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Unit 26: Lines of Curvature




          Now, in any simply connected subset of U, we can always find analytic functions G = x +iy  Notes
                                                                               j   j  j
                   G    F .                
                                      1
                                           3
          satisfying   j  j  We let  X  x , x , x .  Then X is conformal and harmonic. Indeed, x being
                                        2
                                                                                j
          the real parts of complex analytic functions, are harmonic, and hence X is harmonic. Furthermore,
                                                            x
                   x     ,  and by the Cauchy-Riemann equations       y    n . j  Thus, we see
          we have    j  j                                  j      j
                     u                                        v      u
          that
                                                3    2    2  2
                                            v 
                                 X  X   X  X        j     0,
                                                  
                                                       
                                     u
                                         v
                                  u
                                                    j
                                               j 1        
                                                
          and
                                                 3
                                        X  X       0,
                                            v
                                                   j
                                                     j
                                         u
                                                j 1
                                                 
                                                           det
                              3
          and hence, X is conformal.  Since X is real analytic, the zeroes of   X X  j   are isolated. Removing
                                                                i
          the  set Z  of those  zeroes from  U, we  get that  X : U \Z    is  a harmonic  and  conformal
                                                            3
          parametric surface, hence, X is a minimal surface .
                                                  4
          If we carry out this procedure starting with the complex analytic functions f()  =  1  and
          h() = 1/, then X is another parametrization of the catenoid.
          26.3 Surface Area
          In this section, we will give interpretations of the Gauss curvature and the mean curvature. Both
          of these involve the concept of surface area. Before introducing the definition, we first prove a
          proposition which will show that the definition is reparametrization invariant.
          Proposition 5. Let  X : U    be a parametric surface with first fundamental form    ,  and
                                                                               g
                                  3
                                                                                ij
                           3
                      
                                                      
                    
                                                                      g 
           V   U.  Let  X : U    be a reparametrization of X, let  V     1 (V),  and let    be the coordinate
                                                                       ij
          representation of the first fundamental form of  X.  Then, we have:
                                                 
                                                          du du .
                                         du du 
                                    det g     1   2  V   det g  1  2                ...(3)
                                  V    ij               ij
          Proof. Now, we have
                                                    det 
                                      det g  ij  det g ij    j i
                                           
                     i
                         j
                 i
          where     u / u .  Thus, for any open subset V  U, and  V     1 (V),  we have:
                                                          
                        
                 j
                                             det
                                                                    du du
                       det g     1   2  det g    du du i    1    2    V   det g  1  2
                            du du 
                     V    ij        V    ij    j                ij
          Thus, the integral on the right-hand side of (3) is reparametrization invariant. This justifies the
          following definition.
          3  Of course, Y = (y , y , y ) is also conformal.
                           2
                         1
                              3
          4  X is also said to be a branched minimal surface on U. The zeroes of det(g ) are called branched points.
                                                                 ij
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