DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

§3. Surface Theory with Differential Forms 95{x21 + x 2 2 < 1, x 3 =0 } ,vertically project to the southern hemisphere of the unit sphere, andthen stereographically project (from the north pole) back to the unit disk.a. Show that this mapping is given (in terms of polar coordinates) by()Rx(R, θ) =(r, θ) =1+ √ 1 − R ,θ .2Compute that the first fundamental form of the Poincaré metric on D (see Exercise 15),4E =(1 − r 2 ) 2 , F =0,G = 4r 2(1 − r 2 ) 2 ,isgiven in (R, θ) coordinates by Ẽ = 1(1 − R 2 ) 2 ,R˜F =0, ˜G 2= . (Hint: Compute carefully and economically!)1 − R2 b. Changing now to Euclidean coordinates (u, v), show thatÊ =whence you derive1 − v 2(1 − u 2 − v 2 ) 2 , ˆF =uv(1 − u 2 − v 2 ) 2 , Ĝ =Γuu u 2u=1 − u 2 − v 2 , Γ uu v =0,Γuv u v=1 − u 2 − v 2 , Γ uv v u=1 − u 2 − v 2 ,Γ uvv =0,Γ vvv =1 − u 2(1 − u 2 − v 2 ) 2 ,2v1 − u 2 − v 2 .c. Use part b to show that the geodesics of the disk using the first fundamental form Î arechords of the circle u 2 + v 2 =1. (Hint: Show (by using the chain rule) that the equationsfor a geodesic give d 2 v/du 2 = 0.) Discuss the advantages and disadvantages of this model(compared to Poincaré’s).d. Compute the distance from (0, 0) to (a, 0); compare with the formula for distance in thePoincaré model.e. Check your answer in part c by proving (geometrically?) that chords of the circle map byx to geodesics in the hyperbolic disk. (See Exercise 2.1.6.)3. Surface Theory with Differential FormsWe’ve seen that it can be quite awkward to work with coordinates to study surfaces. (Forexample, the Codazzi and Gauss Equations in Section 3 of Chapter 2 are far from beautiful.) Forthose who’ve learned about differential forms, we can given a quick and elegant treatment that isconceptually quite clean.We start (much like the situation with curves) with a moving frame e 1 , e 2 , e 3 on (an open subsetof) our (oriented) surface M. Here e i are vector fields defined on M with the properties that(i) {e 1 , e 2 , e 3 } gives an orthonormal basis for R 3 at each point (so the matrix with thoserespective column vectors is an orthogonal matrix);(ii) {e 1 , e 2 } is a basis for the tangent space of M and e 3 = n.