This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.Lecture 6 Joint Invariants of Cycles: Orthogonality
The invariant cycle product, defined in the previous chapter, allows
us to define joint invariants of two or more cycles. Being
initially defined in an algebraic fashion, they also reveal their rich
geometrical content. We will also see that FSCc matrices
define reflections and inversions in cycles, which extend
Möbius maps.
6.1 Orthogonality of Cycles
According to the categorical viewpoint, the
internal properties of objects are of minor importance in comparison
with their relations to other objects from the same class. Such a
projection of internal properties into external relations was also
discussed at the beginning of Section 4.2. As a
further illustration, we may give the proof of
Theorem 13, outlined below. Thus, we will now look
for invariant relations between two or more cycles.
After we defined the invariant cycle product (3), the
next standard move is to use the analogy with Euclidean and Hilbert
spaces and give the following definition:
Definition 1
Two cycles Cσcs and S
σcs are
called σ
c-orthogonal if their σ
c-cycle product
vanishes:
(a)
(b)
Figure 6.1: Relation between centres and radii of orthogonal circles |
Here are the most fundamental properties of cycle orthogonality:
Exercise 2
Use Exercise 12 to check the following:
-
The σc-orthogonality
condition (1) is invariant under
Möbius transformations.
- The explicit
expression for σc-orthogonality of cycles in terms of their
coefficients is
2σcn′n−2l′l+k′m+m′k=0.
(2) |
- The σc-orthogonality of cycles defined by their e-centres
(u,v) and (u′, v′) with
σ-determinants −r2 and −r′2, respectively, is:
(u−u′)2−σ(v−v′)2−2(σ−σc)vv′−r2−r′2=0.
(3) |
-
Two circles are e-orthogonal if their tangents at an
intersection point form a right angle.
Hint:
Use the previous formula (3), the inverse
of Pythagoras’ theorem and Fig. 6.1(a) for this.
⋄
The last item can be reformulated as follows: For circles, their
e-orthogonality as vectors in the cycle spaces ℙ3 with
the cycle product (3) coincides with their
orthogonality as geometrical sets in the point space
ℝe. This is very strong support for FSCc and the cycle
product (3) defined from it. Thereafter, it is
tempting to find similar interpretations for other types of
orthogonality. The next exercise performs the first step for the case
of σ-orthogonality in the matching point space
ℝσ.
Exercise 3
Check the following geometrical meaning for σ
-orthogonality of σ
-cycles.
-
(e,h)
- Let σ=± 1. Then, two cycles in
ℝσ (that is, circles or hyperbolas) are
σ-orthogonal if slopes S1 and S2 of their
tangents at the intersection point satisfy the condition
The geometrical meaning of this condition can be given either in terms
of angles (A) or centres (C):
-
For the case σ=−1 (circles),
equation (4) implies orthogonality of the
tangents, cf. Exercise 4. For
σ=1, two hyperbolas are h-orthogonal if lines with the
slopes ± 1 bisect the angle of intersection of the
hyperbolas, see Fig. 6.1(b).
Hint:
Define a cycle Cσ by the condition that it
passes a point (u,v)∈ℝσ. Define a second
cycle S
σ by both conditions: it passes
the same point (u,v) and is orthogonal to
Cσ. Then use the implicit derivative formula to
find the slopes of tangents to Cσ and
S
σ at (u,v). A script calculating
this in CAS is also provided.
⋄
- In the cases σ=± 1, the tangent to one cycle
at the intersection point passes the centre of another cycle.
Hint:
This fact is clear for circles from inspection of, say,
Fig. 6.1(a). For hyperbolas, it is enough
to observe that the slope of the tangent to a hyperbola
y=1/x at a point (x, 1/x) is −1/x2 and the slope
of the line from the centre (0,0) to the point (x,1/x)
is 1/x2, so the angle between two lines is bisected by
a vertical/horizontal line. All our hyperbolas are obtained from
y=1/x by rotation of ± 45 ∘ and scaling.
⋄
- (p)
- Let σ=0 and a parabola Cp have two
real roots u1 and u2. If a parabola
S
p is p-orthogonal to Cp,
then the tangent to S
p at a point above one
of the roots u1,2 passes the p-centre
(u1+u2/2,0) of Cp.
Remark 4
Note that the geometric p-orthogonality condition for parabolas is
non-local in the sense that it does not direct behaviour of tangents
at the intersection points. Moreover, orthogonal parabolas need not
intersect at all. We shall see more examples of such non-locality later. The relation 3(p)
is also another example of boundary awareness, cf.
Remark 3—we are taking a
tangent of one parabola above the point of intersection of the other
parabola with the boundary of the upper half-plane.
The stated geometrical conditions for orthogonality of cycles are not
only necessary but are sufficient as well.
Exercise 5
-
Prove the converses of the two statements in Exercise 3.
Hint:
To avoid irrationalities in the parabolic case and make the
calculations accessible for CAS, you may proceed as follows. Define
a generic parabola passing (u,v) and use implicit derivation
to find its tangent at this point. Define the second parabola
passing (u,0) and its centre at the intersection of the
tangent of the first parabola at (u,v) and the horizontal axis. Then,
check the p-orthogonality of the two parabolas.
⋄
- Let a parabola have two tangents
touching it at (u1,v1) and (u2,v2) and these tangents
intersect at a point (u,v). Then, u=u1+u2/2.
Hint:
Use the geometric description of p-orthogonality and note that the two
roots of a parabola are interchangeable in the necessary condition
for p-orthogonality.
⋄
We found geometrically necessary and sufficient conditions for
σ-orthogonality in the matching point space
ℝσ. The remaining six non-matching cases will be
reduced to this in Section 6.3
using an auxiliary ghost cycle. However, it will be useful to study
some more properties of orthogonality.
6.2 Orthogonality Miscellanea
The explicit formulae (2)
and (3) allow us to obtain several simple and
yet useful conclusions.
Exercise 6 Show that:
-
A cycle is σc-self-orthogonal
(isotropic) if and only if it is
σc-zero-radius cycle (11).
-
For σc=± 1, there is no non-trivial cycle
orthogonal to all other non-trivial cycles.
For σc=0, only the horizontal axis v=0 is orthogonal
to all other non-trivial cycles.
- If two real (e-negative) circles are e-orthogonal, then they
intersect. Give an example of h-orthogonal
hyperbolas which do not intersect in ℝh.
Hint:
Use properties of the Cauchy–Schwarz inequality from
Exercise 1.
⋄
- A cycle Cσs is
σ-orthogonal to a zero-radius cycle
Zσs (11) if and only if
σ-implementation of Cσs passes through
the σ-centre of Zσs, or, analytically,
k(u2 − σ v2) − 2⟨ (l,n),(u, σc v)
⟩+m =0.
(5) |
- For σc=± 1, any cycle is uniquely defined
by the family of cycles orthogonal to it,
i.e. (Cσcs⊥)⊥={Cσcs}.
For σc= 0, the set (Cσcs⊥)⊥ is
the pencil spanned by Cσcs and the real line.
In particular, if Cσcs has real roots, then all
cycles in (Cσcs⊥)⊥ have these roots.
- Two σc-zero-radius cycles
are σc-orthogonal if:
-
for σc=−1 and σ=±1, cycles’ σ-centres coincide.
- for σc=0 and σ=±1, cycles’ σ-centres belong to the same
vertical line.
- for σc=1 and σ=±1, each cycle’s σ-centre belongs to
the light cone defined by the other cycle.
(Note, that in the Benz axiomatisation [315]*§ 1
these relations between cycles’ centres are called parallelism.)
Hint:
Use the explicit expression for the cycle product (12).
⋄
The connection between orthogonality and incidence from
Exercise 4 allows us to combine the techniques of
zero-radius cycles and orthogonality in an efficient tool.
Exercise 7
Fill all gaps in the following proof:
Proof.[Sketch of an alternative proof of
Theorem 13]
We already mentioned in Subsection 4.4.1 that
the validity of Theorem 13 for a zero-radius
cycle (11)
with the centre z=x+i y is a straightforward calculation—see
also Exercise 5. This implies the
result for a generic cycle with the help of
-
Möbius invariance of the product (3) (and thus
orthogonality) and
- the above relation 4 between the
orthogonality and the incidence.
The idea of such a proof is borrowed from [65] and
details can be found therein.
The above demonstration suggests a generic technique for extrapolation
of results from zero-radius cycles to the entire cycle space. We will
formulate it with the help of a map Q from the cycle space to conics in the
point space from Definition 10.
Proposition 8
Let T: ℙ
3 → ℙ
3 be an
orthogonality-preserving map
of the cycle space, i.e.
⟨
Cσcs,
S
σcs
⟩=0
⇔
⟨
TCσcs,
TS
σcs
⟩=0
. Then, for
σ≠ 0
, there is a map Tσ: ℝ
σ
→ℝ
σ, such that Q intertwines T
and Tσ:
Proof.
If T preserves orthogonality (i.e. the cycle
product (3) and, consequently, the determinant,
see (9)) then the image
TZσcs(u,v) of a zero-radius cycle
Zσcs(u,v) is again a zero-radius cycle
Zσcs(u1,v1). Thus we can define Tσ by the identity
Tσ: (u,v)↦ (u1,v1).
To prove the intertwining property (6) we need
to show that, if a cycle Cσcs passes through (u,v),
then the image TCσcs passes through Tσ(u,v). However, for σ≠ 0, this is a consequence of the
T-invariance of orthogonality and the expression of the
point-to-cycle incidence through orthogonality from
Exercise 4.
Exercise 9
Let Ti: ℙ3 → ℙ3, i=1,2 be
two orthogonality-preserving maps of the cycle space. Show that, if they
coincide on the subspace of σc-zero-radius cycles, σc≠
0, then they are identical in the whole ℙ3.
We defined orthogonality from an inner product, which is linear in
each component. Thus, orthogonality also respects linearity.
Exercise 10
Check the following relations between orthogonality and pencils:
-
Let a cycle Cσc be σc-orthogonal to two
different cycles S
σc and
G σc. Then Cσc is
σc-orthogonal to every cycle in the pencil spanned by
S
σc and G σc.
-
Check that all cycles σc-orthogonal with σc=± 1 to
two different cycles S
σc and
G σc belong to a single pencil. Describe such a
family for σc=0.
Hint:
For the case σc=0, the family is spanned by an additional
cycle, which was mentioned in
Exercise 2.
⋄
- If two circles are
non-intersecting, then the orthogonal pencil passes through two points, which are
the only two e-zero-radius cycles in the pencil, see the first row
of Fig. 6.2. And vice versa: a pencil
orthogonal to two intersecting circles consists of disjoint
circles. Tangent circles have the orthogonal
pencils of circles which are all tangent at the same point,
cf. Corollary 27.
- The above relations do hold for parabolas and hyperbolas only
for tangent pencils (the middle column of
Fig. 6.2). What are correct statements in
other cases?
Exercise 2 describes two orthogonal
pencils such that each cycle in one pencil
is orthogonal to every cycle in the second. In terms of indefinite
linear algebra, see [113]*§ 2.2, we
are speaking about the orthogonal complement of a two-dimensional
subspace in a four-dimensional space and it turns out to be
two-dimensional as well. For circles, this construction is well known,
see [71]*§ 5.7
[37]*§ 10.10. An illustration for three cases is
provided by Fig. 6.2. The reader may wish to
experiment more with orthogonal complements to various parabolic and
hyperbolic pencils—see Fig. 5.1 and
Exercise 9.
Such orthogonal pencils naturally appear in many circumstances and we
already met them on several occasions. We know from
Exercise 3 and 4 that
K-orbits and transverse lines make coaxial pencils which turn to be
in a relation:
Exercise 11
Check that:
-
Any K-orbit (6) in
ℝσ is σ-orthogonal to any transverse
line (8). Figure 1.2 provides
an illustration.
Hint:
There are several ways to check this. A direct calculation
based on the explicit expressions for cycles is not
difficult. Alternatively, we can observe that the pencil of
transverse lines is generated by K-action from the vertical
axis and orthogonality is Möbius-invariant.
⋄
- Any orbit (13) is
σ-orthogonal to any transverse
line (1) for the same subgroup K,
N′ or , that is for the same value τ.
Figure 3.2 provides an illustration.
The following general statement about pencils and orthogonality is an abstract generalisation of the well-known result on a triangle’s orthocenter.
Exercise 11(a)
Let Ca, Cb, Cc be three pair-wise distinct cycles with at most two of them being orthogonal to each other. We define S
a as the cycle
-
from the pencil spanned by Cb and Cc;
- orthogonal to Ca.
Cycles S
b and S
c are similarly defined. Then, S
a belongs to the pencil spanned by S
b and S
c.
In particular, altitudes of a triangle are concurrent
, that is meet at the same point, which is called the orthocenter
of a triangle.
Hint:
Note that S
a = ⟨ Cc,Ca
⟩ Cb − ⟨ Cb,Ca
⟩ Cc.
⋄
We may describe a finer structure of the cycle space through
Möbius-invariant subclasses of cycles. Three such
families—zero-radius, positive and negative cycles—were already
considered in Sections 5.3
and 5.4. They were defined using the
properties of the cycle product with itself. Another important class
of cycles is given by the value of its cycle product with the real
line.
Definition 12
A cycle Cσcs is called selfadjoint
if it is σ
c-orthogonal with
σ
c≠ 0
to the real line, i.e. it is defined by the condition
⟨
Cσcs,
Rσcs
⟩=0
, where
Rσcs=(0,0,1,0)
corresponds to the horizontal axis
v=0
.
The following algebraic properties of selfadjoint cycles easily
follow from the definition.
Exercise 13
Show that:
-
Selfadjoint cycles make a Möbius-invariant
family.
- A selfadjoint cycle Cσcs is defined explicitly
by n=0 in (1) for both values σc=± 1.
- Any of the following conditions are necessary and sufficient
for a cycle to be selfadjoint:
-
All three centres of the cycle coincide.
- At least two centres of the cycle belong to the real line.
From these analytic conditions, we can derive a geometric
characterisation of selfadjoint cycles.
Exercise 14
Show that selfadjoint cycles have the following implementations in
the point space ℝ
σ:
-
(e,h)
- Circles or hyperbolas with their geometric centres on the real line.
- (p)
- Vertical lines, consisting of points (u,v) such that
| u−u0 |=r for some u0∈ℝ,
r∈ℝ+. The cycles are also given by the
||x−y||=r2 in the parabolic metric defined below
in (3).
Notably, selfadjoint cycles in the parabolic point space were labelled
as “parabolic circles” by
Yaglom—see [339]*§ 7. On the other hand, Yaglom used
the term “parabolic cycle” for our p-cycle with non-zero k and
n.
Exercise 15
Show that:
-
Any cycle Cσcs=(k,l,n,m) belongs to a pencil spanned by a
selfadjoint cycle HCσcs
and the real line
Cσcs=HCσcs+nRσcs,
where
HCσcs=(k,l,0,m).
(7) |
This identity is a definition of linear orthogonal projection
H from the cycle space to its subspace of selfadjoint
cycles. When does HCσcs is a real cycle?
- The decomposition of a cycle into the linear combination of a
selfadjoint cycle and the real line is Möbius-invariant:
Hint:
The first two items are small pieces of linear algebra in an
indefinite product space,
see [113]*§ 2.2.
⋄
-
Cycles Cσcs and HCσcs have the same
real roots.
We are now equipped to consider the geometrical meaning of all nine
types of cycle orthogonality.
6.3 Ghost Cycles and Orthogonality
For the case of σcσ=1, i.e. when geometries of the cycle and
point spaces are both either elliptic or hyperbolic,
σc-orthogonality can be expressed locally through tangents to
cycles at the intersection points—see
Exercise 3(A). A semi-local condition also exists:
the tangent to one cycle at the intersection point passes the centre
of the second cycle—see Exercise 3(C). We may note
that, in the pure parabolic case σ=σc=0, the geometric
orthogonality condition from Exercise 3(p) can be
restated with help from Exercise 1 as
follows:
Corollary 16
Two p-cycles Cp and S
p are
p-orthogonal if the tangent to Cp at its intersection
point with the projection
HS
p (7) of
S
p to selfadjoint cycles passes the p-centre
of S
p.
Hint:
In order to reformulate Exercise 3(p) to the
present form, it is enough to use
Exercises 14(p)
and 3.
⋄
The three cases with matching geometries in point and cycle spaces are
now quite well unified. Would it be possible to extend such a geometric
interpretation of orthogonality to the remaining six (=9−3)
cases?
Elliptic realisations (in the point space) of
Definition 1, i.e. σ=−1, were shown
in Fig. 1.7 and form the first row in
Fig. 6.3. The left picture in this row
corresponds to the elliptic cycle space, e.g. σc=−1. The
orthogonality between the red circle and any circle from the blue or
green families is given in the usual Euclidean sense described in
Exercise 3(e,h). In other words, we can decide on the
orthogonality of circles by observing the angles between their tangents
at the arbitrary small neighbourhood of the intersection point.
Therefore, all circles from either the green or blue families, which are
orthogonal to the red circle, have common tangents at points a
and b respectively.
The central (parabolic in the cycle space) and the right (hyperbolic)
pictures show the the non-local nature of orthogonality if σ≠σc.
The blue family has the intersection point b with the red circle,
and tangents to blue circles at b are different. However, we may
observe that all of them pass the second point d. This
property will be used in Section 6.5 to define
the inversion in a cycle. A further investigation of
Fig. 6.3 reveals that circles from the green
family have a common tangent at point a, however this point does
not belong to the red circle. Moreover, in line with the geometric
interpretation from Exercise 3(C), the common
tangent to the green
family at a passes the p-centre (on the central parabolic
drawing) or h-centre (on the right hyperbolic drawing).
There are analogous pictures in parabolic and hyperbolic point spaces
as well and they are presented in the second and third rows of
Fig. 6.3. The behaviour of green and blue
families of cycles at point a, b and d is similar up to
the obvious modification: the EPH case of the point space coincides
with EPH case of the cycle spaces in the central drawing of the second
row and the right drawing of the third row.
Therefore, we will clarify the nature of orthogonality if the locus of
such points a with tangents passing the other cycle’s σc-centre
are described. We are going to demonstrate that this locus is a
cycle, which we shall call a “ghost”. The ghost cycle is shown by the
dashed lines in Fig. 6.3. To give an analytic
description, we need the Heaviside
function χ(σ):
More specifically, we note the relations χ(σ)=σ if
σ=±1 and χ(σ)=1 if σ=0. Thus, the Heaviside
function will be used to avoid the degeneracy of the parabolic case.
Definition 17
For a cycle Cσc in the σ
-implementation, we
define the associated (σ
c-)ghost cycle
G σc
by the following two conditions:
-
The χ(σ)-centre of G σc coincides
with the σc-centre of Cσc.
- The determinant of G σ1 is equal to
the determinant of Cσσc.
Exercise 18
Verify the following properties of a ghost cycle:
-
G σ coincides with
Cσ if σ σc=1.
- G σ has common roots (real or
imaginary) with Cσ.
-
For a cycle Cσc, its p-ghost cycle
G σc and the non-selfadjoint part
HCσc (7) coincide.
-
All straight lines σc-orthogonal to a cycle pass its
σc-centre.
The significance of the ghost cycle is that the σc-orthogonality between
two cycles in ℝσ is reduced to
σ-orthogonality to the ghost cycle.
Proposition 19
Let cycles Cσc and
S
σc be σ
c-orthogonal in
ℝ
σ and let G σc be the ghost
cycle of Cσc. Then:
-
S
σc and G σc are
σ-orthogonal in ℝσ for seven cases
except two cases σ=0 and σc=± 1.
-
In the σ-implementation, the tangent line
to S
σc at its intersection point with
-
the ghost cycle
G σ, if σ=± 1, or
- the non-selfadjoint part
HCσc (7) of the cycle
Cσc, if σ=0,
passes the σc-centre of Cσc, which coincides with
the σ-centre of G σ.
Proof.
The statement 1 can be shown by
algebraic manipulation, possibly in CAS. Then, the non-parabolic
case 2(a) follows from the first
part 1, which reduces non-matching
orthogonality to a matching one with the ghost cycle, and the
geometric description of matching orthogonality from
Exercise 3. Therefore, we only need to provide a
new calculation for the parabolic case 2(b).
Note that, in the case σ=σc=0, there is no disagreement
between the first and second parts of the proposition since
HCσc=S
σc due
to 3.
Consideration of ghost cycles does present orthogonality in
geometric term, but it hides the symmetry of this relation. Indeed,
it is not obvious that S
σcs relates to the ghost
of Cσcs in the same way as Cσcs relates to
the ghost of S
σcs.
Remark 20
Elliptic and hyperbolic ghost cycles are symmetric in the real line and
the parabolic ghost cycle has its centre on it—see
Fig. 6.3. This is an illustration of the
boundary effect from Remarks 3.
Finally, we note that Proposition 19 expresses
σc-orthogonality through the σc-centre of cycles. It illustrates
the meaningfulness of various centres within our approach which may
not be so obvious at the beginning.
6.4 Actions of FSCc Matrices
Definition 11 associates a 2× 2-matrix
to any cycle. These matrices can be treated analogously to elements of
SL2(ℝ) in many respects. Similar to the SL2(ℝ)
action (4), we can consider a fraction-linear
transformation on the point space ℝσ defined by a
cycle and its FSCc matrix
Cσs: w ↦
Cσs(w)
= | (l + ι sn )w−m |
|
kw+(−l + ι sn ) |
| ,
(9) |
where Cσs is, as usual (5),
Cσs =
| | and
w=u + ι v, σ=ι2.
|
Exercise 21
Check that w=
u+ι
v∈ℝ
σ is a fixed point of
the map Cσ−σ (9) if
and only if the σ
-implementation of Cσ−σ
passes w. If det
S
σ s≠ 0
then the
second iteration of the map is the identity.
We can also extend the conjugated
action (7) on the cycle space from
SL2(ℝ) to cycles. Indeed, a cycle S
σcs in the
matrix form acts on another cycle Cσcs by the
σc-similarity
S
σcs1: Cσcs
↦ −S
σcs1 | | S
σcs1.
(10) |
The similarity can be considered as a transformation of the cycle
space ℙ3 to itself due to the following result.
Exercise 22
Check that:-
The cycle σc-similarity (10)
with a cycle S
σcs, where det
S
σcs≠ 0, preserves the structure of FSCc
matrices and S
σcs1 is its fixed point. In a
non-singular case, detS
σcs≠ 0, the second
iteration of similarity is the identity map.
- The σc-similarity with a σc-zero-radius cycle
Zσcs always produces this cycle.
- The σc-similarity with a cycle (k,l,n,m) is a linear
transformation of the cycle space ℝ4 with the matrix
|
| ⎛
⎜
⎜
⎜
⎝ | km−detCσc | −2 k l | 2 σc k n | k2 |
l m | −2l2−detCσc | 2 σc l n | k l |
n m | −2 n l | 2 σc n2−detCσc | k n |
m2 | −2 ml | 2σc m n | km−detCσc
|
| ⎞
⎟
⎟
⎟
⎠ |
| |
|
|
| = | |
|
Note the apparent regularity of its entries.
Remark 23
Here is another example where usage of complex
(dual or double) numbers is different from Clifford algebras. In order to use commutative hypercomplex
numbers, we require the complex conjugation for the cycle
product (3), linear-fractional
transformation (9) and cycle
similarity (10). Non-commutativity of
Clifford algebras allows us to avoid complex conjugation in all
these formulae—see Appendix B.5. For
example, the reflection in the real line (complex conjugation) is
given by matrix similarity with the corresponding matrix
(
)
.
A comparison of Exercises 21
and 22 suggests that there is a connection
between two actions (9)
and (10) of cycles, which is similar to
the relation SL2(ℝ) actions on points and cycles from
Lemma 20.
Exercise 24
Letting det
S
σcs ≠ 0
, show that:
-
The
σc-similarity (10)
σc-preserves the orthogonality
relation (1). More specifically, if
G σcs and G σcs are
matrix similarity (10) of cycles
Cσcs and G σcs, respectively, with
the cycle S
σcs1, then
⟨ G σcs,G σcs
⟩=
⟨ Cσcs,G σcs
⟩
(detS
σcs)2.
|
Hint:
Note that
S
σcsS
σcs=
−det(S
σcs) I, where I is the identity
matrix. This is a particular case of the Vahlen
condition,
see [100]*Prop. 2. Thus, we have
G σcs | | =
−S
σcs1Cσcs | |
| | · det
S
σcs.
|
The final step uses the invariance of the trace under the matrix
similarity. A CAS calculation is also provided.
⋄
- The image Tσ s=Cσs2
Zσs1Cσs2 of a
σ-zero-radius cycle Tσs1
under the similarity (10) is a
σ-zero-radius cycle Tσs1. The
(s1s2)-centre of Tσcs is the linear-fractional
transformation (9) of the (s2/s1)-centre of
Zσcs.
- Both formulae (9)
and (10) define the same transformation of
the point space ℝσ, with σ=σc≠ 0.
Consequently, the linear-fractional
transformation (9) maps cycles to cycles
in these cases.
Hint:
This part follows from the first two items and
Proposition 8.
⋄
- There is a cycle Cσs such that neither map of the
parabolic point space ℝp represents similarity with
Cσs.
Hint:
Consider S
σs=(1,0,1/2,−1) and
a cycle Cσs passing point (u,v). Then the
similarity of Cσs with
S
σs passes the point
T(u,v)=(1+v/u,v+v2/u2) if and only if either
-
Cσs is a straight line, or
- (u,v) belongs to S
σs and
is fixed by the above map T.
That is, the map T of the point space ℝp serves
flat cycles and S
σs but no others. Thus,
there is no map of the point space which is compatible with the
cycle similarity for an arbitrary cycle.
⋄
(a)
(b)
(c)
(d)
Figure 6.4: Three
types of inversions of the rectangular grid. The initial
rectangular grid (a) is inverted elliptically in the unit circle
(shown in red) in (b), parabolically in (c) and hyperbolically in
(d). The blue cycle (collapsed to a point at the origin in (b))
represent the image of the cycle at infinity under inversion. |
Several demonstrations of inversion are provided in
Fig. 6.4. The initial setup is shown in
Fig. 6.4(a)—the red unit circle and the grid of
horizontal (green) and vertical (blue) straight lines. It is very
convenient in this case that the grid is formed by two orthogonal
pencils of cycles, which can be considered to be of
any EPH type. Figure 6.4(b) shows e-inversion of the
grid in the unit circle, which is, of course, the locus of fixed
points. Straight lines of the rectangular grid are transformed to
circles, but orthogonality between them is preserved—see
Exercise 1.
Similarly, Fig. 6.4(c) presents the result of
p-inversion in the degenerated parabolic cycle u2−1=0. This time the grid is mapped to two orthogonal pencils of parabolas
and vertical lines. Incidentally, due to the known optical illusion, we perceive these vertical straight lines as
being bent.
Finally, Fig. 6.4(d) demonstrates h-inversion in the
unit hyperbola u2−v2−1=0. We again obtained two pencils of
orthogonal hyperbolas. The bold blue cycles—the dot at the origin in
(b), vertical line in (c) and two lines (the light cone) in (d)—will be explained in
Section 8.1. Further details are provided by the
following exercise.
Exercise 25
Check that the rectangular grid in Fig. 6.4(a) is
produced by horizontal and vertical lines given by quadruples
(0,0,1,
m)
and (0,1,0,
m)
, respectively.The similarity with the cycle (1,0,0,−1) sends a cycle
(k,l,n,m) to (m,l,n,k). In particular, the image of the grid
are cycles (m,0,1,0) and (m,1,0,0).
We conclude this section by an observation that cycle similarity is
similar to a mirror reflection, which preserves the directions of vectors
parallel to the mirror and reverses vectors which are orthogonal.
Exercise 26
Let det
S
σcs≠ 0
. Then, for similarity (10) with S
σcs:
-
Verify the identities
where Cσcs is a cycle σc-orthogonal to
S
σc. Note the difference in the signs in
the right-hand sides of both identities.
- Describe all cycles which are fixed (as points in the
projective space ℙ3) by the similarity with the given
cycle S
σcs.
Hint:
Use a decomposition of a generic cycle into a sum
S
σcs and a cycle orthogonal to
S
σcs similar
to (7).
⋄
As we will see in the next section, these orthogonal reflections in the
cycle space correspond to “bent” reflections in the point space.
6.5 Inversions and Reflections in Cycles
The maps in point and cycle spaces considered in the previous section
were introduced from the action of FSCc matrices of cycles. They can also
be approached from the more geometrical viewpoint.
There are at least two natural ways to define an inversion in a cycle:
- One possibility uses orthogonality from Section 6.1.
As we can observe from Fig. 6.3, all cycles
from the blue family passing the point b also meet again at the
point d. This defines a correspondence b ↔ d
on the point space.
- Another option defines “reflections in cycles”. The mirror
reflection z → z in the horizontal axis is a
fundamental operation. If we transform a generic cycle to the real
line, then we can extend the notion of reflection to the cycle.
We can formalise the above observations as follows.
Definition 27 For a given cycle Cσs, we define two
maps of the point space ℝ
σ associated to it:
-
A σc-inversion
in a σ-cycle Cσs sends a point
b∈ℝσ to the second point d of the
intersection of all σ-cycles σc-orthogonal to
Cσs and passing through b—see
Fig. 6.3.
- A σc-reflection
in a σ-cycle Cσs is given by
M−1RM, where M is a
σc-similarity (10) which sends the σ-cycle
Cσs into the horizontal axis and R is the
mirror reflection of ℝσ in that axis.
We are going to see that inversions are given
by (9) and reflections are expressed
through (10), thus they are essentially the
same for EH cases in light of Exercise 1.
However, some facts are easier to establish using the inversion and
others in terms of reflection. Thus, it is
advantageous to keep both notions.
Since we have three different EPH orthogonalities between cycles for
every type of point space, there are also three different inversions
in each of them.
Exercise 28
Prove the following properties of inversion: -
Let a cycle
S
σcs be σc-orthogonal to a cycle
Cσcs=(k,l,n,m). Then, for any point u1+ι
v1∈ ℝσ (ι2=σ) belonging to
σ-implementation of S
σcs, this
implementation also passes through the image of u1+ι v1
under the Möbius transform (9) defined
by the matrix Cσσc:
u2+ι v2
=Cσσc(u1−ι v1 )= | |
(u1−ι v1).
(11) |
Thus, the point u2+ι v2 is the inversion of
u1+ι v1 in Cσcs.
- Conversely, if a cycle S
σcs passes two
different points u1+ι v1 and u2+ι v2 related
through (11), then
S
σcs is σc-orthogonal to
Cσcs.
-
If a cycle S
σcs is σc-orthogonal to
a cycle Cσcs, then the σc-inversion in
Cσcs sends S
σcs to itself.
- σc-inversion in the σ-implementation of a cycle
Cσcs coincides with σ-inversion in its
σc-ghost cycle G σcs.
Note the interplay between parameters σ and σc in the
above statement 1. Although we are speaking
about σc-orthogonality, we take the Möbius
transformation (11) with the imaginary unit
ι such that ι2=σ (as the signature of the point
space). On the other hand, the value σc is used there as the
s-parameter for the cycle Cσσc.
Proposition 29
The reflection 2 of a zero-radius
cycle Zσcs in a cycle Cσcs is given by
the similarity
CσcsZσcsCσcs.
Proof.
Let a cycle S
σcs have the property
S
σcs Cσcs
S
σcs = Rσcs, where
Rσcs
is the cycle representing the real line. Then,
S
σcs Rσcs
S
σcs = Cσcs, since
S
σcsS
σcs=
S
σcsS
σcs = −det
S
σcsI. The mirror reflection in the real line is
given by the similarity with Rσcs, therefore the
transformation described in 2 is a
similarity with the cycle S
σcs Rσcs
S
σcs = Cσcs and, thus, coincides
with (11).
Corollary 30
The σ
c-inversion with a cycle Cσcs in
the point space ℝ
σ coincides with
σ
c-reflection in Cσcs.
The auxiliary cycle S
σcs from the above proof of
Prop. 29 is of separate interest and can be
characterised in the elliptic and hyperbolic cases as follows.
Exercise 31
Let Cσcs=(
k,
l,
n,
m)
be
a cycle such that σ
c det
Cσcs>0
for σ
c≠
0
. Let us define the cycle S
σcs by the
quadruple (
k,
l,
n±√
σc detCσcs,
m)
. Then:
-
S
σcs Cσcs S
σcs
= ℝ and S
σcs ℝ
S
σcs = Cσcs.
-
S
σcs and Cσcs have common
roots.
-
In the σc-implementation, the cycle Cσcs passes
the centre of S
σcs.
Hint:
One can directly observe 2 for real roots,
since they are fixed points of the inversion. Also, the
transformation of Cσcs to a flat cycle implies that
Cσcs passes the centre of inversion,
hence 3. There is also a CAS calculation for
this.
⋄
Inversions are helpful for transforming pencils of cycles to the simplest possible
form.
Exercise 32
Check the following:-
Let the σ-implementation of a cycle
Cσs pass the σ-centre of a cycle
S
σs. Then, the σ-reflection of
Cσs in S
σs is a
straight line.
- Let two cycles Cσs and
S
σs intersect in two points P,
P′∈ℝσ such that P−P′ is not a divisor of
zero in the respective number system. Then, there is an inversion
which maps the pencil of cycles orthogonal to Cσs and
S
σs (see Exercise 2)
into a pencil of concentric cycles.
Hint:
Make an inversion into a cycle with σ-centre P,
then Cσs and S
σs
will be transformed into straight lines due to the previous
item. These straight lines will intersect in a finite point
P″ which is the image of P′ under the inversion. The
pencil orthogonal to Cσs and
S
σs will be transformed to a pencil
orthogonal to these two straight lines. A CAS calculations
shows that all cycles from the pencil have σ-centre at
P″.
⋄
A classical source of the above result in inversive
geometry [71]*Thm. 5.71 tells that an inversion
can convert any pair of non-intersecting circles to concentric ones. This
is due to the fact that an orthogonal pencil to the pencil generated by
two non-intersecting circles always passes two special points—see
Exercise 3 for further development.
Finally, we compare our consideration for the parabolic point space
with Yaglom’s book. The Möbius
transformation (9) and the respective inversion
illustrated by Fig. 6.4(c) essentially coincide with
the inversion of the first kind from
[339]*§ 10. Yaglom also introduces the
inversion of the second kind,
see [339]*§ 10. For a parabola v=k(u−l)2+m, he
defined the map of the parabolic point space to be
(u,v) ↦ (u, 2(k(u−l)2+m)−v),
(12) |
i.e. the parabola bisects the vertical line joining a point and its
image. There are also other geometric characterisations of this map in
[339], which make it very similar to the Euclidean inversion
in a circle. Here is the resulting expression of this transformation through
the usual inversion in parabolas:
Exercise 33
The inversion of the second
kind (12) is a composition of three Möbius
transformations (9) defined by cycles
(1,
l,2
m,
l2+
m/
k)
, (1,
l,0,
l2+
m/
k)
and the real line in the parabolic point space ℝ
p.
Möbius transformations (9) and
similarity (12) with FSCc matrices map cycles
to cycles just like matrices from SL2(ℝ) do. It is natural to ask for
a general type of matrices sharing this property. For this, see
works [65, 100, 289] which deal with
more general elliptic and hyperbolic (but not parabolic) cases. It is
beyond the scope of our consideration since it derails from the
geometry of SL2(ℝ). We only mention the rôle of the Vahlen
condition, CσcsCσcs=
−det(Cσcs) I, used in Exercise 1.
6.6 Higher-order Joint Invariants: Focal Orthogonality
Considering Möbius action (1), there is no need to
be restricted to joint invariants of two cycles and a bilinear form. Indeed, for any
polynomial p(x1,x2,…,xn) of several non-commuting
variables, one may define an invariant joint disposition of n
cycles jCσcs by the condition
trp(1Cσcs, 2Cσcs, …, nCσcs)=0,
(13) |
where the polynomial of FSCc matrices is defined through standard
matrix algebra. To create a Möbius invariant which is not affected
by the projectivity in the cycle space we can either
- use a polynomial p which is homogeneous in every variable xi, or
- substitute variables xi with the cycles’ det-normalised FSCc matrices.
Furthermore, for any two polynomials p(x1,x2,…,xn) and q(x1,x2,…,xn) such that for each variable xi the orders of homogeneity of p and q are equal, the value
| trp(1Cσcs, 2Cσcs, …, nCσcs) |
|
trq(1Cσcs, 2Cσcs, …, nCσcs) |
|
is a Möbius invariant. The simplest and yet most important realisation of this concept is the cycles cross ratio in § 12.3
Let us construct some lower-order realisations
of (13). In order to be essentially different
from the previously considered orthogonality, such invariants may
either contain non-linear powers of the same cycle, or accommodate
more than two cycles. In this respect, consideration of higher-order
invariants is similar to a transition from Riemannian to Finsler
geometry [62, 107]. The latter is based on the
replacement of the quadratic line element gij dxi dxj in the
tangent space by a more complicated function.
A further observation is that we can simultaneously study several
invariants of various orders and link one to another by some
operations. There are some standard procedures changing orders of
invariants working in both directions:
- Higher-order invariants can be built on top of those already defined;
- Lower-order invariants can be derived from higher ones.
Consider both operations as an example. We already know that a
similarity of a cycle with another cycle produces a new cycle. The
cycle product of the latter with a third cycle creates a joint invariant
of these three cycles
⟨ 1Cσcs 2Cσcs 1Cσcs,3Cσcs
⟩,
(14) |
which is built from the second-order invariant
⟨ ·,·
⟩. Now we can reduce the order of this
invariant by fixing 3Cσcs to be the real line,
since it is SL2(ℝ)-invariant. This invariant deserves special
consideration. Its geometrical meaning is connected to the matrix
similarity of cycles (10) (inversion in
cycles) and orthogonality.
Definition 34
A cycle S
σcs is σ
c-focal
orthogonal
(or
f
σc-orthogonal
)
to
a cycle Cσcs if the σ
c-reflection of
Cσcs in S
σcs is
σ
c-orthogonal (in the sense of
Definition 1) to the real line. We
denote it by S
σcs ⊣
Cσcs.
Remark 35
This definition is explicitly based on the invariance of the real
line and is an illustration to the boundary value effect from
Remark 3.
Exercise 36
f-orthogonality is equivalent to either of the following
-
The cycle S
σcs
CσcsS
σcs is a
selfadjoint cycle, see Definition 12.
- Analytical condition:
⟨ S
σcs | | S
σcs,Rσcs
⟩=
tr(S
σcs | | S
σcsRσcs)=0.
(15) |
Remark 37
It is easy to observe the following:
-
f-orthogonality is not symmetric: Cσcs ⊣
S
σcs does not imply S
σcs ⊣
Cσcs.
-
Since the horizontal axis Rσcs and
orthogonality (1) are SL2(ℝ)-invariant
objects, f-orthogonality is also SL2(ℝ)-invariant.
However, an invariance of f-orthogonality under inversion of cycles
required some study since, in general, the real line is not invariant under such
transformations.
Exercise 38
The image G σcs1
Rσcs G σcs1 of the real line under
inversion in G σcs1=(
k,
l,
n,
m)
with s≠ 0
is the cycle
(2 s s1 σc k n, 2 s s1 σc l n, −det(Cσcs1), 2 s s1 σc m n).
|
It is the real line again if det(
Cσcs1)≠0
and either
-
s1 n=0, in which case it is a composition of SL2(ℝ)-action by
() and the reflection in the real line, or
- σc=0, i.e. the parabolic case of the cycle space.
If either of two conditions is satisfied then f-orthogonality
S
σcs⊣
Cσcs is
preserved by the σ
c-similarity with G σcs1.
The following explicit expressions of f-orthogonality reveal further
connections with cycles’ invariants.
Exercise 39
f-orthogonality of S
σcs to
Cσcs1 is given by either of the
equivalent identities
s n (l′2+σc s12n′2− m′ k′) + s1n′(m k′ −2l
l′+ k m′ ) | = | 0 or |
n det(S
σc1) +
n′⟨ Cσc1,S
σc1
⟩ | = | 0,
if s=s1=1.
|
|
The f-orthogonality may again be related to the usual orthogonality
through an appropriately chosen f-ghost cycle, cf. Proposition 19:
Proposition 40
Let Cσcs be a cycle. Then, its
f-ghost cycle
G σcσc =
Cσcχ(σ) ℝ
σcσc
Cσcχ(σ) is the reflection of the real line in
Cσcχ(σ), where χ(σ)
is the
Heaviside function
(8). Then:
-
Cycles Cσc1 and G σcσc
have the same roots.
- The χ(σ)-centre of
G σcσc coincides with the (−σc)-focus
of Cσcs, consequently all straight lines
σc-f-orthogonal to Cσcs pass its
(−σc)-focus.
- f-inversion in Cσcs defined from
f-orthogonality (see
Definition 1) coincides with
the usual inversion in G σcσc.
Figure 6.5: Focal orthogonality of
cycles. In order to highlight both
the similarities and distinctions with the ordinary orthogonality, we
use the same notations as in Fig. 6.3. |
Note the above intriguing interplay between the cycle’s centres and foci.
It also explains our choice of name for focal orthogonality, cf.
Definition 1. f-Orthogonality and
the respective f-ghost cycles are presented in
Fig. 6.5, which uses the same outline and legend
as Fig. 6.3.
The definition of f-orthogonality may look rather extravagant at
first glance. However, it will find new support when we
again consider lengths and distances in the next chapter.
It will also be useful for infinitesimal cycles, cf.
Section 7.5.
Of course, it is possible and meaningful to define other interesting
higher-order joint invariants of two or even more cycles.
Last modified: October 28, 2024.