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Metric spaces which cannot be isometrically embedded in Hilbert space

Published online by Cambridge University Press:  17 April 2009

Yang Lu
Affiliation:
Department of Mathematics, University of Science and Technology of China, Hefei, Anhui, The People's Republic of China.
Zhang Jing-Zhong
Affiliation:
Department of Mathematics, University of Science and Technology of China, Hefei, Anhui, The People's Republic of China.
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Abstract

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Let A1A2A3A4, be a planar convex quadrangle with diagonals A1A3 and A2A4. Is there a quadrangle B1B2B3B4 in Euclidean space such that A1A3 < B1B3, A2A4 < B2B4 but AiAj > BiBj for other edges?

The answer is “no”. It seems to be obvious but the proof is more difficult. In this paper we shall solve similar more complicated problems by using a higher dimensional geometric inequality which is a generalisation of the well-known Pedoe inequality (Proc. Cambridge Philos. Soc.38 (1942), 397–398) and an interesting result by L.M. Blumenthal and B.E. Gillam (Amer. Math. Monthly50 (1943), 181–185).

Type
Research Article
Copyright
Copyright © Australian Mathematical Society 1984

References

[1]Blumenthal, L.M., Theory and applications of distance geometry (Chelsea, New York, 1970).Google Scholar
[2]Blumenthal, L.M. and Gillam, B.E., “Distribution of points in n-space”, Amer. Math. Monthly 50 (1943), 181185.CrossRefGoogle Scholar
[3]Pedoe, D., “An inequality for two triangles”, Proc. Cambridge Philos. Soc. 38 (1942), 397398.CrossRefGoogle Scholar
[4]Lu, Yang and Jing-zhong, Zhang, “A generalisation to several dimensions of the Neuberg-Pedoe inequality, with applications”, Bull. Austral. Math. Soc. 27 (1983), 203214.CrossRefGoogle Scholar
[5]Lu, Yang and Jing-zhong, Zhang, “A high-dimensional extension of the Neuberg-Pedoe inequality and its application” (Chinese), Acta Math. Sinica 24 (1981), 401408.Google Scholar
[6]Lu, Yang and Jing-zhong, Zhang, “A geometric criterion of metric embedding and the skew mapping”, submitted.Google Scholar