Math 5378: News and Announcements, Spring 2013

o 5/13/13: I am done with writing solutions to the Sample Final. I have been writing them sporadically all day long, taking breaks from a Calculus exam that I was giving and grading today. Good luck for tomorrow's exam!

o 5/13/13: Coverage for the Tuesday, May 14, 10:30 a.m - 12:30 p.m. Final (in our regular classroom) is basically all we have studied, except the material on Riemannian geometry covered during the last class meeting. This is Chapters 5-8 (skipping the Involutes and Evolutes section from Chapter 5, the Appendix to Chapter 6, and Chapter 7bis), Chapters 9, 10 (through p. 195), 11 (through the top of p. 210), and the material covered in class on constant-curvature surfaces, non-Euclidean geometries, the hyperboloid, the Poincare disk and upper half-plane models of hyperbolic geometry, smooth manifolds, smooth maps, and diffeomorphisms. (This roughly corresponds to pp. 218, 224-225, 228, 262-269, and 282-287 from Chapters 12, 14, and 15 in the text, but the way we dealt with this material was on many occasions quite different and much simpler.) The best way to get ready for the final would be to try to solve as many problems as you can on your own (with an open textbook, if needed, and no timing) from the sample final. For those problems you cannot solve, it may be a good idea to discuss them with your friends, i.e., your study group, and with me tomorrow. Come to my office hours Monday, May 13, 11 a.m. to 12:30 p.m. Try to solve the real and sample midterm problems once again without consulting your old solutions. Go over homework problems again in a similar way and look for problems from similar classes on the web: search for Differential Geometry. When you know you can get an idea of how you might approach most of the problems right away, you may feel more sure that you will do well on our final. I am also working on posting selected solutions to our sample final, but consult them only after you have tried hard to solve the problems on your own. Good luck!

o 5/13/13: I have just posted a solution to Problem 1 on the Sample Final and plan to post some more, but have to go to bed now and will have little time, being busy giving an exam and then grading for another class today.

o 5/8/13: Sample Final is now posted!

o 5/8/13: Because the material for Problem Z on the last homework will be reviewed briefly during the last class, you do not have to turn that problem in -- just solve it, if you would like to see how well you understand those things. The material on Riemannian geometry will not be covered by the Final, either. However, manifolds, smooth maps, and diffeomorphisms will.

o 5/8/13: I need to move my office hours on Monday, May 13 ahead by 30 minutes. My office hours will be 11:00-12:30.

o 5/7/13: Hint to homework Problem U: Use the equation kg = 0 from the second displayed formula on p. 188 using a parametrization x = t, y = y(t), unless x = const, which needs to be treated separately. The equation should then transform into y y'' + (y')2 = -1 or (yy')' = -1, which integrates directly as yy' = -t + C, which is a separable first-order differential equation and is easy to solve.

o 5/5/13: Some statistics for Exam II: Maximum = 39, Mean = 34.6, Median = 36, which is 90% of the total score. The exam was probably too easy for you, folks. On the other hand, it means you have been learning the material really well!

o 5/3/13: I plan to post a sample final over the weeekend and later post solutions to it. I will also hold extra office hours on Monday, May 13: 11:00-12:30.

o 5/3/13: I have moved the last homework due date to Friday, May 10, but added one problem on Riemannian geometry to it. Make sure that the last problem on Homework 12 is Problem Z.

o 4/26/13: I have just realized that Problem 4 on Sample Test II was too hard for an exam and removed it from the Sample Test. See comments in the new, April 26, version of the test. I kept the same numbering of problems. I apologize about the confusion.

o 4/25/13: I have added a few problems to Sample Test II and will not be adding problems to it anymore. There must be 11 problems total.

o 4/24/13: Reminder: the second midterm exam is coming on Wednesday, May 1. Coverage: Chapters 9, 10 (through p. 195), 11 (through the top of p. 210), and the material covered in class on constant-curvature surfaces, non-Euclidean geometries, the hyperboloid, the Poincare disk and upper half-plane models of hyperbolic geometry. (This roughly corresponds to pp. 218, 225, 228, 262-269 from Chapters 12 and 14 in the text, but the way we dealt with this material was quite different and much simpler.) How to get ready: the more problems you solve before the test, the better you will be prepared. Solve the sample exam, and problems from the past homeworks again (if you feel you do not remember how to do them). Also, look for similar problems from Differential Geometry textbooks, such as Differential Geometry and Its Applications by John Oprea, available in the Math Library, look for old exams on the web for Math 5378 and similar classes at other schools. If you know you need only 3 minutes to get an idea on how to solve at least 50% of random problems you find, then you may be sure you will do well our test.

o 4/24/13: I will be out of town, visiting Purdue University the rest of this week. I will be back on Monday, but I will be passing a citizenship interview and not be able to teach the class, either. I have asked my graduate student Denis Bashkirov to substitute for me on Friday and Monday. I will not hold office hours on Friday, but I will be here for my office hours on Monday morning. I apologize about the inconvenience.

o 4/20/13: Homework 11 is now posted! Sorry about a delay.

o 4/9/13: In Problem M on Homework 9, I forgot to mention explicitly that I was talking about a surface of revolution, even though I used such notions as parallels and the radius. I have corrected that in the latest version of the homework.

o 4/6/13: After you are done with Homework 9, you can add to the characterization of geodesics below (News of 4/5/13) the following:

[energy-minimizing curves] <=> [distance-minimizing curves parametrized proportionally to arc length] => [geodesics parametrized proportionally to arc length] <=> [curves satisfying the geodesic equations].

o 4/5/13: Homework 9 is now posted! It may look long, because it consists of problems not from the text, but they are rather straightforward.

o 4/5/13: Actually, I have made a mistake in that table about geodesics, as we figured out with a few of the students in an after-class discussion. Here is a correction:

[distance-minimizing curves] => [geodesics] <= [geodesics parametrized proportionally to arc length] <=> [curves satisfying the geodesic equations].

I was wrong saying that every geodesic was parametrized proportionally to arc length. That confusion came from a different, inequivalent definition of a geodesic that I used to know.

Let me explain why a geodesic parametrized proportionally to arc length will satisfy the geodesic equations. From the determinant formula for geodesic curvature, we see that a regular curve is a geodesic if and only if the projection prN(α''(t)) of α''(t) to the tangent plane to the surface along the unit normal N to the surface is proportional to the tangent vector α'(t). Now, if a curve is parameterized proportionally to arc length, then α''(t) is orthogonal to the tangent vector α'(t). Therefore, prN(α''(t)), being a linear combination of the normals α''(t) to the curve and N to the surface, will also be orthogonal to the tangent vector α'(t). Thus, prN(α''(t)) could be propotional to the tangent vector only when the coefficient of proportionality is zero or prN(α''(t)) = 0, which is equivalent to the geodesic equations.

o 4/1/13: On Problem 10.8: The tractrix is being rotated about the axis which it does touch, e.g., the tractrix from p. 84 is rotated about the y-axis.

o 3/30/13: One clarification on homework Problem 10.4: the latitude-longitude parametrization is the same as spherical coordinates.

o 3/26/13: I have just learned that there is a free tutor at SMART Learning Commons who can provide help with Math 5378 Differential Geometry, see SMART.

o 3/26/13: I have to cancel my office hours on Friday, March 29. Sorry! If you need to see me, please make an appointment.

o 3/26/13: In Problem 9.7, it should be u1= √ u2- a2 , rather than u1= √ u2+ a2 .

o 3/14/13: Homework 7 is now posted!

o 3/11/13: More hints on Problem 9.1: Use Theorem 9.4 and the computation of the first fundamental form (and thereby the line element) in the example after Corollary 8.9. If you use the arc length parametrization of the generating curve in that example, the formula will be simpler. The rest is to see if you can change your coordinates on the given surface so as to make the line element change from Λ(u) (du2 + dv2) to something like du2 + Λ(u) dv2. See p. 132 on how the components of the metric change with changes of coordinates.

o 3/11/13: I have originally misread Problem 9.1, which is on this week's homework. I have just made corrections to my comments to the original posting of the problem. You need to prove only that if in a coordinate chart the line element can be expressed as in the formula, then this surface is locally isometric to a surface of revolution. And you can ignore the conformal mapping question.

o 3/10/13: Some statistics for Exam I: Maximum = 40, Mean = 28.5, Median = 28.5, which is 71.25% of the total score. Good job!

o 3/7/13: Homework 6 is now posted!

o 3/3/13: You do not need to turn in any homework this week. Concentrate on getting ready to the test.

o 3/2/13: I have added one problem (Make sure you have a copy of the test with 7 problems!) to the sample test and moved it to the Homework page.

o 3/2/13: Reminder: the first midterm exam is coming on Wednesday, March 6. Coverage: Chapters 5-8, skipping the Involutes and Evolutes section from Chapter 5, the Appendix to Chapter 6, and Chapter 7bis. How to get ready: the more problems you solve before the test, the better you will be prepared. Solve the sample exam, and problems from the past homeworks again (if you feel you do not remember how to do them). Also, look for similar problems from Differential Geometry textbooks, such as Differential Geometry and Its Applications by John Oprea, available in the Math Library, look for old exams on the web for Math 5378 and similar classes at other schools. If you know you need only 3 minutes to get an idea on how to solve at least 50% of random problems you find, then you may be sure you will do well our test.

o 2/27/13: I have posted a sample test before the real one which will be in a week, on March 6.

o 2/25/13: I have added the following clarification (and simplification) to the homework problem 8.4: Find the principal curvatures just at the north pole of the ellipsoid ax2 + by2 + cz2 = 1, a, b, c > 0, and at the origin for the cylinder z = ky2, k > 0, and the saddle z = ax2 - by2, a, b > 0.

o 2/21/13: Yesterday late night, I made up HW5, but posted HW4 again instead. Sorry. The flaw is now corrected. Homework 5 is short, as we are getting close to the first test, and I plan to post a sample one soon.

o 2/18/13: Also added the following clarification to Problem 7.9 on HW4. The graph is revolved about the u-axis. The function is assumed to be positive. The problem refers to Example (4) on page 119.

o 2/18/13: I have added the following clarifying and simplifying remark to Problem I on HW4. Use any of your favorite charts for the sphere and assume the point (x0, y0, z0

o 2/18/13: Correction for the textbook: The top formula on page 133 should have the bars on the left-hand side, rather than on the right-hand side.

o 2/17/13: Hint to Problem K on Homework 4: You may assume that (u,v) ↦ xu × xv/||xu × xv|| is a smooth function from the surface to S2.

o 2/16/13: Here is a solution of Problem H from Homework 3. WLOG, you can assume that the first two rows of the Jacobi matrix for parametrization y are linearly independent at and near a point on the surface S2. By the Implicit Function Theorem, it means that the surface S2 near that point may be given as the graph of a function x3 = F(x1,x2). This means that y(u,v) = (y1(u,v), y2(u,v), F(y1(u,v),y2(u,v))), and by the Inverse Function Theorem, the map (y1(u,v), y2(u,v)) has a smooth (i.e., differentiable) inverse (u(x1,x2), v(x1,x2)). The map y-1 on a point (x1,x2,x3) of the surface S2 will then be given by the formula y-1 (x1,x2,x3) = (u(x1,x2), v(x1,x2)). The composition y-1 ◦ f ◦ x will then be (u(f1(x), f2(x)), v(f1(x),f2(x))) and thereby a composition of smooth functions. Another solution may be noticing that in the proof of Proposition 7.5, one might use any smooth, not necessarily regular function in lieu of x, in this case the function f ◦ x. Of course, you have to repeat the proof in your own words, if you write this as a solution of the problem.

o 2/14/13: I have just posted homework for next Wednesday. Reminder: At the beginning of each homework, I give a reading assignment for that homework. Not later than shortly after each class, I outline what has been covered in class and give a reading assignment on the Class Outline page. Some important announcements, including homework hints and study advice are made on this Class News page.

o 2/12/13: Here is a hint to Problem 7.5. Do not try to find x-1, find (x-1 ◦ xbar) without knowing an explicit inverse. I would have hard time looking for x-1, too. I am not sure I would be able to do it with formulas in a reasonable amount of time.

Finding (x-1 ◦ xbar) means solving the equation x(u,v)=xbar(ubar,vbar) for u and v. This is an equation on vector functions x and xbar, so it is actually a system of three equations, one for each coordinate. A solution should produce u and v as functions of ubar and vbar. Beware of the domains of these functions, as you expect them to be the intersection of the two charts expressed in the ubar and vbar parametrization, i.e., xbar-1 of the overlap(s).

The equations x = xbar look awfully complicated, but you may notice some structure. In the first two coordinates, they look like a sin u = b cos ubar, a cos u = - b sin ubar. They have a simple solution, but it is not obvious how to find it. Your life would have been simpler, if the equations were between two sines and two cosines. But this is easy to achieve: sin u = cos (u-π/2) and cos u = - sin (u-π/2). Thus, your equations are

a cos (u-π/2) = b cos ubar,

a sin (u-π/2) = b sin ubar.

Since cos2 + sin2 = 1, these imply that a = + or - b. Recall what a and b are and realize that only one of these options is possible: a = b > 0. Thus we get the polar coordinates of one and the same point in the plane. Therefore, they must be equal, up to a multiple of 2π in the angle:

u-π/2 = ubar + 2π k.

We know that both u and ubar must be between 0 and 2 π. This leaves two options: k = 0 and k = -1, which corresponds to two overlaps between those two charts. Indeed the Moebius band is just a piece of band after all, glued in a funny way at the ends. The two charts you can imagine cover half of the glued band and a bit more, and you expect them to overlap over two separate regions. Think of covering a circle with two intervals!

Now, you have to recall what a and b are and add to the equation a = b the equation which comes from making the z's equal in x (u,v) = xbar (ubar, vbar):

v cos u/2 = vbar cos (π/4 + ubar/2),

which we have ignored so far. You may also wish to plug in the two solutions for u through ubar that you have found above. The two solutions correspond to two different overlaps, with which you will have to work separately. For each overlap this gives a system of two equations, very similar to what we had above, but with u already being expressed through ubar, considerably simpler. Solve it for v, and you will get u and v expressed through ubar and vbar in the two overlaps. Now work with the partials and see whether the Jacobians are positive.

o 2/7/13: Folks, I was too sleepy last night to post the homework and too busy in the morning today. Sorry. I will do that today early afternoon.

o 2/6/13: If you try to use google or other search engines to arrive at our class pages, make sure that you are looking at Math 5378 Spring 2013. The thing is that I also taught this class in 2002, and you probably do not want to do homework from 2002.

o 1/31/13 (after midnight): Homework 2 is now posted!

o 1/28/13: I have made an error computing β''(s), when β(s) = α(t(s)) is the reparametrization of the curve α(t) by its arc length s = s(t), today in class: it should be β"(s) = α" (t(s)) (dt/ds)2 + α'(t(s)) (d2t/ds2), rather than β"(s) = α" (t(s)) (dt/ds) + α'(t(s)) (d2t/ds2). Sorry! This should yield the required computation of the curvature in terms of α(t) as follows:

κ (α; t) = κ (β; s) = det (β'(s)t β''(s)t) = (dt/ds)3 det (α'(t)t α''(t)t) = 1 /||α' (t)||3 det (α'(t)t α''(t)t).

o 1/28/13: A little clarification on the textbook Problem 6 from Chapter 5. Let us reword it as follows: Take any circle of radius R=a with center on the y-axis. Show that the lower half of the circle meets the tractrix at a right angle.

o 1/25/13: You should be able to do all the homework problems, except #'s 4 and 6, over the weekend. We will study the tractrix and oriented (a.k.a. directed) curvature on Monday.

o 1/22/13: I have to move my office hours just for one day, Wednesday, January 23, from 11:15-12:05 to 12:20-1:10, because of a conflict with a special seminar.

o 1/21/13: I recommend the following way to study. Attend each class, take notes, participate in class actively. After each class review your notes and study the corresponding part of the text. You can find out which part of the text at the Class Outlines page. Then do relevant homework problems.

o 1/21/13: If you need to register for this class, please, send a message to Ms. Lawson at ugrad@math.umn.edu and ask for a permission number. If permission is granted, go to OneStop and register.

o 1/21/13: Homework will be collected before each class, starting next Wednesday, January 30, and a small selection of problems will be graded. ("Small," because the grader will allocate only 3 hours per week for our class.)


Last modified: (2014-01-20 16:34:56 CST)