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Definition. Suppose P = X(u10, u20) and let Ω be a neighborhood of (u10, u20) on which X is one-to-one with a continuous inverse X −1 : X(Ω) → Ω. Deﬁne U (u1, u2) to be a unit normal vector to the surface M determined by X at point X(u1, u2) (recall that U = X1 × X2 / X1 × X2 ). Therefore U : X(Ω) → S 2. U is called the sphere mapping or Gauss mapping of X(Ω). The image of X(Ω) under U (a subset of S 2) is the spherical normal image of X(Ω). Example (Exercise 9 (d), page 57). The spherical normal image of a torus (see Example 12, page 34) is the whole sphere S 2 (there is a normal vector pointing in any direction - in fact, the sphere mapping is two-to-one).
Then Eu = Gu = Gv = 0 and the nonzero Christoﬀel symbols are −2r2 cos v sin v Ev = = − tan v = = 2E 2r2 cos2 v 2r2 cos v sin v −Ev 2 = = cos v sin v. Γ11 = 2G 2r2 It is shown (and not trivially) in Exercise 14 that this implies geodesics Γ112 Γ121 are great circles. 8 Note. In Example 19 page 62, it is shown that the Euclidean plane when equipped with polar coordinates (which are orthogonal coordinates) yields geodesics which are lines (as expected). Note. In general, to determine the geodesics for a surface, requires that one solve diﬀerential equations.
Let v = v iXi and w = wi Xi . As in Theorem I-5 (equation (24), page 50) (Lij − k1 gij )v j = 0 for i = 1, 2, and (Lij − k2gij )wj = 0 for i = 1, 2. The ﬁrst of these equations is equivalent to Lij v i = k1 gji v i for j = 1, 2 and since Lij = Lji and gij = gji to Lij v i = k1gij v i for j = 1, 2. (25) The second of these equations implies (Lij − k2gij )v iwj = 0 (we now sum over i = 1, 2). So (Lij v i − k2gij v i )wj = 0 and from (25) we have (k1gij v i − k2 gij v i )wj = 0 or (k1 − k2 )gij v i wj = 0.