Talk:Euler characteristic

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)

In the line, "If $M$ and $N$ are topological spaces, then the Euler characteristic of their product space $M\times N$ is," I'm not sure how to make the M \times N not look gigantic. 00:49, 29 Dec 2004 (CST)

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)

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)

pdf? Why? Is not postscript universally considered superior to pdf? Michael Hardy 23:24, 30 Dec 2004 (UTC)

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)

euler characteristic with compact support; excision

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)

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)

Reorg on 2 Mar 2006

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)

V - E + F

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)

Yes I agree. --Salix alba (talk) 18:02, 6 April 2006 (UTC)

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)

• It seems the Euler formula can be extended to n-dimensional to read:

So-S1+S2-S3+ .... =1,

where S_k is the number of k-dimensional entities. In this notation, V becomes S_o (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=S_o=9, E=S₁=16, F=S₂=10, and S₃=2. It is seen, V–E+F=3 not equal to 2, but S_o–S₁+S₂–S₃=1.

Let us tabulate expansion of the seemingly trivial (1–1)ⁿ⁺¹=0 in the form of S_o–S₁+S₂–S₃+S₄–....=1, and make observations:

For n=0, 1=1, which represents a point with S_o=1.

For n=1, 2–1=1, which represents a line segment with S_o=2, S₁=1.

For n=2, 3–3+1=1, which represents a triangle with S_o=3, S₁=3, S₂=1.

For n=3, 4–6+4–1=1, which represents a tetrahedron with S_o=4, S₁=6, S₂=4, S₃=1.

For n=4, 5–10+10–5+1=1, which represents a 4-D "triangle" with S_o=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 S_o=1.

For n=1, 2–1=1, which represents a line segment with S_o=2, S₁=1.

For n=2, 4–4+1=1, which represents a quadrilateral with S_o=4, S₁=4, S₂=1.

For n=3, 8–12+6–1=1, which represents a cube with S_o=8, S₁=12, S₂=6, S₃=1.

For n=4, 16–32+24–8+1=1, which represents a 4-D "square" with S_o=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)

Indeed. These are the Betti numbers discussed in the article. --Salix alba (talk) 17:32, 15 September 2006 (UTC)

Proposed Merger

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)

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)

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)

Now on AfD. --Salix alba (talk) 20:00, 24 April 2006 (UTC)

Hey

The Euler char of a torus is said to be 0 but what about a cube that has a cube shaped hole going threw it. V=16, E=24, F=10, e(x)=2! right? Try the same with a klien bottle. This make the shapes have edges, vertexis, and more then 1 faces. Joerite 05:24, 1 October 2006 (UTC)

A cube with a smaller cuboid hole cut through it (as a topologically equivalent torus) would have χ=0, if the new faces are properly defined as simple polygons, and exactly two faces per edge. I'd look at it as a cube with a 3x3 grid of squares on each cubic face. Then subtract 2 central squares from opposite faces, and add 4 inner "tunnel" rectangles across the volume center. As a result there's same number of vertices, four new edges, a NET increase of two more faces, so χ will decrease by 2 from the original. Tom Ruen 05:33, 1 October 2006 (UTC)

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)

euler characteristic with compact support; excision

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)

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)

Reorg on 2 Mar 2006

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)

V - E + F

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)

Yes I agree. --Salix alba (talk) 18:02, 6 April 2006 (UTC)

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)

• It seems the Euler formula can be extended to n-dimensional to read:

So-S1+S2-S3+ .... =1,

where S_k is the number of k-dimensional entities. In this notation, V becomes S_o (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=S_o=9, E=S₁=16, F=S₂=10, and S₃=2. It is seen, V–E+F=3 not equal to 2, but S_o–S₁+S₂–S₃=1.

Let us tabulate expansion of the seemingly trivial (1–1)ⁿ⁺¹=0 in the form of S_o–S₁+S₂–S₃+S₄–....=1, and make observations:

For n=0, 1=1, which represents a point with S_o=1.

For n=1, 2–1=1, which represents a line segment with S_o=2, S₁=1.

For n=2, 3–3+1=1, which represents a triangle with S_o=3, S₁=3, S₂=1.

For n=3, 4–6+4–1=1, which represents a tetrahedron with S_o=4, S₁=6, S₂=4, S₃=1.

For n=4, 5–10+10–5+1=1, which represents a 4-D "triangle" with S_o=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 S_o=1.

For n=1, 2–1=1, which represents a line segment with S_o=2, S₁=1.

For n=2, 4–4+1=1, which represents a quadrilateral with S_o=4, S₁=4, S₂=1.

For n=3, 8–12+6–1=1, which represents a cube with S_o=8, S₁=12, S₂=6, S₃=1.

For n=4, 16–32+24–8+1=1, which represents a 4-D "square" with S_o=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)

Indeed. These are the Betti numbers discussed in the article. --Salix alba (talk) 17:32, 15 September 2006 (UTC)

Proposed Merger

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)

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)

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)

Now on AfD. --Salix alba (talk) 20:00, 24 April 2006 (UTC)

Hey

The Euler char of a torus is said to be 0 but what about a cube that has a cube shaped hole going through it. V=16, E=24, F=10, e(x)=2! right? Try the same with a Klein bottle. This make the shapes have edges, vertices, and more than 1 faces. Joerite 05:24, 1 October 2006 (UTC)

A cube with a smaller cuboid hole cut through it (as a topologically equivalent torus) would have χ=0, if the new faces are properly defined as simple polygons, and exactly two faces per edge. I'd look at it as a cube with a 3x3 grid of squares on each cubic face. Then subtract 2 central squares from opposite faces, and add 4 inner "tunnel" rectangles across the volume center. As a result there's same number of vertices, four new edges, a NET increase of two more faces, so χ will decrease by 2 from the original. Tom Ruen 05:33, 1 October 2006 (UTC)

Formula Name?

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)

Roman surface Euler characteristic is wrong

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)

SVG images

13:51, 19 June 2007 Patrick (Talk | contribs) (12,392 bytes) (→Proof of Euler's formula - 
rm Image:V-E+F=2 Proof Illustration.svg, does not work)

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)