These are notes and problems from Rob Kusner's Spring 2010 honors course on Differential Geometry of Curves & Surfaces (Math 563H at UMassAmherst, Copyleft* 2010 by Rob Kusner). The notation "MR" refers to an item in the Montiel & Ros text. Please write up the problems marked # (generally due in class the Tuesday 12-14 days after the date noted): Part 1 Parametrized curves. Arc length and arc-length parametrization. Unit tangent and normal vectors. Frenet and other adapted frames. Curvature and torsion. Global questions for plane and space curves. 19 Jan #Show that any interval I of the real numbers R is diffeomorphic to either [0,1], R_+=[0,\infty) or R; here "diffeomorphic" simply means there is a smooth map from I to one of these (find it) with smooth inverse; and show that it's enough to find a map with nonvanishing derivative (implicit/inverse function theorem in dimension 1). Study the twisted cubic curve X:R -> R^3: t -> (at, bt^2, ct^3) near t=0 for various values of (a, b, c). Note: a=0 is special. #If X(t) is smooth curve with X'(t) nonzero, show that the unit tangent field T(t)=X'(t)/|X'(t)| is also smooth (or, if X is only C^k, then T is C^{k-1}). What about the converse? We talked about (and examined plaster models of) surfaces that minimize their "bending energy". How might you define the bending energy of a curve? How about the stretching energy of a curve (think of Hooke's law from physics)? 21 Jan If X(s) is a C^2 immersed curved in R^n, parametrized by arclength, its velocity vector X'(s)=T(s) is the unit tangent vector, and its curvature vector is its derivative K(s):=T'(s); thus K(s)=X"(s) is the acceleration vector with respect to (any) arclength parameter. Differentiating 1= shows K(s) is perpendicular to T(s). #For any C^2 curve X(s) in R^n, and any C^2 reparametrization s(t), verify that the reparametrized curve Y(t)=X(s(t)) satisfies Y"(t) = s"(t) X'(s) + (s'(t))^2 X"(s). Therefore, if s happens to be an arclength parameter, the acceleration of X(s(t)) with respect to the parameter t is a linear combination of T(s(t)) and K(s(t)) with coefficients (tangential acceleration) and (speed squared). #In particular, to compute the curvature vector K at a point of an immersed curve X(t) as the acceleration X"(t), it suffices that |X'(t)|=1 and |X'(t)|'=0. Check that the second condition holds at t=0 for the twisted cubic, and use this idea to compute K(0) in terms of a, b, c. #More generally, work out an expression for the curvature vector K for an arbitrarily parametrized curve X(t) in terms of X' and X". (Try the case of curves in R^2 and R^3 first; in R^3, the cross product can be used, then specialized to R^2; the special case where X is a graph over its tangent line is also worth trying.) How does curvature behave under scaling? (Show |K| has "dimension" of 1/Length.) For a curve X(s) in R^2, there is a canonical extension of the unit tangent vector field to an orthonormal frame field T, N=JT spanning (the tangent space of) R^2 at any point along X. (Here J is just the 2x2 matrix that rotates R^2 by a quarter-turn.) Thus we can express the curvature vector K=kN for a function k(s) [I called this \kappa in class, but will use lower case here]. What curves have k=0? Which have constant k? What can be determined about X(s) from k(s) in general? (Hint: the ordinary differential system T'(s)=k(s)N(s)=k(s)JT(s) can be solved by integrating k(s) and exponentiating the resulting 2x2 matrix [more next time in class].) We briefly discussed polygonal approximation of a curve, and we promised to say more in these notes. Even for a continuous (C^0) X one can make a partition (see MR 1.6) and associated inscibed polygon; if the supremum of the length of such inscribed polygons exists, we say X is "rectifiable"; MR 1.6 & 1.10 show that a C^1 curve X is rectifiable and that this supremum equals the length of X defined by integrating |X'|. For a C^2 curve, the estimate in MR 1.6 can be improved, as hinted in class: if |P|<\eps then there is a constant C (depending linearly on (|K|^2)_max Length(X) where K is the curvature of X) such that |Length(P)-Length(X)|=0. For a simple (i.e. injective) curve, try to argue the converse. Can you find a non-simple (thus non-convex) plane curve with k>0. (Recall that a curve is convex if it lies to one side of its tangent line; the ideas in MR exercise 1(8) might be helpful.) [Hint: you might do the converse by first showing there is a pair of points p1, p2 on X with a common tangent line, but where the arc of X between p1, p2 is not a segment; then use the previous problem to show (under the assumption of X being simple) the arc has zero total curvature (show the "convexified" curve, gotten by substituting the segment for the arc also has total curvature 2\pi); thus the curvature changes sign on such an arc (contrapositive argument).] 28 Jan For a curve X in R^2 with periodic curvature function, if the total curvature (integral over a full period) is an integer multiple of 2\pi then T is periodic. What additional conditions guarantee X is periodic, i.e. for X to be a closed curve? Frenet-Serret frame (for points on a C^3 curve X in R^3 with |K|=k>0: T, N=K/|K|, B=TxN. For plane curve, B=normal to plane. In general, B(s) normal to osculating plane span(T(s),N(s)). Assume k>0 below. Work out osculating circle at a point. (Second derivatives agree). As usual, taking derivative of inner-products and using product-rule gives B'(s)=t(s)N(s) for some function t, the torsion. Compute the torsion of a (r,q)-helix: t=-q/(r^2+q^2)=constant. Discuss limiting cases. (Later, we will prove the converse: if curvature k=constant and torsion t=constant then X is a helix; see 1.37, 1(26&27).) Note (unfortunate, but MR and many other authors) sign convention: left-handed helix has positive torsion! #Express torsion in terms of X', X" and X"' for an arbitrary C^3 parametrized curve X(u) in R^3 (hint: B = X'x X"/|X' x X"|; now differentiate and find the normal part; the numerator is triple product - explaining computationally why we need C^3 curves for Frenet.) #Compute the torsion of the twisted cubic at 0. #Show that any two C^3 pbal curves X and X* in R^3 "busculate" (agree through third derivatives at X(0)=X*(0)) iff they have the same tangent T(0)=T*(0) and curvature K(0)=K*(0) vectors, and the same torsion t(0)=t*(0). In particular, there is a busculating twisted cubic (or helix, generalizing the osculating circle) at any point of a C^3 curve. [Careful: if only the curvature functions agree (k(0)=k*(0)>0) then the two curves agree only after a rotation in the normal plane about the common tangent direction; and what will happen if k(0)=k*(0)=0 - do we need to exclude this case?!] 02 Feb Frenet formula can be written as columns of 3x3 rotation matrix: [T|N|B]'=[T|N|B]A where A is a transform (note signs) of the skew symmetric matrix with curvature k and torsion t given in class: |0 -k 0| A = |k 0 t| |0 -t 0| The same form holds for any curve of orthonormal 3-frames [E1|E2|E3] where = 1 or 0 iff i=j or not. This is from applying product rule to these inner product equations: '=+ so A_ji = = - = -A_ij. #The above can also be expressed: there is a vector field W(s) along the curve X so that [E1|E2|E3]' = W(s)x[E1|E2|E3] - cf. 1(26); the field W(s) is called the angular velocity; if [E1|E2|E3]=[T|N|B], W(s)=?, and show W(s) constant iff X is helix - cf. 1(27) & below. #The unit tangent vector T(s) of a curve X(s) pbal in R^3 traces out a curve on unit sphere S^2 called the tangent indicatrix. Note that T(s) is not pbal (what's its speed?)! The curve X(s) has total (absolute) curvature \int k(s)ds = \int|K(s)|ds = \int|T'(s)|ds = Length(T). [Fill in the details, and as a challenge, try to prove the following result:] Note that, if X is closed curve in R^3, then T closed on S^2; but not all closed curves on S^2 are tangent indicatrices, whose center of mass (weighted by arclength of X, not T) must be at origin because \int T(s)ds = \int X'(s)ds = 0. This can be used to prove a theorem of W. Fenchel (1940): The total curvature of a closed curve X in R^3 (or in R^n) is at least 2\pi, with equality iff X lies in a plane as a convex curve. (For knotted X, ... > 4\pi - Fary & Milnor.) [This might make a good expository project. A good research project might be to explore how the Frenet frame - viewed as a curve in SO(3) or its double cover S^3 - geometry compares with that of the original curve in R^3 - we might say more about this, especially a formula relating total torsion (or Twist) to the geometry and topology of closed curve - cf. Writhe and Self-Linking Number.] #Exercise 1.37 [Hint: We already computed k and t are constant for a helix; for the converse, we looked at A(s):= X"(s) = kN(s), so (by Frenet formulas) A'(s) = -k^2T-ktB -> |A'|^2 = k^2(k^2+t^2); A"(s) = -k(k^2+t^2)N -> k_A = |A"(s)|/|A'(s)|^2 = 1/k; and also A"'= (some multiple you can compute)A', so triple product and torsion t_A must vanish; thus A is planar curve with curvature k_A=1/k lying on sphere of radius k -> A (great) circle of radius k. Thus can write | cos \sqrt{k^2+t^2}s | X"(s) = A(s) = k| sin \sqrt{k^2+t^2}s | | 0 | which integrates (twice) to give a helix X(s) = (??? in k and t); should get r = k/(k^2+t^2) and q^2=t^2/(k^2+t^2) in old notation.] Back to plane curves: the series of MR exercises 1(9-17) deal with special properties of plane curves Y obtained by perturbing a plane curve X. One nice result (not among these exercises) is a theorem of Archimedes: If X is a convex curve and Y is any closed curve surrounding X then Length(Y)>=Length(X), with equality iff Y = (a reparametrization of) X. This generalizes the fact that straight lines are shortest paths between points (think about the limiting case where X is a back-and- forth line segement). [This topic might be part of a final project.] 04 Feb Parallel (Bishop) frames, holonomy, comparison with Frenet frame: the latter is globally defined as long as k>0, but how to frame a general curve in R^3? Go back to plane curves, but now thought of as curves in R^3 with constant binormal B (so B'=0) and torsion zero (N'=-kT); in particular, the derivatives of normal fields N,B are multiples of T. We call any normal field V (along a pbal curve X(s) in R^3 with unit tangent vector field T=X'(s)) _parallel_ provided V'(s)=f(s)T(s) for some function f. Check (compute ') that a parallel field has constant length. We often express parallel by V'(s)^\perp=0. Let N1(0), N2(0) be an orthonormal basis for normal space of X at X(0). By solving initial value problem V'(s)^\perp=0 we get a parallel normal framing N1(s), N2(s) along X. To compare with Frenet, write N(s)=cos\phi(s)N1(s)+sin\phi(s)N2(s) for some twist angle \phi(s) for [T,N,B] versus [T,N1,N2]; and so B(s)=-sin\phi(s)N1(s)+cos\phi(s)N2(s). Show torsion t(s)=-\phi'(s). In other words, \int |t(s)| is the total twisting of the Frenet frame with respect to a parallel frame. #Let N1, N2 be parallel orthonormal frame of normal fields along X in R^3; the orthonormal frame M1(s)=cos\a(s)N1(s)+sin\a(S)N2(s), M2(s)=-sin\a(s)N1(s)+cos\a(s)N2(s) is parallel iff \a(s)=constant. Now for the bad news: if we follow a parallel field V around a closed curve X in R^3, say with X(0)=X(1), we may not have V(0)=V(1), but instead V(1) will be rotated in the normal plane from V(0). This rotation (or the rotation angle \a) is the _holonomy_ (of parallel translation along X). #Find an explicit closed curve with an abritrary holonomy angle \a. Can you characterize those curves with holonomy \a=0? [Research?] Part 2 Surfaces: parameterized and implicit; embedded (or immersed) in R^3 and R^n. The first and second fundamental forms. Length and area. Mean and Gauss curvature. 09 Feb #A surface patch X:U->R^n for U\subsetR^2 has X homeomoprhism to its image and (if X is C^1) satisfies rank(DX)=2, i.e. X_u, X_v linearly independent; for n=3, this means X_u x X_v nonvanishing, so we can normalize and define the (C^0) Gauss map (unit normal) \nu:U->S^2 by \nu = X_u x X_v/|X_u x X_v|; clearly \nu(u,v) is perpendicular to the tangent plane span(X_u(u,v), X_v(u,v)) to the surface patch at X(u,v); as we did for unit tangent vector for curves, check that if X is C^k then \nu is C^{k-1}. #Find a "cusplike surface patch" X which is C^1 but has rank(DX)=1 at a point (in other words, X_u and X_v are parallel but not both zero); similarly, find a "conelike surface patch" where rank(DX)=0 at a point (i.e. both X_u and X_v must vanish). [Neither of these is a (regular, immersed) surface patch, but sometimes it's good to expand one's point of view. See the torus problem below. Hints: The map X:R^2 -> R^2: (u,v) -> (u^2,v) is C^1 (in fact, smooth) but rank(DX)=1 at (0,0). This isn't the example, but should give some ideas. More directly: How would you parameterize a cone near its vertex? How would you pametrize a surface of revolution made by revolving a circle (other regular smooth curve) that's tangent to the axis of revolution (that point of tangency becomes the cusp)?] The sphere S^2 is covered by a "quilt" (atlas) of 6 surface patches X^N, X^S, X^W, X^E, X^F, X^B which map an open unit disk onto the various open hemispheres (X^N to the northern hemisphere, ... , X^B to the back hemisphere). Each of these patches is graphical over the tangent plane over the correponding "pole" (we have the north, south, west, east, front & back poles, where the standard axes of R^3 meet S^2). 11 Feb #Use sterographic projection to quilt S^2 with 2 patches (this is the minimum number, since S^2 is not region of R^2). [Hint: recall that stereographic projection from a pole P takes the punctured sphere S^2\setminus{P} to R^2, and its inverse gives a patch X^P.] #Find patches covering a torus T^2=S^1xS^1, say, of revolution in R^3 which is a special case of a tube X^\eps around a closed curve: if X*:S^1->R^3 is C^2 and 0<\eps<1/k_max(X*)=maximum curvature of X* then X(u,v)=X*(u)+\eps(cos(v)N1+sin(v)N2), where [N1,N2] is a normal framing (like [N,B] for the Frenet frame) of X*, is a doubly periodic map R^2->R^3 (and thus a map from the torus). The special case where X* is a circle of radius r>0 is worth checking explicitly: X_u and X_v are linearly independent when r>\eps, and when r=\eps there is one "cusp" point where rank[X_u|X_v]=1. What's the least number of patches needed for a torus? [Hint: this will depend on whether the patches are disk-like or annular!] 16 Feb [No class on account of Lincoln and Washington ;-}] 18 Feb Veronese embedding of S^2/{+,-} into R^6 (or R^5 or R^4). Boy's surface in R^3 (see: http://en.wikipedia.org/wiki/Boy's_surface, or http://www.mfo.de/general/boy/, or - for a more familiar face - http://owpdb.mfo.de/detail?photo_id=12131)! #The notion of a fundamental domain ("patch") for a surface. The upper hemisphere is one for the real projective plane; a square in R^2 is one for the torus. Find fundamental domains in R^2 (later we will see that most of these are more naturally viewed as being endowed with hyperbolic rather than euclidean geometry) for other surfaces. #Sketch "movies" of slices of surfaces in space such as the the sphere (in various embedded configurations), the torus and Klein bottle (begun in class), the real projective plane (see websites above) immersed as a Boy's surface, etc. 23 Feb Classifying surfaces of finite or infinite topology (notion of genus for orientable surfaces, number of ends, nonorientability, bands and Mobius bands). #A surface S with boundary \dS has patches modeled on open sets of the (closed) half-plane \R x [0,\infty), with points on \dS coming from points on \d(\R x [0,\infty))=\R x {0}. Sketch pictures of orientable surfaces with boundary for small genus g, ends e, and boundary components b (may want to distinguish bounded components of \dS from unbounded ones). Similarly for non-orientable ones. #Make a series of sketches to indicate why the various examples of 1-ended surfaces with infinite genus (depicted in class) are all diffeomorphic to each other. 25 Feb We indicated how to form the connected sum S # S' for two surfaces (or surfaces with boundary) S and S': remove an open disk from the interior of S, an open disk from the interior of S', and attach these (smoothly) along the common boundary circle. If S and S' are both orientable, orientations can be matched to make the connected sum S # S' orientable. If S or S' is non-orientable, so is S # S'. We can also define (with caveats!) a boundary sum operation in case S and S' each have nonempty boundary by smoothly attaching an arc of \dS to an arc of \dS': the resulting surface is denoted S #_\d S' though some care needs to be taken in case \dS or \dS' have several components (of circle or line (unbounded) type). Both of these operations connected sum operations are commutative up to diffeomorphism, giving the set of surfaces (up to diffeomorphism) the structure of an abelian semi-group under # (or under #_\d with the above caveats!) whose neutral element represented by S^2 (or the by disk D^2 under #_\d). This leads to a decomposition into "prime" pieces. For closed bounded surfaces (without boundary) this semi-group is generated by the torus T=S^1xS^1 (orientable case) and projective plane P=\RP^2 (non-oruentable case) subject to the sole relation 3P=P#P#P=P#T. And if we also include the operations of deleting a point (puncturing, possibly on the boundary) and of deleting an open disk (or a point from boundary), then all finite topology surfaces (with boundary) can be generated in this way. #Check how genus g, number of ends e, number of boundary compenents b=(b_bounded,b_unbounded) behave under each connected sum operation for orientable surfaces. Note that in this case, the genus g is simply the number of T's in the prime decomposition (no P's). For non-orientable surfaces, the number of T's can be taken as 0, and the resulting number p of P's is sometimes referred to as the non- orientable genus (though one must be careful when saying "genus p" out loud ;-). How do g and p behave under # and #_\d? #Trousers means a surface diffeomorphic to a 3-punctured sphere. Show that (except for the sphere S^2, plane \R^2, cylinder S^1 x \R and torus T^2) any orientable, finite topology surface (without boundary) can be decomposed into (how many?) trousers by deleting a finite number (how many?) of simple closed curves. 02 Mar We talked a bit about transversality: if surfaces S and S' meet such that every point p in the intersection has the property that their respective tangent spaces T_pS+T_pS'=R^3, then S intersects S' in a regular curve. This is the foundation for the "movies" we discussed a while back, where S'=S_t is a family of level surfaces of a R-valued function F on R^3 (with non-vanishing gradient), such as the height F(x)= along the direction a: #The surfaces S_t above meet a fixed surface S transversally iff t is a regular value of the restriction f of F to S. Typically, the critical values t are where the level sets of this function f, such as the height function, have a qualitative change. If there are no critical values between a pair of regular values t1 and t2, then the intersections of S_t are equivalent for all t in [t1,t2] (this is the "fundamental lemma of Morse Theory" and is behind the M\"obius classification of surfaces we sketched last week). [End of a topological interlude...] [... finally: back to geometry!] 04 Mar We worked out the derivative D\nu of the Gauss map S -\nu-> S^2 for the case S is the cylinder parametrized by (a fundamental domain in) R^2 -X-> S by the formula (u,v) -> (cos u, sin u, v) and showed how to think of this as a 2-by-2 matrix and as a quadratic form (the second fundamental form) which records the normal curvature in a given direction W in T_pS, i.e., the curvature of the (plane) curve S intresected with the plane spanned by W and \nu(p). #Do the same calculations in case S is i) a plane, ii) a sphere of radius r, iii) the graph (u,v, h(u,v)) of a function h, especially for a quadratic function h(u,v) = au^2 + bv^2 at (0, 0, 0). 09 Mar We worked out (upward) normal for a graph, introducing tangential gradient operator T(h)=\grad(h)/\sqrt{1+|\grad(h)|^2} where \grad(h)=(h_u, h_v). Note that T(h) is the first two components of \nu (up to sign). Showed shape operator A is symmetric since the associated 2nd fundamental form B = = - is symmetric, where Hess(X) is vector of Hessians (second derivative forms) of the coordinate components of X. [Some of you already computed B for the sphere, and - thanks to the confusion I may have caused above - conflated it with the shape operator A, which on the sphere is 1/r times the identity, since the sphere is totally umbilic (with principal curvature 1/r). The clarification needed is discussed below! Although in class, I pointed out that the second fundamental form B(v,w) = can be computed by taking B = -, just a little care needs to be taken when converting this into the shape operator A. Indeed, if we let b denote the matrix B(X_u,X_u) B(X_u,X_v) B(X_v,X_u) B(X_v,X_v) then A = g^{-1}b where g^{-1} is the inverse of the matrix g for the "first fundamental form" Note that when X_u, X_v is an *orthonormal basis*, this confusion goes away, since then g is just the identity matrix. In fact, the trace of a quadratic form B (as we need for defining H, or rather 2H) is always computed this way: either sum B(U,U) + B(V,V), evaluated on any orthonormal basis {U,V}, or take the trace of the associated matrix A=g^{-1}b. I hope this clears things up - it's so fundamental! ;-] #Compute the shape operator and 2nd fundamental form (again) for a graph (u,v, h(u,v)) of h(u,v), expressing in terms of Hess(h) 11 Mar Reviewed some features of moving frames along a curve, clarified angular velocity of such a frame. Compared holonomy of Bishop (parallel) frame (in the normal bundle to a curve) from other notions of holonomy (e.g. in the unit tangent bundle to a surface). #And, now that we are dealing also with surfaces, we discussed the Darboux frame [T|\eta|\nu] adapted to a curve c on a surface S: here T is the unit tangent vector to c, and rotating T by 1/4-turn in the tangent plane gives \eta, i.e. T \cross \eta = \nu (that's our sign or orientation convention). Express the infinitesimal rotation [T|\eta|\nu]'=[T|\eta|\nu]A=W\cross[T|\eta|\nu] of a Darboux frame along c using the shape operator of S. We introduced ideas from integral geometry (Buffon needle problem), especially the "Crofton principle" that the Length(c) of a curve c is proportional to the average number of intersections with standard geometric objects (such as lines in R^2 or great circles in S^2). For curves on S^2 we computed the Crofton Formula proportionality constant by looking at the "test" case where c is a great circle and discovered (in honor of the upcoming Pi Day?!) that it's \pi. We then applied this to complete a key lemma in a proof of Fenchel's theorem about total absolute curvature (from 2 Feb - like Groundhog Day, we keep repeating this ;-), namely that if a curve on S^2 is not contained in any hemisphere it has length at least 2\pi. #Use these ideas to work out Crofton's Formula for a curve c in R^2, equating Length(c) = (univeral constant)(average number of times c intersects a straight line in R^2); in other words, find (univeral constant) for this case. [Hint: it also involves \pi. Bigger hint: use a unit length line segment as your "test" curve.] For fun, check out my friend Bob Palais' article "Pi is Wrong": http://www.math.utah.edu/~palais/pi.html 16 Mar Spring Break (work on your research project! :-) 18 Mar Spring Break (stop procrastinating on the project! ;-) 23 Mar We continued the study of the second fundamental form and proved: Theorem. A surface S is totally umbilic (k1=k2) iff S part of a plane or sphere. #Check that the following are equivalent ways to say S is umbilic at p: 1) k1(p)=k2(p), 2) H^2(p)=K(p), 3) the shape operator A at p is a multiple of the identity linear map from T_pS to itself. #Find the umbilic points on an ellipsoid ax^2+by^2+cz^2=1. You may first want to consider the case of an ellipsoid of revolution (where, say a=b) using symmetry (the general case may be too hard to do explicitly). Sketch the curves whose tangent vectors are aligned with the principal directions associated with the larger and smaller principal curvatures. [Hint: it may help to recall that the unit normal for a level surface {f(x,y,z)=0} is given by \nu = Df/|Df| viewed as a column vector, and then work out A terms of Hess(f) - the formulas you get should specialize to the earlier case of a graph {z=h(x,y)} by taking f(x,y,z)=z-h(x,y).) #Does a torus of revolution have any umbilic points? Again, sketch the curves whose tangent vectors are aligned with the principal directions for the larger and smaller principal curvatures. #One of the steps in proving the above Theorem was showing both principal curvatures are constant fuctions on S. Can you find any non-umbilc surfaces with this property? Can you characterize all such S? 25 Mar We argued: if a surface S has a mirror symmetry plane P, then the intersection curve S \cap P is a principal curve. As any surface of revolution is a union of such curves (congruent to the "generating curve"), the "orbits" of the revolution form the orthogonal family of principal curves. This determines the shape operator of S (the principal directions and curvatures): one principal curvature is (up to sign) the curvature of the plane curve S \cap P; the other (for a surface of revolution) is the curvature of orbit projected to the surface normal (with the new notation c" for the curvature vector of the orbit curve c, using our sign convention: -=-(sin\phi)/r, where \phi is the angle between the axis of revolution and \nu, and where r is the distance to the axis). #In fact, for any pbal space curve c(s) along a surface S we get the normal curvature of S in direction c'(0) at c(0) by computing -. We defined an "asymptotic" direction for S as one with normal curvature 0 and observed (by previous fact) that a ruled surface has asymptotic direction along the rulings (the straight lines which comprise the ruled surface). #Check this for examples of ruled surfaces such as {z=xy} that you can write down explictly. #Check that the Gauss curvature K=det(A) is non-positive at p iff there is an asymptotic direction. More generally, there are 0, 1 or 2 (independent) asymptotic directions if K>0 (resp. =0, <0). #Check the the mean curvature H=trace(A)/2=0 at p iff the (pair of) asymptotic directions at p are orthogonal (so the above "if" is not "iff"). We also remarked about where we expect umbilics to appear on an ellipsoid and began to think globally: associate an index in Z[1/2] to each umbilic, these add to the Euler number (2 for an ellipsoid or any topological sphere), and so forth (how about higher genus?)! 30 Mar We discussed parallel surfaces S^t = S + t\nu (parametrized by X^t(u,v) = X(u,v) + t\nu(u,v) if you like, and more nicely if we take (u,v) to be coordinates along principal curves, possible away from umbilics), and verifed the "area form" dArea^t = X^t_u x X^t_v dudv in these coordinates is (1+2Ht+Kt^2)dArea, so that 2H is the "first variation" of Area while K is "second"! #Redo the above calculation without assuming nice coordinates and try to reach similar conclusions. What about the "first fundamental form" g^t (given by the inner-products , etc. - see note on 09 Mar for more on this), as well as the second fundamental form B^t and shape operator A^t for for the parallel surfaces? 01 Apr [No foolin'! ;-] We discussed the notion of R^3-valued, normal and tangent vector fields on a surface S in R^3, and began to discuss the divergence operators DIV and div. #Fill in details of my claim that the divergence theorem for DIV (in general, not just applied to the coordinate vector fields) is equivalent to the statement that the (outer, unit) normal field integrates to zero for any S=\d\Omega, where \Omega is a (smooth) bounded domain in R^3 and S is its boundary surface. #Consider the shell domain between two surfaces S^t1, S^t2 in any family of parallel surfaces S^t. The unit normals to S^t define a vector field N on this shell domain. Compute DIV N along S^t in terms of the mean curvature of S^t. [Hint: this can be done directly, or by applying the divergence theorem to N and using the area formula from last time.] #Try proving the isoperimetric inequality via the divergence theorem [April Fools - or not ;-]! 06 Apr We continued clarifying some issue raised in the 09 March problems about A and B (when will we get to K? ;-)! We (re)discovered that if g is the matrix for <.,.> with respect to a basis, and if a is the matrix for A in this basis, and b for B= is symmetric then a is "g-symmetric" in the sense that a^t=gag^{-1}. Of course, when g commutes with a (as when we take an orthonormal basis with respect to <.,.>) then a is symmetric in the usual sense. We also defined DIV V and div V as the trace and partial trace of DV for any R^3-valued vector field V on R^3 or a surface S in R^3, respectively. #Show that our formula for DIV V and div V are independent of the choices of orthonrmal frames used to define the traces. 08 Apr We gave a couple arguments why div V = div V^T + 2H, and looked at the consequences for a minimal (H=0) surface in R^3: 1) coordinate functions are harmonic [take V=e_i, so V^T=e_i^T= grad x_i, for i=1,2,3] and 2) the conormal vector \eta integrates to the zero vector for any closed curve c which bounds a domain on a minimal surface (this goes back to Archimedes in the guise of force-balancing, since surface tension in a soap film exerts a constanr force per unit length on the boundary curve c in the conormal direction). #Suppose instead we have a region R of a *constant* mean curvature surface bounded by a curve c. Now argue that the integral of \eta around c must equal (perhaps with the constant factor -2H) the integral of the unit normal \nu over R. This is a again force balancing, except now there is both surface tension (the \eta integral) force and pressure (the \nu integral) force to consider. 13 Apr We showed that force balancing in a cmc surface can be viewed as a "conservation law": homologous curves C and C' have the same force, defined as the integral of \eta over C (the tension force) minus 2H times the integral of \nu over any cap K spanning C (the pressure force). We also used this idea to get a first order ODE for cmc surfaces of revolution. #Analyze this first order ODE: check (by implicit differentiation) that it is equivalent to the second order ODE for the radius from the axis of a cmc surface of revolution; verify that for H=0 the function r = cosh t solves the equation (its graph is a catenary, revolving to the minimal catenoid); what happens as you vary the magnitude m of the force parameter, in case H=0 and for general cmc H? Part 3: Differential forms and the intrinsic geometry of surfaces 15 Apr Here are some problems about exterior algebra and differential forms that were posed to the manifolds class a few years ago. If you go to the bottom, you'll get to the problem I began to outline in class: ----------------------------------------------------------------------- Let V be an n-dimensional vector space over R. Compute the dimension of the exterior algebra V. [Hint: if {e1, e2, ... , en} is a basis for V, show that {1, e1, e2, ... , en, e1e2, ... , en-1en, ... , e1e2e3...en} is a basis for the exterior algebra. ------------------------------------------------------------------------ From the above problem, you should have found the dimension of the degree k vectors (or k-vectors, for short) in the exterior algebra of V to be n \choose k = n!/k!(n-k)! Since n \choose k = n \choose n-k, there is an isomorphism between k- and (n-k)-vectors. Here's a way to define such an isomorphism, "Hodge duality": if {e1, e2, ... , en} is a basis for V, then the n-vectors e1e2...en and -e1e2...en are ORIENTATIONS for V; pick an orientation (say, the + one); then any k-vector is paired with its Hodge dual (n-k)-vector according to this rule for basis vectors: *(ei1ei2...eik) := (+/-) e1...ei1^...eik^...en where the v^ means v is omitted, and where the (+/-) is chosen to make *(x)x a nonnegative multiple of the (+) orientation chosen, i.e. *(ei1ei2...eik)ei1ei2...eik = e1e2...en. Show that, on k-vectors, *^2 = (+/-)Id - how does the sign depend on k? In case V = R^2 and k=1, *^2 = -Id, and * is "rotation by 90 degrees". In case V = R^4 and k=2, *^2 = Id; compute the (+/-)1 eigenspaces for *, i.e. the self- and anti-self-dual 2-vectors on R^4. ------------------------------------------------------------------------ If L:V -> V is a linear map, then L induces a linear map on the exterior algebra. In particular, show that the induced map on the 1-dimensional space of n-vectors is multiplication by det(L). ------------------------------------------------------------------------ If V* is the dual space of V, i.e. the space of all linear maps from V to R, then we can define exterior algebra of V* as well. Show that the k-vectors on V* can be regarded as skew-symmetric k-linear maps from V to R (or k-forms, for short). Show that L as above induces a linear map on k-forms - what is this map? ----------------------------------------------------------------------- We have seen how to define Hodge duality for smooth forms: extend it to commute with multiplication by smooth functions. For example, let \theta = (x dy - y dx)/(x^2 + y^2) be a 1-form on R^2 \ {0}; then *dx = -dy and *dy = dx extend to *\theta = (x dx + y dy)/(x^2 + y^2) = dr/r = d(ln r) where r^2 = x^2 + y^2. Compute d\theta and d(*theta). Interpret \theta in polar coordinates. Do these forms remind you of anything from complex analysis?! ----------------------------------------------------------------------- If we write dz = dx +i dy and d\barz = dx - i dy, then they span the complexified (tensor with C) cotangent space of R^2. The complexified tangent space is spanned by their duals \d_z and \dbar_z. Write \d_z and \dbar_z in terms of \d_x and \d_y. Show that the Cauchy-Riemann equations for f = f1 + i f2 can be expressed \dbar_z f = 0. Also, express the exterior derivative operator d in terms of the complex operators \d and \dbar defined by \df = \d_z f dz and \dbar f = \dbar_z d\barz and conversely. ----------------------------------------------------------------------- The differential operators grad, curl and div from advanced calculus have a nice interpretation in terms of exterior derivative d and Hodge duality *, as follows. Let I, J and K be the standard constant vector fields on R^3. Then any vectorfield W = aI + bJ + cK on R^3 is determined by smooth functions a, b and c. We may identify W with a 1-form \omega = a dx + b dy + c dz via W = #\omega or \omega = &W (here # or "sharp" is what Marcel Berger calls a "musical isomorphism" - its inverse is called "flat": we'll use & for that so &# = #& = Id). Considering the spaces of forms on R^3, the operator d, and Hodge *: <---*---> \Omega^0 -d-> \Omega^1 -d-> \Omega^2 -d-> \Omega^3 <---------------*-----------------> please check the following (up to sign): grad f = #df curl W = #(*d(&W)) div W = *(d*(&W)) and in particular, since d^2 = 0, we have the calculus facts curl(grad f) = 0 and div(curl W) = 0. ----------------------------------------------------------------------- Note that the operator div above is what I've been calling DIV in earlier problems here - it's the R^3 divergence. If you can do this last problem, you've understood much of what one needs to know about differential forms - so please try it.... 20 Apr Tangential projection of D as covariant differentiation (connection) on a surface S in R^3; notion of parallel transport. Connection 1-form (so(2)-valued) \om for an orthonormal (co)frame e1,e2 (\th1, \th2) on S. First structure equations d\th1=\om\wedge\th2 and d\th2=-\om\wedge\th1 depend only on intrinsic information (the metric g=\th1^2+\th2^2) about S. Compute d\om for basic examples: (extrinsic) S^2 in R^3 (get d\om=-\th1\wedge\th2) and hyperbolic plane (upper-half plane with metric g=(dx^2+dy^2)/y^2 where we get d\om=\th1\wedge\th2). Second structure equation give curvature K as ratio of 2-forms: d\om = -K \th1\wedge\th2 = -K da where \th1\wedge\th2 can also be viewed the area form. (Note: even if S is non-orientable, this gives a well-defined K!) Check that this K agrees with determinant of shape operator for S in R^3. This is Gauss's *Theorem Egregium*! Also check that if the o.n.frame is adtaped to a curve c in S (say e1=T is tangent along c) then \om(T)=k_g the geodesic curvature of c in S, which extrinsically is just component of acceleration for (pbal) c tangent to S (see the Darboux frame problem above). A metric on S can always be given as g = e^2h (du^2 + dv^2) where e^2h is the "conforrmal factor" for some function h=h(u,v). Check that g is a flat metric (K=0) iff h is harmonic (h_uu+h_vv=0). More generally, find K in terms of h. 22 Apr What happens if we integrate the second structure equation (using Stokes's Theorem)? We get an even more remarkable theorem (really the same as the Theorem Egregium, rather than "more egregium" as MR suggest, or - a double entendre - *more* of the same ;-): \int_\dR k_g ds \int_\dR \om = \iint_R d\om = -\iint_R K da the Gauss-Bonnet Theorem! (Here we've adapted the (co)frame to the boundary curve \dR.) But we need to be careful, since this doesn't work for a disk in R^2 (k_g=1 and K=0) or hemisphere in S^2 (k_g=0 and K=1)! We've made the same mistake Chern made when he first discovered this (intrinsic) proof: we forgot that the adapted (co)frame may have a singularity in R! When we account for this, we see that for a disk-like region R on S, we have instead: \int_\dR k_g ds + \iint_R Kda = 2\pi which agrees with the disk examples above. More careful accounting (breaking R into geodesic polygons and counting turning angles at vertices as atomic contributions to k_g - this gives both the edge the vertex counts E and V) replaces r.h.s. with 2\pi(F-E+V) where F is the number of faces, i.e. with the Euler number 2\pi\chi(R). For a closed surface S (compact, with empty boundary \dS) in R^3, this says the degree of the Gauss map is \chi(S)/2 in case S is orientable. In case S is nonorientable, the Gauss map goes to RP^2 and we need to be a bit more careful about degree. There's also an nice (extrinsic) proof using Morse Theory of height functions and the change-of-variable formula for the Gauss map to express (1/2\pi times) the total curvature as the (signed) number of critical points averaged over the sphere of directions. ============================================================== Here's the list of final Research Project topics (I took some liberties) and the tentative schedule for the presentations - NOTE THAT WE MEET AT 11 BOTH DAYS (Tuesday in LGRT 119, Thursday TBA): Drew: Knots, Links and their Geometry [Th 10:45] Zhon: Clairut's Theorem for Geodesics [Tu 12:00] Neil: Surfaces as Branched Covers [Th 12:15] Bill: Knots, Links and Biological Applications [Th 11:00] Hugh: Minimal Surfaces [Tu 11:00] Sebastian: Connections and Parallel Transport [Tu 11:15] Keith: Connections and Physics [Tu 11:30] John: Stokes's Theorem for Differential Forms [Th 11:30] Mike: Unicycle Tire Track Geometry [Tu 12:30] Reiter: Morse Theory and Gauss-Bonnet [Th 12:00] Nicky: The Degree of a Smooth Map [Th 11:45] Rich: The Shape of Soap Bubbles [Th 12:30] Derek: Inverse/Implicit Function Theorem [Th 11:15] (possibly 2 more after 12:45 on Thursday) ============================================================== 27 Apr Research Project Presentations (bring lunch :-) Tuesday 11:00 Hugh 11:15 Sebastian 11:30 Keith 11:45 [break] 12:00 Zhon 12:15 Mike 29 Apr More Research Project Presentations (*bring more lunch ;-) 10:45 Drew 11:00 Bill 11:15 Derek 11:30 John 11:45 Nicky 12:00 Reiter 12:15 Neil 12:30 Rich 12:45 [*we are signed up for the cookout at Goodell till 2] 04 May (Get started on your take-home final exam, which is basically a careful - and carefully referenced! - expanded write-up of your presentations, sent as .pdf file to me at robkusner@gmail.com - this is due 14 May! And after that, have a great summer!!! :-) ----------------------------------------------------------- *"Copyleft" means that you have permission to use this material for non-commercial purposes as long as you acknowledge its source. Any use, in whole or in part, must include the "Copyleft 2010 by Rob Kusner" message or its equivalent. Commercial use without prior approval of the copyleft holder is strictly forbidden, and is punishable by methods not even imagined by John Yoo!