Download 1 Appendix to notes 2, on Hyperbolic geometry:

1230, notes 3
1
Appendix to notes 2, on Hyperbolic
geometry:
The axioms of hyperbolic geometry are axioms 1-4 of Euclid, plus
an alternative to axiom 5:
Axiom 5-h: Given a line l and a point p not on l; there are at least
two distinct lines which contain p and do not intersect l:
Theorem: Assuming Euclid’s …rst four axioms, and axiom 5-h, and
given l and p not on l; there are in…nitely many lines containing
p and not intersecting l .
Proof ?
Comment on the “extendability” axiom.
The wording in Euclid is a bit unclear. Recall that I stated it as
follows:
A line can be extended forever.
This is basically what Euclid wrote. Hilbert was more precise,
and adopted an axiom also used by Archimedes:
Axiom: If AB and CD are any segments, then there exists a
number n such that n copies of CD constructed contiguously
from A along the ray AB willl pass beyond the point B .
A model of hyperbolic geometry: Consider an open disk, say of
radius 1, in the earlier model of Euclidean geometry. Thus, let
n
D= ( ; ) j
2
+
2<
o
1
De…nition: A “point” is a pair ( ; ) in this set.
De…nition: A “line”is any diameter of this disk, or the set of points
on a circle which intersects D and which meets the boundary of
D in two places and at right angles to the boundary of D at both
of these places. Notice that a diameter can be considered part of
such a circle of in…nite radius.
However, there a lot to be said before we can claim that this model
satis…es Euclid’s …rst four axioms. In particular, work is required
to de…ne what “congruent” line segments are. It is necessary
to de…ne some sort of distance between points which is di¤erent
from Euclidean distance. With this de…nition, it turns out that
a line segment near the boundary can look very short to us, and
yet according to the de…nition of distance, be congruent (same
length) as a segment near the middle of the disk which looks a
lot longer to us. It is in this sense that a “line” can be extended
inde…nitely (to “in…nity”). This will be discussed in more detail if
time permits later.
1.1
Euclidean solid geometry
Planes are mentioned in some of the 1st 20 de…nitions in chapter
1.
They don’t appear again until chapter 11, where there are new de…nitions (e.g. "solid angle", de…nition 11), but no new postulates.
(Hilbert included several axioms about planes.)
2
Euclid, Book 13 "Platonic Solids”
What are they?
De…nitions:
Although Platonic solids are three dimensional objects, we start
with a de…nition in R2:
1. polygon. (class exercise)
As an example of a polygon we have, for example, triangles,
squares, etc. Please note that I am talking here about the boundary and everything inside. For example, the set
S = f(x; y ) j 0
x
1 and 0
y
1g
is a polygon (called the “unit square”). Its boundary is
(
(x; y ) j (x; y ) 2 S and either x = 0; x = 1;
o9r
y = 0 or y = 1
)
but this set is not itself a polygon, at least not as a polygon is
usually de…ned.
De…nitions:
1. polygon
A polygon in R2 is a closed bounded subset of R2 with the following properties:
(a) It has a non-empty interior.
(b) Its boundary consists of a …nite set of line segments.
(c) Each of these segments intersects exactly two of the others,
one at each of its endpoints. There are no other intersections
between these segments.
(d) The boundary is connected.
A polygon in R3 is a subset of a plane which is a polygon in that
plane.
What “unusual” polygons are there with this de…nition?
2. A set in Rn, is “convex” if each intersection of this set with a
(straight) line is either empty, a single point, or a segment of that
line.
3. A polygon is regular if it is convex, each edge is congruent
to each other edge and all the angles between adjacent edges are
equal.
convex, regular
non-convex
irregular
Now we move to three dimensional objects.
4. A polyhedron is a connected closed bounded region in R3 whose
boundary consists of a …nite set of polygons. These polygons are
called “faces” of the polyhedron. Polyhedra can also be convex
or non-convex.
convex polyhedron j nonconvex polyhedron
Not a polyhedron
But, all de…nitions are arbitrary, and authors often di¤er, according to the setting in which they are working. The de…nition I gave
probably allows for some weird looking polyhedra. Some authors
don’t even attempt to de…ne polyhedron, but simply de…ne “convex polyhedron”, which would just add the conditon that the set is
convex to what I have above, but would very much simplify what
is allowed.
Test of geometric intuition: Start with a cube.
Looking down from the top, cut vertical slices (parallel to the
vertical edges) along the four slanted lines, thus cutting o¤ the
four vertical edges.
Make the same cuts on a second side,
oriented as shown.
Do this again.
Do this again.
What is the result?
De…nition: A Platonic solid, or “regular polyhedron”, is a
convex polyhedron such that all faces are congruent to each
other, as are all solid angles, and each face is a regular
polygon.
Plato: “The …rst will be the simplest and smallest construction,and its element is that triangle which has its hypotenuse twice
the lesser side. When two such triangles are joined at the diagonal, and this is repeated three times, and the triangles rest their
diagonals and shorter sides on the same point as a centre, a single
equilateral triangle is formed out of six triangles; and four equilateral triangles, if put together, make out of every three plane
angles one solid angle , being that which is nearest to the most
obtuse of plane angles; and out of the combination of these four
angles arises the …rst solid form which distributes into equal and
similar parts the whole circle in which it is inscribed.”
See Elements, chapter 11 for de…ntion of "solid angle"
He is describing a tetrahedron, in which three equilateral triangles
meet at each vertex:
sum of angles at vertex: 3
Other possibilities:
60o = 180 < 360:
Four triangular faces meet at a vertex. This gives an octahedron
sum of angles at vertex = 4
60 = 240 < 360:
Five triangular faces meet at a vertex, giving an icosahedron (20
faces):
sum of angles at vertex = 5
60 = 300 < 360
Can six triangular faces meet at a vertex? Angle sum = 6
60 = 360; so the surface would be ‡at and we don’t get a three
dimensional solid. (See Euclid, Book 11, Propositions 20,21.)
How about square faces? The only possibility is a cube – three
faces meet at a vertex.
sum of angles at vertex= 3
90 = 270::
Four faces meeting at a point would give a sum of 360
There is one with pentagonal faces, three meeting at a vertex.
This gives a dodecahedron (12 faces)
sum of angles at vertex: 3
108 = 324 < 360:
That’s all there can be (…ve), because any others would give too
large an angle sum ( 360 )
Exactly what has been proved here?
Theorem: There are no more than …ve platonic solids.
Theorem: There are no more than …ve platonic solids.
But how do we know there are any at all?
The greatest idea of Greek mathematicians: Assertions such as
the existence of any of the platonic solids need to be “proved”.
The greatest idea of Greek mathematicians: Assertions such as
the existence of any of the platonic solids need to be “proved”.
For example, do the Platonic solids exist without the parallel postulate? (This may be the universe in which we live!)
The greatest idea of Greek mathematics: Assertions such as the
existence of any of the platonic solids need to be “proved”.
Earlier example: The "Pythagorean theorem". The result was
discovered many times, in many places, but as far as we know,
without proof before Pythagoras.
Construction of a cube. Proposition 15, Book 13.
To construct a cube, and enclose it in a sphere.
Let the diameter AB of the given sphere be laid on,y and let it have
been cut at C such that AC is double CB.z And Let CD have been
drawn from C at right angles to AB. And let the semi-circle ADB
have been drawn on AB. And let DB have been joined. And let
the square EFGH, having its side equal to DB, be laid out.x And
let EK, FL, GM, and HN have been drawn from points E, F, G,
and H, respctively, at right angles to the plane of square EFGH.{
And let EK, FL, GM, and HN, equal to one of EF, FG, GH, and
HE, have been cut o¤ from EK, FL, GM, and HN, respectively.
And let KL, LM, MN, and NK have been joined. Thus a cube
contained by six equal squares has been constructed.
Question: Did this require the parallel postulate?
y Book
z prop
1, proposition 2
6.10
x proposition
{ prop.
11.12
46, book 1
De…nition: Let P be a convex polyhedron. Then the “dual” of
P is the polyhedron with vertices at the centers of the faces of P;
and edges connecting the centers of any two faces with a common
edge.
Proposition: If P has v vertices, f faces, and e edges, then the
dual of P has f vertices, e edges, and v faces.
Construction of an Octahedron: Book 11, prop 14 or:
The polyhedron whose vertices are the centers of the faces of
a cube is an octahedron. (The "dual" of the cube".) Also,
the polyhedron whose vertices are the centers of the faces of an
octahedron is a cube.)
Construction of a tetrahedron: Proposition 13, Book 13. (pg.
519 in Fitzpatrick edition, 37 lines) or: Prove that an appropriate
set of 4 vertices of a cube form a tetrahedron.
Construction of a Dodecahedron: Proposition 17. (pg. 530,
88 lines)
Construction of an icosahedron: Prop 16, (pg. 512. 85 lines)
“Woodworker’s proof” that a dodecahedron exists
In the slicing method we leave, at the end, one line segment on
each of the original faces of the cube. We assume that these are
of equal length and symmetrically placed.
Consider three of the cut planes as shown:
Here we are assuming that each of the cutting planes goes from a
center line of one face to a point a distance d2 from the center line
of another face, with orientation as shown. We need to calculate
d in terms of the length s of a side of the cube. In fact, d will the
length of an edge of our dodecahedron.
We calculate the intersection of the three planes (yellow dot). This
is a straight forward problem in linear algebra. After some further
algebra it turns out that in order for that point to be the distance
d from the three other points we must have
sd
2
2
s
+ s4 + (s
d)4 = d2 (4s
2d)2
(1)
If we can …nd a positive solution to this equation which is less
than s then we will have proven that a dodecahedron exists. In
fact, some algebra leads to four solutions,
p
!
1+ 5
golden mean,
1:618 s;
d=
s
but we want d < s
2
p
5
1
d=
s (< 0)
2p
3+ 5
d=
s (> 1)
2
0
1
p
only feasible value,
3
5 B
C
d=
s; @ = 2 golden mean A
2
0:382 s
d=
3
p
5
s
2
Here is one of the pentagons (yellow dots) :
The golden mean can be constructed by methods of Euclid (straight
edge and compass; book 2, proposition 11), and so in essence this
method can be said to be Euclidean. But algebra makes it easier
to prove this.
The Icosahedron is the dual of the dodecahedron, and vice-versa.
Better method for constructing a dodecahedron: Start with the
icosahedron, using a construction due to Luca Pacioli, 1509. (See
Stillwell, page 21-22)
First de…ne the "golden ratio" as usual:
1
x
=
x
1 x
p
1+ 5
x=
= 1:618:::::
2
1. Construct a rectangle with this ratio of the longer side to the
shorter. (Book 2, proposition 11, plus book 6, prop 30.)
2. Construct three perpendicular rectangles of this size as shown
(Book 11, prop. 11)
Prove that ABC is an equilateral triangle.
3
Beyond the Platonic Solids
4
The Archimedian solids
Faces are all regular polyhedra, but they are not all the same
polyhedron.
Archimedes proved that there are exactly 13 of these.
4.1
Euler’s formula
e=v+f
2:
Applies to any “simply connected” polyhedron. Really a theorem
in "topology"!
Plausibility argument (??): Consider a polyhedron with only four
vertices. Then this is a tetrahedron (possibly irregular). Show
that if one vertex is added, then another edge is added also, so
v + f e is unchanged. Add an edge. This adds a face, so again
v + f e is unchanged. Check its value for a tetrahedron
Proofs: There are many. For example, see page 471 of Stillwell.
(There are many formulas due to Euler. The best known are this
one and ei = 1:)
This can be used to prove (again) that there are only 5 regular
polyhedra.