By Larry Smith
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Additional info for Algebraic topology: Proc. conf. Goettingen 1984
Therefore, 2 x(s)2 + y(s) − y0 = R2 (x′ (s)2 + y ′ (s)2 ) = R2 , and C is indeed a circle of radius R. 3 1. a. Prove that the shortest path between two points on the unit sphere is the arc of a great circle connecting them. (Hint: Without loss of generality, take one point to be (0, 0, 1) and the other to be (sin u0 , 0, cos u0 ). ) b. Prove that if P and Q are points on the unit sphere, then the shortest path between them has length arccos(P · Q). 32 Chapter 1. Curves 2. Give a closed plane curve C with κ > 0 that is not convex.
S) = x(s), y(s) = x(s), R2 − x(s)2 , s0 ≤ s ≤ L §3. 5) L A= x(s)y ′ (s)ds 0 A = πR2 = − L 0 y(s)x′ (s)ds = − L y(s)x′ (s)ds. 0 Adding these equations and applying the Cauchy-Schwarz inequality, we have L A + πR2 = 0 x(s)y ′ (s) − y(s)x′ (s) ds = L (∗) ≤ x(s), y(s) 0 ′ (y (s), −x′ (s)) L 0 x(s), y(s) · y ′ (s), −x′ (s) ds y ′ (s), −x′ (s) ds = RL, inasmuch as = (x′ (s), y ′ (s)) = 1 since α is arclength-parametrized. We now recall the arithmetic-geometric mean inequality: √ a+b for positive numbers a and b, ab ≤ 2 with equality holding if and only if a = b.
Let C be a C2 closed space curve, say parametrized by arclength by α : [0, L] → R3 . A unit normal field X on C is a C1 vector-valued function with X(0) = X(L) and X(s) · T(s) = 0 and X(s) = 1 for all s. We define the twist of X to be tw(C, X) = 1 2π L 0 X′ (s) · (T(s) × X(s))ds. X∗ a. Show that if X and are two unit normal fields on C, then tw(C, X) and tw(C, X∗ ) differ by an integer. , the twist mod 1) is called the total twist of C. ) 1 τ ds. b. Prove that the total twist of C equals the fractional part of 2π C c.
Algebraic topology: Proc. conf. Goettingen 1984 by Larry Smith