Mathematics 138
Galilean Geometry
Reading: Yaglom, pages 15-25 (especially 23-25) and 33-53.
As a model for Galilean geometry, consider uniform motions along the y axis, described by

y = kt + y0 (k is the velocity, y0 the initial position).
These are invariant under a "change of frame of reference" described by

t' = t + a  ( a = 1 hour when we switch to Daylight Savings Time)
y' = y + vt + b (v is the relative velocity of two frames of reference) 
We can represent such motions graphically.
The points of the t-y plane represent events.
The straight lines represent uniform motions (which happen, according to Newton's First Law, when a particle is subject to no force).
The slope k of a line represents the velocity of the motion.
Vertical lines are special: they are sets of simultaneous events, which remain simultaneous in any frame of reference according to "Galilean relativity."

An isometry of Galilean geometry can be represented by the matrix
,which is the shear  followed by the translation
Galilean geometry is the study of the properties that are invariant under this group of transformations.  Since they are affine transformations, we already now that they
1. are bijections, hence invertible.
2. carry lines into lines and parallel lines into parallel lines.
3. preserve ratios of segments along lines; i.e. map proportional vectors into proportional vectors.
4. preserve area and orientation of figures, since the determinant is +1.
Yaglom takes the trouble to prove all these things on pp 34-36.

We could have considered the larger isometry group that includes matrices like
 (time reversal)
(reflection in the origin of the line on which the motion is taking place)
(both of the above)
In Yaglom, this is left to exercises 4 and IX on p. 31.  By restricting ourselves to the smaller group of "direct isometries" we will be able to define invariant signed distances and angles.
A peculiar feature of Euclidean geometry:

In general, the distance between two lines cannot be defined,
But if the lines are parallel, we can define a "special distance" d, measured along any line perpendicular to the given lines. Since this distance is, like ordinary angle, a number that describes a relationship between line, "special angle" might be a better term.

Now we shall take a purely geometrical view and call the horizontal coordinate x in order to match the notation of Yaglom.  But think of x as being measured in seconds.
Consider points A(x,y) and A'(x',y')
The distance between them is dAA' = x' - x  ( a signed quantity!)
If and only if this distance is zero we can define a special distance dAA' = y' - y
Compare Euclidean geometry: we can define a special distance (or maybe we should call it a special angle) between lines if and only if the angle between them is zero because they are parallel.
We define angle in the "parabolic" manner.  Given two intersecting lines m and m' we construct the special (vertical) line one unit to the right of their intersection and then define the angle between the lines as the special distance along that line. This is again a signed quantity.
As thus defined, the angle between lines can have any positive or negative value.

For parallel lines, we can define a special distance between the lines as the special distance measured along any special line that cuts the two vertical lines.

Finally we define the distance from a point M to a line r as the special distance measured along the vertical line from M to the point P of r that lies on that vertical line.

In both this case and the preceding one, the definition of a quantity that is preserved by any isometry required us to use a "special line" where in Euclidean geometry we would have used a line perpendicular to the line. In general, we can think of a special line as being perpendicular to any ordinary line.

This works (see Supplement B of Yaglom) because the appropriate inner product between vectors v = (x, y) and v' = (x', y') is
<v, v'> = x x'.
So the vector (0, 1), which is the direction vector for any special line, has an inner product of zero with the direction vector for any ordinary line.
But this scalar product is not an adequate replacement for the Euclidean scalar product if we want to characterize Galilean geometry axiomatically.  Yaglom states the two vector axioms that are needed to replace the axiom that any nonzero vector has positive length:
IV5(G) (page 249) There exists a vector a with <a,a> > 0 and a vector o with <o,o> = 0.

IV6(G) (page 252) The set of all vectors such that <o, a> = 0 for all a is a one-dimensional Euclidean vector space (so that any vector between two points on a special line can be assigned a positive length, the magnitude of the special distance)
The connection between Galilean geometry and mechanics:

Geometry				Mechanics
Point					Event
Ordinary line				Uniform motion
Special line				Set of simultaneous events (moment in time)
Ordinary distance d			Time interval
Special distance d			Space interval for simultaneous events
Angle					Relative velocity
Distance between parallel lines		Distance between particles with same velocity,
					measured at any fixed time
Circle of radius r				Events at time r before or after a given event.

Triangles in Galilean geometry

Now a, b, and c are positive lengths and A, B, C are positive angles.
If c is the largest side, then c = a + b, because the length of each side is simply the magnitude of the distance along the x axis.

In terms of slopes ka, kb, and kc of the lines a, b, and c, the angles are
A = kb - kc
B = kc - ka (note that ka is negative in the diagram)
C = kb - ka (in Euclidean terms, this would be an exterior angle).

So if C is the largest angle, then C = A + B.
This is our first example of a pair of theorems that are dual to one another, that can be turned into one another by interchanging "point" and "line."

There is a theorem that is analogous to the Law of Sines: a/A = b/B = c/C.
To prove it, we proceed as in the Euclidean case by constructing the three altitudes, perpendicular to the sides.  But while the three altitudes of a Euclidean triangle are concurrent, these three altitudes are all special lines!

The proof works just like the Euclidean proof, though:
hc = bA = aB   so bcA = caB
hb  = cA = aC   so bcA = abC
ha = bC = cB   so abC = caB
Thus abC = caB = bcA. Divide by abc to get a/A = b/B = c/C.

A corollary: if a triangle is isosceles with equal sides a and b, then it is also isogonal:
the angles A and B are equal, because a/A = b/B and so A/B = a/b = 1.

But this cannot be extended to equilateral triangles. They cannot exist, because for any isosceles triangle the third side must have a length that is the sum of the lengths of the two equal sides.
We can define midpoint as in any affine geometry, and the medians of a triangle are concurrent. This is a theorem of affine geometry, so it is true in Euclidean, Galilean, or Minkowski geometry.

What about angle bisectors?  In contrast to the Euclidean case, we can define a unique angle bisector for a pair of intersecting ordinary lines (we don't need to use rays as in the Euclidean case).

The definition is the same as in the Euclidean case:
Find an isometry (a shear that holds point P fixed) that carries line r into line r'
Take the square root of this isometry (use half the relative velocity)
The square root carries r into the bisector, the bisector into r'
The angle being bisected has measure k'-k
The angle between the bisector and either of the original lines is (k' - k)/2
The angle bisectors of a triangle are not concurrent as in Euclidean geometry, but as we shall soon see, they satisfy another theorem, obtained from the theorem that the medians are concurrent by interchanging "point" and "line" everywhere and thereby turning "midpoint" into "angle bisector."

Congruence of triangles in Galilean geometry:

The "SSS" criterion fails dismally.

a = a'. b = b', and c = c', but the triangles are not congruent. Notice that if we did not require our isometries to preserve special distance as well as ordinary distance, then they would be congruent, since a uniform stretching of the plane along the vertical direction would be an isometry.
The "SAS" and "ASA" criteria are still valid, but only if we require that the directed sides and directed angles of the triangles be equal.

A and A' have equal measure, b and b' have equal (unsigned) length, and c and c' also have equal length.  But it would take a reflection to map ABC onto A'B'C', and our isometry group allows only direct isometries with determinant +1.
Notice that the distances AB and A'B' are equal in magnitude but opposite in sign.
By duality, whatever is true for SAS will also be true for ASA!

Duality in Galilean Geometry
Reading: Yaglom, pp.54-75 and pp. 243-252
From any theorem of Galilean, we can create another theorem (which may turn out to be the same) by the following substitutions:

Point       <------->      Line
Points on same line    <-------> Lines intersecting at same point
Distance between points    <-------> Angle between lines
Parallel points (on same special line)     <------->     Parallel lines
Special distance between parallel points     <-----> Special distance between parallel lines
Midpoint of a segment        <---------->   Bisector of an angle
Some concepts are manifestly self-dual, like "distance from a point to a line".
Others can be proved self-dual, like "isosceles triangle" (the dual is "triangle with two equal angles," which is the same thing.

A simple example with a triangle.
The original theorem:
"In an isosceles triangle, the special line through the vertex bisects the base."
Duals:
isosceles triangle ---> isogonal triangle (same as isosceles triangle).
midpoint of base ---> bisector of angle opposite the base
point parallel to vertex   ----->   line parallel to side opposite vertex
Dual theorem:
In an isogonal triangle, a line f, parallel to the base a, through the vertex A  bisects angle A..

Proving a dual theorem like the one above:
First approach: prove the original theorem, then take the dual of each step.
1 (original). Side AF has length zero (parallel points) so AB and BF have equal length.
1 (dual). The angle between lines a and f is zero (parallel lines) so the angle between a and b equals the angle between b and f.
2 (original). Similarly, AC and CF have equal length.
2 (dual) Similarly, angle ac equals angle fc.
3 (original). AB and AC have equal length because ABC is isosceles.
3 (dual) The angles at B and C are equal because ABC is isogonal.
4 (original). So BF = FC and F is the midpoint of BC
4 (dual) So the angle between b and f equals the angle between f and c and f bisects the angle between b and c.

Second approach: prove that Galilean geometry is self-dual; so that the dual of the proof of any theorem (starting from the duals of the axioms) must be a proof of the dual theorem.

Another pair of dual theorems: 
Original: The diagonals of a parallelogram bisect one another.
(This is a theorem not merely of Galilean geometry but of affine geometry)
To construct the dual, proceed as follows.
Convert the parallelogram whose sides p and r are parallel and whose sides q and s are parallel into a coparallelogram with points P and R parallel and points Q and S parallel.
In the coparallelogram, find the intersection E of a and c (dual to the line joining A and C) and the intersection F of b and d (dual to the line joining B and D).
Construct the line through E and F (dual to the intersection of the diagonals AC and BD)
This line bisects the angles at E and F (dual to the point of intersection of the diagonals bisecting both diagonals).

Proof of the dual theorem:
Just as AD = BC in the parallelogram, so the opposite angles aPd and bRc are congruent in the coparallelogram. So E is the midpoint of UV by the definition of Galilean angle.
Thus angles bFq and dFq are congruent (as are aQb and cSd) and q bisects the angle at F.
In triangles FEQ and FES, the largest angle (at E) is the sum of the other two angles, so
angle aEq = aQb + bFq
angle cEq = cSd + dFq
But aQb = cSd and bFq = dFq, so aEq = cEq and q bisects the angle at E

One last example, with a cotrapezoid:

Another theorem of affine geometry:
In a trapezoid, the line joining the midpoints of the non-parallel opposite sides is parallel to the two parallel opposite sides, and its length is half the sum of their lengths.

The dual theorem concerns a cotrapezoid, a quadrilateral with one pair of parallel points. 
Dual to M as the midpoint of AD we have line m as the angle bisector of ad
Dual to N as the midpoint of BC we have line n as the angle bisector of bc.
Dual to the line joining M and N we have the point of intersection E of m and n, and the theorem states that this point E is parallel to the points S and Q, so that all three lie on the same special line. (For the proof, see Yaglom, p. 60.)

As a special case of the previous theorem, we let the trapezoid degenerate into a triangle.
Original theorem: midline MN is parallel to side AC and the length of MN is half the length of AC.

Dual theorem: Angle bisectors m and n intersect in a point S that is parallel to vertex P, and the angle between them is half the angle at P.

Proof: Since m bisects aRb, it passes through the midpoint S of the altitude PX.
Since n bisects bQc, it also passes through the midpoint S of PX.
Therefore the point of intersection of m and n is S, which is parallel to P.
The angle between m and n at S is the sum of bRm and bQn.
Since m and n are angle bisectors, this angle between m and n is half the sum of bRa and bQc, which is the angle between a and c at P.
(Alternative proof; just note that RS' = 1/2 RP' and compare the angles at S and P.

Now we can see how to formulate and prove a theorem about the angle bisectors of a Galilean triangle.  Since "angle bisector" is the dual of "midpoint," we take the dual of the well-known affine theorem that the medians of a triangle are concurrent.

Proof:
1. Construct M as the intersection of the medians AD and BE.
2. Take N as the midpoint of segment CM.
3. Now E bisects AC and N bisects CM so EN is parallel to AD.
4. Similarly, D bisects BC and N bisects CM so DN is parallel to BE.
5. Therefore DNME is a parallelogram, and
DM = EN = 1/2 MA    Also EM = ND = 1/2 MB.
But the same argument holds for any pair of medians of the triangle, so the intersection point of any two medians divides each median in the ratio 2:1.  SInce M is two-thirds of the way from A to D, it is the point where both the other medians BE and CF intersect AD.

Now we can prove the dual theorem by taking the dual of every step of this proof.
First we must formulate the dual theorem:

Midpoint of a side ----> bisector of the angle opposite a side.
Median(line joining midpoint to vertex) ----> point of intersection of angle bisector with opposite side.
Medians are concurrent -------> points of intersection are collinear
Intersection point of medians divides each median in the ratio 2:1  ----> line formed by the points of intersection of angle bisectors with opposite sides divides the angle at each intersection point in the ratio 2:1

Proof of the dual theorem:
1. Construct line m through the point U where the angle bisector d intersects the opposite side BC and the point V where the angle bisector e intersects the opposite side AC. This line intersects side AB at a point W, which will turn out to lie on the angle bisector f (but we must prove it.)
2. Take n as the bisector of the angle where m intersects the side AB.
3. Consider triangle AWV: angle bisectors n and d intersect at a point R that is parallel to the third vertex V.
4. Similarly, consider triangle BWU: angle bisectors n and e intersect at a point S that is parallel to the third vertex U.
5. Therefore RUSV is a coparallelogram, and
dUm = nSe = 1/2 mUa    Also eVm = nRd = 1/2 mVb.
But the same argument holds for any pair of angle bisectors of the triangle, so the line joining any pair of points where angle bisectors meet the opposite sides divides the angle at the intersection point in the ratio 2:1.  The angle mUa is two-thirds of the angle dUa.  The angle mVb is two-thirds of the angle eVb.  Wherever the bisector f intersects side c (call it W'), the line m' joining that point of intersection to U has the property that m'Ua is two-thirds of dUa.  But the line m is the unique line with that property.  Therefore m' = m and W' = W, so the three points of intersection U, V, W are collinear.

Proving the principle of duality:
If we can show that the fundamental principles of Galilean geometry satisfy the principle of duality, then any consequences of those fundamental principles also satisfy the principle.  That means that the proof of any "dual theorem" can be constructed by taking the proof of the "original theorem" back to first principles and replacing each statement by its dual.  It is then no longer necessary to write out explicitly the steps in the proof of the "dual theorem" since they follow mechanically from the steps in the proof of the original theorem.

Yaglom does this proof in three different ways.  The best one, although the most tedious, is to list all the axioms for Galilean geometry and to show that the dual of each axiom is also a true statement of the geometry.  Then the proof of each theorem from the axioms corresponds to a proof of the dual theorem from the dual axioms.

Since Yaglom's axioms are based on vectors, this approach requires us to invent the dual of a vector.  Yaglom calls this a "doublet."  Two lines determine a doublet just as two points determine a vector.

Think of a line of Galilean geometry as a two-component object  that corresponds to the uniform motion described by y = vt + s.  Then the "difference" between two lines is also a two-component object( a doublet) where v is the relative velocity of the two motions and s is the difference between the "initial positions" of the two motions at time t = 0.
Yaglom proves (pp.71-75) that the system of lines and doublets satisfies exactly the same set of axioms as the system of points and vectors, and he does so in a purely synthetic manner, without assuming that a line and a doublet can be represented as a pair of real numbers.  This proof
is in the notes that follow.

The doublets satisfy the Galilean scalar product axioms.  The scalar product of "ordinary doublets" is
= v1 v2, so that the "length" of an ordinary doublet is just relative velocity.
This scalar product complies with the axiom IV4 on page 246:
 for any doublet a
There are "special doublets" of the form   whose ordinary length is zero.
So the doublet a =  and the doublet o = satisfy axiom  IV5(G) on page 249:
There exists a vector with a2 > 0 and a vector o with o2 = 0.
Finally, the "special scalar product"= s1 s2, which assigns to a special doublet a "special length" equal to the "difference in initial position", satisfies axiom  IV6(G) on page 249:

The set of all vectors o such that <o,a> = 0 for all a is a Euclidean (vector) line.