Lecture Note on Elementary Differential Geometry
                                                                             *
                                                          Ling-Wei Luo
                                                 Institute of Physics, Academia Sinica
                                                           July 20, 2019
                                                               Abstract
                        This is a note based on a course of elementary differential geometry as I gave the lectures in
                    theNCTU-YauJournalClub: InterplayofPhysicsandGeometryatDepartmentofElectrophysics
                    in National Chiao Tung University (NCTU) in Spring semester 2017. The contents of remarks,
                    supplements and examples are highlighted in the red, green and blue frame boxes respectively.
                    The supplements can be omitted at first reading. The basic knowledge of the differential forms
                    can be found in the lecture notes given by Dr. Sheng-Hong Lai (NCTU) and Prof. Jen-Chi Lee
                    (NCTU) on the website. The website address of Interplay of Physics and Geometry is http:
                    //web.it.nctu.edu.tw/~string/journalclub.htmorhttp://web.it.nctu.
                    edu.tw/~string/ipg/.
             Contents
                               2
             1    CurveonE        . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .     1
                              3
             2    CurveinE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .          6
                                        3
             3    Surface theory in E     . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .     9
             4    Cartan’s moving frame and exterior differentiation methods . . . . . . . . . . . . . .               31
                                      2
             1     CurveonE
                                                             n
             Wedefinen-dimensionalEuclidean space E as
                                              n
             a n-dimensional real space R equipped a dot
             product defined n-dimensional vector space.
             Tangent vector       In 2-dimensional Euclidean
                           2
             space, an E plane, we parametrize a curve
                      (          )
             p(t) = x(t),y(t) by one parameter t with re-
             specttoareferencepointowithafixedCartesian
             coordinate frame. The velocity vector at point p
                          ˙       (          )                                       Figure 1: A curve.
             is given by p(t) = x˙(t),y˙(t) with the norm
                        ˙       √˙ ˙        √ 2       2
                       |p(t)| =    p·p= x˙ +y˙ ,              (1)
                 *
                  Electronic address: lwluo@gate.sinica.edu.tw
                                                                   1
                      where x˙ := dx/dt. The arc length s in the interval [a,b] can be calculated by
                                                           ∫               ∫ √               2              2       ∫ b√ 2                  2           ∫ b ˙
                                                   s =           ds =                (dx) +(dy) = a                             x˙   +y˙ dt = a |p(t)|dt.                                              (2)
                      Thearclength can be a function of parameter t given by
                                                                                                           ∫ t ˙         ′        ′
                                                                                              s(t) = a |p(t )|dt .                                                                                     (3)
                      Fromthefundamental theorem of calculus, we have
                                                                                     
                                                                               ds
                                                                                                                                ˙
                                                                               dt ̸= 0             =⇒ s˙(t)=|p(t)| > 0.                                                                              (4)
                      According to the inverse function theorem, we have t = t(s). One can parametrize the curve by arc
                                                         (                   )                                                                                         ′           ( ′             ′      )
                      length s as p(s) = x(s),y(s) . The corresponding velocity vector should be p (s) = x (s),y (s) ,
                      where we have x′ := dx/ds. We can rewrite the derivatives of x and y with respect to s as
                                                                                                    dx          dxdt                dt
                                                                                        ′
                                                                                       x =                 =                =x˙          ,
                                                                                                    ds          dt ds               ds                                                                (5)
                                                                                                       dt
                                                                                        ′
                                                                                       y =y˙ds.
                      Thus, the norm of the velocity vector parametrized by s can be calculated as
                                                              ′             √ ′2             ′2       √ 2              2 dt           ˙   dt         ds dt
                                                          |p (s)| =             x +y = x˙ +y˙ ds =|p|ds = dtds =1,                                                                                     (6)
                      which implies that the velocity vector p′(s) is a unit vector. We can define a unit tangent vector as a
                      velocity vector parametrized by s
                                                                                                  T ≡e1 := p′(s).                                                                                      (7)
                      Normalvector Duetoe ·e =p′·p′ =1,wehave
                                                                      1       1
                                                              ′                      ′                              ′                                     ′
                                                            e ·e +e ·e =0                            =⇒ e ·e =0 =⇒ e ⊥e ,                                                                              (8)
                                                              1      1        1      1                              1      1                              1         1
                                                     ′
                      it indicates that e is a normal vector. The principle normal vector is defined by
                                                     1
                                                                                                                           e′
                                                                                                   N≡e := 1                                                                                            (9)
                                                                                                                2        |e′ |
                                                                                                                             1
                                                                                                                                                                              ′
                      as a unit normal vector at p(s). The curvature of a curve p(s) is given by κ(s) = |e (s)| > 0, which
                                                                                                                                                                              1
                                                                                                                                  ′         ′′
                      can be realized as a norm of the acceleration vector a := e = p . Therefore, we have a relation
                                                                                                                                  1
                                                                                                        ′
                                                                                                      e =κ(s)e2.                                                                                     (10)
                                                                                                        1
                         Remark. If a vector V is an unit vector, |V | = 1, the corresponding derivative vector would be
                         perpendicular to itself, i.e.
                                                                                                         V′ ⊥ V .                                                                                 (11)
                                                                                                                2
           Osculating plane   Theplane is spanned by the vectors e and e is called osculating plane.
                                                               1     2
           Newton’ssecondlaw Inclassicalphysics,wehaveamomentumvectorp = mT = mp′withmass
           m. The force F is defined by Newton’s second law
                                         F = dp =mdT =ma=mp′′                                    (12)
                                               ds      ds
           with respect to parameter s.
           Frame Asetofvector e ,e equipped with a point p calls frame. In such of case, a frame at p is
                                   1  2
           denoted by (p;e ,e ).
                          1  2
           Frenet-Serretformulain2D     Fromtheorthonormalityconditionei·ej = δij (i,j = 1,2), we have
                                  e′ · e + e · e′ = 0                                           (13a)
                                   i  j   i  j
                                   ′          ′           ′
                           =⇒ e ·e +e ·e =κ+e ·e =0                                             (13b)
                                   1  2    1  2        1  2
                                      ′          ′
                           =⇒ e1·e =−κ (e hascomponent −κalonge1direction)                      (13c)
                                      2          2
                           =⇒ e′ =−κe .                                                         (13d)
                                   2       1
           Asaresult, we have the following relations
                             p′ =    +e                   p′     1 0( )
                                        1
                             ′                             ′             e1
                               e =         +κe      =⇒       e   =     0  κ                      (14)
                             1                2            1             e2
                             ′                               ′
                               e = −κe                       e        −κ 0
                                2        1                    2
           called Frenet-Serret formula.
                                 2
             Example (Circle in E ). A circle with radius r can be parametrized by p(t) = (rcost,rsint)
             with 0 ≤ t ≤ 2π.
                                                Figure 2: A circle.
             Thetangent vector is
                                             ˙
                                             p(t) = (−rsint,rcost)                             (15)
             with norm
                                          ˙    √2 2 2 2
                                         |p| =   r sin t,r cos t) := r.                        (16)
                                                       3
             Thearclength s(t) is
                                              ∫ t    ′   ′   ∫ t   ′
                                        s(t) = 0 |p(t )|dt = 0 rdt = rt.                         (17)
             Therefore, the circumference is
                                           ∫ 2π    ′   ′   ∫ 2π    ′
                                       L= 0 |p(t)|dt = 0 rdt =2πr.                               (18)
             Byt=s/r,thecirclep(s)anditstangent vector are
                                             p(s) = (rcos s,rsin s)                             (19a)
                                                           r      r
             and
                                        p′(s) = (−sin s,cos s) = e1 = T                         (19b)
                                                       r     r
             respectively. From (19b), we have
                                          ′      ( 1       s   1    s)
                                         e (s) =   − cos ,− sin         .                        (20)
                                          1           r    r   r    r
             Thecurvature κ can be obtained by
                                            ′   √1      2 s    1   2 s   1
                                      κ=|e | =       cos    + sin      = ,                       (21)
                                            1      r2     r   r2     r   r
             which is the inverse of the constant radius r. The normal vector can be calculated by
                                    ′     (                   ) (                 )
                                   e          1     s   1    s             s     s
                             e = 1 =r − cos − sin                = −cos ,sin        .            (22)
                              2     ′
                                  |e |        r     r   r    r             r     r
                                    1
           Gaussmap GaussmapGisamappingwhich
           globally send all the points p of curve to a unit
           circle S1 (a Gauss circle) centered at c and send
           the corresponding normal vector e to a radius
                                            2
           vector from c pointing to S1, which is shown
           as Fig. 3. Therefore, e2 can be represented as
           a point on S1.
               Let’s consider two normal vectors e (s) and
                                                2
               ′
           e (s ) with respect to two infinitesimal points
            2
           p(s) and p(s′), where s′ = s + ∆s is infinitesi-         Figure 3: The Gauss map G.
                                            ′
           mal close to s. We can expand e (s ) at s:
                                         2
                                         e (s′) = e (s + ∆s)
                                          2       2
                                                          ′
                                               ≈e (s)+e (s)∆s
                                                  2       2
                                               =e (s)+(−κ(s)e (s))∆s
                                                  2             1
                                               =e (s)+(−κ(s)∆s)e (s),                              (23)
                                                  2                 1
                                                        4