By A T Fomenko and Avgustin Tuzhilin, Visit Amazon's A.T. Fomenko Page, search results, Learn about Author Central, A.T. Fomenko, , Avgustin Tuzhilin

This ebook grew out of lectures offered to scholars of arithmetic, physics, and mechanics by means of A. T. Fomenko at Moscow college, below the auspices of the Moscow Mathematical Society. The publication describes smooth and visible facets of the speculation of minimum, two-dimensional surfaces in third-dimensional area. the most subject matters coated are: topological homes of minimum surfaces, solid and volatile minimum motion pictures, classical examples, the Morse-Smale index of minimum two-surfaces in Euclidean area, and minimum motion pictures in Lobachevskian area. Requiring just a general first-year calculus and basic notions of geometry, this publication brings the reader speedily into this attention-grabbing department of recent geometry.

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**Sample text**

A third order system a mn with , m, n = 1, 2, 3 is said to be symmetric in two of its subscripts if the components are unaltered when these subscripts are interchanged. When a a mn = am n =a nm = amn = anm = an m. mn is completely symmetric then Whenever this third order system is completely symmetric, then: (i) How many components are there? (ii) How many of these components are distinct? Hint: Consider the three cases (i) = m = n 22. A third order system b mn (ii) =m=n (iii) = m = n. with , m, n = 1, 2, 3 is said to be skew-symmetric in two of its subscripts if the components change sign when the subscripts are interchanged.

Pk where δsr is the Kronecker delta. 123 eijk = δijk Show (b) Show ijk eijk = δ123 (c) Show ij δmn = eij emn (d) Define rs rsp δmn = δmnp and show (summation on p) rs r s s = δm δn − δnr δm δmn Note that by combining the above result with the result from part (c) we obtain the two dimensional form of the e − δ identity r rn = 12 δmn (e) Define δm 27. Let denote the cofactor of ari a11 in the determinant a21 a31 (a) Show erst Air = eijk asj atk 28. rst δpst = 2δpr (summation on n) and show rst δrst = 3!

59) and obtain the equation r Aqp Bkij ∂xq ∂xs ∂xk ∂xi ∂xj ∂xr ∂xs ∂xm ∂xl ∂xp ∂xs ∂xm . = Arqm Brql l ∂x ∂xp = l Cm Since the summation indices are dummy indices they can be replaced by other symbols. We change l to j, q to i and r to k and write the above equation as ∂xs ∂xj Use inner multiplication by ∂xn ∂xs r Aqp m ∂xq ∂xk k ∂x − A im ∂xi ∂xr ∂xp Bkij = 0. and simplify this equation to the form r δjn Aqp r Aqp m ∂xq ∂xk k ∂x − A Bkij = 0 im ∂xi ∂xr ∂xp or m ∂xq ∂xk k ∂x − A Bkin = 0. im ∂xi ∂xr ∂xp Because Bkin is an arbitrary tensor, the quantity inside the brackets is zero and therefore r Aqp m ∂xq ∂xk k ∂x = 0.