Wikipedia:Reference desk/Archives/Mathematics/2008 July 14

= July 14 =

Vector fields on the torus
Let's consider vector fields on surfaces with unitary vectors. Is it possible that there is only one such vector field without critical points on the torus up to diffeomorphic coordinate changes? (That means there are not two topologically distinct vector fields.)--Pokipsy76 (talk) 10:54, 14 July 2008 (UTC)


 * There are three possible interpretations of the phrase "up to diffeomorphic coordinate change" that I can think of. In order to explain this, it is perhaps better to first introduce some terminology (which will also give you some pointers for further reading). Your vector field is a section of the unit tangent bundle. The group of diffeomorphisms of the surface acts naturally on the unit tangent bundle. So the first interpretation of your question would be that you mean up to such diffeomorphisms. More generally, there are diffeomorphisms of the unit tangent bundle that preserve the structure of the fibre bundle, in the sense that they take fibers to fibers. In other words, to each such diffeomorphism there is an associated diffeomorphism of the surface such that there is a commutative diagram involving these two diffeomorphisms and the projection maps of the fiber bundles. Finally, there are just the self-diffeomorphisms of the fiber bundle thought of as a differential manifold without additional structure. I believe the answers are no for the first interpretation, and yes to the other two. But I will not elaborate until I know which interpretation you are actually interested in. I should say that I'm not an expert on topology in general and fiber bundles in particular, but I have some background. I hope I have not discouraged you with all this terminology. I'll be happy to try to help if it causes too much grief and the links I have provided are not very helpful. Oded (talk) 15:12, 14 July 2008 (UTC)
 * What I had in mind is this: a diffeomorphism from the torus to itself induces a transformation of the vector field acting by multiplication for the differential
 * $$(x,v) \mapsto (\phi(x), D\phi(x)v)$$
 * this is what happens if the tangent vector of a curve is mapped to the tangent vector of the image curve.--Pokipsy76 (talk) 21:26, 14 July 2008 (UTC)


 * In that case, I think the answer is no. The diffeomorphism would take the flow curves of one vector field to the flow lines of the other. If you look at the vector field $$\bigl(\cos(2 \pi x),\sin(2\pi x)\bigr)$$ on the torus $$\R^2/\mathbb Z^2$$, then you can see that most of its flow lines will not be closed. On the other hand, the flow lines of the vector field $$(1,0)$$ are all closed. Oded (talk) 21:57, 14 July 2008 (UTC)

Manhattan distance
In the picture at right, as the number of steps N increases, the distance along that route from A to B stays the same, per Manhattan distance. As N approaches infinity, the distance stays the same, but the route is indistinguishable from a straight line from A to B, which is obviously shorter than the Manhattan distance ($$\sqrt 2$$ vs. 2 for a unit square, say). How does one explain this unintuitive result? Thanks. --Sean 14:51, 14 July 2008 (UTC)


 * I think that's covered in the linked article, which redirects to Taxicab geometry - it's not Euclidean geometry. In Taxicab geometry the straight line also has a length 2.  -- LarryMac  | Talk  15:09, 14 July 2008 (UTC)


 * More generally, "length" is not a continuous function on curves. Suppose that you need to walk 10 miles on a road. If every few steps forward you decide to cross the road, you will need to walk much more than 10 miles, but if you measure the road itself it will appear as if you have traveled just 10 miles. Does this make more sense now? Oded (talk) 15:27, 14 July 2008 (UTC)


 * There was at least one previous discussion of this "paradox" on this desk.. the "user:froth" asked and I seem to remember it getting a bit argumentative (probably my fault)..
 * However should you wish to read it then Reference_desk/Archives/Mathematics/2008_March_5 - as you will see however the answers consists of mostly me arguing with "user:tango"...
 * Still it might help...87.102.86.73 (talk) 16:50, 14 July 2008 (UTC)
 * That did get rather heated, didn't it? No hard feelings? :-) --Tango (talk) 17:34, 14 July 2008 (UTC)
 * Of course not - this used to happen to me a lot - but I'm trying to cut down... Something about the internet being impersonal - I've also made contributions to Lamest edit wars.. (not saying which) do I get a barnstar? 87.102.86.73 (talk) 17:53, 14 July 2008 (UTC)
 * There's also this previous discussion Reference_desk/Archives/Mathematics/2008_February_3 —Preceding unsigned comment added by 87.102.86.73 (talk) 16:52, 14 July 2008 (UTC)  which should be more helpful.87.102.86.73 (talk) 16:54, 14 July 2008 (UTC)


 * I think the bottom line is that the convergence of the stepped line to the straight line isn't uniform which allows strange things to happen. I don't fully understand it myself, so I won't try and explain further. --Tango (talk) 17:34, 14 July 2008 (UTC)


 * Sorry, Tango, but the convergence is uniform when the curves are parameterized properly. The point is that the convergence of the paths is not really relevant here, since length is not a continuous function on curves. Think, for example, of the function f that is 1 on the irrationals and 0 on the rationals. You have a sequence of irrationals xi that converges to $$1$$. Then f(xi)=1, xi converges to zero, but f(0)=0. It is just a fact of life, not all functions are continuous. (That's not completely the end of the story, there are stronger notions of convergence of paths for which length is continuous, but that's probably not going to be helpful to discuss them presently.) Oded (talk) 17:48, 14 July 2008 (UTC)


 * I can recall three separate occasions on which this same question has been asked. Isn't it only logical that someone (who knows a lot more than me) create a separate article for this counterintuitive result?  And from then on, just refer people to that page instead of linking, typing, and arguing everytime.--A Real Kaiser...NOT! (talk) 21:59, 14 July 2008 (UTC)
 * Yes please! --Tango (talk) 23:03, 14 July 2008 (UTC)
 * So I understand it even less than I thought I did... great! --Tango (talk) 23:03, 14 July 2008 (UTC)


 * This seems somewhat related to the issue of not being able to precisely define the length of a coastline, since the more detail you include, the longer the coastline gets. [|StuRat] (talk) 22:15, 14 July 2008 (UTC)
 * We've got an article on that, but this case has the opposite outcome; it stays the same! --Sean 23:31, 14 July 2008 (UTC)
 * Thanks for the responses so far, folks, but I forgot to mention that I'm only 8 years old, and while I can understand the ideas of driving taxi-cab style from A to B and driving straight from A to B, I don't understand how just declaring that these real-world operations are in a different geometry or that length is not continuous on the curves (in my rectilinear diagram!) resolve the paradox. Would it be possible for someone to put it in terms as simple as the problem statement?  Thanks.  --Sean 23:31, 14 July 2008 (UTC)
 * In simplest possible terms, then: Why should the length of the 'taxicab route' have anything to do with the length of the straight line? Less simply: there's really no connection. I could give you a route as close as you wanted to the straight line that was infinitely long. Algebraist 23:41, 14 July 2008 (UTC)
 * And while I'm here: mathematicians' use of the word 'curve' is rather perverse. It certainly doesn't need to be curved. Algebraist 23:44, 14 July 2008 (UTC)


 * You can boil it down to simple arithmatic
 * say to have N steps - so each step goes L/N right and L/N up (the box is L by L)
 * So the distance for eachs step is 2L/N
 * It takes N steps, so the distance is (2L/N) x N =2L
 * ie it's independent of N - the number of steps.
 * Obviously by eye it looks like as N becomes big that the line approximates a diagonal with length sqrt(2) x L
 * But the structure is still there ; because we originally said it was- it just becomes so small that it can't be seen or accurately described (when N tends to infinity)...
 * In other words no matter how small they are if we say the path is composed of up/right rather than diagonal then it always remains up/right..87.102.86.73 (talk) 23:50, 14 July 2008 (UTC)


 * Well... as you saw, for any finite number of steps, the distance remains the same as 2 (for the unit square). However, as mentioned, this uses a different metric of measurement - that of distance = change in x + change in y.
 * Now, suppose that we took a 2-step path. We can change this into a one step path by "pushing out" the indentation. This preserves the distance, as the overall change in x and change in y is unaffected. Thus, we can transform an n-step path to a n-1 step path and vice versa. Thus, repeating this, even if we had an infinite number of steps, we could push out each one, and transform it into a one-step path. Thus, the distance with infinite steps is the same as a single step.
 * But, this manhattan metric is not what we use normally to consider length. We use the Pythagorean Theorem - that square of the direct distance is equal to the sum of the squares of the perpendicular sides. -mattbuck (Talk) 23:59, 14 July 2008 (UTC)


 * Let me attempt to fuel the fire a little bit more. I am going to assume that I am working in a Euclidean plane using ONLY the Euclidean metric to measure distances.  We are not using the taxi-cab metric.  My goal is to get from point a to point b where a and b are the diagonally opposite vertices of a rectangle with dimensions A and B.  I will only be walking in straight lines starting from a.  Let n be the number of right-angle turns I make in my journey from point a to point b and define f(n) to be the total distanace I have traveled from a to b making n turns.  f(1)=A+B because starting from a, I walk along one edge, make a right turn, and walk along the other edge to get to b.  Since the lengths of the two edges (independent of the order) are A and B, I have traveled a total distance of A+B.  Remember that this distance is measured in the Euclidean norm.  The values of f(2)=f(3)=f(4)=f(5)=...=f(k)=A+B for any integer k because the total horizontal distance I traveled and the total vertical I traveled will be the same so the total distance from a to b will always be A+B.  Now the question is why does $$f(\infty)=\sqrt{A^2+B^2}$$ while $$\lim_{n\rightarrow\infty}f(n)=A+B$$?  These two quantities are obviously not the same (unless A or B is zero).   The $$\lim_{n\rightarrow\infty}f(n)=A+B$$ because f(n)=A+B for all n.  But why does $$\lim_{n\rightarrow\infty}f(n)\neq f(\infty)$$.  Can someone please explain that?  We have a constant sequence which is converging to some other number than the values of the terms inthe sequence.  Remember, to measure all distances, I am using Euclidean metric, NOT the taxi-cab metric.  I feel your pain Tango.  I thought I had it down too but I don't think so.--A Real Kaiser...NOT! (talk) 01:07, 15 July 2008 (UTC)


 * But $$f(\infty) \neq \sqrt{2}$$, it is equal to 2. Unless you move up and along simultaneously, which means the function no longer applies, you move in discrete (but infinitessimal) distances up and along an infinite number of times. The root 2 is the displacement, NOT the distance. -mattbuck (Talk) 01:20, 15 July 2008 (UTC)
 * In the limit, you have a straight line. The length of a straight line is what it is, regardless of how you constructed the line. A curve is just a set of points, nothing else matters (you probably need to parametrise it to calculate the length, but I'm pretty sure the length is independent of the choice of parametrisation - if it's not, then this is even weirder that I thought!). This example shows that f is not a continuous function (on the extended real line - it's continuous at any finite point, since it's constant at all finite points, it's only at infinity that things go wrong). Why that happens, I really don't know, but it's clear that it does. --Tango (talk) 01:50, 15 July 2008 (UTC)
 * Surely in the n=infty case, you should be taking infinitely many right-angle turns? It's easy to do this without infinitesimal movements (go up and right 2−n on the nth leg or whatever) and the total distance travelled is 2. More ontopic: I think this is the kind of thing that would have worried 17th (and possibly 18th) century mathematicians, with their naive expectation that behaviour at the limit would nicely reflect finite behaviour. Now we're used to the idea that lots of things (in this case the length operator) don't commute with limits, we shouldn't be very surprised. Algebraist 10:11, 15 July 2008 (UTC)
 * Well, yes, I was assuming each turn was the same size, without that assumption there is nothing strange going on at all. --Tango (talk) 17:01, 15 July 2008 (UTC)

For everyday use, the length of a path is measured simply by a number of steps sufficient to walk that path. Taking shorter steps, x, you expect the number of steps, N(x), to be greater, so the total length of the path is computed as the length of the step times the number of steps, L(x) = x&middot;N(x). Still L(x) does depend a little on the length of each step, because long steps cut corners. Mathematicians use tiny steps. The length of a smooth curve, like a circle arc, does not depend much on the length of the step when the step is tiny compared to the radius of curvature.
 * L = limx&rarr;0L(x) = limx&rarr;0(x&middot;N(x)).

In the example of a zig-zag curve, there is a second number to be taken into account: the distances between corners, = y. The number of steps, N(x,y) and the distance L(x,y)=x&middot;N(x,y) depend on both. If you take the limit of tiny steplength first then you get the Manhattan distance,
 * limy&rarr;0(limx&rarr;0L(x,y)) = limy&rarr;0(LManhattan) = LManhattan

because tiny steps do not cut corners. If you take the limit of tiny cornerdistance first then you get the Euclidean distance,
 * limx&rarr;0(limy&rarr;0L(x,y)) = LEuclid,

because finite steps cut sufficiently small corners. So the moral is that you cannot always switch the order of taking limiting values.
 * limx&rarr;0(limy&rarr;0L(x,y)) is not necessarily equal to limy&rarr;0(limx&rarr;0L(x,y)).

and the intuitive distance lim(x,y)&rarr;(0,0)L(x,y) is not well defined. Bo Jacoby (talk) 13:32, 20 July 2008 (UTC).

Continuous distance
If one defines a sequence of functions f_n: [0,1] → R2 by f_n(t) = ( t, sin(n*t*π)/(n*π) ), then | f_n(t) - f_m(t) | ≤ 1/min(n,m) for all t, and indeed, | f_n(t) - (t,0) | ≤ 1/n for all t. Hence f_n approaches the function t -> (t,0) uniformly in t. Every f_n is smooth.

The arclength of f_n is the integral from t=0 to 1 of sqrt( 1 + f'_n(t)^2 ), which is the integral of sqrt(1+cos(n*t*π)^2) from t=0 to 1. I believe this integral is constant in n, and strictly larger than 1.

However, the limit function t → ( t, 0 ) clearly has arclength 1.

Hence "arclength" as a function from the space of smooth functions from [0,1]→ R2 to the nonnegative reals is not a continuous function when the domain is given the topology of uniform convergence.

However, the derivative of f_n is ( 1, cos(n*t*π) ) which does not converge to the derivative of t → ( t, 0 ), so there is no reason to expect anything defined in terms of derivatives to be continuous. In particular, there is no reason to expect arclength to be continuous. To me at least, this is a typical reason to use sobolev spaces (though with p=&infin;, if you don't mind a little extra continuity, the higher hölder spaces might be simpler).

Question: Is $$\int_0^1 \sqrt{1+\cos(nt\pi)^2}~dt$$ constant in n? Does the constant have a name? Is there a reference for this?

BTW: In case someone dislikes the integral, the same conclusion is reached in a brutish way by using g_n(t) = ( t, sin(n*n*t)/n ). Here the arclength diverges to infinity, rather than remaining constant and inequal to the arclength of the limit function. JackSchmidt (talk) 19:42, 14 July 2008 (UTC)


 * $$\int_0^1 \sqrt{1+\cos(nt\pi)^2}~dt=\frac{2 \sqrt{2} \text{EllipticE}\left[\frac{1}{2}\right]}{\pi }$$
 * Integrals are not something I do for fun. For this, I use Mathematica. Mathematica can evaluate the integral symbolically if n=1,2,3,100 (and gets the same answer, though the calculation for 100 takes longer). So there is some knowlege out there which can evaluate the integral at least when n is integer. But there is really no need to evaluate the integral to see that this is constant when n is integer. You can look at it geometrically. If you scale the path by a factor of 1/2 and put two of these one after the other, you go from n to 2n and the length does not change. The same argument shows that you can go from n=1 to n=k without changing the length. If you do not believe in geometry (I am sure that you do), you can get the same result by substitution. You make the substitution u=n t, and then use the periodicity of cosine. Oded (talk) 21:15, 14 July 2008 (UTC)
 * Thanks! The geometric reason is very clear. I tried the simple substitution before, but didn't change the limits of integration... oopsy.  The periodicity of cosine just rewords your geometric argument, so I like it: n copies of 1/n'th the length.  It also clear both geometrically and basic inequality pushing that the constant is strictly larger than 1, so this solves my problem (self-contained example of arclength not continuous, but not divergent).  JackSchmidt (talk) 22:02, 14 July 2008 (UTC)


 * I played some more and Mathematica can evaluate the indefinite integral:
 * $$\int\sqrt{1+\cos^2t}~dt=\sqrt{2} \text{EllipticE}\left[t,\frac{1}{2}\right].$$
 * Therefore, all that remains for the human to do in order to verify the above equality (if we really care to) is to look up the definition of the ellipticE function and verify by differentiation. Oded (talk) 21:21, 14 July 2008 (UTC)
 * I leave it to the interested reader :) Your simple proof of constancy is enough for me. JackSchmidt (talk) 22:02, 14 July 2008 (UTC)


 * I cannot help taking this opportunity to mention some related current research mathematics that I am excited about. Suppose that you use a (say smooth) curve $$\gamma:[0,t]\to \R^2$$ do drive some differential equation. What does that mean? It means that you are looking at the solution $$u:[0,1]\to\R^n$$ of a differential equation of the form $$u'(t)=F\bigl(t,\gamma(t),\gamma'(t)\bigr)$$ for some smooth F. A nice example is that you have a sphere that you roll around on the plane while keeping the contact point on the curve, and your u consists of the position and rotation of the sphere. The map thus induced from u to $$\gamma$$ is not continuous with respect to the uniform convergence of paths. A British mathematician Terry Lyons had the idea to look at some extra structure on the path which I think is "less" than having the derivative but enough in order to solve all such equations. In this way, the enhanced path carries some structure with it (which I think of as something like a coordinate frame). One point is that you can take the limit of such enhanced curves where the resulting curve (after you forget the extra structure) is highly non-smooth, but still the differential equation makes sense. I don't know much beyond this. Oded (talk) 21:45, 14 July 2008 (UTC)
 * It sounds very interesting. I know very little about ODE, mostly only parabolic and elliptic 2nd order PDE (virtually no smoothness assumptions at least), and that was a while back.  This sounds a lot like the typical gist of a PDE proof: solutions to nice equations carry some extra structure, complete the space they live in, and prove there is a solution to the not nice equation, then prove the resulting solution is a little better than you assumed.  I like the idea that you can do this without even assuming the existence of a whole derivative.  I guess for a path, you basically want the derivative to point in the direction of travel, so if you just had a consistent coordinate frame, you'd at least have some consistent notion of which way you were pointing.  If you know of a survey article, I'd be interested to look through it.  I doubt I could read a research article in ODE, but if that is all there is, it might be a good excuse to learn the basics of ODE. JackSchmidt (talk) 22:02, 14 July 2008 (UTC)


 * Terry's homepage has a link to his list of publications. I assume that you can find the best of what is available there. Here is a nice illustrative example to contemplate. Take your sequence of curves that converge to the straight line segment, take the straight line segment itself, and take another sequence of curves converging to it where the n'th cruve is periodic with period 1/n in the same way that your curves are (periodic is not the precise term here, but you know what I mean) but in each period makes a little clockwise loop of diameter of order 1/n. Then you roll the sphere on the curve in each of these examples, but the curve is so close to the line segment that you cannot visually tell the difference. The ending result of the rotation of the sphere at the end might be different. Thus, on the line segment itself one can put more than one type of these enhanced structures, though only one of these is compatible with its derivative. In that sense, I think it is a bit different than the standard thing that happens in the ODE PDE framework. No? Oded (talk) 22:18, 14 July 2008 (UTC)
 * It seems standard in very (very) broad strokes, but the actual structure being placed on the functions strikes me as quite new, and very natural. I'll let you know when I've had a chance to read through a few of the articles, but it might not be until the weekend. JackSchmidt (talk) 18:17, 15 July 2008 (UTC)