Durham University Pavel Tumarkin Epiphany 207 Dierential Geometry III, Solutions 5 (Week 5 Christoel symbols and Gauss Theorema Egregium 5.. Show that the Gauss curvature K o the surace o revolution locally parametrized by x(u, v = ((v cos(u, (v sin(u, g(v, (u, v U, (or some suitable parameter domain U is given by K = (. 2 + ( /g 2 I the generating curve is parametrized by arc length, show that K = /. Egregium in the latter case. Deduce Theorema Solution: We have already calculated the coeicients o the irst and second undamental orms or a surace o revolution (see e.g. Example 9.3, so we just cite the result here again: E = 2, F = 0, G = 2 + g 2 L = g 2 + g 2, M = 0, N = g g 2 + g 2. I we assume now that (v > 0 everywhere, then we have κ = L E = g ( 2 + g 2 /2, κ 2 = N G = g g (see Prop. 9.2, hence the Gauss curvature is K = κ κ 2 = g ( g g ( 2 + g 2 2 = ( /g g ( ( /g 2 + 2 ( 2 + g 2 3/2 = 2( /g ( /g 2 ( ( /g 2 + 2 = ( 2 ( /g 2 + as desired (we have also implicitly assumed here that (v 0 g (v. I the curve is parametrized by arc length, then 2 + g 2 =, and κ = L E = g, κ 2 = N G = g g. Moreover, dierentiating 2 + g 2 = gives + g g = 0, and we obtain K = κ κ 2 = g ( g g = g g + ( = (g 2 + 2 as desired (this could also be obtained by simpliying the ormula or K obtained above. = Now we can deduce Gauss Theorema Egregium by expressing in terms o the coeicients o the irst undamental orm: = E. Then we have = E v 2 E, = E vve Ev/2 2, 2E 3/2 so that / = E2 v 4E 2 E vv 2E = ( 2 Ev E E v which is a special case (G =, i.e, G u = 0 o the ormula o Example 0.9.
5.2. Let x: U S be a parametrization o a surace S or which E = G = and F = cos(uv (so that uv is the angle between the coordinate curves. Determine a suitable parameter domain U on which x(u is a surace (i.e., where the coordinate curves are not tangential. Show that K = sin(uv. Solution: Suitable parameter domain: The tangent vectors x u and x v are linearly dependent i F = cos ϑ = ±, where ϑ = uv. ( cos ϑ Another way to see this restriction is as ollows: we have to assure that with ϑ = uv is a positive cos ϑ deinite matrix. Its determinant is cos 2 ϑ, and this is positive i cos ϑ ±. Since its trace is always positive (the trace is 2, the matrix is positive deinite i cos ϑ ±. So a maximal parameter domain could be U := { (u, v R 2 uv / πz, or, i you preer a connected domain, another choice could be (choosing just the component 0 < uv < π. U := { (u, v R 2 0 < u < π/v, v > 0. The urther calculations are similar to ones used in Example 0.7 (and in the proo o Theorema Egregium. We amend the order a bit to avoid computations with some zeros, and thus to save time in this way. (a Step : Christoel symbols Γ k ij are unctions deined by x uu = Γ x u + Γ 2 x v + LN x uv = Γ 2x u + Γ 2 2x v + MN x vv = Γ 22x u + Γ 2 22x v + NN (Γ (Γ2 (Γ3 (and we have Γ k 2 = Γ k 2 since x uv = x vu. Beore calculating Γ k ij in terms o E, F and G, let us irst see what we need (to save some time. But we also need the ollowing: Express x uu x u etc. in terms o E, F, G: x uu x u = 2 (x u x u u = 2 E u (= 0 ( x uv x u = 2 (x u x u v = 2 E v (= 0 (2 x uv x v = 2 (x v x v u = 2 G u (= 0 (3 x vv x v = 2 (x v x v v = 2 G v (= 0 (4 x uu x v = (x u x v u x u x uv = F u 2 E v (= v sin(uv (5 x vv x u = (x v x u v x v x uv = F v 2 G u (= u sin(uv (6 (the terms in parentheses correspond to our special case E = G =, F = cos(uv. Multiplying the deining equations or Γ k ij by x u and x v, we obtain equations EΓ + F Γ 2 = 2 E u, EΓ 2 + F Γ 2 2 = 2 E v, EΓ 22 + F Γ 2 22 = F v 2 G u, F Γ + GΓ 2 = F u 2 E v, F Γ 2 + GΓ 2 2 = 2 G u, F Γ 22 + GΓ 2 22 = 2 G v.
Plugging in E = G = and F = cos(uv, we obtain Γ + cos(uvγ 2 = 0, Γ 2 + cos(uvγ 2 2 = 0, Γ 22 + cos(uvγ 2 22 = u sin(uv, cos(uvγ + Γ 2 = v sin(uv, cos(uvγ 2 + Γ 2 2 = 0, cos(uvγ 22 + Γ 2 22 = 0. From this one could easily obtain that Γ = v cos(uv sin(uv, Γ 2 = 0, Γ 22 = u sin(uv, Γ 2 = v sin(uv, Γ2 2 = 0, Γ 2 22 = u cos(uv sin(uv. However, we will see now that we can avoid computations o Γ 22 and Γ 2 22. (b Step 2: Calculate LN M 2 : From the equations above we have LN = (LN (NN = (x uu Γ x u Γ 2 x v (x vv Γ 22x u Γ 2 22x v = x uu x vv Γ 22 x uu x {{ u =0 Γ 2 22 = x uu x vv + (Γ 2 22v + Γ u sin(uv x uu x v {{ = v sin(uv Γ x vv x {{ u Γ 2 x vv x {{ v = u sin(uv =0 + Γ Γ 22E + (Γ Γ 2 22 + Γ 2 Γ 22F + Γ 2 Γ 2 22G + Γ Γ 22 + (Γ Γ 2 22 + Γ 2 Γ 22 cos(uv + Γ 2 Γ 2 22 = x uu x vv + Γ 2 22( Γ + Γ 2 + v sin(uv + Γ 22( Γ + Γ 2 cos(uv + Γ u sin(uv. Note that due to the deining equations on Γ and Γ 2 the irst and second parentheses in the expression above vanish, i.e., LN = x uu x vv + Γ u sin(uv, which means that all we needed is to calculate Γ = v cos(uv sin(uv. Let us now calculate M 2. First we observe that the linear system involving Γ 2 and Γ 2 2 is homogeneous with non-zero determinant, so has a trivial solution only, i.e. Γ 2 = 0 = Γ 2 2. Hence, so that Finally, recall that so that inally, M 2 = (MN (MN = (x uv Γ 2x u Γ 2 2x v (x vv Γ 2x u Γ 2 2x v = x uv x uv, LN M 2 = x uu x vv x uv x uv + Γ u sin(uv = x uu x vv x uv x uv + uv cos(uv. x uu x vv x uv x uv = (x u x vv u (x u x uv v = ( F v G u /2 u (E v/2 v = ( u sin(uv u 0 = sin(uv uv cos(uv, LN M 2 = sin(uv uv cos(uv + uv cos(uv = sin(uv. (c Step 3: Calculate K: The Gauss curvature is as desired. K = LN M 2 EG F 2 = LN M 2 cos 2 (uv = sin(uv
5.3. ( I the coeicients o the irst undamental orm o a surace S are given by E = 2 + v 2, F =, G =, show that the Gauss curvature o S is given by K = ( + v 2 2. Solution: Calculations are similar to the previous exercise. (a Step : Christoel symbols Γ k ij. We have EΓ + F Γ 2 = 2 E u, EΓ 2 + F Γ 2 2 = 2 E v, EΓ 22 + F Γ 2 22 = F v 2 G u, F Γ + GΓ 2 = F u 2 E v, F Γ 2 + GΓ 2 2 = 2 G u, F Γ 22 + GΓ 2 22 = 2 G v. Plugging in E = 2 + v 2 and F = G =, we obtain (2 + v 2 Γ + Γ 2 = 0, (2 + v 2 Γ 2 + Γ 2 2 = v, (2 + v 2 Γ 22 + Γ 2 22 = 0, Γ + Γ 2 = v, Γ 2 + Γ 2 2 = 0, Γ 22 + Γ 2 22 = 0. We see that the equations on Γ 22 and Γ 2 22 have only the trivial solution Γ 22 = Γ 2 22 = 0. For the others, we obtain Γ v = ( + v 2, 2v + Γ2 v3 = + v 2, Γ 2 = v + v 2, Γ2 2 = v + v 2. (b Step 2: Calculate LN M 2 : We have LN = (LN (NN = (x uu Γ x u Γ 2 x v (x vv Γ 22x u Γ 2 22x v = (x uu Γ x u Γ 2 x v x vv = x uu x vv Γ x vv x u Γ 2 x vv x v So we only need x vv x v = G v /2 = 0 and x vv x u = F v G u /2 = 0 in our case here, hence we have Similarly, or M 2 we obtain M 2 = (MN (MN LN = x uu x vv. = (x uv Γ 2x u Γ 2 2x v (x vv Γ 22x u Γ 2 22x v = x uv x uv 2Γ 2 x uv x {{ u 2Γ 2 2 x uv x {{ v =E v/2=v =G u/2=0 + (Γ 2 2 x u x u {{ = x uv x uv 2v2 + v 2 +2Γ 2Γ 2 2 x u x {{ v =E=2+v 2 =F = + v2 (2 + v 2 ( + v 2 2 2v 2 ( + v 2 2 + v 2 ( + v 2 2 = x uv x uv 2v2 + v 2 + v2 ( + v 2 ( + v 2 2 = x uv x uv v2 + v 2 +(Γ 2 2 2 x v x v {{ =G=
Hence Recall again that LN M 2 = x uu x vv x uv x uv + v2 + v 2. x uu x vv x uv x uv = (x u x vv u (x u x uv v ( = F v ( 2 G u 2 E v =, u v so that inally (c Step 3: Calculate K: The Gauss curvature is as desired. LN M 2 = + v2 + v 2 = + v 2 K = LN M 2 EG F 2 = /( + v2 (2 + v 2 = ( + v 2 2 5.4. Let x be a local parametrization o a surace S such that E =, F = 0 and G is a unction o u only. Show that Γ 2 2 = Γ 2 2 = G u 2G, Γ 22 = G u 2 and that all the other Christoel symbols are zero. Hence show that the Gauss curvature K o S is given by K = ( G uu G. Solution: Again, the calculations are similar to the previous exercise. (a Step : Christoel symbols Γ k ij. Since E =, F = 0 and G v = 0, we have which implies Γ = 0, Γ 2 = 0, Γ 22 = 2 G u, GΓ 2 = 0, GΓ 2 2 = 2 G u, GΓ 2 22 = 0, (Γ 2 2 =Γ 2 2 = G u 2G and Γ 22 = G u 2, and all other Christoel symbols are 0, as desired (note that G cannot vanish as the irst undamental orm is positive deinite. (b Step 2: Calculate LN M 2. We have LN = (LN (NN = (x uu Γ x u Γ 2 x v (x vv Γ 22x u Γ 2 22x v = x uu (x vv Γ 22x u = x uu x vv Γ 22 x uu x u {{ =E u/2=0 = x uu x vv.
Similarly, or M 2 we obtain M 2 = (MN (MN = (x uv Γ 2x u Γ 2 2x v (x vv Γ 22x u Γ 2 22x v = (x uv Γ 2 2x v (x vv Γ 2 2x v Hence, Recall again that = x uv x uv 2Γ 2 2 x uv x {{ v +(Γ 2 2 2 x v x {{ v =G u/2 =G = x uv x uv (G u 2 2G + (G u 2 4G = x uv x uv (G u 2 4G LN M 2 = x uu x vv x uv x uv + (G u 2 4G so that inally (c Step 3: Calculate K. The Gauss curvature is Finally, we have x uu x vv x uv x uv = (x u x vv u (x u x uv v ( = F v ( 2 G u 2 E v = 2 G uu, LN M 2 = 2 G uu + (G u 2 4G. K = LN M 2 EG F 2 = G uu/2 + (G u 2 /4G = G uu G 2G + (G u 2 4G 2. ( G uu = ( Gu G G 2 G = ( Guu u G 2 G (G u 2 = G uu 4G 3/2 2G (G u 2 4G 2 = K so that we obtain the desired ormula. Remark. A particular example o such coeicients is given by a surace o revolution with a generating curve parametrized by arc length with u and v interchanged. u v