DIFFERENTIAL GEOMETRY OF CURVES AND SURFACES 7. Geodesics and the Theorem of Gauss-Bonnet

Transcription

1 A P Q O B DIFFERENTIAL GEOMETRY OF CURVES AND SURFACES 7. Geoesics an the Theorem of Gauss-Bonnet 7.. Geoesics on a Surface. The goal of this section is to give an answer to the following question. Question. What kin of curves on a given surface shoul be the analogues of straight lines in the plane? Let s unerstan first what means straight line in the plane. If you want to go from a point in a plane straight to another one, your trajectory will be such a line. In other wors, a straight line L has the property that if we fix two points P an Q on it, then the piece of L between P an Q is the shortest curve in the plane which joins the two points. Now, if instea of a plane ( flat surface) you nee to go from P to Q in a lan with hills an valleys (arbitrary surface), which path will you take? This is how the notion of geoesic line arises: Definition 7... Let S be a surface. A curve : I S parametrize by arc length is calle a geoesic if for any two points P = (s ), Q = (s ) on the curve which are sufficiently close to each other, the piece of the trace of between P an Q is the shortest of all curves in S which join P an Q. There are at least two inconvenients concerning this efinition: first, why i we say that P an Q are sufficiently close to each other?; secon, this efinition will certainly not allow us to o any explicit examples (for instance, fin the geoesics on a hyperbolic paraboloi). Let s see an example, which will at least help us unerstan the efinition. Which are the geoesics on the unit sphere S? They are the great circles, that is, circles with centre at O (rely on your intuition or wait a bit until you get to the rigorous proof below). Let us P ' Q ' Figure. The shortest path between P an Q on the sphere is the (small) piece of the great circle between P an Q. Because this is true for any two points P an Q on the thickene curve, the latter curve is a geoesic. look at Figure. Definition 7... gives two geoesic segments on S which join A an B: the thickene an the otte piece of the great circle. The otte one is the strangest one, because we can fin two points P an Q on it such that the piece of its trace between P an Q is not the shortest curve on S between P an Q ; that s only true if P an Q are sufficiently close to each other. Of course, we coul go back to Definition 7.. an change

2 it, that is, simply remove the wors which are sufficiently close to each other. Without getting into etails, we just mention that we wouln t like the resulting efinition, because it woul be too restrictive. Our next goal is to obtain escriptions of geoesics which can be use in concrete examples. To this en, we will try to unerstan how changes the length of a curve if we vary the curve without changing its enpoints. It is important to note that we will eal with ifferentiable curves efine on close intervals [a, b] (until now, such intervals were always open). By efinition, we say that : [a, b] R is a ifferentiable curve if there exists two numbers c an such that c < a < b <, an a ifferentiable curve on (c, ) whose restriction to [a, b] is. Theorem 7... (i) Let : [a, b] R be a ifferentiable curve parametrize by arc length an let λ : [a, b] R be a family of curves with ǫ < λ < ǫ which epens ifferentiably on λ, such that 0 = (see Figure ). Then we have λ 0l( λ ) = b a λ 0 λ (s) (s)s (ii) Assume that all the above curves have the trace containe in the surface S. Then we have b λ 0l( λ ) = a λ 0 λ (s) (s) T s, where (s) T enotes the orthogonal projection of (s) on the plane T (s) S. λ o λ ( s ) ( s ) ( a ) ( b ) λ ( s ) Figure. The family of curves λ, with ǫ < λ < ǫ is a variation of. The otte curve is λ λ (s) an the vector with tail at (s) is the tangent vector to the latter curve at 0 (s). Proof. We have b λ 0l( λ ) = λ 0 λ (s) s a b = λ 0 λ (s) λ (s)s = = a b a b a 0(s) 0(s) ( λ 0 λ(s)) 0(s))s ( λ 0 λ (s)) (s)s Except 0, which is the same as, the curves λ are not necessarily parametrize by arc length.

3 We also have that λ 0 λ(s) = ( λ 0 λ (s)) because ifferentiation w.r.t. s an λ can be interchange with each other. We use the integration by parts formula an write further b λ 0 λ (s) (s) s=a s=b λ 0 λ (s) (s)s which gives the esire result, as both λ (a) an λ (b) are constant. This is how we prove (i). To prove (ii), we ecompose (s) = (s) T + (s) N where (s) N is perpenicular to T (s) S; we also note that λ 0 λ (s) is in T (s) S, so We euce the following corollary. (s) ( λ 0 λ (s)) = (s) T ( λ 0 λ (s)). Corollary 7... A curve : I S parametrize by arc length is a geoesic if an only if a (s) T = 0, for all s in I (as before, (s) T enotes the orthogonal projection of (s) to T (s) S). Equivalently, for any s in I, the vector (s) is perpenicular to the tangent plane at (s) to S. Note. The corollary is for us the main characterization of a geoesic, which will be use throughout the course. Most textbooks use this as a efinition. Our Definition 7.. is certainly more intuitive, but less useful an less clear (what exactly means P an Q sufficiently close to each other?) The iea of the proof of Corollary 7... If is a geoesic, then we pick two points P an Q on the curve, sufficiently close to each other, an vary the piece of between P an Q; the length of the latter curve is minimal, thus the erivative with respect to the variation parameter λ is equal to 0. From Theorem 7.. (ii), we can euce that (s) T = 0, for all s. The converse (if (s) T = 0 for all s, then is a geoesic) is harer to prove (see for instance [C], Section 4-6.) Examples.. Great circles on a sphere are geoesics. Because if : R S is a parametrization by arc length of such a circle, then for any s in R, the vector (s) is parallel to the raius of (s) (being perpenicular to (s)), thus it is perpenicular to the tangent plane at (s) to S.. One can also etermine the geoesics on a cyliner C (see Chapter 4, Figure 4). To this en we use the local isometry f from the plane to the cyliner escribe in HW no. 4, Question no. 4. The point is that in general, a local isometry between two surfaces maps geoesics to geoesics (see Definition 7.. an remember that a local isometry preserves lengths of curves). Thus geoesics on the cyliner are images of straight lines uner f (the rolling map); it s easy to see that they are just helices on the cyliner. In general, fining the geoesics on a given surface is not easy. In the following we will show how to etermine geoesics containe in the image of a local parametrization, by solving a system of ifferential equations. See Chapter.

4 4 Proposition Let (U, ϕ) be a local parametrization of the surface S an let : I S be a curve parametrize by arc length, whose trace is containe in ϕ(u). Write (s) = ϕ(u(s), v(s)), where s u(s) an s v(s) are real functions of s. Then is a geoesic if an only if () u (s) + Γ (u(s), v(s))(u (s)) + Γ (u(s), v(s))u(s) v (s) + Γ (u(s), v(s))(v (s)) = 0 v (s) + Γ (u(s), v(s))(u (s)) + Γ (u(s), v(s))u (s)v (s) + Γ (u(s), v(s))(v (s)) = 0. Here Γ k ij are the Christoffel symbols associate to (U, ϕ) (see Chapter 6 of the notes). Proof (sketch). In each space T ϕ(q) S we have the basis ϕ u(q), ϕ v(q). By Lemma.., we have (s) = u (s)ϕ u (u(s), v(s)) + v (s)ϕ v (u(s), v(s)). Consequently, if we omit writing (u(s), v(s)), we have () (s) = u (s)ϕ u + u (s)(ϕ uu u (s) + ϕ uv v (s)) + v (s)ϕ v + v (s)(ϕ vu u (s) + ϕ vv v (s)). Now ϕ uu, ϕ uv, ϕ vu an ϕ vv can be expresse by Equations (), Chapter 6. We plug them into () an take into account that the equation (s) T = 0 means that the coefficients of ϕ u an ϕ v are both zero. These give the two equations (). From the general theory of ifferential equations we can euce the following geometric information about geoesics: if P is a point on S an w a vector in T P S, with w =, then there exists a unique geoesic : I S with (0) = P an (0) = w. Here I is a small interval aroun 0. An interesting question is how big can be mae the interval I? An an interesting answer is: if S is compact (with respect to the subspace topology inuce from R ), then we can always take I = R; that is, we can exten the omain of any geoesic to the whole R. We will not aress this point here (this is a consequence of the theorem of Hopf-Rinow, see for instance [C, Section 5-]). 7.. The Theorem of Gauss-Bonnet. The sum of the angles of a triangle is equal to π. Equivalently, in the triangle represente in Figure, we have θ + θ + θ = π. Proof: if you go along the triangle, when you come back to where you starte the trip, you will have rotate yourself with Now let s take a triangle on the sphere S, whose sies are geoesics (that is, pieces of great circles). Let us try to etermine again θ + θ + θ, this time in Figure 4. We can see there that the sphere is partitione in the following five pieces: the thickene triangle, its antipoal image, an the slices of angle θ, θ, an θ. The slice of angle θ has area θ (because the slice of size π is the whole sphere, an has area 4π), an similarly for θ an θ. If A enotes the area of the thickene triangle, then we have A + θ + θ + θ = 4π θ + θ + θ = π A. So the sum of the (insie) angles of the triangle is this time equal to π + A. In this section we will give a formula for the sum of the angles of a geoesic polygon on an arbitrary surface. There is a preparatory result we nee to prove first. We will consier a special kin of a local parametrization, namely orthogonal parametrization. By efinition, this is (U, ϕ) with the property that for any Q in U, the vectors ϕ u (Q) an ϕ v (Q) are It is important to note that () is a nonlinear system of two ifferential equations of orer two with unknowns u(s) an v(s).

5 θ θ θ θ O θ θ θ 5 Figure. θ + θ + θ = π Figure 4. θ + θ + θ < π perpenicular (orthogonal). Equivalently we have F = 0 everywhere on U. Consequently, the vectors () e = ϕ u, e = ϕ v E G are an orthonormal basis in every tangent space. Lemma 7... Let (U, ϕ) be an orthogonal parametrization of a surface an : (a, b) ϕ(u) a geoesic parametrize by arc length, of the form (s) = ϕ(u(s), v(s)), for all s in (a, b). If φ(s) enotes the angle from 4 ϕ u ((s)) to (s), then we have ( ) φ s = E v u EG s v G u. s Proof. First we prove the following claim. 4 The angle φ = φ(s) satisfies 0 φ < π, an is uniquely etermine by = (cosφ)e + (sin φ)e

6 φ e 6 Claim. We have ( ) T s e ((s)) = φ s e ((s)), where, as usually, the superscript T inicates the orthogonal projection to the tangent space. e ' ' e Figure 5. Proof of Lemma 7.. (the plane is T (s) S an the thickene arrow is (e ) T ). To prove the claim, we ifferentiate the relation an obtain e = cos φ (4) e + e = (sin φ)φ e = (sin φ)φ, where we have use that is a geoesic. On the other han, because e =, the vector e is perpenicular to e, so the vector (e )T is parallel to e, that is, (e )T = λe, for some number λ (see also Figure 5). Because equation (4) implies an the claim is prove. e = (e )T = λ e = λ cos(π/ φ) = λ sin φ, λ = φ, From the claim we euce φ s = s (e ((s))) e ((s)). We replace s (e ((s)) = s (e (u(s), v(s)) = (e ) u u s + (e ) v v s, then we take into account of () an euce the esire formula (we skip the computations). We will also nee the notion of surface integral of a function (you may have seen this in the Vector Calculus course). Namely, of (U, ϕ) is a local parametrization of a surface S, f : S R a function, an R a compact subspace of ϕ(u), then, by efinition (5) fσ := f(ϕ(u, v)) ϕ u ϕ v A = f ϕ EG F A R ϕ (R) ϕ (R)

7 Θ Θ where the last two integrals are ouble integrals over the the region ϕ (R) in R. The last equality comes from the fact that ϕ u ϕ v = EG F. We are now reay to state an prove the following version of the theorem of Gauss-Bonnet. Theorem 7... (Gauss-Bonnet local). Let (U, ϕ) be a local orthogonal parametrization of a surface an let R containe in ϕ(u) be a region boune by the geoesics k : I k S, k =,,, with for the numbers s < s in I s < s in I s < s in I (s ) = (s ), (s ) = (s ), an (s ) = (s ), Denote by θ, θ, θ the external angles of the triangle R (for instance, θ is the angle from 5 (s ) to (s ) etc., see Figure 6). Then we have θ + θ + θ = π Kσ, where K is the Gauss curvature. R 7 ( s ) = ( s ) ( s ) = ( s ) Θ ( s ) = ( s ) Figure 6. The region R is boune by (pieces of traces of) the geoesics,, an. By θ we mean the angle between the tangent vectors (s ) an (s ); similarly for θ an θ. Proof (sketch). We use Lemma 7.. for the geoesics i, where i =,,. Denote by φ i the angle between ϕ u an i along i. We have ( ) φ i s = E v u EG s v G u. s 5 It is important, although not enough emphasize in these notes, that the surface is oriente: this means that in any tangent spaces we have a istinguishe sense of rotation, an this allows us to measure the signe angle between two non-opposite vectors in a tangent space to the surface, which is in the interval ( π, π).

8 φ φ φ φ φ φ 8 along ϕ i. If we integrate we obtain (6) φ (s ) φ (s ) = φ (s ) φ (s ) = φ (s ) φ (s ) = If we a the left han sies we obtain s EG s s s s EG s EG ( E v u ( E v u ( E v u ) s v G u s ) s v G u s ) s v G u s (φ (s ) φ (s )) + (φ (s ) φ (s )) + (φ (s ) φ (s )). If we look at Figures 6 an 7, we can easily see that the latter expression is equal to θ θ θ + kπ, where k is an integer. If Figure 6 was in a plane (not on an arbitrary surface) an ϕ u was the same, the sum woul actually be always θ θ θ + π. One can show (an it s not trivial) that even if the triangle is on a surface, the sum equals θ θ θ + π. Figure 7. The thickene arrows are ϕ u three points. at each of the Now let s analyze the sum of the right han sies of Equations (6). This represents the line integral E v C EG u G u EG v, where C is the (close) path in R (with coorinates u, v) obtaine by joining the paths s ϕ ( (s)), s ϕ ( (s)), an s ϕ ( (s)).

9 9 Green s formula says that C fu + gv = In our case, we obtain E v EG u G u EG v = C D ϕ (R) ( g u f ) A. v (( ) E ( ) v G ) + u EG v A. EG u In Chapter 6 we have obtaine a formula for the curvature K in terms of the coefficients E, F, an G of the first funamental form. It turns out that this formula, use in the case F = 0, becomes very simple, namely This implies C K = EG (( ) E ( ) EG v G ) + EG u. v u E v EG u G u EG v = K EGA = Kσ, ϕ (R) R where we have use Equation (5). The theorem is now prove. To unerstan the global version of the theorem of Gauss-Bonnet, we nee a few more notions, as follows: a region R on a surface S is a compact subset of S boune by finitely many curves a triangle is a region boune by three curves a triangulation of a region R is a finite collection of triangles T,...,T n such that. n i= T i = R. If T i T j φ, then either i = j, or T i T j is a common ege of T i an T j, or T i T j is a common vertex of T i an T j. Figure 8 shoul help in unerstaning these notions. A geoesic region is a region boune by geoesics. A geoesic triangle is a triangle whose eges are geoesics. A geoesic triangulation is a triangulation by geoesic triangles. Compact surfaces (like the sphere, the ellipsoi, the torus etc. you are free to think of those as surfaces that have an insie ) are geoesic regions (not boune by any geoesic). If T,...,T n is a triangulation of the region R, then we enote as follows: n the number of triangles (actually n = n) n the number of eges n 0 the number of vertices χ = n n + n 0 We will rely (without proof) on the following two propositions. Proposition 7... (a) Any region on a surface (thus any compact surface) amits a triangulation. Any geoesic region has a geoesic triangulation. Moreover, in the latter case we can refine the triangulation in such a way that any triangle is containe in the image of an orthogonal parametrization. (b) If R is a region, then the number χ = n n + n 0 is the same for any triangulation of R. We call it the Euler-Poincaré characteristic of R, enote χ(r). Proposition Let R be a region in the surface S an T,..., T n a triangulation with the property that each T i is containe in the image of a local parametrization. Let also f be

10 0 Figure 8. A region on a surface an a triangulation of it (with n 0 = 6, n = 5, an n = 8). We take this opportunity to mention that, as explaine in the footnote on page 7, the angles θ an θ are negative (the other external angles are all positive). a ifferentiable function on S. The number n fσ =: T i i= oes not epen on the choice of the triangulation (note that each term of the sum above is well efine in view of Equation (5)). We call this the integral of f on R. We are now reay to state a more general version 6 of the theorem of Gauss-Bonnet. Theorem (Gauss-Bonnet global) Let S be an orientable surface an R a geoesic region in S. Choose the parametrizations of the curves which boun R in such a way that we can go along the contour etermine by them in the sense given by the orientation with increasing parameters; let θ, θ,...,θ n be the external angles (like in Theorem 7.., see also Figure 8). Then we have θ + θ θ n = πχ(r) Kσ. R fσ R This is not the most general version of the theorem, though cf. [C], page 74 or [Gr-Abb-Sa], page

11 Corollary If S is a compact orientable surface, then Kσ = πχ(s). S Rather than the theorem, we will actually prove the corollary, inepenently of the theorem. Of course, the corollary can be euce from the theorem by taking R = S, which implies that the bounary of R is empty an there are no angles θ i to be taken into account: but again, this is not what we re going to o. Proof of Corollary Let us consier a geoesic triangulation T,...,T n of S with the property that each triangle is containe in the image of an orthogonal parametrization whose orientation is compatible with the global orientation of the surface (see Proposition 7.. (a)). As usually, n 0, n, n are the number of vertices, eges, respectively triangles of the triangulation. For each i =,,..., n we consier the external angles θ i, θ i, θ i of the triangle T i (recall that they are in ( π, π) an they epen on the orientation) an the internal angles i := π θ i, i := π θ i, i := π θ i. Theorem 7.. for the triangle T i says that T i Kσ = i + i + i π. We a up all these equalities an obtain n (7) Kσ = (i + i + i ) n π. S The sum of angles at every vertex of the triangulation is π, thus n (i + i + i) = πn 0. Every ege of the triangulation belongs to two triangles, thus Now (7) implies The corollary is prove. S i= i= n = n. Kσ = (n 0 n )π = (n 0 n + n )π = χ(s)π. Examples of compact surfaces are: the sphere, the torus with one or several holes (see Figure 9), an any smooth eformations of those. For instance, we can eform the sphere an obtain ellipsois, but not only there are many shapes you can get by eforming the sphere (use your imagination, you can get a potato, a cucumber, a tomato etc.) The equation state in the corollary is surprising because if we have a compact surface S an we eform it, then (a) in the right han sie, χ(s) oes not change uner (ifferentiable) eformations: for instance, take the triangulate sphere in Figure 0 an make strange shapes out of it (for instance, a cucumber), without making sharp eges or corners an without making self-intersections; the triangulation is preserve uring the eformation, as well as the number of triangles, eges, an vertices.

12 Figure 9. Tori with two, respectively three holes. (b) in the left han sie, K changes uring the eformation: again, for the sphere, you can eform it to a thin an long cucumber in such a way that the top is very curve, so K at that point will become much larger than. Nevertheless, in spite of what we sai at (b), the average of K on S oes not change uner the eformation: an this is unexpecte! Figure 0. A triangulation of the sphere. Let us compute the Euler-Poincaré characteristic of the sphere. One way of oing this is by inscribing a tetraheron in the sphere an joining the four vertices by curves on the sphere. We obtain a triangulation with n = 4, n = 6, n 0 = 4 (see Figure ), so χ(s ) = =. To calculate the Euler-Poincaré characteristic of the torus, we use the triangulation inicate in Figure. We have n = 4, n = 6, n 0 =. This implies χ(t) = = 0. One can show that a torus with k holes has the Euler-Poincaré characteristic equal to k (so it can be negative).

13 Figure. A more lucrative triangulation of the sphere: the thickene points an lines are all on the sphere. Figure. A triangulation of the torus: the figure is incomplete, as each quarilateral has to be ivie in two triangles. Warning: not all triangles you see here are elements of the triangulation (only 4 of them). References [C] M. P. o Carmo, Differential Geometry of Curves an Surfaces, Prentice Hall 976 [Gr-Abb-Sa] A. Gray, E. Abbena, an S. Salamon, Moern Differential Geometry with MATHEMATICA, r eition