Talk:Euler characteristic

message: This article definitely needs more work.   -- Mike Hardy
I added some material. Could you explain the poset case a bit more? AxelBoldt 03:44 Nov 18, 2002 (UTC)
I moved material from the euler formula page to this page. I did some merging but the page still needs some more work. MathMartin 22:27, 2 Aug 2004 (UTC)
replies: text: I agree that that's a problem.  I've fixed it.  But it recurs elsewhere in this article.  Generally on Wikipedia, TeX looks good when "displayed" and often looks terrible when embedded in lines of text.  This is just one of many examples.  Many mathematicians who work on Wikipedia seem unaware of or indifferent to the problem, and won't use non-TeX mathematical notation even when it's pretty easy to do so.  Some mathematicians are aware of the problem and are opposed to efforts to correct it, for reasons best known to themselves. Michael Hardy 01:11, 30 Dec 2004 (UTC)

text: If you're still hanging around I would love to know a proof for the product space Euler characteristic.  I'm not saying I don't believe it, I've just never seen a proof (outside of the finite CW complex case or some other relatively trivial case). Amling 7:03, 17 Dec 2005 (CST)
I think it is problem with soft and writers should not do anything about that hopping that it will be improvedsometime, infact I do not understand why not to choose pdf as the standard for wikipedia. Tosha 02:58, 30 Dec 2004 (UTC)

text: ps is ok, I just do not like how it looks now, and it seems that MathML will  improve it just a bit... Tosha 03:42, 2 Jan 2005 (UTC)
thread_title: untitled

message: Is it not more common to define the euler characteristic in terms of the
compact supported betti numbers? Then of course it is no longer a homotopy
invariant (just invariant under homotopy equivalence between compact spaces)
, but it does give the excision property.
Take for example the real line R. It is homotopically equivalent to
the point, hence has euler characteristic 1. But now split R in three parts:
the positive reals, the negative reals, and zero. As all parts are
homotopic equivalent to the point, we get that the excision property fails
(it would give 1=1+1+1). Using compact support in stead gives -1 as euler
char of R and then the excision property gives -1=-1+1-1. 
Could somebody confirm what the most common definition is? And could we then
add the excission property and some non-compact examples
(Rn for example) to this article?
Also, the euler characteristic with compact support can be defined axiomatically
as being unique with the properties of chi(point)=1, chi(product)=product, excision,
and compact homotopy invariance.
--Lenthe 12:11, 16 February 2006 (UTC)
replies: text: In my experience, the standard definition is definitely as the alternating sum of Betti numbers. I have never seen Euler characteristic with compact support, but I am interested, since it seems to have some nice properties (and lack some others). Is this concept well-established? I mean, can you cite a reference for it? If so, please write a new section about it! Joshuardavis 22:05, 2 March 2006 (UTC)
thread_title:  euler characteristic with compact support; excision 

message: I just did a major reorganization, so I apologize for stepping on any toes. Two notes:
*I removed the following, which appears to be dead wrong: "for the plane we have \chi = 2, counting the outside as a face". If I'm misunderstanding the point, let me know.
*I reworded a lot of Cauchy's proof of Euler's formula. Perhaps I should not have done this, if the proof given was a direct translation from Cauchy, interesting for its historical value. If so, revert away.
By the way, the picture examples on this page are fantastic. Good work. Joshuardavis 17:31, 2 March 2006 (UTC)
thread_title:  Reorg on 2 Mar 2006 

message: I feel very strongly that the polyhedron formula should be V - E + F, not F - E + V, because this agrees with the k_0 - k_1 + k_2 - k_3 + ... formula given for arbitrary CW complexes. Salix alba and Maggu, is this okay with you? Joshua Davis 16:14, 6 April 2006 (UTC)
replies: text: I agree. You both raise good points I didn't think of. I mostly felt that it should be uniform, and faces as a kind of primary characteristic would look better placed first in the tables. --Maggu 21:27, 6 April 2006 (UTC)

text: * It seems the Euler formula can be extended to n-dimensional to read:
 So-S1+S2-S3+ .... =1,
where Sk is the number of k-dimensional entities. In this notation, V becomes So (vertices or points), E becomes S1 (edges or line segments), F becomes S2 (facets or areas), and S3 is the number of 3-D solids, so on and so forth.
Consider an inverted pyramid apex-to-apex on top of another pyramid. We have V=So=9, E=S1=16, F=S2=10, and S3=2. It is seen, V–E+F=3 not equal to 2, but So–S1+S2–S3=1.
Let us tabulate expansion of the seemingly trivial (1–1)n+1=0 in the form of So–S1+S2–S3+S4–....=1, and make observations:
For n=0, 1=1, which represents a point with So=1.
For n=1, 2–1=1, which represents a line segment with So=2, S1=1.
For n=2, 3–3+1=1, which represents a triangle with So=3, S1=3, S2=1.
For n=3, 4–6+4–1=1, which represents a tetrahedron with So=4, S1=6, S2=4, S3=1.
For n=4, 5–10+10–5+1=1, which represents a 4-D "triangle" with So=5, S1=10, S2=10, S3=5, S4=1.
The result  suggest that a 4-D "triangle" has 5 vertices, 10 edges, 10 facets (or triangles) and 5 tetrahedrons.
Likewise, expansion of the seemingly trivial (2-1)n=1 yiels:
For n=0, 1=1, which represents a point with So=1.
For n=1, 2–1=1, which represents a line segment with So=2, S1=1.
For n=2, 4–4+1=1, which represents a quadrilateral with So=4, S1=4, S2=1.
For n=3, 8–12+6–1=1, which represents a cube with So=8, S1=12, S2=6, S3=1.
For n=4, 16–32+24–8+1=1, which represents a 4-D "square" with So=16, S1=32, S2=24, S3=8.
The expansion suggests a 4-D "quadrilateral" would have 16 vertices, 32 edges, 24 facets (or quadrilaterals) and 8 cubes. 209.167.89.139 16:59, 15 September 2006 (UTC)

text: Indeed. These are the Betti numbers discussed in the article. --Salix alba (talk) 17:32, 15 September 2006 (UTC)
thread_title:  ''V'' - ''E'' + ''F'' 

message: It looks like Vertex/Face/Edge relation in a convex polyhedron is a special example of the euler characteristic, and it needs cleanup so merge proposed --PhiJ 16:10, 23 April 2006 (UTC)
replies: text: I don't know that there is anything to be salvaged from Vertex/Face/Edge relation in a convex polyhedron. Everything is already covered here and at Platonic solid. I think that the article should simply be deleted. Its only significant editor was anonymous, so I don't foresee that there will even be an argument about it. Joshua Davis 16:43, 23 April 2006 (UTC)
replies: text: I agree it should be simply deleted. I've now added a prod tag to the article. --Salix alba (talk) 16:51, 23 April 2006 (UTC)
replies: text: Now on AfD. --Salix alba (talk) 20:00, 24 April 2006 (UTC)
thread_title: Proposed Merger

message: So what is the best name for this formula? It's called the Euler Formula on this page, but search Wikipedia for "Euler Formula's" and you'll reach this page: Euler's formula (though you do get a link to this one). Wolfram MathWorld calls it the Poincaré Formula, though references a publication by Coxeter on "Euler's Formula". I've also seen it referred to as the "Euler-Poincaré Formula", e.g. http://www.cs.mtu.edu/~shene/COURSES/cs3621/NOTES/model/euler.html. Any consensus here?
:AFAIK, Euler posed the formula, and Poincaré proved the result. I have also seen the Euler characteristic called the Euler–Poincaré characteristic. Although I can provide no solid reason for which to choose, I do think it would be a good idea to mention Poincaré and the variations in this article. — cBuckley (Talk • Contribs) 18:07, 2 April 2007 (UTC)
thread_title: Formula Name?

message: The table lists its Euler characteristic as 1, but this is wrong.  The Euler characteristic of the Roman surface -- as a subset of 3-space -- is 5.  (One can triangulate it and count 11 vertices, 30 edges, and 24 triangles,
giving χ = 11 - 30 + 24 = 5.)
It is true that the projective plane -- of which the Roman surface is a continuous image -- has Euler characteristic = 1.  But that is a different statement.Daqu 09:12, 29 April 2007 (UTC)
thread_title: Roman surface Euler characteristic is wrong

message: Could somebody enlighten me as to why it does not work anymore? I figured it was a temporary problem with Wikipedia that would go away with time, but that doesn't seem to happen. The image has worked fine in the past after all, and the example on meta:SVG image support still works. Perhaps something related to the naming of the file has changed lately? --Maggu 10:30, 21 June 2007 (UTC)
replies: text: It should work now, I used &action=purge at the end of the image file which manage to restore the image. I've now reinserted the image.into the page. --Salix alba (talk) 14:15, 21 June 2007 (UTC)
replies: text: Thank you. --Maggu 09:24, 2 July 2007 (UTC)
thread_title: SVG images

message: Some points explained in Chapter 23 of Richeson, D.; Euler's gem: the polyhedron formula and the birth of topology, Princeton, 2008.
Poincaré classified higher-dimensional manifolds according to their Betti numbers and torsion coefficients.  A polytope in n dimensions is an (n-1)-manifold and has n Betti numbers, b0 to b(n-1).
The Euler characteristic χ of a polyhedron generalises to the Euler-Poincaré characteristic χ of a higher polytope.
χ can also be obtained as the sum of the n Betti numbers b0 - b(n-1) alternately added/subtracted in a way analogous to the usual alternate addition and subtraction of n-dimensional element counts.
The Poincaré duality theorem describes a symmetry, or duality, of the Betti numbers for a given manifold. For example if a given manifold has four Betti numbers b0 - b3, we need only know the fris two, say b0=1, b1=0, to write the full set b0 - b3 as 1, 0, 0, 1.
This has two immediate consequences:
# The well-known duality of polyhedra.
# For any closed (finite unbounded) manifold of odd dimension (hence, any polytope of even dimension), the Euler-Poincaré characteristic χ = 0.
HTH — Cheers, Steelpillow (Talk) 11:41, 12 January 2011 (UTC)
thread_title: Richeson's explanation

message: Hey, quick note--I notice that the article mentions that the Euler characteristic can be extended to product spaces, but at least in my tiny brain, it is unclear that the notion of product spaces is well-defined.  Even if it is, as I'm sure it must be, are we talking topologically?  Most of the article is suggestive of a geometric interpretation, so I'm a little confused.
Thanks 70.251.0.16 (talk) 23:30, 6 February 2010 (UTC)
replies: text: I have added a brief intro to the Properties section, that attempts to clarify that all of the properties listed are properties of the topological Euler characteristic under operations on topological spaces. Mgnbar (talk) 23:48, 7 February 2010 (UTC)
thread_title:  Product space 

message: The page Euler's polyhedron theorem redirects here.  I guess the "theorem" is that Euler's formula is true, but I've not previously heard it referred to by that name.  Can anyone provide a source for this? or at least confirm that I've guessed correctly? Jowa fan (talk) 07:33, 12 July 2011 (UTC)
replies: text: Here, let me Google that for you. —David Eppstein (talk) 07:42, 12 July 2011 (UTC)
replies: text: Thanks.  I guess I could have visited (oops, can't post a link to the lmgtfy site, it's blacklisted) myself, but it would have been less fun.  I've added the alternative name to the article. Jowa fan (talk) 12:59, 12 July 2011 (UTC)
thread_title:  What is "Euler's polyhedron theorem"? 

message: An anonymous editor just implied that for manifolds of odd dimension, non-orientable examples may have a non-zero Euler characteristic. See the diff.  Does Poincare duality really distinguish between orientable and non-orientable manifolds in this way? — Cheers, Steelpillow (Talk) 12:09, 22 April 2012 (UTC)
replies: text: I'm not sure --- it's been too long since I've studied this stuff --- but as far as I can recall, most textbook formulations of Poincare duality require closedness and orientability (see for example Poincare duality). In that case, you definitely get Euler characteristic 0. I don't know what happens in the non-closed or non-orientable cases. Insofar as we should be conservative, erring on the side of making true but incomplete statements, I guess I agree with the anonymous edit. Mgnbar (talk) 13:23, 22 April 2012 (UTC)

text: Um, and by "true" I of course meant "verifiable". ;) Mgnbar (talk) 13:36, 22 April 2012 (UTC)
replies: text: OK, just got hold of David Richeson's book, Euler's Gem (Princeton, 2008). It is true for all closed manifolds of odd dimension, though the Poincare theorem applies only to orientable ones. I'll clarify the article accordingly. (Sorry, I forgot to specify closure. For non-closed cases, a counter-example is provided by the Euler characteristics χ for a ball and for a solid torus: for a ball χ = +1, while for a solid torus χ = −1). — Cheers, Steelpillow (Talk) 14:18, 22 April 2012 (UTC)
replies: text: Great edit. And yeah, now I realize how easy the non-closed case should have been. Cheers. Mgnbar (talk) 14:50, 22 April 2012 (UTC)
thread_title:  Poicare duality and non-rientability 

message: Just a note to acknowledge that the Euler characteristic for nonconvex polyhedra can be very nonintuitive. This includes shapes that are similar to a sphere in 3D but whose faces may have holes in them: for example, a cube with a cavity on a face, or a cube with a smaller cube stuck onto it. Shapes that are similar to a torus can also be tricky. A torus has characteristic 0, being the product of two circles. A polyhedral approximation to a torus made from wedge-shaped frustums also has characteristic 0; note that each 2D face therein is similar to a disk and has no holes. However by tweaking the angles and merging facets, if two or more 2D faces themselves get holes, then the torus condition doesn't apply any more unless edges are counted differently. Consider a square annulus: 8+8 = 16 vertices, 12+12 = 24 edges, but only 10 faces (4 inside, 4 outside, 1 up, 1 down), giving a characteristic of 16+10-24 = 2. -- 108.122.99.20 (talk) 02:35, 6 April 2015 (UTC)
replies: text: That is not the correct calculation. When faces have holes in them, you have to add a different number than 1 for those faces. It's easier just to require that all faces are disks, but then you would have to subdivide some of them to make that be true. —David Eppstein (talk) 03:03, 6 April 2015 (UTC)
replies: text: OK that's fair. Instead of deleting the edit, could you add them with the correct V/F/E numbers? It'll help me and other non-experts understand the calculation better. -- 108.122.99.20 (talk) 03:08, 6 April 2015 (UTC)
replies: text: Well, we would need a source for the calculation for non-simple faces. It might be in Rote's Geometric Measure Theory, but I can't check that right now (it's in my office and I'm at home). —David Eppstein (talk) 03:18, 6 April 2015 (UTC)
replies: text: Found that http://mathforum.org/library/drmath/view/51815.html covers it somewhat. Using it as a placeholder ref for now. Can improve refs. -- 108.122.99.20 (talk) 03:30, 6 April 2015 (UTC)
replies: text: It would probably be helpful also for you to read Lakatos' Proofs and Refutations, for some background philosophy on why you need to have the definitions carefully nailed down in this problem to avoid going astray. —David Eppstein (talk) 03:41, 6 April 2015 (UTC)
replies: text: Thanks. Will get it from the library. This journey started 6 hours ago for me when I was checking the topology of a CAD drawing for an engineering component I had made. Asking "why is V+F-E not equal to +2 like I was taught?" led me down this rabbit hole of continuing education. The funniest part is that all the (wrong) numbers I entered earlier were from the CAD topology checker. Looks like I need to submit a bug report and point them to this article. In other news, apparently real-world engineering components are being designed with faulty topology... -- 108.122.99.20 (talk) 03:56, 6 April 2015 (UTC)

text: These weird figures are not polyhedra anyway. See for example Richeson; Euler's Gem: The polyhedron Formula and the Birth of Topology. In particular, a face with a hole in is not permissible: it must be subdivided into multiple colanar faces, each of which is itself a topological disc. — Cheers, Steelpillow (Talk) 09:05, 6 April 2015 (UTC)
replies: text: Well, they might or might not be polyhedra depending on exactly what you mean when you define a polyhedron. But a clear definition, and a reliably source that extends the Euler characteristic to things meeting that definition, has to come before adding examples and trying to say what their characteristic is. —David Eppstein (talk) 15:30, 6 April 2015 (UTC)

text: Ok, I looked up the reference I was thinking of (which I wrote incorrectly above: it's Klain and Rota, Introduction to Geometric Probability, section 5.2 The Euler Characteristic, pp. 46–50). They define Euler characteristics for a much more general class of shapes than polyhedra, which they call polyconvex sets; these are finite unions of compact convex sets in Euclidean spaces of any finite dimension. They show that this characteristic is additive: if you decompose a given polyconvex set into a disjoint union of smaller sets, the characteristic of the union is the sum of the characteristics of the smaller sets. They also define a polytope to be a finite union of convex polytopes (which they define to be intersections of finitely many closed halfspaces), and again observe that the Euler characteristic is additive on these shapes. Additionally (unlike for polyconvex sets more generally) the boundary of a polytope is a polytope, so you can calculate Euler characteristics of boundaries, not just of solid polyhedra. So from their point of view all of these examples are perfectly valid, but you have to modify the formula so that instead of adding (–1)dim for a given face, you add the Euler characteristic of the relative interior of the face. This is +1 for a vertex, –1 for a line segment, +1 for a disk-shaped face, 0 for a face with a single disk-shaped hole in its interior, etc. They also define a "system of faces" of a polytope to by a partition of the boundary of the polytope into convex polytopes, and they prove a generalization of Euler's formula for arbitrary polytopes: if you have a system of faces, and add  (–1)dim per face, you get the Euler characteristic. (They don't seem to prove that a system of faces always exists, however.) So, you can generalize the Euler characteristic to the shapes listed above (interpreted as boundaries, not as solids, for consistency with the rest of the table), and it is exactly what one would think it should be (0 for a torus, 2 for a topological sphere). However, it is a mistake to write that the Euler characteristic is or equals the Euler formula V – E + F for these shapes. There is an Euler formula that looks like that, but it doesn't apply to the given vertices, edges, and faces, because they don't form a valid system of faces in the sense of Klain and Rota. —David Eppstein (talk) 19:34, 6 April 2015 (UTC)
replies: text: That sounds a little technical.  I think it would be simpler if we restricted our consideration to simplicial complexes (realized in Euclidean space) or to CW-complexes.  Neither of these is really satisfying (because any convex polytope is as good as a simplex, and a CW-complex doesn't conventionally have straight sides) but they have the advantage that it would be easy to quote results for them and easy to link to other relevant Wikipedia pages.  Ozob (talk) 22:39, 6 April 2015 (UTC)
thread_title:  Nonconvex polyhedra 

message: I visited the page because I forgot the definition of the graded Euler characteristic, but was surprised it wasn't mentioned. Considering its relevance to knot theory (knot Floer homology, Khovanov homology) and other areas of GT, I feel it should at least mentioned on the page. Neuroxic (talk) 01:54, 18 March 2016 (UTC)
thread_title:  no mention of graded Euler characteristic 

message: Substituted at 19:57, 1 May 2016 (UTC)
thread_title: Assessment comment

message: Making a note on correct use of plane versus planar graph, since the article used these terms incorrectly (now corrected). A graph is planar if it can be embedded into the plane, but plane if it is given with such an embedding. Since it is the embedding which determines which the faces of the graph are, the Euler characteristic applies to plane graphs. Any plane embedding of a planar graph would of course have the same number of faces, but that is at best an indirect association of a number of faces with the graph. 130.243.94.123 (talk) 14:17, 8 September 2016 (UTC)
replies: text: Yes, I agree with this correction. —David Eppstein (talk) 16:06, 8 September 2016 (UTC)
thread_title:  Plane graph vs. planar graph 

message: I was about to send a student here who is doing research for an elementary paper about the history of the concept of Euler characteristic, and was dismayed that there's almost nothing here about that history. A short section is justified. I'm hoping someone will write it before I find the time during summer break. Ishboyfay (talk) 20:32, 2 April 2018 (UTC)
thread_title:  History of the concept