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By Luther Pfahler Eisenhart
In Riemannian geometry, parallelism is decided geometrically by means of this estate: alongside a geodesic, vectors are parallel in the event that they make an identical perspective with the tangents. In non-Riemannian geometry, the Levi-Civita parallelism imposed a priori is changed by means of a choice via arbitrary capabilities (affine connections). during this quantity, Eisenhart investigates the most outcomes of the deviation.
Starting with a attention of uneven connections, the writer proceeds to a contrasting survey of symmetric connections. Discussions of the projective geometry of paths persist with, and the ultimate bankruptcy explores the geometry of sub-spaces.
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Extra resources for Non-Riemannian Geometry
L, M, and N are collinear. This is surely a startling result, totally unexpected, easy to grasp, and beautiful to contemplate. In the modern foundations of geometry it is one of the fundamental theorems. If we take other points, say D on f and D' on m, the intersection CD' n C'D lies on the line already constructed. This line will not, we note, usually pass through the intersection of e and m. A P FIG. 340 xxxiv CIRCLES A case where it always does so arises when we choose our points A, A', B, B', and C, C' in a special way, so that the joins AA', BB', and CC' all pass through the same point V.
In the modern foundations of geometry it is one of the fundamental theorems. If we take other points, say D on f and D' on m, the intersection CD' n C'D lies on the line already constructed. This line will not, we note, usually pass through the intersection of e and m. A P FIG. 340 xxxiv CIRCLES A case where it always does so arises when we choose our points A, A', B, B', and C, C' in a special way, so that the joins AA', BB', and CC' all pass through the same point V. The theorem of Pappus still applies, and we obtain a figure almost like a national flag, seen in perspective.
Let B and C be the acute angles of the triangle ABC. On BC, and away from A, describe an equilateral triangle BCD. Then by the extension of Ptolemy's theorem, unless P lies on the circle Fio. BC, or PB + PC > PD, since CD = DB = BC. Therefore PA + PB + PC > PA + PD. 6 Now, unless P lies on AD, we have PA + PD > AD. Hence, unless P is at P' (the other intersection of AD with the circle BCD), we have PA + PB + PC> AD. But if P is at P', both the above inequalities become equalities; so that P'A + P'B + P'U = AD.