Math 3181
Dr. Franz Rothe
May 29, 2014
08SUM\3116_2014h2a.tex
Names:
Homework turned in has to contain p.1 through 11 of this handout.
The homework can be done in groups up to three
due June 4
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Homework
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Theorem 1 (Euler’s formula). For a planar connected graph or multigraph, the number n of vertices, number f of faces, and number m of edges satisfy
n+f −m=2
Note the outer (unbounded) face has to be counted, too.
Definition 1 (Angle defect). For a convex polyhedron, the angle defect at a vertex
is 360◦ minus the sums of the face angles at that vertex. The total angle defect is the
sum of the angle defects at all vertices of the polyhedron.
Theorem 2 (Descartes’s version of the Euler formula). The total angle defect of
a convex polyhedron equal 720◦ ,—which is 4π in arc measurement.
10 Problem 2.1. Check Euler’s formulas for the square pyramid.
10 Problem 2.2. Check Descartes’s version directly for a square pyramid with
four equilateral triangles and a square as its faces.
10 Problem 2.3. We assume that all faces have the same length l. Prove under
this additional assumption that Euler’s formula implies Descartes version.
1
10 Problem 2.4. Euler’s formula implies Descartes version, in the general case
of an arbitrary polyhedron. Complete the proof below.
Proof that Euler’s formula implies Descartes’ version. We calculate the sum S of all
face angles in two ways. Obviously
S = n · 360◦ − D
(2.1)
where D is the total defect. On the other hand, we sum the face angles over all ?
. Let fl denote the number of faces of length l. These faces contribute ?
to the total sum S. Hence the sum of all face angles is
hX
X
X i
(2.2)
S=
fl (l − 2) · 180◦ =
l fl − 2
fl · 180◦ = (m − f ) · 360◦
In the last step, we have used
X
l fl = 2m
and
X
fl = f
which hold by the ?
and the definition of the number
of faces. Since the two expressions (2.1) and (2.2) for the total sum S are ?
S = n · 360◦ − D = (m − f ) · 360◦
Hence ?
implies Descartes’ statement
D = (n + f − m) · 360◦ = 720◦
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2.1
Archimedean polyhedra
10 Problem 2.5. Draw the icosahedron graph in two ways:
• begin near the middle of your page with the star K1,5 , and use as long as convenient
the 5-fold symmetry around the middle vertex;
• begin near the middle of your page with the a triangle C3 , and use as long as
convenient the 3-fold symmetry.
Check the isomorphism between your two graphs. With some numbering or coloring
indicate the isomorphism.
3
10 Problem 2.6. The cuboctahedron is the Archimedean polyhedron for which
all vertices have degree 4 and are surrounded by faces of length (3, 4, 3, 4). Draw the
planar graph for the cuboctahedron and four-color its edges in a proper way.
4
10 Problem 2.7. Draw the planar graph for the snub cube. It is helpful to
remember that the six squares lie on the faces of a surrounding cube. One begins with
the small square in the middle. Later appear four squares on the sides of the cube. A
very large square with much distortion appears only in the end.
The square have their sides rotated against the squares of the surrounding cube. The
entire figure has a left or right handed orientation.
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10 Problem 2.8. Give a reason for Lemma 1 part (i).
10 Problem 2.9. Give a reason for Lemma 1 part (ii).
6
10 Problem 2.10. Give a reason for Lemma 2 part (i).
10 Problem 2.11. Give a reason for Lemma 2 part (ii) and provide a drawing.
7
2.2
The geometric dual
Figure 1: Find the Poincaré duals.
10 Problem 2.12. Draw the Poincaré dual of each of the four plane drawings of
planar, connected graphs, given in the figure on page 8.
10 Problem 2.13. For the graphs G and H from the last problem, count the
vertices and faces of all degrees and length. Write down their numbers ni and fl , as well
as fi∗ and n∗l .
8
10 Problem 2.14. Convince yourself that each pyramid is isomorphic to its dual
and provide a few drawings for illustration.
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2.3
Dual Archimedean polyhedra
10 Problem 2.15. The rhombus-dodecahedron is the dual Archimedean polyhedron for which all faces have length 4 and are surrounded by vertices of degrees (3, 4, 3, 4).
Draw the planar graph for the rhombus-dodecahedron and four-color its edges in a proper
way.
10
10 Problem 2.16. Let the truncated tetrahedron be graph G.
• Produce planar drawings of G, and of its dual G∗ .
• Count the vertices and faces of all degrees and length.
• Write down their numbers ni and fl , and fi∗ and n∗l .
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2.4
Five series of polyhedra with a main rotational axis
Here are three-dimensional geometric descriptions for these polyhedra. The figures on
page 13 and 14 contain their automorphism groups, too.
• An n-sided pyramid consists of a regular n-gon as base, and a top vertex lying on
the main rotational axis at a higher level. In the notation introduced earlier, it is
the graph Cn ∨ N1 .
• An n-sided prism consists of two regular n-gons lying in parallel planes. Their
corresponding vertices are connected by parallel edges. This is the graph Cn × P2 .
• An n-sided doubleprism consists of a regular n-gon, and two extra vertices t and
b on the main rotational axis. The vertices of the n-gon are connected by edges
two both the top vertex t and the bottom vertex b. It is the graph Cn ∨ N2 .
• To obtain the n-sided antiprism, we imagine that one of the n-gon of a prim is
rotated by an angle 360◦ /(2n). Afterwards the vertices of the two polygons are
connected by a path with vertices alternating between the two polygons. It turns
out that the three-sided antiprism is isomorphic to the octahedron.
• To obtain the n-sided antipyramide, we take a regular 2n-gon with vertices alternating in two parallel planes, and put two extra vertices t and b on the main
rotational axis. Each vertex of the 2n-gon is connected by an edge either to the
top vertex t or the bottom vertex b, alternatingly as one circulates the polygon.
It turns out that the three-sided antipyramid is isomorphic to the cube.
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Figure 2: Pyramid, prism and doublepyramid
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Figure 3: Antiprism and antipyramid
2.5
Archimedean polyhedra
To obtain a complete list of the Archimedean polyhedra, we use Descartes’s version
of the Euler formula Theorem ?? together with a few observations gathered in the
Lemmas 1, 2 and 3 below.
Lemma 1. For a 3-regular Archimedean polyhedron with vertex star (a, b, c) and n
vertices
(i)
2 1
1 1 1
+ + = + ;
a b c
n 2
(ii) If a 6= b then c is even.
Lemma 2. For a 4-regular Archimedean polyhedron with vertex star (a, b, c, d) and n
vertices
(i)
2
1 1 1 1
+ + + = + 1;
a b c d
n
(ii) If a = 3 then b = d.
Lemma 3. For a 5-regular Archimedean polyhedron with vertex star (a, b, c, d, e) and n
vertices
(i)
1 1 1 1 1
2 3
+ + + + = + ;
a b c d e
n 2
(ii) hence there are only three possibilities: a = b = c = d = 3 and e = 3, 4 or 5.
Theorem 3. The complete list of Archimedean polyhedra is given in the table below,
and consists of
(a) the five Platonic bodies;
(b) the n-sided prisms for n = 3 and n ≥ 5;
(c) the n-sided anti-prisms for n ≥ 4;
(d) and the extra polyhedra listed below. Only the latter are depicted in the figure on
page 18.
In each instance, the Archimedean graph G and its automorphism group Aut(G) have
the following properties:
(i) For any two vertices v and w there exists an isomorphism g ∈ Aut(G) such that
gv = w. Thus the automorphism group acts transitive on the vertices. In everyday
words: all vertex figures, (also called ”clown hats”) look the same.
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(ii) With the exception of the pseudo-rhombi-cuboctahedron, snub-cube and snub-dodecahedron, all other Archimedean bodies can be obtained by truncation of either vertices or edges from a Platonic body. In some cases, in a second step repetition of
this process is necessary. The Archimedean body has the same the automorphism
group as the ancestral Platonic body.
The rhombi-cuboctahedron exists in two different non-isomorphic version which have the
same vertex star (3, 4, 4, 4), but different graphs and different automorphism groups. The
automorphism group of the rhombi-cuboctahedron is the cube-group with 48 elements, but
the automorphism group of the pseudo-rhombi-cuboctahedron is the semidirect product
D4 ×s S2 and has the order 16.
Theorem 3 is proved simply by drawing the appropriate graphs and checking their
properties. With a bid of spherical geometry, one gets the following more stringent
Theorem 4 about the metric properties of the Archimedean polyhedra.
Theorem 4. To each Archimedean graph corresponds exactly one Archimedean polyhedron ,—except for the snub-cube and the snub-dodecahedron which exist in two mirror
symmetric versions.
Proposition 1. The Archimedean polyhedra have the following properties:
(i) All vertices lie on a sphere, and the midpoints of the edges lie on a concentric sphere.
(ii) The automorphism group consists of rotations in the group SO3 , and as many
reflections or rotational reflections. There are two exceptions: The automorphism
group of the snub-cube is the group of rotations mapping a cube to itself, and has
24 elements. The automorphism group of the snub-dodecahedron is the group of
rotations of an icosahedron and has 60 elements.
(iii) All vertex figures (”clown hats”) are congruent.
(iv) Let fl be the number of faces with l sides and l be the defect of its angle sum over
the Euclidean value (l − 2) · 180◦ . The total angle defect is
X
fl l = 720◦
l
Remark. To really obtain semi-regular polyhedra in the case of the n-sided prisms, and
the n-sided anti-prisms, one has to choose their heights appropriately: for the n-sided
prism such that its sides are squares, for the n-sided anti-prisms such that its sides are
equilateral triangles.
Remark. The snub-cube and the snub-dodecahedron exist in two different versions.
These are isomorphic as graphs, but mirror images of each other.
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The situation for the snub-cube, as well as the snub-dodecahedron, can be put into
the language of group theory: The reflection R taking the snub-cube to its mirror image
is an outer automorphism of the automorphism group Aut(Snub), but not an inner
automorphism g 7→ sgs−1 . This happens since the reflection R ∈
/ Aut(Snub).
Too, as an outer automorphism, the reflection R leads to more inequivalent irreducible representations of the groups Aut(Snubcube). The situation of the snub-dodecahedron is similar. In this way, one gets extra representations with non-integer characters
of the alternating group A5 , which is isomorphic to Aut(Snub − dodecahedron).
name
tetrahedron
cube
octahedron
dodecahedron
icosahedron
vertex star
(3, 3, 3)
(4, 4, 4)
(3, 3, 3, 3)
(5, 5, 5)
(3, 3, 3, 3, 3)
v
4
8
6
20
12
n-side prism, n 6= 4
n-faced antiprism, n 6= 3
cuboctahedron
icosidodecahedron
(4, 4, n)
(3, 3, 3, n)
(3, 4, 3, 4)
(3, 5, 3, 5)
2n
2n
12
30
truncated
truncated
truncated
truncated
truncated
truncated
truncated
(3, 6, 6)
(4, 6, 6)
(3, 8, 8)
(5, 6, 6)
(3, 10, 10)
(4, 6, 8)
(4, 6, 10)
12
24
24
60
60
48
120
tetrahedron
octahedron
cube
icosahedron
dodecahedron
cuboctahedron
icosidodecahedron
f3
4
f4
f5
f10
fn
6
12
20
n
2n
8
20
4
8
2
2
6
8
12
6
6
6
4
8
6
12
20
20
12
12
30
8
20
24
24
60
8 18
8 18
20 30 12
snub-cube
snub-dodecahedron
24
60
32
80
17
f8
8
rhombi-cuboctahedron
(3, 4, 4, 4)
pseudo-rhombi-cuboctahedron (3, 4, 4, 4)
rhomb-icosidodecahedron
(3, 4, 5, 4)
(3, 3, 3, 3, 4)
(3, 3, 3, 3, 5)
f6
6
12
6
12
Aut(G) |Aut|
T ×s S2
24
C × S2
48
C × S2
48
I × S2
120
I × S2
120
Dn × S2
Dn ×s S2
C × S2
I × S2
4n
4n
48
120
T ×s S2
C × S2
C × S2
I × S2
I × S2
C × S2
I × S2
24
48
48
120
120
48
120
C × S2
D4 ×s S2
I × S2
48
16
120
C
I
24
60
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