Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20,

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1 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, Gauss curvature This section gives an introduction to curvature of manifolds. It begins with a survey of classical notions of curvature of two dimensional surfaces embedded in Euclidean three-space, followed in the next section by a generalization to general Riemannian manifolds. Subsequently, a very concise and elegant characterization in terms of connections is given. The emphasis is on the geometric character of the notion of curvature, i.e. its independence of choices of coordinates and imbeddings in ambient spaces. Following the historical development it is natural to build on familiar concepts of curvature of plane and space curves to develop useful notions of curvature of surfaces. This is an opportune time to think about why one may want a quantitative notion of curvature, and what its properties should be: Exercise 4.34 Make lists of desirable properties of a notion of curvature, and of possible purposes. [[Suggestions: How should curvature of a surface relate to curvature of curves, consider e.g. cylinders. Should curvature be a scalar function or what else? How should it transform, if at all, under coordinate changes? Under what other operations should it be invariant? What manifolds should be distinguished from each other by curvature? ]] The natural start point is to consider smooth hypersurfaces M 2 R 3 (with the induced metric. At any point p M one may consider the curvatures κ σ (0 of all smooth curves σ : M 2 with σ(0 = p. The first significant step is to go from this large collection of all possible curves to those that are intersections of M 2 with various planes. One might consider (suitably parameterized curves of intersection of M 2 with planes Π j = {q M 2 : x j (q = x j (p} through p that are parallel to the coordinate planes. Alternatively, in the case that the surface is the graph of a function x 3 = f(x 1, x 2 one might consider the curves of intersection with all planes through p that contain the line l(t = (x 1 (p, x 2 (p, t x 3 (p. Or consider all normal planes through p, i.e those that contain the normal line ν(t = p + tn(p where N(p is a normal vector to M 2 at p i.e. in the case that M 2 is the graph of a function f one might take N = ( (D 1 f, (D 2 f, 1 evaluated at (x 1 (p, x 2 (p, with the usual identification of R 3 with T p R 3. Exercise 4.35 Which of the above approaches is not geometric? Before reading on, try to come up with explicit formulas for the curvatures of the intersection curves with the various planes suggested above. Explore whether any information is lost when only considering the curvature with curves of intersection with planes as opposed to all curves. Explore if either one of the suggested alternatives allows one to recover the information of any of the others e.g. if all the curvatures at p of the curves of intersection with the normal planes are known, can one use these to calculate all the other curvatures mentioned above. Instead of considering all normal planes, might it be enough to consider two orthogonal planes, or any two different normal planes. [[Utilize the MAPLE worksheets to get a better feeling for these curves of intersection.]] The historical developments over the last 250 years are highly recommended for reading. There are many publications which discuss excerpts the following notes are very close to Spivak II which provides an enlightening discussion contrasting and comparing the original works, with modern terminologies and tools. For notational convenience and clarity, the following survey of classical result liberally shall identifies at any point p R 3 the tangent spaces T p R 3 with R 3, and also frequently suppresses any mention of inclusion maps such as ı p : T p M T ı(p R 3. The first major theorem is due to Euler (about 1760.

2 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, Theorem 4.23 Suppose M 2 R 3 is a smooth surface (2-manifold, p M and {Y 1p, Y 2p, ν(p} is a positively oriented orthonormal basis for T p R 3 such that {Y 1p, Y 2p } is a basis for T p M. For each θ R let σ θ : ( ε θ, ε θ M (for suitable ε θ be the unique curve parameterized by arclength with σ θ (0 = p and such that σ θ (s, sin θ Y 1p + cos θ Y 2p 0 for all s. (Technically, σ θ (s is parallel-transported back to T pm using the Euclidean connection. Then there exist κ min, κ max R and θ 0 R such that the signed curvature κ θ of σ θ at t = 0 satisfies κ θ = cos 2 (θ θ 0 κ max + sin 2 (θ θ 0 κ min. (228 In general it makes no sense to talk about signed curvatures of curves in R 3 but here each curve σ θ lies in a plane that inherits a natural orientation by requiring that the basis {cos θy 1p + sin θy 2p, ν p } is positively oriented. This classical result identifies two principal curvatures κ min and κ max together with two orthogonal principal directions that determine the curvatures κ θ of the curves σ θ in all possible directions. This situation hints that there is a symmetric quadratic form the second derivative (of something to be identified lurking behind. In linear algebra language, the principal directions should be the eigenvectors, and the principal curvatures the eigenvalues associated with this form. Naturally, one expects the trace and the determinant of associated with this quadratic form be natural invariants indeed they will later be called mean curvature and Gaussian curvature. However, this theorem is really a statement about curves that lie in a surface. Its proof is a straightforward calculation requiring little more than multi-variable calculus provided one judiciously chooses a good set of coordinates. (The subsequent discussion of Gauss and Riemann s work will recover these curvatures from a geometric viewpoint that really describes curvature of surfaces as opposed to curvature of curves in a surface. Proof. Choose coordinates (x, y, z for R 3 such that p = 0 and ν p = D 3,(0,0,0 = (0, 0, 1. Note that this can always be achieved by using only translations and rotations this is important since the curvature of a space curve is invariant under such coordinate changes, compare exercise (1.10. Furthermore, the rotation allows enough freedom such that if near p the surface is given as the graph of a function z = f(x, y and D 1 D 2 f(p = 0. Exercise 4.36 Suppose that f C (R 2 satisfies f(0 = (D 1 f(0 = (D 2 f(0 = 0. Explicitly calculate the coordinate change (rotation in R 2 such that in the new coordinates D 1 D 2 f(p = 0. In these coordinates the curve σ θ is given as some reparameterization of γ(t = (t cos θ, t sin θ, f(t cos θ, t sin θ (229 Use that (D 1 f(0 = (D 2 f(0 = 0 to see that γ θ (0 = (cos θ, sin θ, 0. Similarly, (D 1D 2 f(0 = 0 (and the induced orientation on the plane containing the image of the curve yields that κ θ = γ θ (0 = (0, 0, cos2 θ(d 2 1f(0 + sin 2 θ(d 2 2f(0. (230 A closely related theorem of Meusnier in 1776 obtains the curvatures of the curves of intersection of the surface with arbitrary planes. Theorem 4.24 Suppose p M 2 R 3 and ν p is a normal vector to M at p. Suppose Π φ and Π are planes that intersect in a line through p that is perpendicular to ν p, and such that ν p in contained in Π and the angle between (the normals of the planes is φ. If κ φ and κ are the curvatures of the curves of intersection of M with Π φ and Π at p, respectively, then κ φ cos φ = κ. (231

3 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, For a simple, modern proof of this theorem, and to prepare for the subsequent discussions it is useful to introduce the notion of a smoothly varying unit normal vector field: Lemma 4.25 Suppose M n R m+1 is an oriented immersed hypersurface. Then there exists a unique smooth function ν : M T R n+1, called a unit normal vector field, with π ν = ı, (i.e. ν p T ı(p R 3 for all p M and ν( 1, such that whenever {X 1p,... X mp } T p M is a positively oriented basis, then {ı p X 1p,... ı p X mp, ν p } T ı(p R m+1 is a positively oriented basis. This existence is an immediate consequence of the basic linear algebra theorem that any basis for a subspace can be extended to a basis for the whole space. The smoothness is immediate, e.g. from working in a chart. (One also may argue using the Gram-Schmidt algorithm. The following discussions almost entirely address only (imbedded smooth surfaces in R 3, and it is convenient to be a little less precise: Throughout most of the following discussions we shall omit mention of maps such as ı and ı and conveniently identify e.g. p M with ı(p R M+1, X p T p M with ı p X p T ı(p R m+1 and the Euclidean inner product, ı(p on T ı(p R m+1 with its pullback ı (, ı(p to T p M. Moreover, if τ : R m+1 R m+1 is a translation τ(p = p + q for some fixed q R m+1, then also identify X p T p R m+1 with τ p X p T τ(p R m+1. Check carefully: The sequel uses Euclidean connection. Should this be first introduced in section on geodesics and Γ k ij? Exercise 4.37 Explain what this latter translation has to do with connections and parallel transport, i.e. how the identifications described above make use of the trivial Euclidean connection Di D j 0 for all i, j m + 1 on R m+1. Suppose M 2 R 3 is an immersed smooth surface and σ : ( ε, ε M is a smooth curve parameterized by arc-length (i.e. σ 1 with σ(0 = p. If necessary, working locally, assume that M has an orientation and let ν be the associated unit normal field. Since the curve lies in M it follows that σ, ν σ 0. Differentiating this identity and evaluating at t = 0 yields, in modern notation σ (0, (ν σ(0 = σ D (0, (ν σ (232 t=0 [[Here D dt is the covariant derivative along σ associated to the Euclidean connection on R3, i.e. if σ (t = a i (td i and ν(q = b j (qd j then D dt (ν σ = i,j (ai (0 (D i b j (pd j. t=0 The following pages frequently will use the language of connections, but it will always be the trivial Euclidean connections: It provides a very concise notation even for various expressions arising in classical calculations, and it is meant to suggest the obvious generalizations to not necessarily imbedded manifolds. ]] The right hand side is uniquely determined by the value of σ at t = 0, and writing Q(σ (0 for this expression, it defines a function Q: T p M R. Proof (of Meusnier s theorem. Using the conventions and notation of the preceding discussion and of the theorem statement suppose that σ : ( ε, ε M Π and σ φ : ( ε, ε M Π φ are smooth curves parameterized by arc-length such that σ (0 = σ φ (0 = p and X p = σ (0 = σ φ (0. The signed curvature of either curve is the magnitude of the second derivative σ (0 or σ φ (0. Since σ (t Π for all t it is clear that σ (0 is a multiple of ν p, and hence dt κ (0 = σ (0, ν p = Q(X p. (233

4 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, Similarly σ φ (t Π φ for all t implies that σ φ (0 = κ φ(0n φ is a multiple of a vector N φ Π φ characterized by X p, N φ = 0 (and respecting the orientation convention. Since X p P Π φ this means that N φ, ν p = cos φ, and thus σ φ (0, ν p = κ φ (0 cos φ The result follows from combining this with (232 κ (0 = σ (0, ν p = Q(σ (0! = Q(σ φ (0 = σ φ (0, ν p = κ φ (0 cos φ. (234 The constructions leading to the proof given above are mostly elementary except for the key observation that the right hand side of (232 only depended on the single tangent vector σ (0. While one may establish this fact using only basic calculus, we consider this as a sign post leading to the modern view point in terms of covariant derivatives. The first step is to recognize the utility of the normal field ν, or rather its rate of change to give a new characterization (definition! of the (or rather a curvature of a surface. Define the Gauss map, and, using modern terminology its tangent map/differential: Definition 4.11 Suppose that M m R m+1 is an oriented smooth hypersurface and ν : M T R m+1 is the unit normal vector field (as constructed above that provides the compatibility of the orientations of M m and R m+1. With the natural identification, considered as a map ν : M S 2 this map is called the Gauss map. Its tangent map (or differential, see below considered at each p M as a map IL p = ν p : T p M T p M is called the Weingarten map. Pictorially there are no problems with these maps, but it takes a little patience to sort out all of the details if one wants to be precise about the natural identifications. Clearly they are all about the trivial Euclidean connection, which allows one to identify tangent vectors X p T p R m+1 X q T q R m+1 at different points p, q R m+1 with each other, and even with points in R m+1. While not essential, it is a useful, and somewhat challenging exercise to walk through the details: To begin with, ν p T ı(p R m+1 has unit norm and the unit sphere in T ı(p R m+1 is identified with the sphere S m R m+1 via a suitable restriction Φ of the linear map Φ : T p R m+1 R m+1 to the set {X ı(p T ı(p R m+1 : X ı(p = 1} that takes the tangent vector j aj D j,ı(p to the point (a 1, a 2,... a m+1 S m R m+1. The tangent map ν p takes the tangent space T p M to the tangent space T ν(p (T R m+1. The tangent map of Φ then serves to identify T ν(p (T R m+1 with T (Φ ν(p S m. It remains to identify T (Φ ν(p S m back with T p M. For this, note that T p M ν p T ı(p R m+1 means that the image of X p T p M satisfies (Φ ν p (X p T (Φ ν(p S 2 (Φ ν(p R m+1, where in this last step (Φ ν(p is identified with a tangent vector in T (Φ ν(p R m+1. An alternative viewpoint considers dν, or more accurately d(φ ν p as a map from T p M to R m+1 : If ν(p = j νj (pd j T ı(p R m+1 and X p = X i pd i,p T p M then dν(x p = (dν 1 (X p, dν 2 (X p,..., dν m+1 (X p = (X p (ν 1, X p (ν 2,... X p (ν m+1 = X p (ν 1 D 1 + X p (ν 2 D X p (ν m+1 D m+1 (235 def = Xp ν may again be identified with a tangent vector first in T ı(p R m+1, but actually also with a tangent vector in T p M because dν(x p ν(p in T ı(p R m+1 as is seen from differentiating the constant function q ν q 2 = ν q, ν q 1 (as a function in C (M, i.e. 0 = X p (1 = X p ν, ν = ν p, X p (ν 1 D X p (ν m+1 D m+1. (236

5 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, These technical details are not essential when only considering curvature of smooth hypersurfaces in Euclidean spaces but they are helpful for the development of notions of curvature in general manifolds, not necessarily Riemannian. The key is to keep in mind which parts of the construction made above are only possible in Euclidean spaces, and which ones may be generalized to general manifolds using connections. Intuitively the rate of change of the normal vector field ν provides a measure of curvature. It is worthwhile to first get a better intuitive understanding of this map, i.e. consider the images ν(u S m of subsets U M m. Exercise 4.38 Identify the images of familiar surfaces under the Gauss-map: Verify that on the unit sphere the map ν is the identity. Find the images of the cylinder x 2 +y 2 = 1, of the cone z = α x 2 + y 2 (how does the image depend on α?, of any plane ax + by + cz = k. Continue with describing the images of the paraboloid z = x 2 + y 2, of the saddle z = x 2 y 2, and of the graph of z(1 + x 2 + y 2 = α (how does the image depend on α?. [[Use the provided MAPLE worksheet to confirm your descriptions and to enjoy the images... ]] Definition 4.12 Define the total curvature of an oriented smooth hypersurface M m R m+1 to the volume (area of the image ν(m m S m. Exercise 4.39 Calculate the total curvatures for each of the surfaces of exercise (4.38. A natural proposal for a quantitative measure of the curvature at a point p M m is to consider sequences of neighborhoods U k M m that shrink [?] to the point p and propose k(p def? (signed area of = lim U p area of U ν(u (237 This is basically what Gauss idea was. Compare this to defining curl and divergence in multivariable calculus as curlf def? (p N = lim R F ds and divf R p ( area of R def? R F (p = lim dn (238 R p ( volume of R where e.g. in the first case R is a planar surface element with normal N, and in the second case dn is the outward normal on the boundary of the region R. The main advantage is that such definitions provide immediate understanding of what these objects measure, and they make the associated integral theorems rather trivial: With these definitions (? Stokes theorem (in its narrow sense and as the divergence theorem are immediate, and similarly will be a theorem that relates the total curvature to the curvature at a point via an integral. On the other hand it takes substantial, and rather unpleasant, efforts to make these into real definitions. One way is to restrict the subsets U or regions R to be of a very special form (but if this is too restrictive, the proofs of the integral theorems immediately become much harder. What is even less attractive is the necessarily rather arbitrary character of what one calls admissible for U or R. The alternative is to characterize quantities like the curvature k(p, curl and divergence in algebraic terms, and show that the quantities so defined may be interpreted in the above way. However, in all fairness, the algebraic formulas do not just fall out of the sky, but rather ought to be inspired by the proposed preliminary definition given above. Historically one has informally taken the above as a preliminary definition, and under whatever suitable niceness conditions

6 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, work out what a candidate algebraic definition ought to be... Where Gauss worked with infinitesimal surface elements, the modern machinery provides an elegant alternative that is at the same time precise and leads to all algebraic formulas while at the same time being close to the intuitive ideas. The key innovation is to consider of subsets of the tangent space T p M and their images instead of small neighborhoods U M and their images ν(u S 2. Using the Riemannian metric on M 2 induced from the Euclidean metric on R 3 the area of the parallelogram spanned by X p and Y p is area( {λx p + µy p T p M : λ, µ [0, 1]} = det ( Xp, X p M p X p, Y p M p Y p, X p M p Y p, Y p M p (239 Similarly, using the Riemannian metric on S 2 induced from the Euclidean metric on R 3 the area of the parallelogram spanned by ν p (X p and ν p (Y p is det ( ν p X p, ν p X p S2 ν(p ν p Y p, ν p X p S2 ν(p ν p X p, ν p Y p S2 ν(p ν p Y p, ν p Y p S2 ν(p (240 To simplify notation we note that the determinants are nothing else than the volume forms on the Riemannian manifolds M 2 and S 2, with metrics inherited from their immersions into R 3, and therefore may write dv M and dv S2, respectively. To further simplify notation use the pullback of dv S2 by ν and formally define: Definition 4.13 Suppose : M 2 R 3 is a smooth surface. The Gaussian curvature at p M is k(p = ν p(dv S2 dv M p (241 Note that despite its strange appearance this division makes sense as every 2-from on M 2 is a scalar multiple of the natural volume form dv M, and thus k C (M. Exercise 4.40 Verify that the Gaussian curvature is independent of the orientation on M 2. It is appropriate to take a short detour and obtain more explicit formulas for graphs of functions in R 3. One key simplification is based on the observation that since the tangent plane T p M is parallel to the tangent plane T ν(p S 2 the ratio of the areas of the respective parallelograms is equal to the ratio of the areas of their respective projections onto the coordinate planes! Hence if M 2 is the graph of a function g : R 2 R 3, a natural choice of two independent tangent vectors at p = F (s, t M 2 R 3 is X p = F (D 1 = D 1 + (D 1 gd 3 and Y p = F (D 2 = D 2 + (D 2 gd 3. The area of the projection of the parallelogram spanned by these vectors onto the (x 1, x 2 coordinate plane is 1. On the other hand ν p X p = ν F (D 1 = (ν F (D 1 = (D 1 (ν 1 F D 1 + (D 1 (ν 1 F D 2 + (D 1 (ν 1 F D 3 ν p Y p = ν F (D 2 = (ν F (D 2 = (D 2 (ν 1 F D 1 + (D 2 (ν 1 F D 2 + (D 2 (ν 1 F D 3 (242 Exercise 4.41 Identify where each of the expressions, functions and tangent vectors, in the preceding calculations is to be evaluated.

7 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, The area of the projection of the parallelogram spanned by these tangent vectors to S 2 onto the (x 1, x 2 coordinate plane, and hence the Gaussian curvature at p is equal to k(p = D 1 (ν 1 F D 2 (ν 2 F D 2 (ν 1 F D 1 (ν 2 F (243 evaluated at F 1 (s, t. Finally, explicitly calculate the unit normal ν p from X p and Y p as (ν p F = (D 1gD 1 (D 2 gd 2 + D 3 (1 + (D 1 g 2 + (D 2 g (244 Exercise 4.42 Combine the expressions in the equations (243 (244 to derive by direct calculation (use a computer algebra system as needed: k(p = (D2 1 g (D2 2 g (D 1D 2 g (1 + (D 1 g 2 + (D 2 g 2 2 F 1 (p (245 Briefly return to Euler s theorem: In the first proof we chose the coordinates so that (D 1 g, (D 2 g, and (D 2 D 1 g all were zero at F 1 (p, and the formula (245 immediately yields that the Gaussian curvature is the product of the extremal curvatures k min and k max of Euler s theorem. Exercise 4.43 Use the formula (245 to explicitly calculate the Gaussian curvatures of a plane z = mx + ny, the paraboloid z = ax 2 + by 2, the saddles z = x 2 y 2 and z = xy, and the graphs of z(1 + x 2 + y 2 = 1 and z = Re(x + 1y 3. In each case identify all regions where the curvature is positive, zero, or negative. [[Compare with the visual images in the MAPLE worksheet gaussmap.mws.]] Exercise 4.44 Use the expression (245 to prove that the curvature k(p of the graph of every polynomial converges to zero as p grows to infinity. Exercise 4.45 Verify by direct calculation (computer algebra system that the expression (245 is invariant under rotations and reflections. Exercise 4.46 Develop an explicit formula for the Gaussian curvature on a smooth parameterized surface F : (s, t (F 1 (s, t, F 2 (s, t, F 3 (s, t (not necessarily F 1 (s, t = s, F 2 (s, t = t. Exercise 4.47 Verify by direct calculation (computer algebra system that the expression obtained in exercise (4.46 is invariant under rotations and reflections. Exercise 4.48 Verify by direct calculation (computer algebra system that the expression obtained in exercise (4.46 yields constant curvature for the sphere. Also calculate the Gaussian curvatures of a cone and cylinder using cylindrical coordinates, and of the torus F : (s, t ((R + r cos(s cos(t, (R + r cos(s sin(t, r sin(s. In each case identify all regions where the curvature is positive, zero, or negative. [[Utilize a computer algebra system as appropriate, and compare with the visual images in the MAPLE worksheet gaussmap.mws.]]

8 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, Returning to the more theoretical discussion, we now make use of the prior identifications, considering the tangent map or differential of the Gauss map ν as the Weingarten map IL p = (dν p : T p M T p M. There is an elegant way to express some classical terminology in terms of the Weingarten map, and thereby obtain new insights about the character of the Gauss curvature as an magnification factor of infinitesimal area, or rather area in the tangent spaces. Definition 4.14 Suppose M m R m+1 is an oriented smooth manifold (with the induced metric. Define the 1 st, 2 nd and 3 rd fundamental forms I, II, III pointwise by I p, II p, III p : T p M T p M R, I p (X p, Y p = X p, Y p II p (X p, Y p = IL p (X p, Y p (246 III p (X p, Y p = IL 2 p(x p, Y p In the case of surfaces M 2 in R 3 the classical notation for the components of the fundamental forms in a chart ((x, y, M 2 uses the letters F, G, H and l, m, n for the components of I and II: I = F dx dx + Gdx dy + Gdy dy + Hdy dy II = ldx dx + mdx dy + mdy dy + ndy dy. (247 Exercise 4.49 Use the precise definition in terms of dν to verify that the map II is bilinear over C (M. Demonstrate that I and II depend on the orientation on M (hence II/I does not. It is straightforward to recognize an interpretation of the second fundamental form in terms of the curvatures of the curves of intersection of M with normal planes as discussed earlier: If Π is the plane in R 3 through p M that is spanned by ν p and a unit vector X p T p M, and σ : ( ε, ε M Π is a smooth curve parameterized by arc length, with σ (0 = X p, then the signed curvature (with the same orientation convention of σ at t = 0 equals κ(0 = σ (0, ν p = II σ(0 (σ (0, σ (0 (248 Before further analyzing the Weingarten map we carry out some basic technical constructions which demonstrate that, locally, vector fields along a submanifold may be smoothly extended. We check some basic properties of such extensions, which will allow for some elegant proof of important properties of the Weingarten map. Suppose ı: M m N n is an imbedded submanifold, and (u, U, ū, Ū are charts (about p of M and N, respectively such that U = Ū M = {q Ū : um+1 (q =... = u n (q = 0} and u i = ū i U for i = 1, 2,..., m. W.l.o.g. assume that ū(ū = Bn 0 (r is an open ball in Rn so that the map π : U U, π(q = u 1 (u 1 (q, u 2 (q,..., u m (q, 0,... 0 is well-defined. If X C (M, T N such that π T N X = ı is a smooth vector field along M, then it may be smoothly extended to a vector field X Γ (Ū by defining X q = m i=1 (Xui π(q ū i q. (249 Lemma 4.26 Suppose M, N, p, U, Ū are as above and ı: U Ū is the inclusion map. and Z Γ (U and Z Γ (Ū are tangent vector fields. Then Z is an extension of Z if and only if ( Z f U = Z( f U for every Z C (Ū. Moreover, if X and Ȳ extend X, Y Γ (U to Ū, then [ X, Ȳ ] extends [X, Y ] to Ū.

9 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, Proof. If p U, Zp = ı Z p, and f C (Ū, then ( Z f(p = Z p f = Zp ( f ı = (Z(f U (p. Conversely, if ( Z f U = Z( f U then Z p f = ( Z fp = Z p (f u = Z p ( f ı = (ı Z p f. Now suppose p U, f C (Ū and that X, Ȳ extend X, Y Γ (U to Ū. Then using the first part, the following shows that [ X, Ȳ ] extends [X, Y ]. [ X, Ȳ ] p f = X p (Ȳ f Ȳp( X f = X p (Y ( f U Y p (X( f U = [X, Y ](f U. (250 Proposition 4.27 Suppose M m R m+1 is a smooth oriented hypersurface and p M. The Weingarten map IL p : T p M T p M is self-adjoint. Proof. The preceding lemma allows for an elegant proof that uses the Euclidean connection. Suppose X p, Y p T p M. First extend to tangent vector fields X, Y in a neighborhood on p M, then to vector fields X, Ȳ on a neighborhood of ı(p Rm+1. Calculate X, IL(Y p IL(X, Y p = X, Y ν p X ν, Y p = X, Ȳ ν p X ν, Ȳ p = Ȳ X, ν p + ν, XȲ p (251 = ν, [ X, Ȳ ] p = ν p, [X, Y ] p = 0 using in the third step that X p ν, Ȳ = Ȳp Ȳ, ν = 0 since ν, Y X, ν 0. In a nutshell the self-adjointness of the Weingarten map may be seen as an immediate consequence of the compatibility of the Euclidean connections (directional derivatives with the Euclidean metric (standard inner product. Note that Euler s theorem about principal curvatures now becomes a simple corollary: As a self-adjoint operator IL: T p M T p M has real eigenvalues and is orthogonally diagonalizable. The principal curvatures and directions are mere consequences of elementary linear algebra. Exercise 4.50 Fill in the details to derive Euler s theorem as a corollary of IL being self-adjoint. Theorem 4.28 The Gaussian curvature k(p at p M 2 R 3 is the determinant of the Weingarten map IL: T p M T p M. k(p = det IL p (252 Corollary 4.29 Suppose p M 2. If X p, Y p T p M are linearly independent or orthonormal, then the Gaussian curvature at p is, respectively equal to ( IIp (X det p, X p II p (X p, Y p ( II p (Y p, Y p II p (Y p, Y p IIp (X k(p = ( and k(p = det p, Y p II p (X p, Y p (253 Ip (X det p, X p I p (X p, Y p II p (Y p, Y p II p (Y p, Y p I p (Y p, Y p I p (Y p, Y p Lemma 4.30 Suppose that M 2 R 3 is an immersed manifold with the metric inherited from R 3. Then in each coordinate chart (u, U of M 2 the Gaussian curvature may be written as a smooth function of the components g ij =, of the Riemannian metric and its first and u i u j second derivatives g ij and 2 g ij. u k u l u k

10 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, Proof (outline. This outline follows Opreah, but more highlighting the role of the connection, and with details of the calculations provided in the MAPLE worksheet egeregium.mws. (For a different approach using determinants and vector cross products see Spivak II, pp Suppose (u, U is a chart of M with p U. Let ν : U R 3 be the unit normal vector field along U that provides the compatibility of the orientation between U and R 3. Then the Euclidean connection defines the coefficients in: After reorganization: How much of Euclidean connection need to be defined before? u 1 u 2 u 2 = Γ 1 u Γ 2 u l ν u 2 = Γ 1 u Γ 2 u m ν u 2 = Γ 1 u Γ 2 u n ν u 2 using the symmetry of as a Riemannian (or Levi-Civita connection: (254 Γ k 12 = Γ k 21 since u 2 u 1 u 2 u 1 = [ u 1, u 2 ] 0. (255 In this basis the Weingarten map L: T p M T p M may be expanded in the form IL( u 1 = u 1 ν = w 11 u 1 + w 12 IL( u 2 = u 2 ν = w 21 u 1 + w 22 u ν = (l u 2 EG F ν = (m u 2 EG F 2 u 1 + m u 1 + n u 2 (256 u 2 Each of these identities may be differentiated in particular, the equality of the mixed partial derivative (this is the Euclidean connection! provides two different expressions from equating the derivatives ( ( (257 u 2 u 1 = u 2 u 2 u 1 As shown in the MAPLE worksheet egregium.mws, it is straightforward to differentiate the corresponding right hand sides, re-using the identities (254 and (256 to rewrite all derivatives of the vector fields, and ν again as linear combinations of these vector fields. The crucial u 1 u 2 feature is that while now also derivatives of the coefficients l, m and n appear, these only appear in the coefficients of ν. Consequently, one obtains two equations from equating the coefficients of each of, in (257 to zero. The surprising feature is that each of these will yield an u 1 u 2 equation that only depends smoothly on the combination (ln m 2, but otherwise is independent of l, m and n. This allows one to solve for (ln m 2 in terms of E, F, and G and their first and second order derivatives (in the directions and. (Alternatively, these may also u 1 u 2 be written in terms of the g ij and their first and second order derivatives, or in terms of the Christoffel-symbols Γ k ij and their first derivatives (in the directions and. u 1 u 2 The resulting formulas look unappealing (actually they are a preview of the components of the Riemann curvature tensor!. But what matters is that these formulas express the Gaussian curvature entirely in terms of the intrinsic metric and its (covariant derivatives (on the manifold M. This proves that even though the Gaussian curvature was defined in terms of the normal vector field ν, it is independent of the imbedding into the ambient space:

11 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, Theorem 4.31 (Gauss, 1827: Theorema Egregium If Φ, Ψ: M 2 R 3 are immersions such that Φ, = Ψ, then for all p M the Gaussian curvatures of Φ(M and Ψ(M at Φ(p and Ψ(p, respectively, are equal. A few remarks about geodesic triangles and the Gauss Bonnet theorem. With the tools developed so far it is basically a just a lengthy exercise, involving some trickier calculations, to connect the total curvature (the integral of the Gauss curvature to the angle sum in a triangle: Theorem 4.32 Suppose M 2 is a Riemannian manifold, and U U M are neighborhoods of p M such that every pair of points in U is joined by a unique geodesic that lies entirely in U, and such that U is contained in the image of exp p, compare theorem (4.18. Suppose q, q U and R U is a geodesic triangle whose sides are the geodesics connecting the three points p, q, q, respectively. If the interior angles are α i, i = 1, 2, 3, then k dv = α 1 + α 2 + α 3 π. (258 R In the standard proof one uses the map exp p to pull back the volume form (area form dv from U to T p M, thereby reducing the integration to an integral over a subset of the linear space T p M. Note that the preimages under exp p of the geodesics connecting p to q and q, respectively, are straight rays emanating from 0 p T p M. Thus the integral is particularly easy to evaluate using polar coordinates (basically these are the pullbacks of the geodesic coordinates on U to T p M. Via a simple application of Green s theorem it reduces to a single integral involving the direction of the velocity vector of the preimage of the geodesic connecting q to q as a function of the angular variable. So far everything beyond the pullback via exp p is basic calculus. What remains is to explicitly calculate the pullback of k in the geodesic, or polar coordinates. Thus write (r, θ for the standard polar coordinates on T p M and u = (r, θ exp 1 for are the associated geodesic coordinates on U M. By definition of the geodesics it is clear that in these coordinates g 11 = E =, = 1. Gauss lemma (4.19 implies that g u 1 u 1 12 = g21 = F = 0. Consequently in these coordinates the metric has the form, = du 1 du 1 + G du 2 du 2 for some smooth positive function G: U (0,. Exercise 4.51 For the metric as given above verify that k = 1 2 G G. u 1 u 1 On the other hand one needs properties of the geodesics connecting q to q. It is straightforward to write down the geodesic equation given the data for g ij as above. Exercise 4.52 Given the metric in coordinates as above, calculate the Christoffel symbols Γ k ij of the second kind and write out the geodesic equations in these geodesic coordinates. [[The coefficients will be in terms of derivatives of the given function G = g 22 ]]. To complete the proof it remains to find a convenient expression for the integral along the third side, the geodesic connecting q to q. It appears most convenient to rewrite (reinterpret the geodesic equation to obtain a differential equation for the rate of change of the angle between its velocity vector and either of the coordinate directions. There are several technical details involved and thus leave it as an extended project:

12 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, Project 4.53 Fill in the missing details into the outline of the proof of the theorem. As reference consult e.g. Spivak II.pp in progress Definition 4.15 A triangulation of what?? is union... intersections are edges or vertices... in progress The final highlight of this classical development is without doubt: Theorem 4.33 (Gauss-Bonnet : Suppose M 2 is a compact, orientable Riemannian manifold, k is the Gaussian curvature, and χ E (M is the Euler characteristic of M. Then k dv = 2π χ E (M (259 M The Euler characteristic is a fundamental topological invariant. In the case that M is a triangulated C 0 manifold (think of a polyhedron, and subdivide the faces if necessary the Euler characteristic is equal to the alternating sum χ E (M = V E + F of the number of vertices, edges, and faces. Exercise 4.54 Calculate the Euler characteristic of the regular polyhedra (tetrahedron, cube, octahedron, dodecahedron, and icosahedron. Are they the same (as topologically manifolds, i.e. are they homeomorphic? Are they homeomorphic to the sphere? Given that the Euler characteristic is a topological invariant calculate the Euler characteristics of a torus, of a double-torus (figure eight shaped tube, and a bretzel-like surface with three holes. [[Construct, simply draw, a polyhedron that is homeomorphic, e.g. start with a solid rectangular block and remove one, two, or three appropriate rectangular blocks. Count the faces, edges, and vertices of the surface of the solid.]] Does it matter for this count whether the faces are allowed to have any polygonal shape or whether they must first be fully triangulated? The Euler characteristics belongs into the realm of algebraic topology the desire for a solid proof may be taken as a good motivation to develop the tools of cohomologies. Arguably the closest differential geometric analogue is the de-rham cohomology which starts with an algebraic description of in how many ways can closed differential forms on the manifold fail to be exact (technically these are the dimensions of the quotient spaces of closed differential k-forms by the exact differential k-forms. [[Look under Betty-numbers in most any advanced book on the interfaces of differential and algebraic, geometry and topology.]] What makes the Gauss Bonnet theorem so intriguing that on one side is the integral of an important geometric invariant, but on the other side is a purely topological invariant. This has very profound consequences: E.g. given that χ E (S 2 = 2 it follows immediately that the total curvature of the sphere is 4π no matter what Riemannian metric S 2 (or, for that, any other compact simply connected 2-manifold is equipped with! Similarly, if the Gaussian curvature is negative at all points on a compact manifold M 2 then M cannot be simply connected. A complete proof of the Gauss Bonnet theorem requires the development of a substantial machinery of algebraic topology. But if one takes the topological invariance of the Euler characteristic as a given, then modulo one technical detail the proof is amazingly simple:

13 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, Exercise 4.55 Suppose that M 2 is a Riemannian manifold that is triangulated into a finite number of geodesic triangles. Use the theorem (4.32 to derive the Gauss-Bonnet theorem for this special case. [[Sum the left and right hand sides of the identity in theorem (4.32 over all triangles and note that at each vertex all the interior angles sum to 2π. However, in a case of general polygons, and also for the generalization to Not necessarily geodesic triangles it is better to replace the interior α i angles with the change of direction ( σ i = lim t ti + σ (t lim t ti σ (t = π α i. Note that this angle only involves the inner product in T σ(ti M 2. ]] Thus aside form the topological background (that χ E is indeed an invariant the main missing part is the existence of a (finite! triangulation by geodesic triangles. The general result that every C -manifold has a triangulation is a difficult theorem (cf. Spivak I. p.579, for a proof see e.g. Munkres, Elementary Differential Topology. In our case the compactness hypothesis makes things much easier, but still nontrivial. On the other side it is relatively straightforward to weaken the demand that the sides of the triangles are geodesics in that case one merely has to account for the change of direction of the velocity vector along the edges this is amounts to integrating the geodesic curvature of the edges. This may be done using classical formulations (calculus in Euclidean spaces, but modern approach in terms of connections, covariant derivatives along the edges and parallel transports appears much more appealing. Project 4.56 Work out an outline to fill in the steps to the more general version of the Gauss- Bonnet formula given in Schaum s Outline (see below and the modern approach in terms of connections. Suppose C is a C 2 curvi-linear polygon and a C 3 manifold, which is positively oriented and its interior is simply connected. Then κ g ds + kdv = 2π α i (260 R i where κ g is the geodesic curvature along C, R is the union of C with its inte- -rior [[??]], k is the Gaussian curvature, and α i are the exterior angles of C. There are many closely related theorems, some almost immediate corollaries of the Gauss Bonnet theorem. We just mention a few. Theorem 4.34 (Puiseux, 1848 (compare Spivak II, p.129 Let C(r be the circumference of the geodesic sphere (circle with radius r about p. Then k(p = lim r 0 12 πr2 A(r πr 4 (261 Theorem 4.35 (Diquet, 1848 (comp. Spivak II, p.131 Let A(r be the area of the region inside the geodesic sphere (circle with radius r about p. Then 2πr C(r k(p = lim 3 r 0 πr 3 (262 The following two result from relating the Gaussian curvature to the minimality of geodesics. Theorem 4.36 (Bonnet (comp. Sternberg p.292 If M 2 is a connected Riemannian manifold with k 1/C 2, then for any two points p, q M the distance d R (p, q is at most πc.

14 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, Theorem 4.37 (compare Sternberg p.292 If M 2 is a connected Riemannian manifold such that k 0, then no geodesic on M has any conjugate points, and if M is simply connected then M is diffeomorphic to the plane R 2. (Here a conjugate point along a geodesic is defined as the first point along the geodesic where the tangent map of the map exp: T p M M becomes singular.

15 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, Riemann curvature tensor Since Gauss the concept of curvature has been generalized to apply to general manifolds, not necessarily hypersurfaces or Riemannian. The most prominent notion is due to Riemann who in 1861 asked the simple question stated here in modern language Riemann s question: When is a Riemannian manifold flat, i.e. locally isometric to R m? Find conditions for the existence (and an algorithm for the construction of local coordinates (u, U such that in these coordinates g ij =, = δ u i u j ij. The following exposition loosely follows Spivak s discussion of what basically was Riemann s original approach. However, we shall use modern terminology throughout, and sprinkle in some reference to connections which eventually will suggest a perfectly elegant definition in terms of Koszul connections. Thus suppose (M m,, is a Riemannian manifold, and (v, V is a chart about a point p M. Let g ij =, denote the components of the metric in these given coordinates. The v i v j objective is to find conditions for the existence of a local coordinate change u 1 that transforms the given g ij into, = δ u i u j ij. The basic strategy is to derive a set of partial differential equations for the coordinate change, and massage them so that eventually we may apply standard integrability theorems (some version of Frobenius theorem. The transformation formulas for the components of the metric provide the basic conditions. Write them out in both directions. Note the ranges of the summations. δ ij = u i, g ij = v i, u j = v j = k,l=1 k=1 v k u i vl u j g kl (263 u k v i uk v j (264 Interpret these as a set of 1 2m(m + 1 quadratic, first order inhomogeneous partial differential equations for m functions u k : U V R m. Basically the only operation one can always use is to differentiate [[From the late Robert Gardener I learnt the phrase apply the Pavlovoperator, i.e. take the exterior derivative.]] Here we take partial derivatives in the directions of the coordinate vector fields (clearly, working with d comes to mind as a more algebraic alternative. In order to obtain familiar formulas that eventually will lead to linear partial differential equations, add and subtract the following, carefully noting the cancelations. ( + ( + ( g il v j g jl v i = g ij v l = m k=1 k=1 = m k=1 ( 2 u k v j v i uk v l + uk v i 2 u k v j v l ( 2 u k v i v j uk v l + uk v j 2 u k v i v l ( 2 u k v l v i uk v j + uk v i 2 u k v l v j

16 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, [i j, l ] = 1 2 ( gjl v i + g il v j g ij v l = k=1 2 u k v i v j uk v l (265 The plan is to use the identity (264 in order to move the first derivative uk to the left side, v l noting that a product of two such terms yields the g ij and thus raises the indices, leading to Christoffel symbols of the second kind on the left hand side. Specifically use: Γ k ij = [ i j, l ] g kl = l=1 [ i j, l ] vl u r vk u r (266 l, r=1 Multiply both sides by us and sum over k. Reuse the preliminary result (265, and contract v k the sums as usual: k=1 Γ k ij us v k = = = k, l, r=1 l=1 k, l=1 = 2 u k v i v j [ i j, l ] vl u r vk u r us v k [ i j, l ] vl u s 2 u k v i v j uk v l vl u s (267 Consider these as a set of m 3 first order explicit (i.e. solved for the highest derivative linear homogeneous partial differential equations for m 2 unknown functions ui, i.e. think of them as v j ( u k v i v j = l=1 Γ l ij uk v l (268 Note that these are actually the same partial differential equations for each index k the eventual objective is to find (conditions for the existence of m linearly independent solutions u k satisfying e.g as boundary conditions u k (p = 0 and p = p. The Frobenius integrability u k v k test requires the equality of mixed partial derivatives, i.e. for all i, j, k, s = 1,..., m ( ( 2 u k 2 u k v i! v v s v j = i v j v s i.e. (Γ lij v s uk! v l = (Γ lis v j uk v l (269 l=1 Differentiate the products and reuse the partial differential equations (268 to express every l=1

17 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, second order derivative of the u k as a linear combination of first order derivatives. (! Γ l ij 0 = v s uk v l Γl is v j uk v l + Γ l 2 u k ij v s v l Γ l 2 u k is v j v l l=1 ( Γ l ij = v s uk v l Γl is v j uk v l + Γ l ij Γ r sl uk v r Γ l is l=1 r=1 r=1 ( Γ l ij = v s Γl is v j uk ( v l + Γ l ij Γ r sl Γl is Γ r jl l=1 l,r=1 ( Γ l ij = v s Γl is v j + Γ r ij Γ l sr Γ r is Γ l jr l=1 r=1 }{{} Rsij l Γ r jl uk v r uk v r (270 uk v l (271 So much for an exposition of the historical origin of the Riemann curvature tensor... Theorem 4.38 (Riemann, 1861 A Riemannian manifold (M m,, is flat if and only if in every coordinate chart if Rjkl i 0 for all i, j, k, l = 1,... m, The prior calculations essentially establish the necessity. For sufficiency, including a construction of the new coordinates u k see e.g. Spivak II, pp (referred to as the test case. Theorem 4.39 The functions R i jkl as above are the components of a tensor of type ( 3 1. A purely mechanical proof checks that these functions transform as required, i.e. suppose that (u, U and (v, V are coordinate charts with functions Γ k ij, Ri jkl and Γ t rs, R r stq, respectively. Then on U V it is to be verified that for r, s, t, q = 1,... m R r stq = i,j,k,l=1 Rjkl i v r u i uj v s uk v t ul v q (272 This is a straightforward calculation that uses the transformation formula for the Christoffel symbols of the second kind derived earlier Γ t rs = Γ k v t ij u k ui v r uj m v s + 2 u q v r v s vt u q (273 i,j,k=1 Thus the Riemann curvature tensor is a map q=1 R: Γ (M Γ (M Γ (M Γ (M (274 and in a chart (u, U there are functions R i jkl C (U such that R( u k, u l u j i = R i jkl u i (275

18 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, ( 4 For convenience one also introduces a tensor of type 0, as the map and in a chart (u, U R: Γ (M Γ (M Γ (M Γ (M C (M (276 R ijkl = g is Rjkl i s=1 = R( u k, u l u j, u i (277 In the case of a two dimensional manifold M 2 on recovers the Gaussian curvature as k(p = R(X p, Y p Y p, X p X p, Y p 2 (278 where X p, Y p T p M are linearly independent and X p, Y p is the area of the parallelogram in T p M spanned by X p and Y p. In the special case that X p = p are Y u 1 p = p orthonormal u 2 one obtains k(p = R 1221 = R(,, (279 u 1 u 2 u 2 u 1 Exercise 4.57 Verify the assertion that k = R 1221 by direct calculation. [[Write Rjkl i of Γ k ij, then in terms of g ij or E, F, G and their derivatives.]] in terms Exercise 4.58 Calculate the components of the Riemann curvature tensor for the special case that M is the graph of a function z = f(x, y coordinatized by u = (x, y. [[ Use a computer algebra system first do it yourself, then compare the results with the tensor-package.]] Exercise 4.59 Use the results of the previous exercise to calculate the components of the Riemann curvature tensor for familiar graphs such as a plane, a paraboloid, a saddle, the graph of (x, y 1. [[ Use a computer algebra system first do it yourself, then compare the 1+x 2 +y 2 results with the tensor-package.]] Exercise 4.60 Calculate the components of the Riemann curvature tensor for the special case that M is a parameterized surface (x, y, z = F (s, t coordinatized by u = (s, t [[ Use a computer algebra system first do it yourself, then compare the results with the tensor-package.]] Exercise 4.61 Use the results of the previous exercise to calculate the components of the Riemann curvature tensor for familiar surfaces such as spheres and tori of different radii. [[ Use a computer algebra system first do it yourself, then compare the results with the tensor-package.]] Exercise 4.62 Calculate the components of the Riemann curvature tensor for familiar surfaces such a plane, a cone, a cylinder, and a paraboloid using polar / cylindrical coordinates. Especially, check that in these coordinates still R = 0 for the cone and plane. The following symmetries of Riemann curvature tensor are very useful.

19 Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 20, Proposition 4.40 The Riemann curvature tensor is antisymmetric in the first two, and in the last two entries, respectively, but it is symmetric in pairs and satisfies an analogue of the Jacobi identity, called the Bianchi identity. Specifically, for X, Y, Z, W Γ (M R(X, Y Z = R(Y, XZ (280 R(X, Y Z, W = R(X, Y W, Z (281 R(X, Y Z, W = R(W, ZX, Y (282 Exercise 4.63 Prove the assertions made in the proposition. 0 = R(X, Y Z + R(Y, ZX + R(Z, XY. (283 While we found that in a chart on a two-dimensional manifold the component R 1221 recovers the Gaussian curvature, it is far from easy to understand the geometric roles of all the other components. The following quite surprising property helps substantially: Proposition 4.41 The Riemannian curvature tensor is completely determined by the values, called sectional curvatures, of the quadratic forms (X p, Y p R(X p, Y p, Y p, X p (284 Exercise 4.64 Prove the proposition: Suppose that R, S : V V V V R are multi-linear maps satisfying the identities of proposition (4.40 suitably rewritten. Show that then R = S. The next section will obtain the following more appealing formula for the Riemann curvature tensor: R(X, Y Z = X ( Y Z Y ( X Z [X,Y ] Z (285