs where doesn't matter if it is planar or not. It is an independent topological space. If it is not planar, then when we draw it on a sheet of paper, then there will appear line crossings that "do not count", i.e, that arent a vertex. Of course we can draw planar graphs also in an entangled form where the lines cross not only at vertices. Still it is planar. But I think, that we can talk about "faces" only when we embed a ${\displaystyle G}$ planar graph in the sphere. If the embedding is the ${\displaystyle f:G\to S^{2}}$ function, then the faces are the connected parts of ${\displaystyle S^{2}\setminus f(G)}$. Generally, the faces depend on ${\displaystyle f}$ too, not only on ${\displaystyle G}$. But how do you define faces in the case of a nonplanar graph? 89.135.8.194 (talk) 05:51, 2 October 2015 (UTC)[reply]

You choose some cycles in the graph and call them faces. If you like, you can also associate each chosen cycle with a topological disk and glue the disks together along the edges, but that step is not necessary for understanding the collection of vertices, edges, and cycles as an abstract polyhedron. —David Eppstein (talk) 06:24, 2 October 2015 (UTC)[reply]

Is there a difference between your "abstract polyhedron" and CW complexes? 89.135.8.194 (talk) 06:31, 2 October 2015 (UTC)[reply]

Is there a difference between your "abstact polyhedron" and CW-complexes? 89.135.8.194 (talk) 06:31, 2 October 2015 (UTC)[reply]

Yes. There are abstract polytopes which are not CW-complexes and there are CW-complexes which are not abstract polytopes (although in three dimensions, all abstract polyhedra are CW-complexes). At a foundational level, abstract theory is overtly set-theoretic in nature, so the two theories express themselves rather differently. In terms of the structures allowed, a CW-complex requires all cells (of any dimension) to be topologically simple, while abstract theory is more general in allowing toroidal and other non-simple cells (or j-faces). As it happens, planar faces are always simple so for 3D polyhedra this distinction is trivial. On the other hand CW-complexes are more general in that they do not require cells to be assembled into a higher polytope. I don't know much about CW-complexes but as I understand it, say an n-ball attached to a 0-dimensional CW-complex (0-complex) together comprise a valid n-complex, but this is certainly not a valid abstract polytope. Also, a CW-complex need not "fill in the gaps", for example a graph need not highlight any particular cycles as cells or faces. A skeletal polyhedron is abstractly "unfaithful" or incomplete but, understood as a graph, is a valid CW-complex in its own right. This can be significant, for example consider a skeletal regular icosahedron. Abstractly we may have identified triangular cycles as bounding 2-faces of a convex icosahedron, or we might have identified pentagonal cycles bounding a great dodecahedron: we have to have made the choice. But the CW-complex of the skeleton is sufficient to itself and does not need to choose. (One can of course make the choice anyway and construct a distinct, higher-dimensional CW-complex.) — Cheers, Steelpillow (Talk) 08:52, 2 October 2015 (UTC)[reply]

This is very interesting, thank you. I'm starting to understand you. 89.135.8.194 (talk) 06:46, 3 October 2015 (UTC)[reply]

There has been a bit of editing back-and-forth about duality. I'm inclined to prefer David Eppstein's version (though actually I might even prefer something like this: "For every convex polyhedron there exists a dual polyhedron, having .... (Abstract polyhedra also have abstract polyhedral duals, with the same properties. However, some definitions of non-convex geometric polyhedra may not have duals meeting the same definition.)"). There is one minor thing that concerns me about it: at that stage of the article, the class of convex polyhedra has not been introduced yet. (Possibly this is an indication that convexity should be mentioned earlier.) --JBL (talk) 18:29, 11 February 2017 (UTC)[reply]

Oh, and I should mention: the center of this disagreement is the discussion here. --JBL (talk) 18:31, 11 February 2017 (UTC)[reply]

I will not enter further discussion with an editor who accuses me of lying and shouts at me in his edit comments. By rights I should be taking this straight to WP:ANI. I will say here that I have cited the case for nonconvex polyhedra in both articles - Wenninger's is a popular and well-regarded text from a reputable academic publisher. — Cheers, Steelpillow (Talk) 18:46, 11 February 2017 (UTC)[reply]

If you want to take David Eppstein to ANI, go for it, but it has nothing to do with choosing among a few options for how to word a sentence. --JBL (talk) 18:58, 11 February 2017 (UTC)[reply]

Yes, that's exactly the problem. — Cheers, Steelpillow (Talk) 19:01, 11 February 2017 (UTC)[reply]

[edit conflict] Since this response was not particularly informative, let me describe what I think Steelpillow's position is (he can obviously correct me if he would like to actually participate in the discussion rather than blustering). It is that the theory of abstract polyhedra provides a valid form of duality for almost all instances of what people call polyhedra (true) and therefore that all uses of the word "polyhedra" in our articles (unless otherwise qualified) should be assumed to mean abstract polyhedra. Under this interpretation, the sentence "all polyhedra have duals" is true, because what it really means is "all abtract polyhedra have duals". My own position, on the other hand, is that most readers are likely to come to the article with a naive conception of what it means to be a polyhedron (involving something embedded into Euclidean space with flat sides), and that the sentence "all polyhedra have duals" is likely to seriously mislead these readers into thinking that all non-convex Euclidean things with flat sides have dual Euclidean things with flat sides, something that generally isn't true. If we say things that we can reasonably expect to lead to false beliefs in our readers, we are lying to them. (This, by the way, is what Steelpillow thinks of as incivility: pointing out situations where what we write may cause readers to have false beliefs.) To avoid lying to the readers, I would prefer to qualify the statement by saying which kinds of polyhedra have duals: convex polyhedra have convex duals, and abstract polyhedra have abstract duals, but other kinds of polyhedra may not have duals within those other classes of polyhedra. —David Eppstein (talk) 19:02, 11 February 2017 (UTC)[reply]