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Complex Analysis and Differential Geometry




                    Notes          Due to its importance, we state the Theorema Egregium of Gauss.
                                   Theorem 5: Given a surface X and a point p = X(u, v) on X, the Gaussian curvature K at (u, v) can
                                   be expressed as a function of E, F,G and their partial derivatives. In fact
                                                                        1 G  1 G        1 E  1 G
                                                                     v
                                                              C     F  2  u  2  v  0   2  v  2  u
                                                             1                     1
                                                (EG – F ) K =   E                  E
                                                      2 2
                                                             2  u      E      F    2  v  E    F
                                                               1                   1
                                                            F   2 E v  F     G    2 G u  F   G
                                                            u
                                   where
                                                                  1
                                                               C   (–E vv    2F  G ).
                                                                                uu
                                                                           vv
                                                                  2
                                   Proof. Way of proving theorem is to start from the formula
                                                                       LN – M 2
                                                                   K =
                                                                       EG – F 2
                                   and to go back to the expressions of L, M, N using D, D’, D” as determinants:

                                                      L =   D    ,  M =   D'  ,  N =   D"  ,
                                                           EG – F 2     EG – F 2    EG – F 2
                                   where

                                                   D = (X , X , X ),
                                                              uu
                                                           v
                                                        u
                                                   D’ = (X , X , X ),
                                                           v
                                                              uv
                                                        u
                                                  D” = (X , X , X ).
                                                           v
                                                        u
                                                              vv
                                   Then, we can write
                                                    (EG – F ) K = (X , X , X ) (X , X , X ) – (X , X , X ) ,
                                                                                            2
                                                          2 2
                                                                          u
                                                                      uu
                                                                   v
                                                                                        v
                                                                 u
                                                                                           uv
                                                                             v
                                                                                vv
                                                                                     u
                                   and compute these determinants by multiplying them out. One will eventually get the expression
                                   given in the theorem!
                                   It can be shown that the other two equations, known as the Codazzi-Mainardi equations, are the
                                   equations:
                                                              2
                                                           
                                              Mu – Lv =  2 11 N (   1 11 )M   1 12 L,
                                                              12
                                                              2
                                                           
                                             Nu – Mv =  2 12 N (   1 12 )M   1 22 L.
                                                              22
                                   We conclude this section with an important theorem of Ossian Bonnet. First, we show that the
                                   first and the second fundamental forms determine a surface up to rigid motion. More precisely,
                                   we have the following lemma:
                                   Lemma 6. Let X :   E  and Y :   E  be two surfaces over a connected open set . If X and Y
                                                                  3
                                                      3
                                   have the same coefficients E, F, G, L, M, N over , then there is a rigid motion mapping X()
                                   onto Y ().
                                   The above lemma can be shown using a standard theorem about ordinary differential equations.
                                   Finally, we state Bonnet’s theorem.
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