These are notes and problems for Rob Kusner's Spring 2021 honors course on Differential Geometry of Curves & Surfaces (Math 563H at UMassAmherst, Copyleft* 2010-21 by Rob Kusner), and are always under revision. [Rob may eventually prepare some .tex/.pdf Problem Sets, and revise a .tex/.pdf version of notes taken by his S'13 TA, Dugan Hammock]. Homework is assigned and due weekly, but is retrospective by one week; in practice, this means the problems marked ## are to be written up and submitted as .pdf to Rob and our TA by 11:59pm on the Saturday 10-12 days after the date noted – e.g. the first hw is due 13 Feb and consists of the ## problems from the first week of class. Intro Geometry.... Topology.... And the Universe!!!!!!!! 01 Feb Introduce ourselves. Review course logistics. Imagine how models for the physical universe lead to the interplay the between local geometry (curvature) and global topology (which, for a surface can be measured by its Euler number \chi=F-E+V, where F=#faces, E=#edges, V=#vertices in any cell-decomposition). ##Compute the Euler numbers of CUBE, TETRAHEDRON, SPHERE, TORUS and PRETZEL surfaces with g>1 handles. See how subdviding may change F, E or V, but not change \chi. Explore other ways to compute \chi, like cutting out a pair of disjoint disks (reducing F by 2) and gluing their boundary circles together (increasing g by 1); this only gives surfaces with even \chi ≤ 2; what about odd \chi?! Discuss surfaces that minimize "bending energy" (plastic panels and metal plates) or some other geometric or physical quantity like area or "stretching energy" (possibly with an enclosed-volume constraint - soap films and bubbles minimize this). Future: how might one define stretching or bending energy of a curve in the plane or space (think of Hooke's law from mechanics)? 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. 03 Feb ##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. (Draw pictures, projecting to the various coordinate planes. How smooth is it, and how do you reconcile this with the pictures?) ##How smooth is f:R -> R where f(t)=e^{-1/t} for t>0 and 0 otherwise? ##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 onto one of these (find it) with smooth inverse; and show it's enough to find such a map with nonvanishing derivative (implicit/inverse function theorem in dimension 1). ##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? ##Explicity parametrize the path of an "abrupt right turn" (the union of the negative x2-axis and the positive x1-axis, with the "turn" at the origin X=(x1,x2)=(0,0)) so that it's a C^k-smooth parametrization. How about a C^\infty parametrization? (Compare with the a=0 case – the "cusp" – of the "twisted" cubic.) 08 Feb Closed (periodic) curves X:R —> R^n with X(t)=X(t+1) can be viewed as curves defined on the (abstract) circle S^1 = [0,1]/0~1. Discuss: the length of a polygon (sum of edge lengths) & inscribed polygons for a C^0 curve (index increases with parameter). Define length of a C^0 curve as supremum of lengths of inscribed polygons. If the supremum of the lengths over all such inscribed polygons exists, we say X is "rectifiable" an define its length as this sup. MR 1.6 & 1.10 show that a C^1 curve X is rectifiable and that its sup length equals integral of its speed |X'| over the domain of X. Or... Define length of C^1 curve X as the (Riemann) integral of |X'|. ##Tedious exercise: Show these definitions agree for C^1 curves, first by estimating how close the length of an inscribed polygon approximates the corresponding Riemann sum for the integral of the speed |X'|, and then by estimating how well Riemann sum approximates this integral (your estimate may be easier and more explicit if you assume X is C^2 with an explicit bound on the acceleration |X"|; see MR sections 1.2 and 1.3, especially MR items 1.10, 1.11). [Note: good students (JMN20) dropped by to ask about this exercise above, which seems tricky and not-easy, as well as tedious! After some discussion and false starts, we decided that a good strategy would be to show: (A) Given any inscribed polygon \P, we can find a polygon \P_N with vertices at X(k/N) (for k=0,1,...,N) such that ($) length(\P_N)≥length(\P)-\eps(N), where \eps(N)—>0 as N—>\infty (the idea is that the vertices of \P are well-approximated by a subset of vertices for \P_N, which works even for a C^0 curve X); since (why?!) length(\P_N)≤sup(length(\P))=length(X([0,1])), and assuming it's finite, ($) implies length(\P_N)—>length(X([0,1])). (B) For any k, the intermediate value theoremm for each component x_i' of X' yields a t_i in [k/N,(k+1)/N] with ($$) x_i'(t_i)/N= x_i((k+1)/N)-x_i(k/N) (here we need X is C^1); then argue (again use X is C^1, or for a more explicit estimate, assume X is C^2 so that x'_i oscillates≤c/N over [k/N,(k+1)/N] where c=max_[0,1]|X"|) if we move all the t_i to an endpoint of this interval, say t=k/N, we can estimate ||X((k+1)/N)-X(k/N)|-|X'(k/N)|/N|≤c_n/N^2 (we use ($$) n times since X(t) in R^n, so c_n depends on c & n); now sum from k=0 to N-1 to see that the Nth Riemann sum is within c_n/N of length(\P_N).... Another (AW20) student suggested in class to use the intermediate value th'm, but one must use care since it's NOT TRUE that we can find t in [a,b] so that (b-a)X'(t)=X(b)-X(a).) ?!?!?!?MAYBE ONE OF YOU CAN FIND AN EVEN BETTER ARGUMENT?!?!?!?] ##Easy exercise: compute the length L of the cusp X:[0,1] —> R^2 where X(t)=(t^3,t^2); more generally, compute its length function s(t) := length(X([0,t])) and its derivative s'(t).... ##Tricky exercise: ... you should see s'(t)>0 for t in (0,1]; and since s=s(t) maps [0,1] onto [0,L], the inverse function theorem shows there's an inverse function t=t(s); can you explicitly find t(s) and the arclength reparametrization Y:[0,L]—>R^2, Y(s)=X(t(s)) for the cusp? (Extra credit: How smooth is t(s) at s=0?) Discuss the Koch snowflake K. It's a limit of polygonal loops K_n, where K_0 is an equilateral triangle of sidelegth 1, and where K_{n+1} is obtained from K_n by replacing the middle third of each edge of K_n by two more edges of an "outward" equilateral triangle. Observe that K is not rectifiable - not even locally! [For a Koch animation: en.wikipedia.org/wiki/Self-similarity] Quick discussion of measuring length(K) by covering K with balls of a given radius and counting how many are needed as this radius tends to 0; this number grows like 1/radius if K is a rectifiable curve, and otherwise leads to notion of Hausdorff measure & dim - a possible project topic! (Try this for the Koch snowflake K.) 10 Feb Oriented curve is a parametrized curve modulo orientation-preserving diffeomorphism (reparametrization) of the domain intervals. Length of a C^1 curve is independent of C^1 reparametrization (chain rule and change of variables formula for interval). Same is true for a rectifiable C^0 curve and C^0 reparametrization (homeomorphism of the domain intervals). ##Check the second statement above [it requires no calculation, just the fact that the ordering of vertices of an inscribed polygon is preserved (or reversed) if the homeomorphism is orientation-preserving (or reversing) — and the definition of length as the sup...]. Proposition: Immersed C^1 curve X(t) has unit speed parametrization Y(s)=X(t(s) with C^0 unit tangent vector T, unique up to orientation and choice of starting point t(s=0)=0. Proof sketch: "Solve" the differential equation s'(t)=|X'(t)| [just integrate the given function |X'(t)|]. Apply inverse function th'm [since s'(t)>0] to get t(s). Then compute Y'(s) [via chain rule]. [Discussion with a student suggests a simpler way to think about "up to orientatation..." above: if S(t) is another arclength parameter for the curve, then S'(t)=±s'(t); so writing S(t(s))=S(s) via the inv fcn thm, the chain rule gives S'(s)=±1, and thus S=±s+S(0).] Corollary: The "cusp" fails to be immersed at (0,0) with repsect to any parametrization (because T changes sign there). ##Check the details of the above three things! 15 Feb Finish discussion of length (at long last ;-)! ##[Optional challenge question] Decide what extra conditions on a C^k curve X(t) make the arclength function s(t) also C^k. [It's true for an immersion, since |X'(t)|=s'(t)>0, and IFT can be applied. A student (AW20) warned about an s(t) with s'(t)=|t| so that s(t) is C^1 but not C^2; another (SS20) noted the curve X(t)=(t^2/2,0,...,0) gives such an example....] ##Work out the u.s.p (unit speed parametrization) for the helix X(t)=(r cost, r sint, qt). The unit tangent vector defines a curve T:I -> S^{n-1} for curves X:I -> R^n. Can recover X (up to translation) from T, again by integration. If T is periodic, X may not close, but have a "period" (the "translation" or "constant vector of integration"). ##Check that helix above is an example. What's its period vector? Give other examples of curves X in R^2 and in R^n with periodic T but nonvanishing periods. Helices (and their limits, circles and lines) are important curves in R^3 because they are orbits of Euclidean motions (1-parameter subgroups of the Euclidean group E(3)). Discuss in class how E(3) is a group generated by reflections in planes, and how to represent via certain 4x4 matrices with [0 0 0 1] in the bottom row, with the rotation/reflection part is 3\times3 block above the 3 zeros, and translation part A=(a1, a2, a3) is first 3 rows of the 4th column. 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). ##Use a u.s.p. for helix above to compute its curvature vector K(s) and discuss various limits as r or q approach 0. [We saw in class that |K| is a constant depending on r and q; if both approach 0, their relative rate is important to see how |K| behaves.] ##For any C^2 curve X(u) in R^n, and any C^2 reparametrization u(t), verify that the reparametrized curve Z(t)=X(u(t)) satisfies Z"(t) = u"(t) X'(u) + (u'(t))^2 X"(u). Therefore, if u=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 (at, bt^2, ct^3); use this idea to compute K(0) in terms of a, b, c. ##Express the curvature vector K for an arbitrarily parametrized curve X(t) in R^n in terms of X' and X" (it involves projection of X" to the hyperplane normal to X'). Try the case of curves in R^3 first, then specialize to the R^2 case. (In R^3, the cross product can be used to express projection of a vector W to the plane perpendicular to nonzero vector V via proj_{V^\perp}(W) = (V \timesW \times V)/|V|^2.) 17 Feb All of the above means that the curvature is "geometric" - i.e. K is independent of the parametrization. How does K change under rotation or translation in space? Under scaling? Observe |K| has "dimension" of 1/Length. The osculating circle \O_p at a point p=X(0) on a C^2 curve X(s) makes this "geometric" idea concrete: \O_p is the "best fitting" circle at p, meaning it passes through p=X(0) with same unit tangent vector T=X'(0) and curvature vector K=X"(0). Let \L_p=\{p+tT:t\in\R\} be the tangent line at p, and \H_p=\{\L_p+uK:u≥0\} the halfspace containing \L_p & K; then \O_p is the circle of radius 1/|K| centered at p+K/|K|^2 in \H_p (note that K=0 iff \O_p=\L_p=\H_p). ##Work out the osculating circle at any point on the helix, and also at the point (0,0,0) of the twisted cubic. [Later we noted these are tilted, and that the \H_p slices the cylinder in an ellipse, whose osculating circle in \H_ must agree with \O_p for the helix. Even later we mentioned a century-old theorem of A. Kneser (the osculating circles of a plane curve with monotone signed curvature "foliate" a neighborhood of the plane — in particular, \O_p is disjoint from \O_q if p≠q), and wondered if this could be used to prove something about the collection of osculating circles on helix using Cici's nice view (bird's-eye or worm's-eye, depending on your taste ;-) — yet another great topic to explore if you're more "research-oriented"!] If the curve is less smooth, osculating circles can be defined using triples of distict points on the curve: any such triple defines a circle or line in \R^n, and as triple of points converge to a point p on the curve, any limit circle (or line) is an osculating circle at p. FROM NOW ON (unless stated otherwise), WE ASSUME A CURVE IS IMMERSED! ##Describe the set of osculating circles at the point (0,0) of the C^{1,1} curve X(t)=(t,t|t|)=(t,±t^2) if ±t≥0. [Hint: there are two "extreme" osculating circles (that are limits of the unique osculating circles at X(t) for t<0 or t>0, where X is C^2) and more "inbetween"!] 22 Feb A C^1 curve X(s) is a C^1 graph over its tangent line \L_p at p=X(0) for some small interval (s_-,s_+) around s=0 where T(s)•T(0)>0; if X is C^2 (or C^{1,1}) with a curvature bound |K(s)=T'(s)=X"(s)|≤c, we argued (using the geometry of the unit sphere) that T(s)•T(0)>0 is true for at least |s|<π/2c. ##Compute the curvature vector when X(x) is expressed as a normal graph over its tangent line \L_p at p=X(0). [Hint: without loss of generality (by translation and rotation), assume X(x)=(x,x_2(x),...,x_n(x))=(x,y(x)) with y(0)=y'(0)=0 in \L_p^{perp}=\R^{n-1}, so p=X(0)=0, T=X'(0)=e_1, and \L_p is the x_1-axis; x=x_1 is an arclength parameter along axis/tangent line, but not necessarily a u.s.p. for the curve X(x); the unit tangent vector T(x)=X'(x)/|X'(x)|=(1,y'(x))/\sqrt{1+|y'(x)|^2}, so (by the chain rule and the fact that dx/ds=1/(ds/dx)=1/\sqrt{1+|y'(x)|^2}) K(x)=dT/ds=T'(x)dx/ds=((1,y'(x))/\sqrt{1+|y'(x)|^2})'/\sqrt{1+|y'(x)|^2}) =(0,y"(x))/{1+|y'(x)|^2}-(1,y'(x))y'(x)•y"(x)/{1+|y'(x)|^2}^2 which you should check, simplify, note K(x)•T(x)=0, and observe that two terms in your more general formula (n>2) will cancel when n=2, giving a formula you should recognize from calculus in case of a graphical curve X(x)=(x,y(x)) in R^2 (as y(x) takes values in R^{n-1}=R^{2-1}=R^1=R).] Schur's Lemma sharpens the normal graph estimate above: a curve of a given length L≤π with |K|≤1 must have endpoints at distance at least 2\sin(L/2) with equality iff the curve is a great circle arc. [This might be part of a good topic for a project!] 24 Feb [Well-being Wednesday — if folks want to discuss the above....] 01 Mar 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 c.c. by a quarter-turn.) Thus we express the curvature vector K=kN for a function k(s)=K(s)•N(s) [some may call this \kappa, but I'll use lower case here], the signed curvature of a plane curve. Unlike the curvature vector K(s), the sign of k(s) flips when the orientation of the parametrization flips (making the signs of T(s) and N(s) flip)! We saw that the total curvature \int k(s) of a closed curve in R^2 is a multiple (namely, 2π) of its turning number w, the number of times the unit tangent vector T winds around the circle S^1. Note that the sign of w flips if the orientation of the curve flips. 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? ##Draw pictures of (oriented) plane curves illustrating the various values of w (among small integers, including w=0, which is the most interesting case). 03 Mar ##Which plane curves have k=0? Which have constant k≠0? 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)=\theta'(s) to get \theta(s) (up to a constant) and T(s)=(\cos\theta(s),\sin\theta(s)) (up to rotation); and X(s) can be recovered (up to translation) by integrating T(s)=X'(s).] This can also be done using the moving frame for X(s) in R^2, namely the 2\times2 matrix U(s)=[T(s)|N(s)] as follows: ##Verify the 2x2 matrix ODE for the frame U'(s)=Jk(s)U(s), where J is the usual c.c. quarter-turn matrix. Solve this first order ODE. [Hint: you should get U(s)=exp(J \int k(s) ds)U(0), integrating over the interval [0,s]). Recall the exponential of any square (2\times2 here) matrix A is the usual power series exp(A)=I+A+...+A^k/k!+....] Remark: in class we said U'(s)=U(s)Jk(s), and this is also correct (why?)! [Hint: J commutes with U(s) – this is special for n=2!] In fact, since we want to rewrite the derivatives of frame vectors as linear combinations of the frame vectors, we really want to use *right* multiplication which acts on columns (not left multiplication, which acts on rows). Back to periods: we saw above that we have k(s)=\theta'(s) where T(s)=\cos\theta(s)e_1+\sin\theta(s)e_2 is unit tangent vector; we can recover T (up to initial angle \theta(0)) from integrating k(s), and recover X (upto translation by initial position X(0)) from integrating T. ##If X(s) is a closed curve – or even a periodic curve with nonzero translation period – then the tangent T(s) curve on S^1 in R^2 and the signed curvature function k(s) are also periodic; what about the various converses (please give proofs and examples)? 08 Mar "Two wrongs don't make a right, but three lefts do!" [Some above material, and much remaining material, will be further revised as we do our best to make this an on-line course: over the break I set up a home studio in order to make short YouTube videos (so far about 30 URLs have been shared "privately" with 563S20 students) under the "Coronavirus University" (TM and copyleft 2021 by Rob Kusner ;-) and a 563videos.html page with links to these videos is now under development (Mar 2021).] As noted last week, in the context of modern highways and the turning number w (which could also stand for Whitney, who proved a very nice theorem about turning numbers and topology of immersed curves in R^2 — another great project topic!), this slogan is almost true.... Since you folks asked for another break, here are the turning number of plane curves videos I've posted at YouTube to memorialize my views on these ideas: Total curvature & turning number of plane curves: https://youtu.be/UYp7xVHAImU [~17 min] The "easy half" of the Whitney-Graustein Theorem: https://youtu.be/6-X2tnrCXHQ [~11 min] The next 4 are on the "hard half" of Whitney-Graustein: https://youtu.be/Y1QZROune-s [draft* intro <2 minutes] https://youtu.be/keCi79TRC1U [main part ~19 minutes)] https://youtu.be/p3KTl7cGzQc [supplement <6 minutes] https://youtu.be/bG9Zpn4LpZQ [supplement^2 ~6 minutes] 10 Mar In the first video, you were asked to compare the (oft-fractional) turning numbers (or total curvature \int k(s) ds) for various ways to make a "right turn" on a full "cloverleaf" highway intersection: ##Please do that, and also decide what other turning numbers would correspond to the other possible "turns" (including the "no turn"). How would this change if driving in the UK instead of the US (this should help you intuit the idea of "orientation" for both a curve and a surface)? If X is closed curve in R^2, we saw the possible values for total curvature \int k(s) ds mut be integer multiples of 2π: ##What about the possible values for a simple (injective from S^1) closed curve? (Try the case of a convex plane curve first, then "dissect" the region bounded by a simple curve into convex pieces.) [A possible way to do this is to approximate X with a circumscribed simple polygon: First show, for fine enough approximation, the total curvature of X between adjacent points of tangency is simply the turning angle between adjacent endges of this polygon. Next, we need a way to account for the sums of these turning angles. That can be done by triangulating the interior of the polygon, then clipping away triangles from the boundary, checking that the total turning angle doesn't change. In the end we have a single triangle, with total turning angle 2\pi (if oriented counter-clockwise).] The unit tangent T(s) of u.s.p. curve X(s) in R^n (n>2) traces a curve on unit sphere S^{n-1} (the tangent indicatrix). Note that T(s) is not u.s.p (what's its speed?)! The curve X(s) has total (absolute) curvature \int|K(s)|ds = \int|T'(s)|ds = length(T). For n=2 we have the (signed) curvature k(s)=K(s)•N(s)=T'(s)•JT(s) so |\int k(s)ds| ≤\int |k(s)|ds = \int|K(s)|ds = length(T) since there will be length cancellations as T(s) sweeps back (clockwise, k<0) and forth (counterclockwise, k>0) around the "direction" S^1: this means \int k(s)ds measures the "signed" length for the tangent indicatrix as T(s) wraps around S^1. Another way to think of this involves signed-length(T), which we called "happiness" in class: if T(s) points "east" (toward e1) and k(s)>0 ("smiling") then this contributes positively to signed-length(T); if k(s)<0 ("frowning") it contributes negatively to signed-length(T); and (by the wrapping- around-S^1 picture, or the change of variable formula) we see that length(T)=2π(average over e \in S^1 number of X(s) with T(s)=e) and signed-length(T)=2π(signed-average of this number). ##Fill in the details of the above, and as a challenge, try to prove the following result (for "π day" we may do this in class next week): 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 of 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 (below).] Let's finish class with a pre-π-day experiment using chopsticks and floorboards – please try this yourselves at home on π-day this Sunday and we'll pool the results next week to compute π (and develop a bit of the theory behind this too)! ======Watch my "Coronavirus University" videos as they appear on YouTube====== Mar 15 It's the Ides of March, so please get started on thinking about and choosing a topic for your final projects! ==================reviewing our experiments from last week=================== [Something I started back in 2010 to help celebrate Pi(π)Day] 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). [Some of my videos are about this appear at the end of Part I.] For curves on S^2 we compute the Crofton Formula proportionality constant by looking at the "test" case where c is a great circle and discover (in honor of the upcoming Pi Day?!) that it's \pi. We then apply this to complete a key lemma in a proof of Fenchel's theorem about total absolute curvature: 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.] And for fun, check out my friend Bob Palais's article "Pi is Wrong": http://www.math.utah.edu/~palais/pi.html ============================================================================== [Sadly, we don't have Spring Break this year :-((((] ======Watch my "Coronavirus University" videos as they appear on YouTube====== A k-frame in R^n is an ordered basis for k-dimensional subpace of R^n (think of an n\times k matrix, whose columns are the basis vectors); the k-frame defines an orientation of the subspace (in case k=n this is the sign of the determinant, but for k R^2 has k≥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).] You might also try 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 closed 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.] ======Watch my "Coronavirus University" videos as they appear on YouTube====== 17 Mar Frenet-Serret frame along C^3 curve X in R^3 with |K|=k>0 defined by: 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. 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, you 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's 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. I would have picked the opposite sign!) ##Compute the torsion of the twisted cubic at 0. ##Show that any two C^3 u.s.p curves X, 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) and first derivative of curvature k'(0)=k*'(0) [thanks to M. Feller (2012) for pointing out this last necessary (and sufficient) condition]. In particular, at least at critical point for curvature (k'(0)=0) [and perhaps more generally?], there is a busculating twisted cubic (or helix, generalizing the osculating parabola or circle) at any point of a C^3 curve. [Careful: if only the curvature functions and first derivatives agree (k(0)=k*(0)>0, k'(0)=k*'(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?!] 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 weird sign convention would be fixed if a right-helix had 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. Another way to see this uses the matrix transpose: if U=[E1|E2|E3] is a path in SO(3) and U* is its transpose, then differentiating the identity U*U=I and using U*'=U'* gives U'*U+U*U'=0, that is the matrix U*U'=-U'*U=-(U*U')* is skew symmetric. [Recall (XY)*=Y*X* and Z**=Z!] ##The above can also be expressed: there is a vector field W(s) along the curve X so that [E1|E2|E3]'=W(s)\times[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 vector W(s) is often named after Darboux....) More on helices' curvatures and torsions; limiting cases of circles and lines. Pitch p=q/r dimensionless; sign of q gives handedness. Helices are orbits of screw motions (the generic 1-parameter group of Euclidean symmetries). ##Exercise 1.37 [Hint: We already computed k and t are constant for a helix; for the converse, one looks at K(s):= X"(s) = kN(s), so (by Frenet formulas) K'(s) = -k^2T-ktB => |K'|^2 = k^2(k^2+t^2); K"(s) = -k(k^2+t^2)N => k_K = |K"(s)|/|K'(s)|^2 = 1/k; and also K"'=(some multiple you can compute)K', so triple product and torsion t_K must vanish; thus K is planar curve with curvature k_K=1/k lying on sphere of radius k => K (great) circle of radius k. Thus can write | cos \sqrt{k^2+t^2}s | X"(s) = K(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.] ======Watch my "Coronavirus University" videos as they appear on YouTube====== 22 Mar 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 u.s.p 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. ##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?] Parallel (or Bishop) frame rotates about the binormal B - if we think of osculating plane span{T,N} as the "equatorial" plane, then N1 and N2 are tangent to "parallels" ("latitudes" or "poles") - in fact, we know Ni'=f_iT, so these point in same direction as T (i.e. both Ni' are normal to the same "longitude" as well). [This is my attempt to say in words what I tried to draw on the board in class today! ;-] Compare now 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) ds is the total (signed) twist of the Frenet frame with respect to any parallel frame. ======Watch my "Coronavirus University" videos as they appear on YouTube====== 24 Mar Wrap-up on curves and their global properties. We already mentioned the Fenchel-Fary-Milnor theorem: a closed curve in R^3 has total curvature >= 2\pi, with equality only for convex planar curves, and knotted curves have total curvature > 4\pi (the latter result already brings us to the "edge" of surface theory - sorry for the pun - since one way to characterize unknotted closed curves is to say they are the boundaries of embedded 2-disks. This result dates to the 1940s, though as recently as 2 decades ago there was new work on this by Eckholm-BWhite-Wienholz (soap films...) and Cantarella-Kuperberg- Kusner[yours truly ;-]-Sullivan (nonempty second hulls...). What about total torsion? This is the theorem of Calugareanu-JWhite- Fuller: Self-link = Twist + Writhe. What do these mean? Again, most are best understood using surfaces, but we can say a bit about each: Twist = (1/2\pi)\int t ds is the total torsion (a single integral), Writhe = (1/4\pi)\iint /|X(s)-X(r)|^3 ds dr is the double integral of this (normalized) triple-product, and Self-link = the linking number Lk(X,X+\epsN) where X is the original curve and X+\epsN is a "push-off" of X by small \eps>0 along Frenet normal N to a nearby, disjoint embedded closed curve (so this depends on the choice of framing: we've made the standard assumption k>0 to ensure the Frenet normal framing N,B is defined globally) and Lk(X,Y) is defined as the number of times X crosses over Y in a right-handed way minus the number of left-handed crossings (here we assume X and Y are oriented simple closed curve, and when Y=X+\epsN we orient it the same way we oriented X; we also assume a planar projection in which we can depict the under- and over-crossings). ##Sketch some links X \union Y with Lk(X,Y) small. ##Check that Lk(X,Y) doesn't change as we deform X and Y so that they remain disjoint and embedded. In particular, by flipping the plane of projection, we get Lk(X,Y)=Lk(Y,X). (Gauss worked out a double- integral formula for Lk(X,Y) which is essentially the same as that for Writhe, but with Y(r), Y'(r) in place of X(r) and X'(r).) ============Homework Hiatus: *no* homework due this Sat 27 Mar!!!============= To memorialize more of my thoughts on (space) curves and frames (and to give you something to do during – or after – a Homework Hiatus ;-) videos I've posted at YouTube: https://youtu.be/EdoABRZpC6U [Pi Day+Pi(e) 2:39] https://youtu.be/rErnMJWhCag [Crofton Principle 19:37] https://youtu.be/j4vE88PzLhg [Fenchel-Fary-Milnor 24:49] https://youtu.be/hnODmkwg1uE [Normal Framings... I 15:23] https://youtu.be/wQxryEXBW6E [Normal Framings... II 13:08] https://youtu.be/HVTmjF9WBRE [Normal Framings... III 29:20] And some Al Fresco "very-shorts" on Twist + Writhe = Self Link: https://youtu.be/5Xy1j9is-fQ [T+W I: Hose 0:37] https://youtu.be/qrzEDJNl-zQ [T+W II: Hose Theory 2:08] https://youtu.be/z-43u9jyt8I [T+W III: Wire Theory 5:40] https://youtu.be/i8kanwUsaAU [T+W IV: Wire 1:26] https://youtu.be/zVuUqvzRut0 [T+W V: Coil 1.59] ======Watch my "Coronavirus University" videos as they appear on YouTube====== 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. 29 Mar Just as we focused on immersed curves, the notion of a surface being regular (or immersed) is very important, and I want you to explore it in the following material: A surface patch X:U->R^n for U\subset R^2 is a homeomoprhism to its image and (if X is C^1) satisfies rank(DX)=2, i.e. X_u, X_v linearly independent. For general n this means some 2\times2 minor in the n\times2 matrix DX has nonzero determinant; for n=3, this means the cross product X_u\times X_v is nonvanishing, so we can normalize and define the (C^0) tangent map (Gauss map, unit normal) \nu:U->S^2 by \nu = X_u\times X_v/|X_u\times 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)?] ##Find patches covering a torus T^2=S^1\times S^1 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!] Recall, for a C^1 surface patch X:U–>R^n, its derivative matrix DX matrix has 2\times 2 minors M_ij indexed by pairs of rows, giving rise to (n choose 2) determinants µ_ij. The (ordered) set µ of these generalizes the cross product X_u \times X_v, and when n=3 (µ_23, µ_31, µ_12)/|"| gives the unit normal \nu. ##Work out µ for the torus S^1\timesS^1 in R^4=C^2 defined by X(u,v):= (cos u, sin u, cos v, sin v) = (e^2πiu, e^2πiv). [Here we "cheat" and use the universal covering space R^2 of the torus, rather than a quilt of patches.] ======Watch my "Coronavirus University" videos as they appear on YouTube====== 31 Mar 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). ##What's the fewest hemispherical patches needed to cover S^2? (This is a discrete geometry problem: clearly neither 1 nor 2 are enough; are 3?) If two coordinate patches overlap on the immersed surface, they rise to a change-of-coordinates between domains in R^2. This is the foundation for the idea of "intrinsic" surface (or manifold). [For more on this, watch my "Idea of a Riemann Surface" videos.] ##Use stereographic projection to quilt S^2 with 2 patches X^N and X^S (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. (In class we point out that this works for S^n and that everything follows from the n=1 picture. By using similar triangles, we found the formula for the projection; for example, we have X^N:R^1->S^1:t->(2t,t^2-1)/(t^2+1) - this rational map parametrizes rational points on S^1 a.k.a. "Pythogorean triples"!) For n>1, think of t as a vector, e.g. t=(u,v) for a surface.... What you need to check is that both X = X^N and X^S are immersions.] In the lecture we began computing X_u\times X_v — you should finish that and check that it is a multiple of X itself!] Here's a good reason for us to understand stereographic projection better - it lets us go back and forth between R^n and S^n to study the "bending" or Willmore energy W for surfaces: Veronese embedding of S^2/{+,-} into R^6 (or R^5 or S^4 or R^4 via stereographic projection) uses quadratic functions of linear coords on S^2 (see !Newsflash! on my webpage for its importance). Also 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)! ============================================================================ For more on Boy's surface, see this on the arXiv: ============================================================================ arXiv:1303.6448 Date: Tue, 26 Mar 2013 12:05:10 GMT (2195kb) Title: Make your Boy surface Authors: Eiji Ogasa Categories: math.GT Comments: 14pages, 15 figures \\ This is an introductory article on the Boy surface. Boy found that RP2 can be immersed into R3 and published it 1901. (The image of) the immersion is called the Boy surface after Boy's discovery. We have created a way to construct the Boy surface by using a pair of scissors, a piece of paper, and a strip of scotch tape. In this article we introduce the way. \\ ( http://arxiv.org/abs/1303.6448 , 2195kb)\\ ============================================================================== Perhaps it could be part of a project?!? ============================================================================== ##Check that the composition of the patches {X^N,X^S}, {X^E,X^W} and {X^F, X^B}: D\subset R^2 -> R^3 for S^2 (u,v) —> (x,y,z) [in my YouTube lecture (u1,u2) —> (x1,x2,x3)] and the quadratic map R^3—>R^5 X=(x,y,z) —> Q=(q1=xy, q2=x^2-y^2, q3=yz, q4=y^2-z^2, q5=xz) is an immersion, and if we consider as a map on antipodal pairs in S^2, is actually 1-1. (We could also include a q6=z^2-x^2 in Q, so QX would lie in the 5-dimensional hyperplane {q2+q4+q6=0} of R^6.) [The above is one version of the "Veronese" surface, but clearly we can adjust coefficients and change the shape a bit. I may say more about these immersions later.] Finish discussing how to think of RP^2=S^2/{+,-} "intriniscally": it's built from 3 patches, each corresponding to the pairs of patches {X^N,X^S}, {X^E,X^W} and {X^F, X^B} for S^2. ##Fill in the details of the picture we made that RP^2\{N,S} is a Möbius band built from the first two patches {X^E,X^W},{X^F, X^B}, and thus the third patch {X^N,X^S} "attaches" a disk along the "border" of this band to yield all of RP^2. ============================================================================== [The stuff below is from an earlier version of my 563H course.... A potential source for project ideas (or future YouTube videos)!?!] ===================================\/\/\/\/=================================== We begin discussing classification of surfaces of finite or infinite topology, orientable or non-orientable. The former embed as closed subsets of R^3, while the latter have self-intersections. #Sketch examples of 1-ended surfaces of infinite genus embedded in R^3. Make a series of sketches to indicate why these various examples are diffeomorphic to each other. Classifying surfaces of finite or infinite topology (notion of genus for orientable surfaces, number of ends, nonorientability, bands and Mobius bands). ?#Sketch "movies" of slices of surfaces in space such as the sphere (in various embedded configurations), the torus (begun in class) and Klein bottle, the real projective plane (see websites above) immersed as a Boy's surface, etc. We indicate how to perform "surgery" on a surface S: remove a pair of disjoint open disks (i.e. S^0 x D^2) from the interior of S, and attach these (smoothly) along their boundary circles (or "sew in" an annulus D^1 x S^1). When S is connected and orientable, we see how this increases the genus by 1 (and the reverse process decreases the genus, eventually disconnecting the surface if carried out far enough). ?#Analogously, form the connected sum S#S' for two connected 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 (sum and boundary sum) 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-orientable 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. We discuss (using Stanford-red-sweatshirt double-cover) covering projections and spaces, as well as "deck transformations" on the universal cover of a surface. We also explain what "smooth" maps between surfaces means, in particular, the notion of diffeomorphic surfaces (we'll come back to smooth maps when we study the "Gauss" or normal map in more detail next month). We check that S^2 -> S^2/{+,-} = RP^2 is a covering projection and that Veronese surface "factors through" this. Also we see (with paper models) that an annular neighborhood of a great circle projects to a M\"obius band. The antipodal map x -> -x is the deck transformation, interchanging the two "decks" or "layers" which lie over a given neighborhood of a point [x~-x] in RP^2. We also introduce 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. But we saw there are many more f.d.s (puzzle pieces assebled by deck transformations - in case of the torus, translations by the subgroup Z^2 of R^2). ?#Find fundamental domains in R^2 for other surfaces (later we will see that most of these are more naturally viewed as being endowed with hyperbolic rather than euclidean geometry) and see if you can describe the deck transformations. We introduce (mod 2) homology of closed curves on a surface S: a and a' are homologous (a~a') if there is "subsurface" R of S with boundary \dR = a+a'. This led to a Z/2 vector space H_1(S,Z/2) or an abelian group (if we work over Z and keep track of orientations). There is also a Z/2-quadratic form on this group, given by counting intersection number (mod 2) of curves. (The subsurface R defining a homology can be "singular" i.e. not immersed or embedded.) #?Check (by drawing pictures) that for a genus g surface S_g (connected sum of g tori) we have dim(H_1(S_g,Z/2))=2g and that the intersection form has matrix (with respect to a basis a_1,...,a_g,b_1,...,b_g of curves where a_i meets b_i once and otherwise disjoint): | 0_g I_g | | I_g 0_g | (and, over Z, one of the I_g blocks has a minus sign - use the disk with bands picture of S_g \ disk introduced in class). #?Check (by drawing more pictures) that a nonorientable surface N_p (connected sum of p RP^2s) has dim(H_1(N_p,Z/2))=p and that the intersection form has matrix I_p (in a basis c_1,...,c_p of curves whose neighborhoods are M\"obius bands - use the disk with half- twisted bands picture of N_p \ disk to see c_i meets c_i once). We introduce the fundamental (Poincare) group \pi_1(S) of a surface S (with basepoint p): this the group generated by (oriented) loops starting and ending at p with the relation that loops a and a' are homotopic if the product (concatenation) a(a')* (in class used "bar" instead of "*" to reverse orientation of the curve) is boundary of an annulus A mapped into S (the mapping of A needn't be immersion...). We sketch (last time, for case of the torus) why \pi_1(S) acts as deck transformations on the universal covering space, translating one fundamental domain to another. We also sketch why H_1(S,Z) is the abelianization of \pi_1(S). ?#Check that \pi_1(S^2)=\pi_1(R^2) is trivial group, \pi_1(RP^2)=Z/2, and \pi_1(A)=\pi_1(M)=Z where A is annulus and M is M\"obius band. ?#Check \pi_1(S_g)=group generated by a_1,...,a_g,b_1,...,b_g with one relation [a_1,b_1]...[a_g,b_g]=1 (product of commutators, where [a,b]=aba*b* - used "bar" instead of * in class, to denote inverse); this is non-abelian if g>1. ?#Can you figure out \pi_1(N_p) for p>1 (it's also non-abelian; the case p=2 is Klein bottle, so may be useful to think of square with edges identfied by deck transformations to work out \pi_1(N_2))? [The preceding three problems could be a final project instead!!!] Deck transformations of torus are translations z -> z + b of C=R^2 for b in a lattice Z^2=Z+iZ (or more general parallelogram) of C. More general Euclidean motions are of the form z -> az + b, |a|=1. We sketch a synthetic picture of hyperbolic plane (as unit disk or upper-half-space \H in \C with "lines" as circles meeting boundary orthogonally) and asserted that fractional linear maps z->(az+b)/(cz+d) for a,b,c,d real, ad-bc=1 give hyperbolic motions of \H which take "lines" to "lines." In this way, the octogon fundamental domain for S_2 is realized in \H and the nonabelian group of deck transformations can be represented by 2x2 real matrices (SL(2,R) or PSL(2,R)=SL(2,R)/I~-I) as above. (More general representations into PSL(2,C) treated in the book "Indra's Pearls" by Mumford, Series and Wright lead to the study of surfaces as complex curves or "Riemann surfaces" - also a good project!!!) 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. We talk 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 Möbius classification of surfaces we sketched last week). ===================================/\/\/\/\=================================== [That concludes the topological interlude above: maybe we can find final project ideas, especially among the "unproblems" marked "?#"!] ============================================================================== ======Watch my "Coronavirus University" videos as they appear on YouTube====== 05 Apr [The notes above record a 2013 foray into the topology of surfaces: & they could guide a final project, but in class this term, I'll stick 07 Apr (mostly) to geometry! We touched a bit on these this week....] ======Watch my "Coronavirus University" videos as they appear on YouTube====== 12 Apr We explored how a "geometrically obvious" fact — that the unit normal \nu is (up to ± sign) is well-defined — also follows from the chain rule. And we observed that this meeans \nu can be viewed as a map to the unit sphere S^2 if our surface can be given a consistently oriented quilt of patches (or atlas of charts), and ±\nu always gives a map to \RP^2. The unit normal to the unit sphere or unit cylinder is easy to work out synthetically or via formulas, and you should use both methods for these: ##Work out the shape operator of a surface \S — the derivative D\nu of the Gauss map \nu: \S——> S^2 — for the case \S is the unit cylinder parametrized by (a fundamental domain in) R^2 -X-> \S by the formula (u,v) -> (cos u, sin u, v) and show 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_p\S, i.e., the curvature of the (plane) curve S intersected 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). ======Watch my "Coronavirus University" videos as they appear on YouTube====== 14 Apr Wellbeing Wednesday Redux (next week we meet MTW!!!) Here another suite of my sweet videos (on surfaces)!!!!! A pair of videos "Getting to (the) Surface(s)": https://youtu.be/Kb_Xc_Pve2Q [Surfaces I, 17:22] https://youtu.be/vS7LCFOqb8E [Surfaces II, 20:27] Another pair on embedded surfaces in R^n (n>3): https://youtu.be/oPNF-nMHRhU [Surfaces in R^n Veronese, 12:18] https://youtu.be/HS1sRQuRsOs [Surfaces in R^n Plücker, 13.52] A “triptych" on holomorphic (C-analytic) immersed surfaces in R^2=C and "the idea of a Riemann surface" as an intrinsic (abstract, ideal) surface that isn't immersed anywhere: https://youtu.be/l9m6J6FKd_U [Holomorphic immersions to C, 13:28] https://youtu.be/pH7Rqw_W4kI [Idea of a Riemann surface, 17:42] https://youtu.be/ekay2sM6otE [Review & examples, 13:29] An introduction to Covering Maps and Classification of Surfaces (“slides" from a public lecture I gave long ago while wearing several T-shirts and pulling off one layer at a time – you had be there... :-): https://youtu.be/7zVDn2Wwp-g [(Un)covering...Surfaces, 2:44] which is the "hyperbolic geometry bridge" which you can watch before AND after my (very) long version of the Covering Maps & Classification of Surfaces video: https://youtu.be/2wcq50y5VOE [(Under)covering...Surfaces, 32:51] After some major technical issues (failure of the device to store the recorded video for lack of memory), I re-made what had been among the nicest lectures so far (echos of Fermat complaining his marvelous proof couldn't be recorded for lack of margins?! ;-) in my Coronavirus U series: https://youtu.be/amI0D99ouH8 [A bit more on Classifying Surfaces, 22:09] Please forgive (but figure out how to correct!) all the glitches that arise in my reprise – alas, the first time is always better! Finally, I was once asked about a figure drawn on the whiteboard; the answer is my last Coronavirus University video: https://youtu.be/4UY7yUU5rYc [Glimpse of Gauss-Bonnet, 15:18] Hope you enjoy these as much as I enjoyed making them, – Rob ======Watch my "Coronavirus University" videos as they appear on YouTube====== 19 Apr The shape operator A is symmetric (with respect to the metric <,>) & since associated 2nd fundamental form B(.,.) = = - 20 Apr is symmetric (i.e. B(v,w)=B(w,v) for any tangent vectors v,w), where & Hess(X) is vector of Hessians (2nd derivative quadratic forms) of the 21 Apr coordinate components of X. [Some of you already computed B for the sphere of radius R - and thanks to some 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 clarifications needed are discussed below! Although in class, I pointed out that the second fundamental form B(v,w) = can be computed by taking B = -, a bit more care needs to be taken when converting this into the shape operator A. Indeed, if we let b denote the 2x2 matrix B(X_u,X_u) B(X_u,X_v) B(X_v,X_u) B(X_v,X_v) then the matrix for A (with respec to the ordered basis X_u, X_v for the tangent plane) is a = g^{-1}b where g^{-1} is the inverse of the matrix g for the "first fundamental form" in this ordered basis 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! ;-] Review some linear algebra of linear maps and their eigenstuff. Symmetric bilinear and quadratic forms. The main fact we need: if a 2x2 matrix r s A = s t is symmetric, then it has real eigenvalues and a corresponding basis of orthonormal eigenvectors. Eigenvalues k1, k2 (the principal curvatures in case A = a = b is the matrix for second fundamental form taken with respect to an <,>-orthonormal basis, i.e. g=I_2) are roots of the characteristic polynomial c_b(t) = t^2 - (r+t)t + (rt - s^2) whose disciminant (r+t)^2 - 4(rt - s^2) = (r-t)^2 + 4s^2 >=0 and so the roots are real. (When k1 \ne k2, corresponding eigenvectors u1, u2 must be orthogonal [consider 0=-=(k2-k1)], and when k1=k2 we have b is a multiple of I_2 and so any vector is an eigenvector....) ======Watch my "Coronavirus University" videos as they appear on YouTube====== For the rest of the semester, we'll de-focus on homework and lectures to re-focus on final projects – please watch your email for more information! ======Watch my "Coronavirus University" videos as they appear on YouTube====== =========================!!!being revised below!!!========================= =========================!!!being revised below!!!========================= [Extra remarks on Part 1 (stuff on the cutting room floor ;-)] Curvature and length. ?#Use Cauchy-Schwarz ≤ to show line segments minimize length. Try giving alternate proof via (iterated) triangle inequality [MR 1.9]. (Generalize to Archimedes Theorem about curves surrounding a convex curve?!) Discuss the possible domains I for "complete" and "incomplete" u.s.p. curves (this uses Cauchy sequences, i.e. {x_i} in I with the property |x_i - x_j| is arbitrarily small for i,j arbitraily large; completeness means {x_i} has a limit in I also). Any closed interval of R is complete, as is R itself, but a bounded (half) open interval is NOT complete. We agree I=S^1(L)=[0,L]/0~L (circle of arclength L) also qualifies as a complete domain curve. Are there any others (why or why not)? This is, in effect, The Classification of (connected) 1-Manifolds [see J. W. Milnor, Topology from the Differentiable Viewpoint]! Recall the definitions of length. For a C^2 curve, the estimate in MR 1.6 can be improved, as we hint in class: if |P|<\eps 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)| 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 show 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 intersected 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). Show shape operator A is symmetric (with respect to the metric <,>) since the associated 2nd fundamental form B = = - is symmetric (i.e. B(v,w)=B(w,v) for any tangent vectors v,w), 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 the matrix for A (with respec to the ordered basis X_u, X_v for the tangent plane) is a = g^{-1}b where g^{-1} is the inverse of the matrix g for the "first fundamental form" in this ordered basis 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! ;-] Review some linear algebra of linear maps and their eigenstuff. Symmetric bilinear and quadratic forms. The main fact we need: if a 2x2 matrix r s A = s t is symmetric, then it has real eigenvalues and a corresponding basis of orthonormal eigenvectors. Eigenvalues k1, k2 (the principal curvatures in case A = a = b is the matrix for second fundamental form taken with respect to an <,>-orthonormal basis, i.e. g=I_2) are roots of the characteristic polynomial c_b(t) = t^2 - (r+t)t + (rt - s^2) whose disciminant (r+t)^2 - 4(rt - s^2) = (r-t)^2 + 4s^2 >=0 and so the roots are real. (When k1 \ne k2, corresponding eigenvectors u1, u2 must be orthogonal [consider 0=-=(k2-k1)], and when k1=k2 we have b is a multiple of I_2 and so any vector is an eigenvector....) Apr If we keep trying to clarify this subtle distinction between A and B, when will we get to K? ;-) Another way to think of this: if g is the matrix for <.,.> with respect to a basis, if a is the matrix for A, and if the matrix b for B= is symmetric (b=b^t), then a is "g-symmetric" meaning 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 (a=a^t). Apr We work 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). #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) Apr We continue the study of the second fundamental form and prove: 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 is 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? Apr Flat surfaces (k1=0) include cylinders and cones on, and tangent ribbon of, space curves - are these all possibilities? We discuss a bit about ribbon/ruled surfaces in general (more later). Apr In the past we argued (using linear algebra): If a surface S has a mirror symmetry plane P, then the intersection curve S \cap P is a principal curve. This time we argue (using synthetic geometry) a stronger result: If a surface S is cut by a plane P, such that the surface (*) normal along the intersection curve S \cap P lies in P, then S \cap P is a principal curve. Any surface of revolution is a union of such curves (congruent to the "generating curve"), and 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 u.s.p. space curve c(s) along a surface S we get the normal curvature of S in direction c'(0) at c(0) by computing -. Apr ?#A generalization of (*) is a version of Joachimsthal's Theorem: If along S \cap P the unit normal of S make a constant angle with P, then S \cap P is a principal curve on S. In fact, this a special case of an even more general verion: The intersection S \cap S' is a principal curve on both S and S' iff they meet and constant angle and (at least) S \cap S' is a principal curve on at least one of S or S'. [As I noted in class, one must assume the intersection curve is principal on at least one of the surfaces, because there are (counter-)examples of surfaces S and S' - take a pair of helicoids with common axis but rotated about this axis - which meet at a constant angle but where the intersection curve is not principal - indeed, their axis is an "asymptotic" curve. Perhaps there's a further generalization of J's theorem which deals with that case?] Apr Review some features of moving frames along a curve, clarify angular velocity of such a frame. Compare 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 discuss 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. Apr We define an "asymptotic" direction for S as one with normal curvature 0 and observe (by previous fact) that a ruled surface has asymptotic direction along the rulings (the straight lines which comprise the ruled surface). For a "concrete" application of asymptotic directions and ruled surfaces: one of the first steps in finishing concrete "flat" work is to make it flat by "screeding" - rubbing a long straight board back and forth across its surface. This will produce a ruled surface in general, but not necessarily a planar surface (with shape operator A=0 everywhere). Is it enough to have two independent rulings through each point on a surface to get this? ?#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 remark 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?)! Apr We discuss 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 verify 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 ?? Mar for more on this), as well as the second fundamental form B^t and shape operator A^t for for the parallel surfaces? Apr In class we discuss the notion of the R^3-valued, normal and tangent vector fields on a surface S in R^3, and begin 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. Apr ?#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 We define 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 formulas for DIV V and div V are independent of the choices of orthonrmal frames used to define the traces. We give 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. We show 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? Very Late April Part 3: Differential forms and the intrinsic geometry of surfaces 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.... 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 |\th1| | 0 \om| |\th1| (or the matrix equation d| | = | |\wedge| |) |\th2| |-\om 0 | |\th2| depend only on intrinsic information (the metric g=\th1^2+\th2^2) about S. Compute d\om for basic examples: sphere (extrinsic) S^2 in R^3 (\th1=cosv du, \th2=du in usual longitude, latitude coord's) we 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 gives 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. 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. ===================================================================== May Geometry Research Project Presentations [10AM till 5PM, in LGRT ???? - liquid refreshment & pizza possible, and if the weather is nice, afterward we can all hike over to my place and build a roaring fire in the outdoor hearth to celebrate the end of the semester!] ===================================================================== Here's the list of final Research Project topics (where I took some liberties) and a tentative schedule for the presentations: ???day ???day 10:00 10:30 11:00 11:30 12:00 12:30 13:00 [Pizza delivery!?!] 14:00 14:30 15:00 15:30 16:00 16:30 [Possibly over two days?!] ***************************************************************************** ***************************************************************************** ***************************************************************************** NOTA BENE: Please bring a printed draft of your presentation/paper for our ???&???day sessions, and send the final paper (preferably in .tex or .doc format, but ALSO in .pdf) as an email attachment to my usual address: profkusner@gmail.com BEFORE 11:59PM ?????????????? (and please email there with any questions). ***************************************************************************** ***************************************************************************** ***************************************************************************** ----------------------------------------------------------- *"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 2010-21 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 (remember him?)!