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4.5 a Formula for Gaussian Curvature

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4.5 a Formula for Gaussian Curvature 48 4.5 A Formula for Gaussian Curvature The Gaussian curvature can tell us a lot about a surface. We compute K using the unit normal U, so that it would seem reasonable to think that the way in which we embed the surface in three space would affect the value of K while leaving the geometry of M un- changed. This would mean that the Gaussian curvature would not be a geometric invariant and, therefore, would not be as helpful in studying surfaces. If we can find a formula for K which does not depend on U, we would then show that the value of K does not depend on how M is situated in space. We will give a formula for K when depends only on E, F , and G. These three quantities E, F, G are called the metric of the surface. I will give the { } more general formula later, but we will derive this for the case that F = xu xv = 0. Note that this means that the u and v-parameter curves form perpendicular families· of curves. Theorem 4.2 (Gauss’ Theorem Egregrium) The Gaussian curvature depends only on the metric E, F , and G, 1 ∂ Ev ∂ Gu K = + . (4.1) −2√EG ∂v √EG ∂u √EG where ∂ ∂ ∂ ∂ Ev = E = (xu xu) and Gu = G = (xv xv) ∂v ∂v · ∂u ∂u · Proof: This is nothing more than finding the coefficients of a vector with respect to a particular basis. Since we assumed that our patch is regular, we know that xu, xv,U forms a basis for 3 { } R . Now, we need to eliminate U from our formula for Gaussian curvature and l = xuu U, · m = xuv U, n = xvv U. Expand xuu, xuv, and xvv in terms of this basis. · · u v xuu = Γuuxu + Γuuxv + lU u v xuv = Γuvxu + Γuvxv + mU (4.2) u v xvv = Γvvxu + Γvvxv + nU Our job is to find the Γ’s. While they are just the coefficients in the basis expansion, they are traditionally known as Christoffel symbols. u xuu xu = Γuuxu xu + 0 + 0 · u · = ΓuuE likewise v xuu xv = Γ G · uu u so if we can compute xuu xu we can find Γ . · uu 4.5. A FORMULA FOR GAUSSIAN CURVATURE 49 E = xu xu so by taking the derivative with respect to u, we get that • · Eu = xuu xu + xu xuu = 2xu xuu. · · · Therefore, Eu u Eu xuu xu = and Γ = . · 2 uu 2E xu xv = 0 so differentiating with respect to u gives • · 0 = xuu xv + xu xuv or xuu xv = xu xuv · · · − · E = xu xu and differentiating with respect to v gives Ev = 2xu xuv. So, • · · Ev = xu xuv = xuu xv. 2 · − · Thus, xuu xv Ev xuv xu Ev Γv = · = and Γu = · = uu G −2G uv E 2E G = xv xv, so Gu/2 = xuv xv so • · · xuv xv Gu Γv = · = uv G 2G 0 = xu xv so differentiating with respect to v and following the same technique as • above will· give us xvv xu Gu Γu = · = . vv E −2E Finally, xv xv = G so xvv xv = Gv/2 and • · · xvv xv Gv Γv = · = . vv G −2G We end up with the following formulas Eu Ev xuu = xu xv + lU 2E − 2G E G x = v x + u x + mU uv 2E u 2G v Gu Gv xvv = xu + xv + nU −2E 2G l m Uu = xu xv −E − G m n Uv = xu xv − E − G In order to complete our proof, it is necessary to look at certain mixed third partial derivatives. We know that the mixed partials are equal, regardless of the order in which we 50 take the partials. Thus, xuuv = xuvu, or xuuv xuvu = 0. This means that when we write − xuuv xuvu in terms of xu, xv,U all of the coefficients are zero. We will concentrate on − { } the xv term for our result. Other terms give other results. Eu Eu Ev Ev xuuv = xu + xuv xv xvv + lvU + lUv 2E v 2E − 2G v − 2G Expand xuv, xvv and Uv by their basis expansions. Eu EuEv EvGu lm EuGu Ev EvGv ln xuuv = + 2 + xu + 2 xv 2E v 4E 4EG − E 4EG − 2G v − 4G − G mEu nEv + + lv U 2E − 2G Ev Ev Gu Gu xuvu = xu + xuu xv xuv + muU + lUu 2E u 2E − 2G u − 2G 2 Ev EuEv EvGu lm EvEv Gu GuGu m = + 2 + xu + + + 2 xv 2E u 4E 4EG − E − 4EG 2G u 4G − G lEv mGu + + + mu U 2E 2G Now, the xv coefficient of xuuv xuvu must be zero, so − 2 EuGu Ev EvGv ln EvEv Gu GuGu m 0 = 2 + + 2 4EG − 2G v − 4G − G − − 4EG 2G u 4G − G 2 ln m EuGu Ev EvGv EvEv Gu GuGu − = 2 + 2 G 4EG − 2G v − 4G 4EG − 2G u − 4G 2 ln m EuGu 1 Ev EvGv EvEv 1 Gu GuGu − = 2 2 + 2 2 EG 4E G − E 2G v − 4EG 4E G − E 2G u − 4EG EuGu 1 Ev EvGv EvEv 1 Gu GuGu K = 2 2 + 2 2 4E G − E 2G v − 4EG 4E G − E 2G u − 4EG We have to check that right hand side of the above equation is the same as the right hand side of Equation 4.1. That means we must compute the derivatives. 1 ∂ Ev ∂ Gu + = − 2√EG ∂v √EG ∂u √EG 1 1/2 1 1/2 1 Evv√EG Ev( (EG) (EvG EGv) Guu√EG Gu( (EG) (EuG EGu) − 2 − − + − 2 − − ! −2√EG EG EG Evv Ev(EvG EGv) Guu Gu(EuG EGu) = + − + − −2EG 4E2G2 − 2EG 4E2G2 1 = [GEuGu 2EGEvv + EEvGv + GEvEv 2EGGuu + EGuGu] −(2EG)2 − − EuGu 2GEvv 2EvGv EvGv EvEv 2GGuu 2GuGu GuGu = − + − = K 4E2G − 4EG2 − 4EG2 4E2G − 4EG2 − 4EG2 Thus, we have developed the Gaussian curvature as a quantity involving only E, G, and (implicitly) F . 4.6. SOME EFFECTS OF CURVATURE 51 Lemma 4.4 In general the Gaussian curvature is given by Evv + F Guu Eu F Ev 0 Ev Gu 1 2 uv 2 2 u 2 2 2 − Gu− − Ev K = Fv EF EF . ((EG F 2)2 − 2 − 2 Gv FG Gu FG − 2 2 4.6 Some Effects of Curvature Recall that a point on a surface is an umbilic point if the principal curvatures a p are equal. Theorem 4.3 A surface M consisting entirely of umbilic points is contained in either a plane or a sphere. A surface in R3 is compact if it is closed and bounded. Here bounded means that it is contained in a sphere of sufficiently large, but finite, radius. Closed means that every sequence of points on the surface converges to a point on the surface. Theorem 4.4 On every compact surface M R3 there is some point p with K(p) > 0. ⊂ Corollary 5 There are no compact surfaces in R3 with K 0. In particular, no minimal surface embedded in R3 is compact. ≤ Theorem 4.5 (Liebmann) If M is a compact surface of constant Gaussian curvature K, then M is a sphere of radius 1/√K..
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