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CCSMath120: Calculus on Manifolds Simon Rubinstein–Salzedo Spring 2004 0.1 Introduction These notes are based on a course on calculus on manifolds I took from Professor MartinScharlemannintheSpringof2004. Thecoursewasdesignedforfirst-yearCCS math majors. The primary textbook was Michael Spivak’s Calculus on Manifolds. A recommendedsupplementary text was Maxwell Rosenlicht’s Introduction to Analysis. 1 Chapter 1 Basic Analysis and Topology Wedefine Rn ={(x1,...,xn) | xi ∈ R}. For example, (π,e,√2) ∈ R3. Rn is a vector space. This means that we have two operations + : Rn×RRn → Rn and · : R × Rn → Rn satisfying the following properties: 1. x+y =y+x. 2. x+(y+z)=(x+y)+z. 3. There exists 0 ∈ Rn so that for all x, 0 + x = x. 4. For all x ∈ Rn, there exists −x ∈ Rn so that x + (−x) = 0. 5. α(x+y) = αx+αy for all α ∈ R. 6. (α +β)x = αx+βx. 7. α(βx) = (αβ)x. 8. 1x = x. If x = (x1,...,xn) and y = (y1,...,yn), we define x +y = (x1 +y1,...,xn +yn). We assume the following properties of R: For α,β ∈ R, |α| = |β| iff α2 = β2, and |α| ≥ |β| iff α2 ≥ β2. 2 There is a norm on Rn: | · | : Rn → R with useful properties. It is defined by 1 n p 1 2 n 2 |(x ,...,x )| = (x ) +···+(x ) . Proposition. |·| satisfies, for all α ∈ R and x,y ∈ Rn, the following properties: 1. |x| ≥ 0 and (|x| = 0 iff x = 0). P i i 2. | x y | ≤ |x| |y|, with equality iff x and y are linearly dependent. 3. |x + y| ≤ |x| + |y|. 4. |αx| = |α| |x|. Proof. 2 P i 2 i P i 2 1. |x| = 0 iff |x| = 0 iff (x ) = 0. If all x = 0, then (x ) = 0. If some i i 2 P j2 i 2 i 2 x 6= 0, then (x ) > 0, so j6=i(x ) + (x ) ≥ (x ) > 0. P i i 2. If x and y are linearly dependent, then | x y | = |x| |y|. [Digression: Suppose v1,...,vm are vectors in a vector space. They are said to be linearly dependent iff there exists α ,...,α , not all zero, so that Pα v = 0. Hence αx+βy = 0 1 m i i for some α,β not both zero. Suppose for example that α 6= 0. Then x+ βy = 0. α Thustheyare linearly independent iff there exists a λ so that x = λy or y = λx. To see this, suppose α 6= 0. Take λ = −β. If α = 0, then βy = 0, so y = 0.] α Suppose without loss of generality that y = λx, i.e. yi = λxi for all i. Then X 2 X 2 xiyi = xiyi ! n 2 = Xxiλxi i=1 2 X i2 2 =λ (x ) 2 X i2 X i2 =λ (x ) (x ) X i 2X i 2 = (x ) (λx ) 2 2 =|x| |y| , 3
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