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欧拉定理的证明 2011-05-21 17:21:39

先叙述一下欧拉定理,应该说是欧拉定理之一,以欧拉名字命名的数学名词有900多项。

任何一个多面体(不论凹凸), 棱数(L),顶角数(M),面的数目(N),满足如下关系:

N + M - L =2.。

先拿掉一个面,然后把其表面“摊”平。在这过程中,面上每个多边形的几何尺寸会改变,但点面棱的拓扑关系保持不变。

先假定这摊平的多边形的外围有个三角形,我们把这个三角形拿掉。这样L-2,N-1, M-1。所以欧拉定理如果成立,拿掉这个三角形后还是成立。所以我们可以把所有三角形拿掉,不改变欧拉定理的正确性。

我们把所有的K边形(K〉3)添对角线变成两个多边形。这个添加过程中,L+1, N+1, M不变,所以欧拉定理继续成立。这样可以把K边形变成 K-2  个三角形而不改变欧拉定理的正确性。

我们把所有的三角形(原来的和自己制造的)全部拿完,到最后一个时

L=3,M=3, N=1

这时

N + M - L = 1

记住在第一步我们拿掉了一个多边形。这使M-1,L和N不变。把这个面加上去就是欧拉定理。

这个近乎完美工艺品的数学证明,我是在1981年聆听费鹤良老师在上海师院作科普讲座时知道的。他现在当然已经是费鹤良教授了,是《十万个为什么》数学分册的编委。

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作者:wtxwtx 留言时间:2015-05-25 13:25:09
1. 你的命题有错:
“任何一个多面体(不论凹凸), 棱数(L),顶角数(M),面的数目(N),满足如下关系:

N + M - L =2.。”

应为“任何一个凸多面体, 棱数(L),顶角数(M),面的数目(N),满足如下关系:

N + M - L =2.。”

2. 哈佛教授给的证明明显比你的好!我读了你的,没懂,为此,上网查到了哈佛教授给出的一个证明,我懂了,而且懂得更多。

3. 我还是很喜欢阅读你的作品。

http://people.fas.harvard.edu/~ivogt/Exposition/EulerChar_PrimesTalk.pdf

AN INTRODUCTION TO THE EULER CHARACTERISTIC
ISABEL VOGT
1. The Euler Characteristic of a Polyhedron
In this talk we will work almost exclusively with the Euler characteristic of a polyhedron. A
polyhedron is a solid in three dimensions that consists of straight edges and faces. Two distinct
faces intersect in an edge and two distinct edges intersect in a vertex.
Definition 1.1 (The Euler Characteristic). Let P be a polyhedron with V vertices, E edges, and
F faces. Then we define the Euler characteristic to be
χE(P) = V − E + F.
Notice that the Euler characteristic is the alternating sum of the “building blocks” of our polyhedron
in each dimension: the vertices are dimension 0 and get a + sign, the edges are dimension
1 and get a − sign, and the faces are dimension 2 and get a + sign.
The Euler characteristic is best understood by example.
Example 1.2 (Cube). Let C be the cube in 3-dimesions:

• •


• •

There are 8 dots in the above diagram of the cube, so V = 8. Simiarly, E = 12 and F = 6. So
χE(C) = V − E + F
= 8 − 12 + 6
= 2.
Example 1.3 (Tetrahedron). Let T be the tetrahedron in 3-dimensions:




Here V = 4, E = 6, F = 4, so
χE(T) = 4 − 6 + 4 = 2.
From these examples it seems that the Euler characteristic of polyhedra also satisfy the seemingly
mysterious relation the χE = 2 seen for connected planer graphs. However, the above polyhedra
have a very special property: they can be drawn on the surface of a sphere. Imagine that one of
the above polyhedra were made of an elastic rubber material like a balloon. Then you could image
“blowing up” the polyhedron through an imaginary hole at one of the vertices, and the figure would
lie on the face of a sphere!
1
2 ISABEL VOGT
From here it is not hard to see why the Euler characteristic of these examples must be 2, given
the theorem for planer graphs. Given a graph on the circle, we could puncture a tiny hole in one of
the faces of the graph, grab the sphere through that whole, and “splay” it out so that it lies flat,
so that the boundary of our tiny puncture becomes the surrounding boundary of the... connected
planer graph!
There are, however, polyhedra that cannot be drawn on the surface of a sphere. Consider the
following examples.
Example 1.4. Let D be the cube with a central shaft removed:

• •


• •

• •
• •
• •
• •
In this case, V = 16, E = 32, F = 16. So in total
χE(D) = 16 − 32 + 16 = 0 6= 2!
If we imagine doing the same procedure of “blowing up” to the polyhedron D above, we would
end up with something that looked like a donought, called a torus:
(insert image of a torus)
Clearly, for this polyhedron D lying on the surface of a torus, the Euler characteristic is 0. But
is this true of all polyhedra that can be drawn on the surface of a torus?
2. A Topological Invariant
In order to understand if it should be true that the Euler characteristic “reads” the information
of what kind of surface a certain polyhedron can be drawn on, we must understand what kinds of
changes to a graph leave the Euler characteristic unchanged.
Lemma 2.1. The Euler characteristic of a polyhedron does not change when we subdivide a face
or an edge.
Proof. As an example, consider the following triangle that may be part of the underlying graph of
a polyhedron:
• •

Now, if we subdivide the face of triangle, this involves adding a single vertex, and then 3 edges
to connect this vertex to the other 3 vertices of the trianglular face:
• •


AN INTRODUCTION TO THE EULER CHARACTERISTIC 3
In general, a face might be an n-gon. We still add a single vertex, so the number of vertices V
changes by ∆V = 1. This vertex must connect to all outer vertices on the n-gon, of which there
are n. This means that E increases by n, so ∆E = n. These new edges divide the face into n
compartments (subfaces) where there used to just be 1. Thus F has increased by ∆F = n + 1. So
in total the Eular characteristic changes by
∆χE = ∆V − ∆E + ∆F
= 1 − n + n − 1
= 0,
so the process of subdividing a face does not change the global Euler characteristic.
Similarly, if we “subdivide” an edge we introduce one new vertex in the middle of the edge, so
∆V = 1. We also have divided one edge into two, so ∆E = 1. Thus in total
∆χE = ∆V − ∆E + ∆F = 1 − 1 + 0 = 0,
and again this process does not affect the global Euler characteristic.

Lemma 2.2. The Euler characteristic of a polyhedron does not change when we delete an edge
bordering distinct faces, or a vertex bordering 2 distinct edges.
Remark. It is necessary to have these qualifications on to which edges and vertices the lemma applies,
in order to disqualify the following examples:

vertex meets more
than 2 edges

loop around
a single vertex
edge bounds
same face
for which the vertex (respectively edge) is essential for the “integrity” and information inherent in
the graph.
Proof. As in the previous lemma, we will argue by looking at the local changes to the Euler characteristic
caused by our deletion. Upon deleting an edge, clearly ∆E = −1. But, additionally, as
long as the edge bordered 2 distinct faces, the number of faces decreases by 1 as well, so ∆F = −1.
So in total:
∆χE = ∆V − ∆E + ∆F = 0 + 1 − 1 = 0.
Similarly, if we delete a vertex bordering 2 distinct edges then ∆V = −1, ∆E = −1 and ∆F = 0,
so
∆χE = ∆V − ∆E + ∆F = −1 + 1 + 0 = 0.

This last lemma suggests that the underlying graph of a polyhedron drawn on some surface S
can be “stripped down” to its essentials - that is vertices that do not border distinct edges and
edges that do not border distinct faces - without changing the Euler characteristic.
Using both of these lemmas, we can now answer our question posed originally,
Do all polyhedra that can be drawn on the same
surface have the same Euler characteristic?
with the following theorem.
Theorem 2.3. The Euler characteristic of a polyhedron depends only on the surface on which it
can be drawn; in other words, it is constant for polyhedra that can be drawn on the same surface.
4 ISABEL VOGT
Sketch of Proof. Any graph can be transformed into any other graph by subdividing faces and edges
and deleting edges and vertices. By Lemmas 2.1 and 2.2 these do not affect the Euler characteristic.
We could use this theorem to give a definition of the Euler characteristic of a surface as the
Euler characteristic of the class of polyhedra that can be drawn on its surface. There is a more
sophisticated way to define the Euler characteristic of a topological space, but for the surfaces we
are considering, these definitions agree.
Thus the Euler characteristic is something we might call a crude topological invariant. Knowing
the Euler characteristic of a surface, or polyhedron tells you something about its shape, its topology,
irrespective of stretching, skewing, “blowing up”, or similar nondestructive deformations.
We have already seen that the Euler characteristic of a sphere is 2 and of a torus is 0. What
about other surfaces? It turns out that for a donought with g holes, what we call a g-hole torus,
the Euler characteristic has a very nice form.
Theorem 2.4. Let Mg be a g-hole torus. Then we have
χE(Mg) = 2 − 2g.
Remark. Note that in the cases we already know, this theorem agrees with what we have calculated.
When g = 0, the surface is just the sphere, and so
χE(M0) = 2 − 2 · 0 = 2.
And when g = 1, the surface is the torus so
χE(M1) = 2 − 2 · 1 = 0.
3. Applications of the Euler Characteristic
3.1. Brussel Sprouts. In the course of playing Brussels Sprouts, you create a geometric object
known as a planer graph.
The game of Brussels Sprouts goes as follows:
(1) Draw some number n of four-pointed crosses. In the game below, n = 2.
(2) On each round, players alternate joining crosses by drawing a line from one point of one
cross, to another point of another cross, without intersecting any the other lines already
drawn. Then adding a dash in the middle of the line, so as to create another cross in the
middle (a cross that is already occupied by in two positions).
(3) The game ends when it is no longer possible to add any lines without there being an intersection
somewhere.
Here is an example of a game starting with 2 crosses. As you will see, there are 8 rounds and
player 2 wins.
AN INTRODUCTION TO THE EULER CHARACTERISTIC 5
In fact, a game with 2 crosses will always be a win for the second player, and you can prove this
using the Euler characteristic
3.2. Platonic Solids. The Platonic solids are polyhedra built out of regular polygons (like squares,
pentagons, triangles) such that the number of edges meeting each vertex is constant (the same for
each vertex).
The Euler characteristic allows us to prove that there are only 5 platonic solids, in other words,
this list is exhaustive.
3.3. Fixed Point Theory. Let S be a surface. Then in an appropriate sense that I cannot make
less vague here, χE equals the sum of the fixed points of S (with appropriate coefficients) under an
infinitessimal deformation of S.
Consider the case that S is just a sphere. Then one such “infintessimal deformation” of S is just
to draw a line through the sphere, passing through the north and south poles and rotate the sphere
about this axis. This map has 2 fixed points: the north and south poles where the line forming the
axis of the rotation met the sphere.
6 ISABEL VOGT
We can visualize this idea by thinking of a “hairy” sphere. Then by combing the hairs in some
particular way gives us a map of the sphere: send each point to the corresponding point at the end
of the ahir. A fixed point would correspond to a hair pointing straight up, ie a cowlick of hair.
The fact that χE of a sphere is 2 means that any “combing” of the sphere always has at least one
cowlick! This is also known as the hairy-ball theorem.
Department of Mathematics, Harvard University
E-mail address: ivogt@college.harvard.edu
URL: www.people.fas.harvard.edu/~ivogt
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