Csaba Vincze (2013)

University of Debrecen

In this section we are going to present more explicit examples of circular conics (conics with acircle as the set of foci) in the coordinate space of dimension three. Conics in the coordinate plane with squares as the set of focuses will be also considered. In both cases we use integration to compute the average distance. Integration of the Euclidean distance over subsets in the space is crucial for finding** alternatives of Euclidean geometry**. Let O(n) be the group of linear isometries in the Euclidean space and consider a subgroup H in the orthogonal group together with a compact convex subset K containing the origin in its interior such that

(i) it is not an ellipsoid (ellipsoid problem),

(ii) it is invariant under the elements of H and

(iii) its boundary is a smooth hypersurface (regularity condition).

By the second condition H is a subgroup of the linear isometry group O'(n) of the Minkowski space induced bythe Minkowski functional of K. The first condition shows that O'(n) is a real subgroup in the orthogonal group. In other words the Minkowski geometry based on the functional associated to K is an alternative of the Euclidean geometry[12] for the subgroup H. One of the main applications of the generalized conics' theory is to present convex bodies satisfying conditions (i) - (iii).

**Definition **The subgroup H is called
transitive on the Euclidean unit sphere if any two elements of the unit sphere can be transported into each other by a transformation from H.

If the subgroup H is transitive on the Euclidean unit sphere then the Euclidean geometry is the only possible one for H. The list of transitive subgroups in O(n) are

and

For the classification see [11], [12] and [29]. In what follows we present alternatives of the Euclidean geometry for non-transitive subgroups of the orthogonal group [57]. Theoretically we have two different cases: reducible and irreducible subgroups.

If the group is reducible then there exists a non-trivial invariant linear subspace of dimension 0 < k < n in the coordinate space under the elements of the subgroup. This subspace cuts a (k - 1)- dimensional sphere S from the unit sphere in the embedding space. In this case S plays the role of the set of foci. To avoid the theoretical and technical difficulties of higher dimensional coordinate spaces we restrict ourselves to the space of dimension three (circular conics) [56]. Let

$w:\left[\mathrm{0,2}\pi \right]\to {E}^{3},w\left(t\right):=(\mathrm{c}\mathrm{o}\mathrm{s}t,\mathrm{s}\mathrm{i}\mathrm{n}t,0)$ |
(10.7) |

be the unit circle in the (x, y)-coordinate plane and

$F(x,y,z):=\frac{1}{2\pi}{\int}_{0}^{2\pi}\sqrt{(x-\mathrm{c}\mathrm{o}\mathrm{s}t{)}^{2}+(y-\mathrm{s}\mathrm{i}\mathrm{n}t{)}^{2}+{z}^{2}}dt.$ |

The surface of the form

$F(x,y,z)=\frac{8}{2\pi}$ |
(10.8) |

is a generalized conic with foci S(1) in the Euclidean space. According to the invariance of the set of foci under the rotation around the z-axis 10.8 is a revolution surface with generatrix

${\int}_{0}^{2\pi}\sqrt{{\mathrm{c}\mathrm{o}\mathrm{s}}^{2}t+(y-\mathrm{s}\mathrm{i}\mathrm{n}t{)}^{2}+{z}^{2}}dt=8$ |
(10.9) |

in the (y, z)-coordinate plane.

**Lemma 10.2.1 **
*The generalized conic 10.8 is not an ellipsoid. *

**Proof **It is enough to prove that the generatrix

${\int}_{0}^{2\pi}\sqrt{{\mathrm{c}\mathrm{o}\mathrm{s}}^{2}t+(y-\mathrm{s}\mathrm{i}\mathrm{n}t{)}^{2}+{z}^{2}}dt=8$ |

is not an ellipse in the (y, z)-coordinate plane. If y=0 then we have that

$z=\pm \sqrt{(\frac{8}{2\pi}{)}^{2}-1}.$ |

On the other hand, if z=0 then the solutions of the equation

${\int}_{0}^{2\pi}\sqrt{{\mathrm{c}\mathrm{o}\mathrm{s}}^{2}t+(y-\mathrm{s}\mathrm{i}\mathrm{n}t{)}^{2}}dt=8$ |

are just y=+1 or - 1. Therefore the only possible ellipse has the parametric form

$y\left(s\right)=\mathrm{c}\mathrm{o}\mathrm{s}s\text{}and\text{}z\left(s\right)=\sqrt{(\frac{8}{2\pi}{)}^{2}-1}\mathrm{s}\mathrm{i}\mathrm{n}s.$ |
(10.10) |

The figure shows the generatrix (pointstyle) and its approximating ellipse. Consider the auxiliary function

$v\left(s\right):{=\int}_{0}^{2\pi}\sqrt{{\mathrm{c}\mathrm{o}\mathrm{s}}^{2}t+\left(y\right(s)-\mathrm{s}\mathrm{i}\mathrm{n}t{)}^{2}+{z}^{2}(s)}dt.$ |

Then v(0)=v(π/2)=8 but

$v\left(\frac{\pi}{3}\right)=\frac{2}{\pi}\sqrt{2}\sqrt{3}\sqrt{8+{\pi}^{2}}E\left(\frac{2\sqrt{3}\pi}{3\sqrt{8+{\pi}^{2}}}\right),$ |

where

$E\left(r\right):={\int}_{0}^{\frac{\pi}{2}}\sqrt{1-{r}^{2}{\mathrm{s}\mathrm{i}\mathrm{n}}^{2}t}dt$ |

is the standard elliptic integral. Using that

$E\left(r\right)\ge \frac{\pi}{2}(\frac{1+(r\mathrm{\text{'}}{)}^{\frac{3}{2}}}{2}{)}^{\frac{2}{3}},\text{}where\text{}r\mathrm{\text{'}}=\sqrt{1-{r}^{2}}$ |
(10.11) |

(Vuorinen's conjecture, for the proof see [2]) the inequality

$\sqrt{3}\sqrt{2}\sqrt{8+{\pi}^{2}}(\frac{1}{2}+\frac{1}{18}\sqrt{3}(9-\frac{12{\pi}^{2}}{8+{\pi}^{2}}{)}^{\frac{3}{4}}{)}^{\frac{2}{3}}>8$ |
(10.12) |

shows that v(s) is not a constant function. ▮

**Corollary 10.2.2 **
*The generalized conic 10.8 induces a non-Euclidean Minkowski functional l such that the Euclidean isometries leaving the set of foci 10.7 invariant form a subgroup of the linear isometries with respect to l. *

**Proof **It is clear that generalized conics together with the induced Minkowski functionals heritage all symmetry properties of the set of foci. Onthe other hand the previous lemma shows that L can not be arised from an inner product.

In general the group of linear isometries of a non-Euclidean Minkowski space is trivial. Results like 10.2.2 give examples on geometric spaces having more and more rich linear isometry group up to the Euclidean geometry.

**Definition **A transformation group H is closed if for any point p theorbit

$P\left(p\right)=\left\{\phi \right(p\left)\right|\phi \in H\}$ |

is a closed subset.

**Example **The group of rotations in the plane around the origin with rational magnitudes is not a closed subgroup.

**Excercise 10.2.3 **
*Prove that each non-transitive closed subgroup in the orthogonal group of the Euclidean plane is finite. *

Hint. The proof is a kind of nearest-point-type argumentation. Take a point u on the unit circle and consider its orbit P(u) under H. Since H is not transitive there exists a point u' on the unit circle which is not in P(u). By the closedness we can consider the closest points of P(u) to u' into the two possible directions. They form a circular arc C' of positive length such that its (relative) interior is disjoint from P(u). The nearest-point property implies that the relative interiors of the circular arcs P(C') must be pairwise disjoint. Then H must be finite because the sum of the lengths is obviously bounded.

**Remark **The result says that the alternative geometries of dimension two for non-transitive closed subgroups in O(2) always can be realized by Minkowski functionals induced by polyellipses in the plane. A similar theorem can be formulated in case of the coordinate space of dimension three because of Wang's theorem [62] stating that the dimension of a non-transitive closed subgroup inO(3) is just 0 or 1. In case of a one-dimensional subgroup the unit component is actually a one-parameter family of rotations around a line. Choose a finite "unit component" - invariant system of points on this line we can consider the image of the system under mappings from different connected components (it is enough to choose only one mapping from each component). Since the number of the components is finite
we have a finite collection of points which is invariant under the whole group.

In case of higher dimensional spaces the convex hull of non-trivial orbits[13] will play the role of the set of foci.

**Lemma 10.2.4 **
*Let H be a closed subgroup in the orthogonal group. It is irreducible if and only if the origin is the interior point of the convex hull of any non-trivial orbit under H. *

Proof First of all note that the convex hulls of the orbits are obviously invariant under H. If H is irreducible and the origin is not a point of the convex hull of a non-trivial orbit then we can use a simple nearest-point-type argumentation to present a contradiction as follows: taking the uniquely determined nearest point of the convex hull to the origin it can be easily seen that it must be a
fixed point of any element of H. This contradicts to the irreducibility. If the origin is not in the interior of the convex hull of a non-trivial orbit P(u) we can consider the common part T of supporting hyperplanes at the origin. Since** 0 **is not in P(u) it can not be an extreme point of the convex hull which means that T is at least a one-dimensional linear subspace. On the other hand it is invariant under H which is a contradiction. Therefore the origin must be in the interior of the convex hull of any non-trivial orbit. The converse of the statement is trivial.

**Excercise 10.2.5 **
*Prove that invariant ellipsoids must be Euclidean balls in case of any irreducible subgroup of the orthogonal group. *

Hint. Suppose that H contains orthogonal transformations with respect to different inner products. For the sake of simplicity let one of them be the canonical inner product and consider another one given by a symmetric matrix M. If Mv=λv for some nonzero vector v then for any transformation h in H we have

$wMh\left(v\right)={h}^{-1}\left(w\right)Mv=\lambda {h}^{-1}\left(w\right)v=\lambda wh\left(v\right)\Rightarrow Mh\left(v\right)=\lambda h\left(v\right)$ |

because both h and its inverse are orthogonal transformations with respect to both M and the canonical inner product. Therefore eigenvectors with eigenvalue λ form an invariant linear subspace of H which must be the whole space according to the irreducibility. Thus Mv=λv for all vectors and the balls with respect to these inner products coincide.

**Example **Consider the group of symmetries of the square

$[-\mathrm{1,1}]\times [-\mathrm{1,1}]$ |
(10.13) |

centered at the origin in the Euclidean plane. The convex hull of any non-trivial orbit is a convex polygon having singularities at the vertices. For example the orbit

$P\left(\right(-1,-1\left)\right)=\left\{\right(-1,-1),(1,-1),(\mathrm{1,1}),(-\mathrm{1,1}\left)\right\}$ |

induces the supremum norm

$\left|\right(x,y\left)\right|:=\frac{1}{\sqrt{2}}max\left\{\right|x|,|y\left|\right\}.$ |

To avoid the singularities at the vertices consider the function

$F(x,y):=\frac{1}{4}{\int}_{-1}^{1}{\int}_{-1}^{1}\sqrt{(x-t{)}^{2}+(y-s{)}^{2}}dsdt.$ |

Curves of the form F(x,y)=const. are just generalized conics with the square 10.13 as the set of foci. In what follows we investigate the level curve C passing through the point (2,1).

**Excercise 10.2.6 **
*Prove that C is not a circle: recall that invariant ellipses under the symmetry group of the square must be circles. *

Hint. According to the symmetric role of the variables x, t and y, s we cancalculate the coordinate functions

${D}_{1}F(x,y)=\frac{1}{4}{\int}_{-1}^{1}{\int}_{-1}^{1}\frac{x-t}{\sqrt{(x-t{)}^{2}+(y-s{)}^{2}}}dsdt,$ |

${D}_{2}F(x,y)=\frac{1}{4}{\int}_{-1}^{1}{\int}_{-1}^{1}\frac{y-s}{\sqrt{(x-t{)}^{2}+(y-s{)}^{2}}}dsdt$ |

of the gradient vector field:

${D}_{1}F(x,y)=-\frac{1}{8}\left[\right(s-y)\sqrt{(x-1{)}^{2}+(y-s{)}^{2}}+$ |

$+(x-1{)}^{2}\mathrm{l}\mathrm{n}((s-y)+\sqrt{(x-1{)}^{2}+(y-s{)}^{2}})+$ |

$(s-y)\sqrt{(x+1{)}^{2}+(y-s{)}^{2}}+(x+1{)}^{2}\mathrm{l}\mathrm{n}((s-y)+$ |

$+\sqrt{(x+1{)}^{2}+(y-s{)}^{2}}\left){]}_{-1}^{1}\text{}and\text{}{D}_{2}F\right(x,y)={D}_{1}F(y,x).$ |

Using these formulas consider the auxiliary function

$v(x,y):=y{D}_{1}F(x,y)-x{D}_{2}F(x,y)$ |

to measure the difference between the gradient vectors of the family of generalized conics and circles. We have

$v\left(\mathrm{2,1}\right)=-2\sqrt{13}+\frac{9}{2}\mathrm{l}\mathrm{n}3-\frac{9}{2}\mathrm{l}\mathrm{n}(-2+\sqrt{13})+\frac{1}{2}\mathrm{l}\mathrm{n}(-2+\sqrt{5})-8\mathrm{l}\mathrm{n}2+$ |

$+4\mathrm{l}\mathrm{n}(-3+\sqrt{13})+4\mathrm{l}\mathrm{n}(\sqrt{5}+1)+8$ |

which is obviously different from zero.

The general case is discussed via the following theorem of the alternatives.

**Definition **Let z be a fixed element of the unit sphere S in the coordinate space of dimension n and consider its orbit under H. The minimax point of the orbit is such a point z* on the sphere where the minimum

$a:=\underset{\parallel w\parallel =1}{\mathrm{m}\mathrm{i}\mathrm{n}}\underset{\gamma \in P\left(z\right)}{\mathrm{m}\mathrm{a}\mathrm{x}}d(w,\gamma )$ |

is attained at.

Consider the function

$f:R\to R,f\left(t\right):=\left\{\begin{array}{cc}0& ift\le a\\ (t-a){e}^{-\frac{1}{t-a}}& ift>a.\end{array}\right.$ |

By the help of the standard calculus [38] it can be seen that it is a smooth convex function on the real line. Define

$g\left(t\right):=t+f\left(t\right)$ |

and take the functions

$F\left(p\right):={\int}_{convP\left(z\right)}\gamma \mapsto d(p,\gamma )d\gamma $ |

and

${F}^{\mathrm{*}}\left(p\right):={\int}_{convP\left(z\right)}\gamma \mapsto g\left(d\right(p,\gamma \left)\right)d\gamma .$ |

It is clear that

$c:=F\left({z}^{\mathrm{*}}\right)={F}^{\mathrm{*}}\left({z}^{\mathrm{*}}\right).$ |

On the other hand one of the hypersurfaces F(p)=c and F*(p)=c must be different from the sphere unless the mapping

$w\in S\mapsto \underset{\gamma \in {P}_{z}}{\mathrm{m}\mathrm{a}\mathrm{x}}d(w,\gamma )$ |

is constant. It is impossible because H is not transitive.

**Excercise 10.2.7 **
*Prove that if the mapping *

$w\in S\mapsto \underset{\gamma \in P\left(z\right)}{\mathrm{m}\mathrm{a}\mathrm{x}}d(w,\gamma )$ |

is constant then

$P\left(z\right)=S$ |

and H is transitive on the Euclidean unit sphere.

Hint. Taking any element in P(z) its antipole shows that the constant is just the diameter of the sphere. Finally we have the following theorem of the alternatives.

**Theorem 10.2.8 **
*(Theorem of the alternatives) If H is non-transitive on the unit sphere, closed and irreducible, z in S and c is the common value of the functions F and F* at the minimax point z* then at least one of the hypersurfaces *

${\int}_{convP\left(z\right)}\gamma \mapsto d(p,\gamma )d\gamma =c\text{}or\text{}{\int}_{convP\left(z\right)}\gamma \mapsto g\left(d\right(p,\gamma \left)\right)d\gamma =c,$ |

induces a non-Euclidean Minkowski functional l such that H is the subgroup of the linear isometries with respect to l.

Finally we slightly modify the rate of the level in such a way that c* > c. A continuity-type argumentation shows that the generalized conic belonging to the level c* induces a non-Euclidean regular Minkowski functional for the subgroup H. Regularity follows easily because the set of foci is containedin the interior of these convex hypersurfaces. The theorem of the alternatives motivates the following definition.

**Definition **Let Γ be a bounded orientable submanifold in the coordinate space of dimension n with finite positive measure. If g is a strictly monotone increasing convex function on the non-negative real numbers with initial value g(0)=0 and

${F}_{g}\left(p\right):=\frac{1}{vol\mathrm{\Gamma}}{\int}_{\mathrm{\Gamma}}\gamma \mapsto g\left(d\right(p,\gamma \left)\right)d\gamma $ |
(10.14) |

then hypersurfaces of the form F(g)(p)=const. are called generalized conics with distorsion g.

**Excercise 10.2.9 **
*Prove that 10.14 is a convex function satisfying the growth condition 10.6 *

Hint. Use that

$\underset{r\to \mathrm{\infty}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{g\left(r\right)}{r}>0.$ |

**Theorem 10.2.10 **
*If H is a non-transitive closed subgroup of the orthogonal group then the alternative geometry for H always can be realized by Minkowski functionals associated to generalized conics in the space. *