Ugrás a tartalomhoz

Convex Geometry

Csaba Vincze (2013)

University of Debrecen

14. fejezet - References

14. fejezet - References

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[1] In opposite case the complements of B(k)'s form an open cover of the space, especially, they form an open cover of B(1). Choosing a finite subcover and taking the complement again we have that the intersection of a finite subfamily of sets B(k)'s is in the complement of B(1). At the same time it is a subset in B(1) which is obviously a contradiction.

[2] Affine subspaces mean translates of linear subspaces, see also section 1.3

[3] The relative boundary of 2.10 means its boundary in its affine hull.

[4] Taking the sum of the coordinates is a linear operator and thus the sum of the coordinates in a linear combination is the linear combination of the sum of the coordinates.

[5] If t is less than 60 then the vertex is the foot of the perpendicular line from A to the corresponding tangent line 4.5. As they are rotating into the clockwise direction the path of the vertex will be the pedal curve of the arc AC with respect to A. In other words the lower line is not tangential to the full circle belonging to the lower arc AB under the critical value of the parameter.

[6] Hungarian translation: p megfigyelési pontja q-nak

[7] If H would bound the interior of K then it would support K at the same time. The existence of interior points in both sides of H causes an (n - 1)-dimensional intersection.

[8] Hungarian translation: k pontra feszíthető.

[9] It is clear that one of the sets A and B is non-empty; if (for example) A is the empty set the separation means the choice of a point not in the convex hull of B.

[10] Steinitz's comparison lemma states that if we have two polygons A(1)A(2) ... A(n) and B(1)B(2) ... B(n) in the plane such that the sides A(i)A(i+1) and B(i)B(i+1) are equal for i=1, 2, ..., n - 1 but the angles of the first polygon are less than or equal to the angles of the second polygon with at least one strict inequality then A(1)A(n) < B(1)B(n). The proof is based on the induction with respect to the number of vertices. All of Euclidean axioms are working except the parallel postulate. Therefore the lemma holds in spherical geometry too. Consider now a vertex figure and suppose in contrary that the sign changes at most two times (the number of changes of sign is obviously even including zero change too). If we have exactly two changes then there is a diagonal cutting the spherical polygon into two parts. One of them contains only - vertices and the other only + vertices. Applying the spherical version of Steinitz's comparison lemma to the - side we obtain that the diagonal is greater than the corresponding diagonal in the vertex figure belonging to the second polyhedron. Using the other side we have just the opposite conclusion which is impossible.

[11] As explicite examples we can consider curves, surfaces or compact domains in the space.

[12] In case of differentiable manifolds with Riemannian structures the subgroup H is the holonomy group of the Lévi-Civita connection and the alternative geometry is called Finsler geometry: instead of the Euclidean spheres in the tangent spaces, the unit vectors form the boundary of general convex sets containing the origin in their interiors (M. Berger).

[13] The only trivial orbit is that of the origin -- it is a singleton.

[14] For a more precise formulation of determination, verification and successive determination see [26]

[15] The Hausdorff distance of sets is determined by the distances between their points and, because of the invariance of T under H, the possible distances between the points of the boundary of Γ and T are the same as the possible distances between the points of E and T.