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