By David Bao, Robert L. Bryant, Shiing-Shen Chern, Zhongmin Shen
Finsler geometry generalizes Riemannian geometry in precisely an analogous method that Banach areas generalize Hilbert areas. This booklet offers expository debts of six very important issues in Finsler geometry at a degree compatible for a distinct themes graduate path in differential geometry. The members give some thought to concerns regarding quantity, geodesics, curvature and mathematical biology, and comprise numerous instructive examples.
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Extra info for A sampler of Riemann-Finsler geometry
In the same way, we may define the integral of a k-density over a k-chain. For the rest of the chapter, we associate to a given k-volume density φ on a vector space X the functional N −→ φ, N and investigate the relationship between the behavior of the functional and certain convexity properties of φ. The easiest case is when φ is an (n−1)-volume density in an n-dimensional vector space. 2. Convexity of (n−1)-volume densities. This case is special because every (n−1)-vector in an n-dimensional vector space, X, is simple and we may impose the condition that an (n−1)-volume density be a norm in Λ n−1 X.
C. THOMPSON The first step in solving the isoperimetric problem in a normed space X is to find a centrally symmetric convex body I such that x = V ([x], I), for all x ∈ X. Of course, I will also depend on the choice of Lebesgue measure λ used to define the mixed volume. However, given a volume definition the body I will be defined intrinsically in terms of the norm. The construction of I is extremely simple: Let B be the unit ball of X and let Ω be the volume form on X that satisfies Ω(x ∧ y) = λ(x ∧ y) for all positive bases x, y of X (we are forced to take an orientation of X at this point, but the result will not depend on the choice).
THOMPSON By letting a3 , . . , an tend to zero in the above inequality we obtain the triangle inequality φ(a1 + a2 ) ≤ φ(a1 ) + φ(a2 ), and, therefore, φ is a norm. 9. Consider the tetrahedron in the normed space 3∞ with vertices (0, 0, 0), (−1, 1, 1), (1, −1, 1), (1, 1, −1), and show that the mass of the facet opposite the origin is greater than the sum of the masses of the three other facets. Hint. Use the definition of the mass 2-volume density in terms of minimal circumscribed parallelograms.