The Curvature of the Torus

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1 The Curvature of the Torus MLI Home Mathematics The Torus Curvature The Curvature of the Torus Let's get bent We begin with the parameterized surface: x( u, v) = x = ( c + a cos v)cos u y = ( c + a cos v)sin u z = a sin v Take the partial derivatives of this parameterization and compute inner products to find the coefficients of the first fundamental form: E = ( c + a cos v) 2, F = 0, G = a 2. This gives us the line element ds 2 = ( c + a cos v) 2 du 2 + a 2 dv 2, from which we read off the metric: g ij = ( c + a cos v) a 2 Some straightforward and boring computation yields the nonzero Christoffel symbols of the second kind: Γ u uv = Γ u vu = a sin v ( c + a cos v) v Γ uu = 1 a sin v ( c + a cos v ) Another two pages of index juggling and basic algebra gives the nonzero components of the Riemann tensor: 1 of 3 10/07/ :17 AM

2 The Curvature of the Torus u u R = vuv R vvu = a cos v ( c + a cos v) v R uvu v = R uuv = 1 a cos v ( c + a cos v ) Contract to get the Ricci tensor: 1 R ij = R m a = imj cos v ( c + a cos v ) 0 0 a cos v ( c + a cos v) Finally, contract with the upper form of the metric to get the Ricci scalar (a.k.a. the curvature scalar): R = g ij R ij = 2cos v a( c + a cos v) The result is twice the Gaussian curvature, as expected. What does the Gaussian curvature tell us about the torus? Since c > a the denominator is always positive, so the sign of the curvature is determined only by cos v. The illustration shows regions of different curvature: on the outside of the torus curvature is positive (blue), on the inside it's negative (red), and at the top and bottom circles it's zero (grey). Understanding the torus's curvature will help us in our search for the torus's geodesics. Last updated 25 April All contents released into the public domain by Mark L. Irons 2 of 3 10/07/ :17 AM

3 The Curvature of the Torus MLI Home Mathematics The Torus Curvature 3 of 3 10/07/ :17 AM

4 The Geodesics of the Torus MLI Home Mathematics The Torus Geodesics The Geodesics of the Torus Here be monsters, or at least varmints The Geodesic Equation Part of our computation of the torus's curvature was the computation of the Christoffel symbols of the second kind. They also appear in the geodesic equation x a a b + Γbc ẋ ẋ c = 0 Plugging in our results, we get a pair of differential equations. u v + 1 a 2a sin v u v = 0 c + a cos v sin v ( c + a cos v )u 2 = 0 Using the substitution w = c + a cos v and integrating (with a trick or two) gives a solution in terms of u and v : u = v = ± k ( c + a cos v) 2 k 2 where k and l are constants of integration. a 2 ( c + a cos v) + l 2 We'd like to integrate these equations to get u as a function of v or vice versa. However, the constant of integration l throws a monkey wrench into the works, and we can't integrate them. That's not the only bad news: most (k,l) pairs don't yield geodesics, and without a way to relate one constant to the other, we have no way to decide whether a given pair yields a geodesic or not. (Thought question: why are there two constants of integration, not one? Only one parameter a direction is required to choose a unique 1 of 8 10/07/ :18 AM

5 The Geodesics of the Torus geodesic through a given point, not counting reversal of direction along the chosen geodesic.) The same problem dashes our hopes of finding a unit speed parameterization. We've hit a dead end; does the Clairaut parameterization help? The Clairaut Parameterization The Clairaut parameterization of a torus treats it as a surface of revolution. From it, we obtain a formula for du dv. du dv = ± ah ( c + a cos v) ( c + a cos v) 2 h 2 Using this difference equation (which, like the solution to the geodesic equation, can't be integrated), we can find the families of geodesics on the torus by varying the parameter h, which is the "slant" of a geodesic. It governs the angle ϕ at which a geodesic crosses parallels. h = ( c + a cos v)sin ϕ For a given geodesic h is constant, which means that as the geodesic passes through regions of different curvature, the angle it makes with parallels changes to compensate. For the torus, this means that as v changes, so must ϕ. Returning to the formula for du, the term under the radical, dv ( c + a cos v) 2 h 2, must be positive. This bounds h: h c + a. Armed with this fact, we can now categorize the geodesics of the torus by their h value. The Five Families of Geodesics [Notes: in this section, we consider only the absolute value of h, not its sign. Changing the sign of h yields a mirror image of the original geodesic. Also, to be precise, this section considers pregeodesics, since we're not providing unit speed parameterizations.] h = 0. These geodesics are simple: as v changes u doesn't, so these are the meridians. 2 of 8 10/07/ :18 AM

6 The Geodesics of the Torus [An intuitive way to see that meridians are geodesics is to realize that the torus has a mirror symmetry through meridians. Anything that would push the geodesic off a meridian in one direction has a balancing mirror image counterpart, so a geodesic that starts on a meridian cannot leave it. A similar argument can be made for both the inner and outer equators, hence they must be geodesics also.] 0 < h < c a. The geodesics now diverge from the meridians, and loop around the torus in the v direction, crossing both the inner and outer equators. Since a geodesic can pass through any point on the surface, we call these unbounded geodesics. Note in the figure above the difference in slant of the geodesic as it crosses the inner and outer equators. This is a consequence of the definition of h. The closer h is to zero, the more tightly wound are the geodesic's coils. As h increases, the coils spread out and the angles at which the geodesic crosses the equators decrease. Since the crossing angle at the inner equator is always less than the crossing angle at the outer equator, the former will reach zero sooner. This is the next case. h = c a. When h reaches this value, the only angle a geodesic can make with the inner equator is zero. Hence a geodesic starting on the inner equator must remain on it, so the inner equator is a geodesic. 3 of 8 10/07/ :18 AM

7 The Geodesics of the Torus The geodesics with this h value that don't start on the inner equator are asymptotic to it. They spiral ever closer to the inner equator without touching it. (This is hard to illustrate without the ability to draw infinitely thin lines. Use your imagination.) This geodesic is unique barring rotation around the z axis and reflection through the xy plane. These geodesics are the edge case of the next family, the bounded geodesics. c a < h < c + a. Things start getting interesting here. A consequence of the Clairaut parameterization is that a geodesic can't leave the region where ( c + a cos v) 2 h 2. This is trivially satisfied for the preceding cases where h c a, but when h > c a (as it now is), something surprising happens: there's a region of the torus which a geodesic can't reach. Instead, geodesics are restricted to the outer region of the torus between two parallels, known as barrier curves. (The pink curves in the illustration.) These geodesics cross the outer equator, but when they touch a barrier curve they reverse v-direction and bounce off it. We call these the bounded geodesics. When h = c a the barrier curves coincide at the inner equator. As h 4 of 8 10/07/ :18 AM

8 The Geodesics of the Torus increases, the barrier curves sweep out from the inner equator, and converge on the outer equator as h c + a. h = c + a. The last geodesic is the limit of the bounded geodesics, the outer equator. When h > c + a, there are no real solutions to the formula for du. dv This completes our survey of the families of a torus's geodesics. To summarize: Slant h Geodesics 0 meridians 0 < h < c a c a c a < h < c + a c + a unbounded geodesics, which alternately cross inner and outer equators the inner equator, and geodesics asymptotic to it bounded geodesics, which cross the outer equator but bounce off barrier curves the outer equator A Gallery of Geodesics The geodesics in the illustrations above were carefully chosen. Most aren't as photogenic; they're aperiodic and cover either the entire torus (for unbounded geodesics) or the entire region between the barrier curves (for bounded geodesics). If we define a geodesic's period as the number of times it circles the z axis before returning to its starting point, then the photogenic geodesics are the unbounded geodesics of period 1, and bounded geodesics of period 1 or 2. The unbounded geodesics of period 1 cross each equator n times (n 1). 5 of 8 10/07/ :18 AM

9 The Geodesics of the Torus n = 1 n = 5 The interesting bounded geodesics fall into two groups. Those of period 1 do not self-intersect. (For bounded geodesics, n denotes how many times the geodesic touches each barrier curve.) n = 1 The geodesics of period 2 intersect themselves n times (n odd, of course). n = 1 n = 3 For bounded geodesics the allowed values of n depend on the ratio c/a. Unbounded geodesics are not affected by c/a. Open Questions The influence of c/a on bounded geodesics 6 of 8 10/07/ :18 AM

10 The Geodesics of the Torus The kinds of bounded geodesics one can find on a particular torus are determined not only by h, but also by the ratio c/a. For instance, given a torus with c/a=3/1, there is no period 1 bounded geodesic which touches each barrier curve more than once. Yet for a torus with c/a=8/1, there is a period 1 bounded geodesic which touches each barrier curve twice, and another which touches each barrier curve three times. This raises a question: what is the range of c/a of the tori that contain bounded period 1 geodesics which touch each barrier curve exactly n times, as a function of n? How about for different periods? There is no analytic apparatus I know of with which we can approach the problem. Calculation appears to be the only way to go. Note that this restriction applies only to bounded geodesics. There is no corresponding restriction for unbounded geodesics; by choosing an appropriate h, one can find a period 1 geodesic which crosses both equators as often as one pleases. Questions about critical values of h Another open question concerns the values of h which yield crowd-pleasing geodesics with periods 1 and 2. As c/a changes, so does the value of h which yields a particular pleasing geodesic (say, a period 1 geodesic which crosses both equators three times). Is there a simple relation between these two quantities? A similar question exists for values of h for a given c/a. Define h p as the value of h which yields a period 1 geodesic which crosses both equators p times. As p increases, at what rate does h p converge to 0? Is this governed by a simple rule? What about period 2 bounded geodesics? Lessons Learned Irons' First Law of Examples About the illustrations. I've attempted to make this page's illustrations as accurate as possible, but there's an inevitable tradeoff between illustrative power and accuracy. If you're curious, you can read about how I created them. Last updated 3 March All contents released into the public domain by Mark L. Irons 7 of 8 10/07/ :18 AM

11 The Geodesics of the Torus MLI Home Mathematics The Torus Geodesics 8 of 8 10/07/ :18 AM

12 Parallel Transport on the Torus MLI Home Mathematics The Torus Parallel Transport Parallel Transport on the Torus Because it really is all about the torus, baby After reading about the torus s curvature, shape operator, and geodesics, you re probably asking yourself how parallel transport on the torus affects tangent vectors. Good question! We start as always with our standard parameterization of the surface: x( u, v) = x = ( c + a cos v)cos u y = ( c + a cos v)sin u z = a sin v This parameterization is particularly nice, since it s an orthogonal patch ( F = x u x v = 0). This lets us easily compute the associated frame field E 1,E 2 : E 1 = x u E E 2 = x v G = ( sin u,cos u,0 ) = ( cos u sin v, sin u sin v,cos v ) Let s check that E 1 and E 2 are orthogonal by computing their dot product. Also, the cross product should be normal to the surface: E 1 E 2 =sin u cos u sin v sin u cos u sin v + 0 = 0, check x y z E 1 E 2 = sin u cos u 0 = cos u cos v, sin u cos v, sin 2 u sin v + cos 2 u sin v cos u sin v sin v sin u cos v Taking partial derivatives of E and G, we have an equation that gives the connection form ω 12 : 1 of 6 10/07/ :19 AM

13 Parallel Transport on the Torus ( ω 12 = E ) v ( d u + G ) u d v = a sin v d u + 0=sin vd u G E a ω 12 tells us how a vector rotates as it is parallel transported. It encodes pretty much everything about the space s curvature. What does this tell us about parallel transport on the torus? First, the d u term tells us that parallel transport along lines of constant u (longitude lines) doesn t affect vectors: We could have predicted this from the symmetry of the torus; along a line of longitude, the neighborhoods to the left and right are mirror images, so there s no preferred direction for a vector to rotate. Second, the sinv term tells us that parallel transport along lines of constant v (latitude lines) causes vectors to rotate through the angle 2π sin v during their journey back to their starting point. Let s look at some specific cases, starting with parallel transport along the outer equator: Here v = 0, so sin v is also zero, and the vectors don t rotate at all. This is also true at the inner equator. 2 of 6 10/07/ :19 AM

14 Parallel Transport on the Torus At the top of the torus (v = π ), sin v = 1, so a vector rotates through a full 2π during 2 its journey. Note how the angle between the blue vector and its path (red) changes as the vector is parallel transported: Something else interesting is happening simultaneously: the vector s origin is also rotating through 2π. These rotations cancel, leaving the blue vector pointing in the same direction in the embedding space. So someone living on the torus would say the vector rotates as it is parallel transported, while someone living outside the surface would not. (Note that at v = π the torus s Gaussian curvature is zero, so it s 2 not surprising that vectors parallel transported along that path don t appear to rotate in the embedding space.) Parallel transport along other lines of latitude causes vectors to rotate varying amounts (2π sin v). In the next illustration, four frames aligned with the u and v axes (at v = 0, π, π, and π ) are parallel transported widdershins around the torus The amount of rotation is a function of their v coördinate. By the time these frames return to their starting point, each will have rotated 2π sin v around its origin. For example, the second line of latitude from the bottom 3 of 6 10/07/ :19 AM

15 Parallel Transport on the Torus is at v = π, so vectors parallel transported along it will have rotated through an 6 angle of 2π sin π = 2π 1 = π, as indeed they have. 6 2 Putting all of this together, here s how a whole bunch of frames rotate while being parallel transported widdershins along lines of latitude, starting at the red longitude line: When creating these images, one thing that surprised me was how quickly the value of 2π sin v changed near v = 0. Then I remembered that d sin v = cos v, which d v has extremes at v = k π, k an integer. So the rate of change of the effect of parallel transport along lines of latitude is most extreme at the outer and inner equators. 4 of 6 10/07/ :19 AM

16 Parallel Transport on the Torus Finally, parallel transport on the bottom half of the torus is the same except for direction of rotation, since sin v is negative there. Students of differential geometry may have noticed that sin vd u is also the ω 12 of the sphere. The difference is that for the sphere, the range of v is π, π, while for 2 2 the torus it s [ π,π]. Indeed, you can see that parallel transport on the outer half of the torus mirrors parallel transport on the sphere. Bonus: computing the Gaussian curvature from the patch Given an orthogonal patch, we can directly compute the Gaussian curvature K from E and G: K = 1 EG ( G) u E u ( E) v + = G v 1 a( c + a cos v) a sin v ( sin v) v = a v a( c + a cos v) = cos v a( c + a cos v) This agrees with the value for the Gaussian curvature we computed from the shape operator, but unlike that calculation this one doesn t require a normal to the surface. Thus, if we lived on the torus, we could compute our space s Gaussian curvature directly from measurements made within our space, and we don t have to assume the existence of an embedding space. That s the beauty of differential geometry. Last updated 18 November of 6 10/07/ :19 AM

17 Parallel Transport on the Torus All contents released into the public domain by Mark L. Irons MLI Home Mathematics The Torus Parallel Transport 6 of 6 10/07/ :19 AM

18 The Shape Operator of the Torus MLI Home Mathematics The Torus Shape Operator The Shape Operator of the Torus What's normal here may not be normal there The shape operator describes how the normal to a surface changes as we move around on the surface. Pretty much everything you could want to know about a surface's curvature is locked up in the shape operator. But, as usual, the torus's shape operator is neglected in the literature. To find it, we begin with the parameterized surface x: x( u, v) = x = ( c + a cos v)cos u y = ( c + a cos v)sin u z = a sin v The normal to this surface is N = ( cos u cos v,sin u cos v,sin v). Taking the partial derivatives of N with respect to u and v gives the shape operator in those directions: S( x u ) = N u = ( sin u cos v,cos u cos v,0) S( x v ) = N v = ( cos u sin v, sin u sin v,cos v) Comparing these to the partial derivatives of the parameterization, x u = ( ( c + a cos v)sin u, ( c + a cos v)cos u,0) x v = ( a cos u sin v, a sin u sin v,a cos v) we find the former are multiples of the latter: 1 of 4 10/07/ :18 AM

19 The Shape Operator of the Torus S( x u ) = cos v c + a cos v x u S( x v ) = 1 a x v The shape operator is linear, so we can find it not only for the x u and x v directions, but for any direction θ (measured from x v ): S( x) = cos v c + a cos v x u cos θ 1 a x v sin θ From the shape operator, we obtain the Gaussian curvature K (the determinant of the shape operator) and the mean curvature H (the trace of the shape operator). cos v 0 c + a cos v K = 0 1 = a cos v a( c + a cos v) H = c + 2a cos v 2a( c + a cos v) The Geometric Significance of the Shape Operator For a circle with a normal pointing outward, the numeric coefficient of the shape operator is 1, where r is the circle's radius. We can use this r to check our solution. Consider the how the normals change in the following illustration. 2 of 4 10/07/ :18 AM

20 The Shape Operator of the Torus The blue normals lie on a meridian (u constant, v changing), hence they change at a constant rate of 1. This is precisely the negative reciprocal a of the minor radius of the torus, so that's correct. The green normals are on the uppermost circle of the torus, where v = π 2. The numeric coefficient of the shape operator in the u direction is a multiple of cos v = 0, so the shape operator on this circle in the u direction is zero. As the illustration demonstrates, these normals don't change direction as we move around the circle; they all point straight up (discounting the illustration's perspective, of course). This makes sense for another reason. The curvature of the torus on its uppermost and lowermost circles is zero. Hence locally along these circles, the torus looks like a cylinder. Since the shape operator of a cylinder in a direction parallel to its axis is zero, we'd expect the shape operator along these circles to be zero as well. The brown normals lie on the circle v = 0. Working out the shape operator in the x u direction, we find its numeric coefficient to be 1 c +a, the negative reciprocal of the outer equator's radius, as expected. The last set of normals to check are the red ones on the inner equator. Here v = π, so instead of being negative, the shape operator's numeric coefficient is positive. We can see the difference: instead of diverging like the normals on a meridian or the outer equator, the tips of these normals are closer than their bases. This is a sign that we're on the inside of a circle, where the numeric coefficient of the shape operator is positive. 3 of 4 10/07/ :18 AM

21 The Shape Operator of the Torus And, being 1, it is. c a Four checks, four sensible results. Looks like a winner. If you enjoyed this, you might try verifying that the shape operator takes the form 1 v (for some constant p) along the torus's Villarceau circles. p That's a level of masochism I'm not prepared to descend to. Last updated 4 October All contents released into the public domain by Mark L. Irons MLI Home Mathematics The Torus Shape Operator 4 of 4 10/07/ :18 AM

Parallel Transport on the Torus

MLI Home Mathematics The Torus Parallel Transport Parallel Transport on the Torus Because it really is all about the torus, baby After reading about the torus s curvature, shape operator, and geodesics,