CS195H Homework 2 The Whitney Graustein Theorem

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1 CS195H Homework 2 The Whitney Graustein Theorem Due: February 13th, 2014 This third homework is about (a) definitions, homotopy proofs, etc, (b) an interesting mesh that we ll be looking at for the remainder of the semester, and (c) the Whitney-Graustein Theorem. Some of it is to be done alone, but much of it is to be done in pairs. The code should be submitted by in the form of a zip file, with an informative name, such as jfh-pkz-02.zip, indicating that it represents work by jfh and pkz on homework 2 It should contain A README.txt document saying whose work this is, and clarifying anything unusual you did. A bunch of.m files, one for each procedure you wrote. The PDF for your written portion of the assignment, if you ve chosen to hand it in this way. Introductory material In class, we defined Ω(X, x 0 ) to be the set of all loops in the space X, based at the point x 0 ; a loop in X at x 0 is a continuous function γ : [0, 1] X such that γ(0) = γ(1) = x 0. We also defined an equivalence relation on loops ( is homotopic to ), written γ 1 γ 2, and defined π 1 (X, x 0 ) = Ω(X, x 0 )/. That defined π 1 as a set, but it happens to be a group, with the operation being concatenation of loops. We didn t actually finish proving that in detail, which is what this first section is about. Math Part Problem 1 [10 pts] We defined an operation,, from pairs of loops to loops, namely follow one loop after the other. Explicitly, if γ 0, γ 1 Ω(X, x 0 ) we defined (γ 0 γ 1 )(t) = { γ 0 (2t) 0 t 1 2 γ 1 (2t 1) 1 2 t 1 1

2 I claimed that was associative up to homotopy, i.e., that (γ 0 γ 1 ) γ 2 is homotopic to γ 0 (γ 1 γ 2 ), where by homotopic I mean homotopic as loops. You ll recall that the diagram for this proof is the one below, which may help guide you. Remember to prove that the homotopy you construct is in fact continuous. Problem 2 [10 pts] I also claimed that had an identity element ɛ, defined by ɛ : [0, 1] X : t x 0, and that (γ ɛ) γ for any loop γ Ω(X, x 0 ). (The constant loop ɛ is also a left-identity, but we won t bother proving that, since the proof s almost identical.) Prove this claim. The mnemonic diagram for the proof is shown below. Problem 3 [10 pts] The following picture describes an abstract polyhedron, X something that s algebraically like a polyhedron, because it s made up of faces and edges and vertices that meet the way that you expect: each edge has two distinct vertices, each face has three, each edge is shared by two faces, and so on. The one tricky thing is that in this picture, some edges appear more than once. That s because the polyhedron described by this abstraction can t be put simply in the plane (or in 3-space). That s OK. We can still think about it and reason about it just fine. To make things clear (I hope), I ve taken the double edges and marked directions on them, so that you know what s glued to what else, and in which direction. (The outer edge of the yellow face, for instance, is glued to the outer edge of the orange face. More correctly: The yellow face has vertices 0, 2, and 3; the orange face has vertices 0, 2, and 4, so they share the edge 02.) Look at the shape, and get to know it. You re going to try to compute the fundamental group π 1 (X, x 0 ). 2

3 Notice that X has exactly 6 vertices. How many faces and edges does it have I ve marked one vertex in red to act as x 0 for this problem, and I ve drawn a dark red curve that s continuous in the polyhedron, but not in the plane. (You should feel free to pick a different point as x 0 for your computations if it makes things easier for you.) The dark red curve starts at vertex 3, goes up through the beige triangle, onto the pale purple one near the 0 vertex, back out of it near the middle of edge 01, into the beige triangle, then the dark green, light blue, dark blue, and back to vertex 3. That s a typical element of Ω(X, x 0 ). Your challenge is to try to come up with a description of the fundamental group of X. To do so, you may assume that every curve can be deformed into a grid curve one that runs along the edges only, and can be described by a sequence of edges and that all homotopies can be thought of as grid-homotopies (which include insertion/deletion of hairpins, and replacing any one side of a triangle by the other two, and vice versa). With that in mind, try to describe one representative of each nontrivial class of loops, and explain why every curve is homotopic either to one of your representatives or to the constant loop. Once you ve done that, you also need to describe the group structure: what happens when you combine your representative loops with one another using? Try to prove that your answers are correct. Problem 4 [10 pts] Recall that the turning number for a regular 1 path γ : S 1 R 2 is defined by constructing the map γ : S 1 S 1 : t γ (t) γ (t) and observing that this map has some degree (i.e., winding number around the origin). That degree is called the turning number. An analogous definition holds for a polygonal curve, at least one for which there are no length-zero edges, and for which there are no reflex edges edges whose direction is the opposite of the direction of the preceding or following edge. (We ll call such a polygon nice.) For a nice polygon, there s an exterior angle (positive or negative, strictly between π and π) defined at every vertex (see the figure below), and the turning number is the sum of the exterior angles, divided by 2π. 1 A path γ is regular if γ (t) > 0 for all t. 3

4 Suppose a nice polygon has n edges and turning number k. What s the largest possible value for k in terms of n? Can a polygon with turning number 0 have fewer than four vertices? Explain. Computing Part Introductory material You ll recall the proof of the Whitney-Graustein theorem from class. The theorem said that if two constant-nonzero-speed curves had the same turning number, they were in fact regularly homotopic. ( Regularly homotopic implies same turning number is relatively easy.) There was a little preliminar discussion to show that any nonzero-speed curve was regularly homotopic to one that s constant speed, so this really means that any two nonzero-speed curves with the same turning number are regularly homotopic. The proof went like this: First, suppose that both curves start at the origin (if not, translate both until they do; that s a regular homotopy). Further, suppose that both start out headed in the positive-x direction. (If not, rotate each until they do; that s again a regular homotopy). So now we only have to prove the theorem in the case of loops at the origin that start out headed in a positive multiple of the (1, 0) direction. To a curve γ : I R 2, starting at the origin, heading right, associate a curve γ : I S 1 defined by γ(t) = γ (t) γ (t), and a velocity v = γ (0). Note that γ(0) = (1, 0). (Why?) This defines a map from plane-loops to (circle-loop, length) pairs. There s a map in the other direction. Given a function φ : I S 1 with φ(0) = (1, 0), and a speed v, we can define a curve α(t) = t 0 v(cos φ(u), sin φ(u)) du The curve α starts at the origin headed right, and at time t, its tangent vector is v(cos φ(u), sin φ(u)), so its unit tangent vector is (cos φ(u), sin φ(u)). 4

5 It s possible that α(1) α(0), but we know that α (1) = α (0), because φ(1) = φ(0), because φ is a loop in S 1. Let w = α(1) α(0). Then let C φ,v (t) = wt + t 0 v(cos φ(u), sin φ(u)) du Now C φ,v is a closed loop at the origin that starts out headed to the right. There remains the question of whether C φ,v in fact has nonzero derivative everywhere. It turns out that unless φ is constant, C φ,v does in fact have nonzero derivative everywhere. The constant-φ case arises when we interpolate two curves whose direction-angles are exact opposites at every time, so that their average ends up being a constant zero. This happens only for turning number zero only. And the fix is relatively easy: you just have to slightly deform one of the two curves. Now to prove the theorem, we start with two curves γ 1 and γ 2, with the same turning number, and find their associated curves γ 1 and γ 2 and velocities v 1 and v 2. We now interpolate these by defining and let v(s) = (1 s)v 1 + sv 2 (1) φ s (t) = (1 s) γ 1 (t) + s γ 2 (t) (2) K(s, t) = C φs,v s (t). It then turns out that K provides a regular homotopy between the two curves. The next few questions are all about implementing these ideas, but in a discretized form: instead of smooth curves, we ll have polygonal ones with no reflex vertices and no zero-length edges. That means that the associated direction curve will no longer be continuous; instead it will be piecewise constant. Fortunately, if we interpolate between piecewise constant curves, we get a piecewise constant curve (possibly with more discontinuities). Everything else from the smooth case works in the discretized form as well. To handle the special case of turning number zero, you need to verify that your starting and ending curves do not have exact opposite direction curves. One way to ensure this is to check that (a) the initial segments of the two curves have different lengths and (b) the second segments have different directions from the first segment. If that s not the case, you do a preliminary regular homotopy of the first curve to alter either the length of its first segment or the direction of its second segment, or both. Throughout this section, a polygon in the plane will be represented by a n 2 array of points, with the first and last points equal. The angle of a polygon edge will be the angle it makes with the positive x-axis, which, for a single edge could be 45 degrees, or = 405 degrees or = 765 degrees, and so on. In other words, it s only defined up to multiples of 360 (or, because we ll work in radians, multiples of 2π). The problems build sequentially. Problem 1 [10 pts] Use the TA-provided getpoly to get a polygon from the user. Call that p1. Flip p1 about the x-axis to get a polygon p2. Suppose that p1 has n points. Then you can regard it as a mapping from [0, n] to the plane. Build (and plot) a homotopy (but not a regular homotopy!) from p1 to p2. I recommend the straight line interpolation, in which the intermediate curve is created by making a weighted average of the points of the first and second curves, with the relative weights determined by 0 s 1, which indicates how far along from the first to the second curve we have come. 5

6 See the TA-supplied code for how to draw a homotopy in several different ways. Your procedure should be called function plotinversion(p1), taking the polygon p1 as its sole argument. Problem 2 [10 pts] a Given a real number angle, adjust it, by adding a multiple of 2π, to lie within π of a target angle target. Do so by writing a procedure function Y = angleadjust(x, target) The result, Y, of this procedure should have two properties: Y x should be a multiple of 2π, and y target should be no more than π. Hint: You can do this with mod if you work carefully. A former TA tried it and said it was easier just to do it myself. b Modify your code, if necessary, to allow the user to call angleadjust(x, target) where x is an n 1 array of numbers; the result should be an n 1 array of values Y. Problem 3 [10 pts] Write code to check whether a polygon has any short edges or nearly reflex vertexes. Your function should have the form answer = goodpoly(poly, shortlimit, anglelimit), where any edge shorter than (or equal to) shortlimit is declared short, and any vertex whose interior angle is within anglelimit of zero is declared a reflex vertex. If there are no reflex vertices or short edges, return true; return false otherwise. Remember to check the angle at the first vertex! Your procedure may allow default values for the two limits. I suggest a shortlimit of about 0.1 (suitable for polygons whose typical edges have length one) and an anglelimit of about five degrees, which is about 0.08 radians. Problem 4 [10 pts] Given a polygon or polyline as an n 2 array of points, write [angles, lengths] = anglesandlengths(p) to produce a list of n 1 angles and n 1 lengths, one for each edge of the polygon, in order, provided the polygon has no short edges or too-small interior angles. (If it has these problems, your code should exit by calling error( Bad polygon );.) So angles(i) and lengths(i) are the angle and length of the polygon edge from p(i, :) to p(i+1, :). Use your work from problem 6

7 2 to make certain that each angle differs from the previous one by no more than π. Thus if you traverse the unit square twice, your angle sequence will be 0, π/2, 2π/2, 3π/2, 4π/2, 5π/2, 6π/2, 7π/2. Hint: you can do this with the procedure you wrote in step 2, applying it about n 1 times, or you can be clever and apply it only once, to a sequence of angles. Think about angle-differences rather than angles (i.e., think about exterior angles for a polygon, rather than the angle between each edge and the x-axis). Problem 5 [10 pts] a Write a program function p = integrate(start, lengths, angles) that takes a starting point (represented as a 1 2 array, a length-sequence (an n 1 array), and an angle sequence (an n 1 array), and creates a polyline (an (n + 1) 2 array) (not necessarily closed!) whose length and angle sequences are the ones given, and which starts at the specified starting point. b Test your program by allowing a user to input a polygon, computing its length-and-angle sequence, integrating these with a new starting point slightly offset from the original polygon, and then drawing both the original and the reconstructed polygons in a single graph. Problem 6 [10 pts] Now write a revised version of the previous problem, which does the computation just described, converting a length-and-angle sequence into a curve. Call the starting and ending points of the curve A and B, and the total curve-length L. Each point on the curve has an associated length-from-the-start (which you can compute with cumsum; watch out for off-by-one errors). Calling the points (x i, y i ), call that length L i. Now alter the curve by moving point i by adding to it L i (A B). L For the first point, this will add 0(A B), i.e., nothing. For the last point, it ll add Ln L (A B) = 1(A B) = (A B). Since that point is B, this will result in B + (A B) = A, i.e., we ll return to the start. Call your procedure wgintegrate for Whitney-Graustein integration. It should have signature function p = wgintegrate(start, lengths, angles) with the same argument and output types as above. 7

8 Problem 7 [10 pts] Given a polygon, use your angle computation to find its turning number (assuming it s nice, i.e, has no short edges or reflex vertices). Test your computation on some simple models of polygons with turning numbers 1, 0, 1, and 2. Do this in the form of a matlab function function n = turningnumber(p) where p is an n 2 array, representing the n 1 vertices of the polygon, with the first and last rows of p being identical. Problem 8 [10 pts] Write a procedure standardpolygon(n, k) that produces an n-vertex representative of a polygon with turning number k. (Of course, k will have to be large enough for this to work; use your results from the third problem in the Math section to check for valid input.) Problem 9 [10 pts] Put it all together in one of two ways (you get to choose; they re about equally difficult). The first is great for debugging. The second requires only a slight modification to the first once you ve got everything working. First way: Given a polygon of n vertices and turning number k, construct a regular homotopy from it to the standard polygon of n vertices and turning number k by i. converting to the length and angle representation at each end, ii. linearly interpolating between these two representations, and iii. applying Whitney-Graustein integration to the result. You should produce a function poly = wginterpolate(poly1, poly2, s, t) that interpolates s of the way from poly1 to poly2, and then evaluates the resulting interpolated polygon at the point t (where t is a number between 0 and n). In this situation, poly2 will be the standard polygon that you constructed. As usual, your procedure should handle the case where t is a vector of values rather than a single value. (You need not, however, handle a vector of s-values). There s a question of whether the value t is restricted to integers or allowed to be a real number. For an integers, like t = 2 the function should return the third vertex of the polygon. If you want, you can also define it for non-integer values, so that at t = 2.25, the function returns a point one quarter of the way from the third to the fourth vertex. (Recall that the first vertex of the polygon corresponds to t = 0, hence the apparent off-by-one discrepancy in the previous sentence.) Second way: Start with two user-defined polygons, check that they have the same turning number, and then interpolate between them as described above. Note that in this second approach, you ll need to make certain that the polygons have the same number of vertices. One way to do this is to take the one with fewer vertices and insert new vertices along edges either randomly or in such a way as to divide the longest remaining edge in half. There are lots of other possible choices as well, such as inserting all the new vertices on the first or the last edge, but these tend not to look very nice. 8

9 Problem 10 [10 pts] Write a program that does the whole thing: the user inputs a polygon, which you then show gradually changing to the standard polygon without ever having a reflex vertex. We ll provide some sample code that shows how to draw a sequence of things so that they appear nicely. Alternatively, ask the user to input two polygons and then show the interpolation. In either case, you need to check, if the turning number is zero, that the angle-and-length sequences for your two polygons are not identical but with opposite angle-signs. If they are, you need to adjust one of them, perhaps during the first half of the homotopy, and then interpolate the adjusted polygon to the other, during the second half. The adjustment can be as simple as moving the third vertex a small distance along the angle bisector at the third. (That will alter the second and third angles in the sequence.) Extra coolness: Figure out, for the interpolation, how large the largest axis has to be to show everything (perhaps by doing the interpolation once without showing it) and then show each stage of the interpolation in a window that large, rather than allowing Matlab s default auto-resizing. Problem 11 [20 pts] Solo (i.e., no collaboration) problem. The Whitney-Graustein Theorem (part I) said (roughly) that any two curves in the plane with the same turning number are regularly homotopic. The first five pages of Whitney s paper on the subject describe the proof I gave in class, albeit with more details and fewer pictures. I want you now to think about the Whitney-Graustein theorem for grid loops. If you have two grid-loops in the plane, each with its starting point at the origin, and initially heading east, and the two loops have the same turning number (easy to compute, right?), are they grid-regularlyhomotopic? (That means something like grid homotopic, but no hairpin insertions/deletions. ) Note that the two curves have to be regular curves (i.e., correspond to nonzero-derivative at each point). Probably the right grid analog for this is the constant loop isn t allowed. As you know from the earlier parts of this assignment, Whitney s proof involves drawing the angle curve for each loop, interpolating angle-curves, integrating to get a non-closed curve, and then adjusting to close it up. The angle curve for a grid loop is easy: it s just the starting-pointand-sequence-of-directions form of the curve. But what about interpolating? And what about curves that have different numbers of edges? (To deal with that latter problem, I suggest that you allow the bump move, in which one side of a square is replaced by the other three sides, as long as it doesn t create any hairpins.) And how, even if you can interpolate between the angle curves in some rational way, are you going to do the adjusting to close it up? What I want you to do is think about this problem, draw examples, experiment, toss around ideas, and say what you ve come up with. Your first challenge is to give clear definitions of the notions of a regular grid curve, regular grid homotopy, turning number of a regular grid curve, and a clear statement of the potential theorem that corresponds to the smooth Whitney Graustein Theorem. Then you have to think about whether the theorem is true. Perhaps there s some pair of turning-number-one curves that are NOT grid-regular-homotopic. Perhaps there s a proof that s strongly analogous to Whitney s proof, but with some fiddling to deal with the extra-edges problem. Perhaps neither. Maybe it s hard to solve on a square grid, but on an equilateral-triangle grid it s easy to work out. Your job is to play around with ideas and see what you come up with. Perhaps 9

10 the ideas you learned about when I talked about Thurston corrugations will help you. I honestly do not know. I ve played around with this at my whiteboard briefly, but have neither conclusively proved the theorem nor created a counterexample. Success on this problem can come in many forms, but the key elements of a good answer are (1) evidence that you thought carefully about the question, and (2) evidence that your careful thought led you to some valid conclusions. You don t have to prove a theorem, or come up with a counterexample, but you have to show that you re on the way towards doing one or the other. Note: For this problem, writing up your progress in LaTeX or other math-formatting tools could be a huge amount of work. If you prefer, you may hand in a neatly handwritten answer. I emphasize neatly. If I can t read it easily, I won t read it at all. Problem 12 [0 pts] Tell us how long you spent on this problem set.) 10