Indefinite Product Space of Cycles

In the previous chapter, we represented cycles by points of the projective space ℙ³ or by lines of 2× 2 matrices. The latter is justified so far only by the similarity formula (7). Now, we will investigate connections between cycles and vector space structure. Thereafter we will use the special form of FSCc matrices to introduce very important additional structure in the cycle space. The structure is an indefinite inner product, which is particularly helpful to describe tangency of cycles, see [101] [163]*Ch. 4 and Sec. 5.5.

5.1 Cycles: an Appearance and the Essence

Our extension (5) of FSCc adds two new elements in comparison with the standard one [65]:

Such a possibility of an extension exists because elements of SL₂(ℝ) are matrices with real entries. For generic Möbius transformations with hypercomplex-valued matrices considered in [65], it is impossible.

Indeed, the similarity formula (8) does not contain any squares of hypercomplex units, so their type is irrelevant for this purpose. At the moment, the hypercomplex unit ι_c serves only as a placeholder which keeps components l and n separated. However, the rôle of ι_c will be greatly extended thereafter. On the other hand, the hypercomplex unit ι defines the appearance of cycles on the plane, that is, any element (k,l,n,m) in the cycle space ℙ³ admits its drawing as circles, parabolas or hyperbolas in the point space ℝ_σ. We may think on points of ℙ³ as ideal cycles while their depictions on ℝ_σ are only their shadows on the wall of Plato’s cave . More prosaically, we can consider cones and their (conic) sections as in Fig. 1.3.

Of course, some elliptic shadows may be imaginary , see Exercise 5, but, in most cases, we are able to correctly guess a cycle from its σ-drawing.

Exercise 1 Describe all pairs of different cycles C_{σ_c}^s and S _{σ_c}^s such that their σ-drawing for some σ are exactly the same sets. Hint: Note that the vertical axis is the p-drawing of two different cycles (1,0,0,0) and (0,1,0,0) and make a proper generalisation of this observation. ⋄

Figure 1.5(a) shows the same cycles drawn in different EPH styles. Points c_σ=(l/k, −σ n/k) are their respective e/p/h-centres from Definition 3. They are related to each other through several identities:

From analytic geometry, we know that a parabola with the equation ku²−2lu−2nv+m=0 has a focal length, the distance from its focus to the vertex, equal to n/2k. As we can see, it is half of the second coordinate in the e-centre. Figure 1.5(b) presents two cycles drawn as parabolas. They have the same focal length n/2k and, thus, their e-centres are on the same level. In other words, concentric parabolas are obtained by a vertical shift, not by scaling, as an analogy with circles or hyperbolas may suggest.

We already extended the definition of centres from circles and hyperbolas to parabolas. It is time for a courtesy payback: parabolas share with other types of cycles their focal attributes.

Definition 2 A σ_r-focus , where σ_r is a variable of EPH type, of a cycle C=(k,l,n,m) is the point in ℝ_σ

f_{σ_r}=

⎛
⎜
⎜
⎝

detC_{σ_r}^s

2nk

⎞
⎟
⎟
⎠

or, explicitly, f_{σ_r} =

⎛
⎜
⎜
⎝

mk−l²+σ_r n²

2nk

⎞
⎟
⎟
⎠

. (2)

We also use names e-focus , p-focus , h-focus and σ_r-focus, in line with previous conventions.

The focal length of the cycle C is n/2k.

Note that values of all centres, foci and the focal length are independent of a choice of a quadruple of real numbers (k,l,n,m) which represents a point in the projective space ℙ³.

Figure 1.5(b) presents e/p/h-foci of two parabolas with the same focal length. If a cycle is depicted as a parabola, then the h-focus, p-focus and e-focus are, correspondingly, the geometrical focus of the parabola, the vertex of the parabola , and the point on the directrix of the parabola nearest to the vertex.

As we will see, cf. Propositions 19 and 40, all three centres and three foci are useful attributes of a cycle, even if it is drawn as a circle.

Remark 3 Such a variety of choices is a consequence of the usage of SL₂(ℝ)—a smaller group of symmetries in comparison to the all Möbius maps of ℝ_σ. The SL₂(ℝ) group fixes the real line and, consequently, a decomposition of vectors into “real” (1) and “imaginary” (ι) parts is obvious. This permits us to assign an arbitrary value to the square of the hypercomplex unit ι.

The exceptional rôle of the real line can be viewed in many places. For example, geometric invariants defined below, e.g. orthogonalities in Sections 6.1 and 6.6, demonstrate “awareness” of the real-line invariance in one way or another. We will call this the boundary effect in the upper half-plane geometry. The famous question about hearing a drum’s shape has a counterpart:

Can we see/feel the boundary from inside a domain?

Remarks 19, 4, 20 and 35 provide hints for positive answers.

Exercise 4 Check that different realisations of the same cycle have these properties in common:

All implementations have the same vertical axis of symmetry.
Intersections with the real axis (if they exist) coincide, see points r₀ and r₁ for the left triple of cycles in Fig. 1.5(a). Moreover, all three EPH drawings of a cycle have common tangents at these intersection points.
Centres of circle c_e and corresponding hyperbolas c_h are mirror reflections of each other in the real axis with the parabolic centre as the middle point.
The σ_c-focus of the elliptic K-orbits , see Exercise 1, is the reflection in the real axis of the midpoint between the σ_c-centre of this orbit and the inverse of its h-centre.

Exercise 14 gives another example of similarities between different implementations of the same cycles defined by the Equation (6).

5.2 Cycles as Vectors

Elements of the projective space ℙ³ are lines in the linear space ℝ⁴. Would it be possible to pick a single point on each line as its “label”? We may wish to do this for the following reasons:

However, the general scheme of projective spaces does not permit such a unique representation serving the whole space. Otherwise the cumbersome construction with lines in vector spaces would not be needed. Nevertheless, there are several partial possibilities which have certain advantages and disadvantages. We will consider two such opportunities, calling them normalisation of a cycle.

The first is very obvious: we try to make the coefficient k in front of the squared terms in cycle equation (1) equal to 1. The second normalises the value of the determinant of the cycle’s matrix. More formally:

Definition 5 A FSCc matrix representing a cycle is said to be k-normalised if its (2,1)-element is 1 and it is det_{σ_c}-normalised if its σ_c-determinant is equal to ±1.

We will discuss cycles which do not admit either normalisation at the end of Sec. 8.1.

Exercise 6 Check that:

Element (1,1) of the k-normalised C_{σ_c}¹ matrix is the σ_c-centre of the cycle.
Any cycle (k,l,n,m) with k≠ 0 has an equivalent k-normalised cycle. Cycles which do not admit k-normalisation correspond to single straight lines in any point space.
det-normalisation is preserved by matrix similarity with SL₂(ℝ) elements, as in Theorem 13, while k-normalisation is not.
Any cycle with a non-zero σ_c-determinant admits det_{σ_c}-normalisation. Cycles which cannot be det_{σ_c}-normalised will be studied in Section 5.4. What are cycles which cannot be det_{σ_c}-normalised for any value of σ_c?

Thus, each normalisation may be preferred in particular circumstances. The det-normalisation was used, for example, in [163] to obtain a satisfactory condition for tangent cycles , cf. Exercise 2. On the other hand, we will see in Section 7.1 that det C_{σ_c}^s of a k-normalised cycle is equal to the square of the cycle radius.

Remark 7 It is straightforward to check that there is one more cycle normalisation which is invariant under similarity (7). It is given by the condition n=1. However, we will not use it at all in this work.

A cycle’s normalisation is connected with scaling of vectors. Now we turn to the second linear operation: addition. Any two different lines define a unique two-dimensional plane passing through them. Vectors from the plane are linear combinations of two vectors spanning each line. If we consider circles corresponding to elements of the linear span, we will obtain a pencil of circles, see [71]*§ 2.3 [37]*§ 10.10. As usual, there is no need to be restricted to circles only:

Definition 8 A pencil of cycles is the linear span (in the sense of ℝ⁴) of two cycles.

Figure 5.1 shows the appearance of such pencils as circles, parabolas and hyperbolas. The elliptic case is very well known in classical literature, see [71]*Figs. 2.2(A,B,C); 2.3A, for example. The appearance of pencils is visually different for two cases: spanning cycles are either intersecting or disjoint. These two possibilities are represented by the left and right columns in Fig. 5.1. We shall return to these situations in Corollary 27.

Exercise 9 Investigate the following:

What happens with an elliptic pencil if one of spanning circles is imaginary ? Or if both spanning circles are imaginary?
How does a pencil look if one spanning cycle is a straight line? If both cycles are straight lines?
A pair of circles is symmetric about the line joining their centres. Thus, circular pencils look similar regardless of the direction of this line of centres. Our hyperbolas and parabolas have a special orientation: their axes of symmetry are vertical. Thus, a horizontal or vertical line joining the centres of two hyperbolas/parabolas make a special arrangement, and it was used for Fig. 5.1. How do hyperbolic pencils look if the line of centres is not horizontal?

Exercise 10 Show the following:

The image of a pencil of cycles under a Möbius transformation is again a pencil of cycles.
A pencil spanned by two concentric cycles consists of concentric cycles.
All cycles in a pencil are coaxial , see Exercise 3 for the definition. The joint radical axis , see Exercise 2, is the only straight line in the pencil. This is also visible from Fig. 5.1.
Let two pencils of cycles have no cycle in common. Then, any cycle from ℙ³ belongs to a new pencil spanned by two cycles from two given pencils.

This section demonstrated that there are numerous connections between the linear structure of the cycle space ℙ³ and the geometrical property of cycles in the point space ℝ_σ.

5.3 Invariant Cycle Product

We are looking for a possibility to enrich the geometry of the cycle space through the FSCc matrices. Many important relations between cycles are based on the following Möbius invariant cycle product.

Definition 11 The cycle σ_c-product of two cycles is given by

⟨ C_{σ_c}^s,S _{σ_c}^s ⟩ = −tr(C_{σ_c}^s

S _{σ_c}^s

), (3)

that is, the negative trace of the matrix product of C_{σ_c}^s and hypercomplex conjugation of S _{σ_c}^s.

In the essence the above product is Frobenius inner product of two matrices [137]*Ch. 5. And as we already mentioned, an inner product of type (3) is used, for example, in the Gelfand–Naimark–Segal construction to make a Hilbert space out of C^*-algebra. This may be more than a simple coincidence since FSCc matrices can be considered as linear operators on a two-dimensional vector space. However, a significant difference with the Hilbert space inner product is that the cycle product is indefinite, see Exercise 4 for details. Thus, cycles form a Pontrjagin or Krein space [113] rather than Euclidean or Hilbert. A geometrical interpretation of the cycle product will be given in Exercise 3.

Exercise 12 Check that:

The value of cycle product (3) will remain the same if both matrices are replaced by their images under similarity (7) with the same element g∈SL₂(ℝ).

The σ_c-cycle product (3) of cycles defined by quadruples (k,l,n,m) and (k′,l′,n′,m′) is given by

−2l′l+2σ_c s²n′n+k′m+m′k.

(4)

More specifically, and also taking into account the value s=± 1, it is

−2n′n−2l′l+k′m+m′k,			(5)
−2l′l+k′m+m′k,			(6)
2n′n−2l′l+k′m+m′k			(7)

in the elliptic, parabolic and hyperbolic cases, respectively.

Let C_{σ_c} and S _{σ_c} be two cycles defined by e-centres (u,v) and (u′, v′) with σ-determinants −r² and −r′², respectively. Then, their σ_c-cycle product explicitly is
⟨ C_{σ_c},S _{σ_c} ⟩= (u−u′)²−σ(v−v′)²−2(σ−σ_c)vv′−r²−r′². (8)
The cycle product is a symmetric bilinear function of two cycles. It is indefinite in the sense that there are two cycles C_{σ_c} and S _{σ_c} such that
⟨ C_{σ_c},C_{σ_c} ⟩>0 and ⟨ S _{σ_c},S _{σ_c} ⟩<0.

Hint: It is easy to show symmetry of the cycle product from the explicit value (4). This is not so obvious from the initial definition (3) due to the presence of hypercomplex conjugation. ⋄
Check, that the change of variables (k,l,n,m)↦ (t,l,n,s), where k=t+s and m=t−s, transforms the elliptic inner product (5) to the metric of the Minkowski space [163]*§ 4.1:
−2n′n−2l′l+k′m+m′k ↦ −2n′n−2l′l−2s′s+2t′t.

A simple, but interesting, observation is that, for FSCc matrices representing cycles, we obtain the second classical invariant (determinant) under similarities (7) from the first (trace) as follows:

Therefore it is not surprising that the determinant of a cycle enters Definition 2 of the foci and the following definition:

Definition 13 We say that a C_{σ_c} is σ_c-zero-radius , σ_c-positive or σ_c-negative cycle if the value of the cycle product of C_{σ_c} with itself (9) (and thus its determinant) is zero, positive or negative, respectively.

Exercise 14 Show that the zero-radius, positive and negative cycles form three non-empty disjoint Möbius-invariant families. Hint: For non-emptiness, see Exercise 4 or use the explicit formula (9). ⋄

The following relations are useful for a description of these three classes of cycles.

Exercise 15 Check that:

For any cycle C_{σ_c}^s there is the following relation between its determinants evaluated with elliptic, parabolic and hyperbolic unit ι_c:
detC_e^s ≤ detC_p^s ≤ detC_h^s.
The negative of the parabolic determinant, −detC_p^s, of a cycle (k,l,n,m) coincides with the discriminant of the quadratic equation
ku²−2lu+m=0, (10)

which defines intersection of the cycle with the real line (in any EPH presentation).

We already illustrated zero-radius cycles in Fig. 1.6. This will be compared with positive and negative cycles for various values of σ_c in Fig. 5.2. The positive or negative aspect of cycles has a clear geometric manifestation for some combinations of σ and σ_c.

Exercise 16 Verify the following statements:

Circles corresponding to e-positive cycles are imaginary . All regular circles are e-negative.
Parabolas drawn from p- and e-negative cycles have two points of intersection with the real axis, and those drawn from p- and h-positive cycles do not intersect the real axis.
Hyperbolas representing h-negative cycles have “vertical” branches and those representing h-positive cycles have “horizontal” ones. Hint: Write the condition for a hyperbola defined by the Equation (1) to intersect the vertical line passing its centres. ⋄

Of course, manifestations of the indefinite nature of the cycle product (3) are not limited to the above examples and we will meet more of them on several other occasions.

5.4 Zero-radius Cycles

Due to the projective nature of the cycle space ℙ³, the absolute values of the cycle product (3) on non-normalised matrices are irrelevant, unless it is zero. There are many reasons to take a closer look at cycles with a zero-value product—zero-radius cycles, for example:

To highlight that a certain cycle is σ_c-zero-radius we will denote it by Z_{σ_c}^s. A justification of the chosen name for such cycles is given in Exercise 1—further connections will be provided in Section 7.1.

As often happens in our study, we again have nine different possibilities—for cycles drawn in three EPH types in the point space (parametrised by σ) there are three independent conditions detC_{σ_c}^s=0 in the cycle space (parametrised by σ_c). This is illustrated in Fig. 1.6.

Exercise 17 Show that:

Any σ_c-zero-radius cycle admitting k-normalisation can be represented by the matrix

Z_{σ_c}^s=

⎛
⎜
⎝

z	−zz
1	−z

⎞
⎟
⎠

⎛
⎜
⎝

z	z
1	1

⎞
⎟
⎠

⎛
⎜
⎝

1	−z
1	−z

⎞
⎟
⎠

⎛
⎜
⎝

u₀+ ι_c s v₀	u₀²−σ_c v₀²
1	−u₀+ι_c s v₀

⎞
⎟
⎠

, (11)

for some z=u₀+ ι_c s v₀ being its centre and s=± 1. Compare with (10).

Describe all σ_c-zero-radius cycles which cannot be k-normalised.
The σ_c-focus of the σ_c-zero-radius cycle belongs to the real axis—see Fig. 1.6.
Let Z_{σ_c}^s and T_{σ_c}^s be two k-normalised σ_c-zero-radius cycles of the form (11) with e-centres (u₀,v₀) and (u₁,v₁). Then
⟨ Z_{σ_c}^s,T_{σ_c}^s ⟩= (u₀−u₁)²−σ_c(v₀−v₁)². (12)

As follows from (11), the class of σ_c-zero-radius cycles is parametrised by two real numbers (u,v) only and, as such, is easily attached to the respective point of z=u+ι v∈ ℝ_σ, at least in the elliptic and hyperbolic point spaces. In ℝ_p, a connection between a σ_c-zero-radius cycle and its centre is obscure.

Exercise 18 Prove the following observations from Fig. 1.6:

The cycle Z_{σ_c}^s (11) with det Z_{σ_c}^s=0, σ_c=−1 drawn elliptically is just a single point (u₀,v₀), i.e. an (elliptic) zero-radius circle.
The same condition detZ_{σ_c}^s=0 with σ_c=1 in the hyperbolic drawing produces a light cone originating at point (u₀,v₀)
(u−u₀)²−(v−v₀)²=0,

i.e. a zero-radius cycle in hyperbolic metric, see Section 7.1.
p-zero-radius cycles in any implementation touch the real axis, see Fig. 1.6. This property defines a horocycle in Lobachevsky geometry [70]*§ 16.6. Hint: Note, that the parabolic determinant of a cycle (k,l,n,m) coincides with the discriminant of the quadratic equation ku²−2lu+m=0. See also Exercise 4. ⋄
An h-zero-radius cycle in ℝ_σ with the e-center (a,b) intersects the real axis at points (a± b,0) with slopes ±1, see Fig. 1.6. Furthermore, the h-centre of this cycle coincides with its e-focus. Hint: For slopes, use implicit derivation. Also, for a parabola, it follows from the Exercise 3 and the classical reflection property of parabolas, cf. App. B.2. ⋄
The e-centre of the transformation gZ_{σ_c}^sg⁻¹, g∈SL₂(ℝ) of the σ_c-zero-radius cycle (11) coincides with the Möbius action g· z in ℝ^σ_c, where z is the e-centre of Z_{σ_c}^s. Hint: The result can be obtained along the lines from Subsection 4.4.1. ⋄

Remark 19 The above “touching” property (3) and intersection behaviour (4) are manifestations of the already-mentioned boundary effect in the upper half-plane geometry, see Remark 3.

The previous exercise shows that σ_c-zero-radius cycles “encode” points into the “cycle language”. The following reformulation of Exercise 5 stresses that this encoding is Möbius-invariant as well.

Lemma 20 The conjugate g⁻¹Z_{σ_c}^s(y)g of a σ_c-zero-radius cycle Z_{σ_c}^s(y) with g∈SL₂(ℝ) is a σ_c-zero-radius cycle Z_{σ_c}^s(g· y) with centre at g· y—the Möbius transform of the centre of Z_{σ_c}^s(y).

Furthermore, we can extend the relation between a zero-radius cycle Z_{σ_c} and points through the following connection of Z_{σ_c} with the power of its centre.

Exercise 21 Let Z_σ be the zero-radius cycle (11) defined by z=u−ισ v, that is, with the σ-centre at the point (u, v). Show that, in the elliptic and hyperbolic (but not parabolic) cases, a power of a point, see Definition 18, (u,v)∈ℝ_σ with respect to a cycle C_σ, is equal to the cycle product ⟨ Z_σ,C_σ ⟩, where both cycles are k-normalised.

It is noteworthy that the notion of a point’s power is not Möbius-invariant even despite its definition through the invariant cycle product. This is due to the presence of non-invariant k-normalisation. This suggests extending the notion of the power from a point (that is, a zero-radius cycle) to arbitrary cycles:

Definition 22 A (σ,σ_c)-power of a cycle C with respect to another cycle S is equal to ⟨ C_{σ_c},S _{σ_c} ⟩, where both matrices are norm-normalised by the conditions ⟨ C_σ,C_σ ⟩=± 1 and ⟨ S _σ,S _σ ⟩=± 1.

Alternatively, (σ,σ_c)-power is equal to 1/2⟨ C_{σ_c},S _{σ_c} ⟩ (one half their cycle product), if both matrices are det_σ-normalised. Since both elements of this definition—the cycle product and normalisation—are Möbius-invariant, the resulting value is preserved by Möbius transformations as well. Of course, the (e,e)-variant of this notion is well known in the classical theory.

Exercise 23

Show that the (e,e)-power of a circle C with respect to another circle S has an absolute value equal to the inversive distance between C and S , see [24]*Defn. 3.2.2 [27]*§ 4 [71]*§ 5.8 and Thm. 5.91:
d(C, S )= ⎪
⎪
⎪
⎪
⎪
⎪ c₁−c₂ ⎪
⎪ ²−r₁²−r₂²

2r₁r₂

⎪
⎪
⎪
⎪ . (13)

Here, c_1,2 and r_1,2 are the e-centres and radii of the circles. For other two cases, Exercise 8.
Check for σ=± 1, that the (σ,σ)-power of two intersecting σ-cycles is the σ-cosine (3) of the angle between the tangents to cycles at an intersecting point.
For σ=± 1, let C and S be two given σ-cycles and α and β be two acute angles. Consider the collection T of all σ-cycles, such that the the intersection angle with C is α and the intersection angle with S is β. Then, any σ-cycle from the pencil spanned by C and S has the same intersection angle with σ-cycles from T (see also Exercise 29 and Fig. 5.3).

As another illustration of the technique based on zero-radius cycles, we return to orbits of isotropy subgroups, cf. Exercise 2.

Exercise 24 Fulfil the following steps:

Write the coefficients of the σ-zero-radius cycle Z_σ(ι) in ℝ_σ with e-centre at the hypercomplex unit ι=(0,1).
According to Exercise 5, Z_σ(ι) is invariant under the similarity Z_σ(ι) ↦ hZ_σ(ι)h⁻¹ with h in the respective isotropy subgroups K, N′ and of ι. Check this directly.
Write the coefficients of a generic cycle in the pencil spanned by Z_σ(ι) and the real line. Note that the real line is also invariant under the action of the isotropy subgroups (as any other Möbius transformations) and conclude that any cycle from the pencil will also be invariant under the action of the isotropy subgroups. In other words, those cycles are orbits of the isotropy subgroups . Check that you obtained their equation (13).

We halt our study of zero-radius cycles on their own, but they will appear repeatedly in the following text in relation to other objects.

5.5 Cauchy–Schwarz Inequality and Tangent Cycles

We already noted that the invariant cycle product is a special (and remarkable!) example of an indefinite product in a vector space. Continuing this comparison, it will be interesting to look for a role of a Cauchy–Schwarz–type inequality

First of all, the classical form (14) of this inequality failed in any indefinite product space. This can be seen from examples or an observation that all classical proofs start from the assumption that ⟨ x+ty,x+ty ⟩≥ 0 in an inner product space. In an indefinite product space, there are always pairs of vectors which realise any of three possible relations:

Some regularity appears from the fact that the type of inequality is inherited by linear spans.

Exercise 25 Let two vectors x and y in an indefinite product space satisfy either of the inequalities:

⟨ x,y ⟩⟨ y,x ⟩ < ⟨ x,x ⟩ ⟨ y,y ⟩ or ⟨ x,y ⟩⟨ y,x ⟩ > ⟨ x,x ⟩ ⟨ y,y ⟩.

Then, any two non-collinear vectors z and w from the real linear span of x and y satisfy the corresponding inequality:

⟨ z,w ⟩⟨ w,z ⟩ < ⟨ z,z ⟩ ⟨ w,w ⟩ or ⟨ z,w ⟩⟨ w,z ⟩ > ⟨ z,z ⟩ ⟨ w,w ⟩.

The equality ⟨ x,y ⟩⟨ y,x ⟩ = ⟨ x,x ⟩ ⟨ y,y ⟩ always implies ⟨ z,w ⟩⟨ w,z ⟩ = ⟨ z,z ⟩ ⟨ w,w ⟩.

Exercise 26 Check the following:

Let C_σ and S _σ be two cycles defined by e-centres (u,v) and (u′, v′) with σ-determinants −r² and −r′². The relations
⟨ C_σ^s,S _σ^s ⟩² ⪋ ⟨ C_σ^s,C_σ^s ⟩ ⟨ S _σ^s,S _σ^s ⟩

guarantee that σ-implementations of C_σ and S _σ for σ=±1 are intersecting , tangent or disjoint , respectively. What happens for the parabolic value σ=0? Hint: Determine the sign of the expression ⟨ C_σ^s,S _σ^s ⟩ −√⟨ C_σ^s,C_σ^s ⟩ ⟨ S _σ^s,S _σ^s ⟩ using the formula (8). For the parabolic case, Exercise 2.p will be useful. ⋄
Let two cycles C_σ and S _σ be in det_σ-normalised form. Deduce from the previous item the following Descartes–Kirillov condition [163]*Lem. 4.1 for C_σ and S _σ to be externally σ-tangent (that is externally tangent in σ-implementation):
⟨ C_σ+S _σ,C_σ+S _σ ⟩=0 and ⟨ C_σ,S _σ ⟩>0. (15)

Since the first identity is equivalent to det (C_σ+S _σ)=0, C_σ+S _σ is a σ-zero-radius cycle. Show that the centre of C_σ+S _σ is the tangent point of C_σ and S _σ. Hint: The identity in (15) follows from the inequality there and the relation | ⟨ C_σ,S _σ ⟩ | =√⟨ C_σ^s,C_σ^s ⟩ ⟨ S _σ^s,S _σ^s ⟩=1 for tangent cycles from the first item. For the last statement, use Exercise 21. ⋄
How shall we modify Descartes–Kirillov condition (15) for two cycles internally σ-tangent?
Show the following. If det_p C_σ^s=0 then det_σ(C_σ^s−R_σ^s)=0, where C_σ^s is det_σ-normalized and R_σ^s is the det-normalized cycle (0,0,1,0) representing the real line. Thus, C_σ^s is σ-tangent to the real line, i.e. is σ-horocycle, cf. Exercise 3. How to interpret the external/internal touching in this case? Hint: The condition det_p C_σ^s=0 defines a cycle C_σ^s=(1,u,v,u²) with det_σC_σ^s=−σ v². ⋄

Corollary 27 For a given pencil of cycles, see Definition 8, either all cycles are pair-wise disjoint, or every pair of cycles are tangent, or all of them pass the same pair of points.

Zero-radius cycles form a two-parameter family (in fact, a manifold) in the three-dimensional projective cycle space ℙ³. It is not flat, as can be seen from its intersection with projective lines—cycle pencils . The Cauchy–Schwarz inequality turns out to be relevant here as well.

Exercise 28 A pencil of cycles either contains at most two σ_c-zero-radius cycles or consists entirely of σ_c-zero-radius cycles. Moreover:

A pencil spanned by two different cycles cannot consist only of e-zero-radius cycles. Describe all pencils consisting only of p- and h-zero-radius cycles. Hint: Formula (12) will be useful for describing pencils consisting of (and thus spanned by) σ_c-zero-radius cycles. Orbits of subgroup N′ shown on the central drawing of Fig. 3.1 are an example of pencils of p-zero-radius cycles drawn as parabolas. You can experiment with σ-drawing of certain σ_c-zero-radius pencils. ⋄
A pencil spanned by two different cycles C_{σ_c}^s and S _{σ_c}^s, which does not consist only of σ_c-zero-radius cycles, has exactly two, one or zero σ_c-zero-radius cycles depending on which of three possible Cauchy–Schwarz-type relations holds:
⟨ C_{σ_c}^s,S _{σ_c}^s ⟩² ⪋ ⟨ C_{σ_c}^s,C_{σ_c}^s ⟩ ⟨ S _{σ_c}^s,S _{σ_c}^s ⟩.

Hint: Write the expression for the cycle product of the span tC_{σ_c}^s+S _{σ_c}^s, t∈ℝ with itself in terms of products ⟨ C_{σ_c}^s,S _{σ_c}^s ⟩, ⟨ C_{σ_c}^s,C_{σ_c}^s ⟩ and ⟨ S _{σ_c}^s,S _{σ_c}^s ⟩. ⋄

Tangent circles may be treated as having zero intersection angle, thus we continue theme from Exercise 3.

Exercise 29 For σ=±1, let C and S be two given σ-cycles and α and β be two acute angles. Consider the collection T of all σ-cycles, such that the the intersection angle with C is α and the intersection angle with S is β. Let

a=cos_σα, b=cos_σβ, c=⟨ C,C ⟩, d=⟨ S ,S ⟩, p=⟨ C,S ⟩.

Assume (a b+p)²−(a²+c ) (b²+d )≥ 0, then, any σ-cycle from the family T is tangent to cycles C_1,2=x_1,2C+S from the pencil spanned by C and S , where:

x_1,2=

−a b−p ±

√

(a b+p)²−(a²+c ) (b²+d )

a²+c

 . (16)

See Fig. 5.3 for an illustration. Hint: Let C_t∈ T and C_x=xC+S be a σ-cycle from the pencil spanned by C and S . Then, the tangency condition ⟨ C_x,C_t ⟩²=⟨ C_x,C_x ⟩ ⟨ C_t,C_t ⟩ and Exercise 2 lead to a quadratic equation with roots (16). ⋄

Remark 30 Note that the tangency condition (15) is quadratic in parameters of a cycle C_σ. However, n such conditions with a single unknown cycle C can be reduced to the single quadratic condition ⟨ C,C ⟩=1 and n linear conditions like ⟨ C,S _(i) ⟩=λ_i. This observation was implemented in the C++ library figure [211]. A similar technique was also used in [101].

Recall that straight lines are cycles which pass the infinity, that is orthogonal to the zero-radius cycle at infinity. Here is tangency property for lines:

Lecture 5 Indefinite Product Space of Cycles

5.1 Cycles: an Appearance and the Essence

5.2 Cycles as Vectors

5.3 Invariant Cycle Product

5.4 Zero-radius Cycles

5.5 Cauchy–Schwarz Inequality and Tangent Cycles