Sides of the M?bius Strip

Sides of the M�obius StripThomas Randrup and Peter R�genJune, 1995AbstractWe present the necessary and su�cient conditions for a curve to bethe center curve of an analytic and at embedding of the M�obius strip (oran orientable cylinder) into euclidean 3-space. Using these conditions weextend an example by G. Schwarz into a continuous family of analytic and at M�obius strips. This family is split into two connected components.We give a topological argument that explains this behaviour.1 IntroductionThe works [5] and [6] of G. Schwarz focus on at and analytic embeddings of theM�obius strip into euclidean 3-space. The article [6] presents an analytic, alge-braic, and at embedding of the M�obius strip into euclidean 3-space. Throughan analysis of his construction we have found necessary and su�cient condi-tions for a curve to be the center curve of an analytic and at embedding of theM�obius strip (or the orientable cylinder) into euclidean 3-space. By a M�obiusstrip respectively an orientable cylinder we mean a geometric object obtainedas follows: Take a strip of paper and \twist" it, tie a knot on it, and glue itsends together. From the necessary and su�cient conditions we can provide acontinuous family of analytic and at M�obius strips given by trigonometric poly-nomials. Furthermore, we will give a topological argument using the \twist" ofM�obius strips explaining why such families never are connected. Finally, we willpoint out how the work [1] of C. Chicone and N.J. Kalton on at and analyticM�obius strips is connected to the construction we consider below.2 Necessary and Su�cient ConditionsIn the following we will consider isometric embeddings of the M�obius stripinto euclidean 3-space. Consequently, all geodesic curves on an embedding arestraight lines on an unfolding of the embedding. Consider a rectangular unfold-ing of a M�obius strip. On this unfolding there are two types of closed straightlines, namely, those which close after two traversions and exactly one whichcloses after one traversion. We will call this unique shortest closed geodesiccurve on a at M�obius strip the center curve of the strip.1

2 Thomas Randrup and Peter R�genLet the Frenet frame (t;n;b) along a curve r : I ! R3 with r0 � r00 6= 0satisfy the cross product relationst = n� b; n = b� t; and b = t� nand the Frenet formulas t0 = �nn0 = ��t+ �bb0 = ��nwhere � > 0 is the curvature and � is the torsion of r. This determines the signof the torsion1.Given a curve with non-vanishing curvature there exists one unique at ruledsurface on which this curve is a geodesic curve. This at surface is the so calledrectifying developable. For reference see [3], p. 71. A short calculation showsthat given a curve r : I ! R3 satisfying r0� r00 6= 0 its rectifying developable isgiven by ~r(s; t) = r(s) + t(�t+ �b); s 2 I; t 2 R:If the interval I is compact there exists an � > 0 such that the restriction~rjI�]�";"[ is an embedding. If one instead chooses the ruling ��t+b the parametert is the geodesic distance to the center curve.The rectifying developable has the same normal vector as the principal nor-mal vector of the center curve along the center curve. If the center curve hasnon-vanishing curvature it has a Frenet frame in every point. So, the rectifyingdevelopable will be an orientable surface.To obtain a smooth surface it is only necessary to have a projectively well-de�ned normal vector �eld. This leads toDe�nition 1 Let r : I ! R3 be a regular curve and let t0 2 I. Assume thelimits n�(t0) = limt!t�0 n(t) exist.� If n�(t0) = n+(t0), then r(t0) will be called a frame preserving point.� If n�(t0) = �n+(t0), then r(t0) will be called a frame switching point.Points with non-vanishing curvature are frame preserving. But a point withvanishing curvature can be a frame preserving point, frame switching point, ora point where the principal normal cannot be projectively well-de�ned.In order to obtain a M�obius strip it is necessary that its center curve hasan odd number of frame switching points and the rest frame preserving. Todescribe frame preserving and switching points we giveTheorem 2 Let r : I ! R3 be a Cn-parametrization of a regular curve, n � 2,so that the points with zero curvature are isolated. Set J = (r0 � r00)�1(0). Iffor t0 2 J the natural number (the order of the zero for the curvature)N (t0) = min0�j�n�2(j ����� jXi=0 r(i+1)(t0)� r(j�i+2)(t0)i!(j � i)! 6= 0)1In [1] they use the opposite sign on the torsion.

Sides of the M�obius Strip 3exists, then r(t0) is a frame preserving point if N (t0) is even and r(t0) is aframe switching point if N (t0) is odd.Proof: Let r : I ! R3 be as in the theorem. We may assume that t0 = 0. So,r0 � r00(0) = 0 and r0 � r00(t) 6= 0 in a punctured neighbourhood of zero. If theparametrization is Cn it has the Taylor seriesr(t) = nXi=0 ri(0)i! ti + o(tn)and the jth derivative has the Taylor seriesr(j)(t) = n�jXi=0 ri+j(0)i! ti + o(tn�j):Cauchy multiplication givesr0 � r00(t) = n�2Xj=0 jXi=0 r(i+1)(0) � r(j�i+2)(0)i!(j � i)! ! ti + o(tn�2):From the cross product relation n = b � t it follows that n+(0) = �n�(0)are equivalent to b+(0) = �b�(0). As b = r0�r00jr0�r00j it follows that the leadingterm in the Taylor series of r0 � r00 gives the sign in the right hand side ofb+(0) = �b�(0). Denoting the order of the leading term in the Taylor seriesof r0 � r00 by N (0), i.e.,N (0) = min(j 2 f0; 1; : : : ; n� 2g ����� jXi=0 r(i+1)(0) � r(j�i+2)(0)i!(j � i)! 6= 0)we have b+(0) = (�1)N(0)b�(0) if N (0) exists. Hence,n+(0) = (�1)N(0)n�(0): �Conditions for pieces of rectifying developables to �t together in frame pre-serving and switching points are given inTheorem 3 Let r : I ! R3 be a Cn-parametrization of a regular curve, n � 3,so that the points with zero curvature are isolated. Set J = (r0 � r00)�1(0). Iffor every t0 2 J the natural numbersN (t0) = min0�j�n�2(j ����� jXi=0 r(i+1)(t0) � r(j�i+2)(t0)i!(j � i)! 6= 0)

4 Thomas Randrup and Peter R�genand M (t0) =min0�k�n�38<:k ������ kXj=0 1(k � j)! jXi=0 r(i+1)(t0)� r(j�i+2)(t0)i!(j � i)! ! � r(k�j+3)(t0) 6= 09=;exist, then the rectifying developable of r exists if and only if for every t0 2 Jthe inequality M (t0) � 3N (t0) is ful�lled.If the parametrization r : I ! R3 is C1 then M = +1 is allowed.Remark 4 Observe that N is the order of the zero for the curvature and thatM is the order of the zero for the torsion multiplied by the squared curvature.Hence, the inequality M (t0) � 3N (t0) expresses geometrically that the order ofthe zero for the torsion has to be bigger than or equal to the order of the zerofor the curvature.Proof: On a ruled surface the rulings only have to be projectively well-de�ned.We will now analyse the situation where the ruled surface is the rectifyingdevelopable of a regular Cn-curve which has an isolated zero for its curvature.We may assume that this happens for t = 0. For t 6= 0 we haveqjqj = �b+ �tp�2 + �2= 1jr0j3 r0�r00jr0�r00 j + [r0r00r000]jr0�r00 j3 r0jr0jq 1jr0j6 + [r0r00r000]2jr0�r00 j6 :As r0 6= 0 the rulings q are projectively well-de�ned iflimt!0+ r0 � r00jr0 � r00j = � limt!0� r0 � r00jr0 � r00j ) limt!0+ [r0r00r000]jr0 � r00j3 = � limt!0� [r0r00r000]jr0 � r00j3 :Suppose thatN (0) exists, then 3N (0) is the order of the leading term in jr0�r00j3.Let M (0) be the order of the leading term in[r0r00r000](t) = n�3Xk=00@ kXj=0 1(k � j)! jXi=0 r(i+1)(0)� r(j�i+2)(0)i!(j � i)! ! � r(k�j+3)(0)1A tk+ o(tn�3):Then the limits for [r0r00r000]jr0�r00j3 exist if and only ifM (0) � 3N (0). If M (0) = 3N (0)the signs in the above implications are correct else limt!0 [r0r00r000]jr0�r00j3 = 0, whereby,the implications trivially are ful�lled. �In connection to the above proof note that, if we cut a M�obius strip alongthe ruling through r(t0) and unfold it, then the unfolding of the ruled surface

Sides of the M�obius Strip 5will be a rectangle exactly when limt!t0 [r0r00r000]jr0�r00j3 = 0, i.e., M (t0) > 3N (t0), elsethis unfolding is a trapezium.If qjqj are the rulings before a frame switching point, r(t0), it is necessary tochange to � qjqj as rulings after this frame switching point2. This procedure en-sures that the rectifying developable of an analytic curve is analytic because theruling qjqj is � t�t0jt�t0j�N(t0) times an analytic vector �eld. Note, that analyticityof a non-straight curve ensures that the points with zero curvature lie discreteon the curve, whereby, Theorem 3 applies.The next theorem classi�es the closed curves described in Theorem 3.Theorem 5 Let the conditions in Theorem 3 be ful�lled by a closed curve withno self intersections. If the number of frame switching points� is even, then the rectifying developable is an orientable, at surface.� is odd, then the rectifying developable contains a at M�obius strip.If the parametrization is analytic these conditions are both necessary and suf-�cient for the curve to be the center curve of a at and analytic M�obius striprespectively an orientable, at, and analytic surface.Proof: Let r be a curve that ful�lls the conditions in Theorem 3. In a frameswitching point the rulings have opposite limits if and only if the integers Nand M ful�ll the conditions in Theorem 3. So, if the number of frame switch-ing points is even respectively odd the rectifying developable is an orientablerespectively unorientable, at surface.Consider an analytic closed curve. As already noticed, this curve only hasisolated zeros for its curvature. These zeros are of �nite order since else thecurve has constant zero curvature. Thus, the integer N exists in each of theisolated zeros for the curvature.If an analytic curve has M = +1, then [r0r00r000] is identical equal to zero.Hence, the curve lies in a plane, and will, therefore, carry an orientable, atsurface. Consider the curve as a curve in the plane. In a frame switching pointthe curvature of the plane curve changes sign. Since the curvature of the planecurve is periodic, the number of frame switching points is even.By these arguments we can conclude that the conditions in Theorem 5 areboth necessary and su�cient for an analytic curve to be the center curve of a at and analytic M�obius strip or of an orientable, at, and analytic surface. �Remark 6 It is noteworthy, that as well atness of surfaces as the constantsN and M from Theorem 3 are preserved under a�ne transformations. Thus,we are only interested in �nding center curves of M�obius strips modulo a�netransformations.2On a M�obius strip this gives a two fold covering.

6 Thomas Randrup and Peter R�genThroughout we will consider at least three times continuous di�erentiableparametrizations r : I ! R3 of regular closed curves. Hence, they all haveFourier expansions. Given a parametrization, r : I ! R3, of a regular closedcurve we say that it is of degree n (+1 is allowed) if the Fourier expansion ofr is of the form r(t) = a0 +Pnm=1 (am cosmt + bm sinmt). By the degree of aclosed curve we mean the minimumof the degrees of its parametrizations whereagain +1 is allowed.Theorem 7 Given the center curve, , of a M�obius strip. Let p be a projectionof into a plane, p. And let l be a projection of into a line, l. Thendeg( ) � 3, deg( p) � 2, and deg( l) � 1.Proof: It is not possible for a curve, , parametrized by a trigonometric poly-nomial of degree 2 to be the center curve of a at M�obius strip. This is veri�edby using Theorem 5. Since, after some calculation we �nd that already thedemands N (0) � 1 and M (0) � 3 force the curve to lie in a plane, whereby, itsrectifying developable is orientable. Hence, deg( ) � 3.Let p be a projection of into a plane, p. If the degree of p is one, then p is an ellipse. Hence, p and, thereby, also has non-vanishing curvature.Therefore, carries an orientable surface.Let l be a projection of into a line, l. If deg( l) = 0 then l is a point,i.e., lies in a plane, whereby, its rectifying developable is orientable. �3 A Family of M�obius StripsWe will illustrate how Theorem 5 can be used to �nd center curves for at andanalytic M�obius strips. We want to �nd degree 3 curves which each has a degree2 plane projection and a degree 1 line projection. It is, according to Theorem7, necessary that the line, l, is contained in the plane, p. Hence, there exists acoordinate system (x; y; z) such that l is the x-axis and p is the xy-plane.According to Theorem 5 there has to be at least one point on the curve rwith zero curvature. Consider the projection, rxy, of the space curve r into thexy-plane. If r has zero curvature in a point, then rxy has zero curvature in theprojection of this point. We can assume that there is zero curvature for t = 0,and that rxy(0) = (0; 0), r0xy(0) = (1; 0), and r00xy(0) = (?; 0). Solving theseequations we have the expressionrxy(t) = �a (1� cos t) + sin t; b�1� 43 cos t+ 13 cos 2t�+ c�sin t� 12 sin 2t�� :The only a�ne transformation left over is that the y-coordinate can be scaled.So, we might choose b2+ c2 = 1. By construction we now have found all degree1,2 projections of center curves of at and analytic M�obius strips up to a�netransformation.

Sides of the M�obius Strip 7To simplify calculations and to include G. Schwarz's example we take a =0, b = 0, and c = 1. As the third coordinate of the curve r we choose atrigonometric polynomial of degree 3. So, we consider the following family ofcurves: r(t) = (sin t; sin t(1� cos t); z(t))where z is a trigonometric polynomial of degree 3, i.e.,z(t) = a0 + 3Xm=1 (am cos (mt) + bm sin (mt)) for t 2 [0; 2�[:Theorem 8 A curve r from the above family of curves with zero curvature fort = 0 is the center curve of a at and analytic M�obius strip if and only ifr(t) = (sin t; sin t(1� cos t); z(t))withz(t) = a2�53 � 52 cos t + cos 2t� 16 cos 3t�+ b1 sin t + b2 sin 2t+ b3 sin 3twhere a2 2 Rnf0g and b1; b2; b3 2 R.All other center curves of at and analytic M�obius strips from the abovefamily of curves are a�ne transformations of those with zero curvature for t = 0.Proof: The projection, rxy, into the xy-plane has only zero curvature for tequal to zero or �. This means that the curvature of the space curve r only canbe zero for t equal to zero or �. Since,1Xi=0 r(i+1)(0)� r(1�i+2)(0)i!(1� i)! = r0(0) � r000(0) + r00(0) � r00(0)= r0(0) � r000(0)= (1; 0; b1+ 2b2 + 3b3)� (�1; 3;�b1 � 8b2 � 27b3)= (?; ?; 3) 6= 0and1Xi=0 r(i+1)(�)� r(1�i+2)(�)i!(1� i)! = r0(0)� r000(0)= (�1;�2;�b1 + 2b2 � 3b3) � (1; 5; b1� 8b2 + 27b3)= (?; ?;�3) 6= 0it is only possible for N to equal zero and one in r(0) and in r(�) (cf. Theorem3). According to Theorem 5 we can only obtain M�obius strips if (N (0); N (�))

8 Thomas Randrup and Peter R�genequals (1; 0) or (0; 1). The two cases (0; 0) and (1; 1) give orientable surfaces.We will start with the case (N (0); N (�)) = (1; 0). Now, N (�) = 0 requiresr0 � r00(�) = (�2z00(�); z00(�); 0) 6= 0 , z00(�) = a1 � 4a2 + 9a3 6= 0: (1)In order to use Theorem 5 we need to know the �rst four derivatives of r fort = 0. r0 = (1; 0; z0(0)) = (1; 0; b1+ 2b2 + 3b3)r00 = (0; 0; z00(0)) = (0; 0;�a1 � 4a2 � 9a3)r000 = (�1; 3; z000(0)) = (�1; 3;�b1 � 8b2 � 27b3)r(4) = (0; 0; z(4)(0)) = (0; 0; a1+ 16a2 + 81a3)For N (0) to be greater than or equal to one there has to be zero curvaturefor t = 0. Hence,r0 � r00 = (0;�z00(0); 0) = 0 , z00(0) = a1 + 4a2 + 9a3 = 0: (2)The statements (1) and (2) implya2 6= 0 ^ a1 + 4a2 + 9a3 = 0: (3)Note, that (2) is equivalent to the statement r00(0) = 0. Requiring N (0) to beequal to one demands that M (0) is bigger than two (cf. Theorem 3). This givesthe following constrains for t = 0.M (0) 6= 1: 1Xj=0 1(1� j)! jXi=0 r(i+1) � r(j�i+2)i!(j � i)! ! � r(1�j+3)= [r0r00r(4)] + [r0r000r000] + [r00r00r000]= 0:M (0) 6= 2:2Xj=0 1(2� j)! jXi=0 r(i+1) � r(j�i+2)i!(j � i)! ! � r(2�j+3)= 12[r0r00r(5)] + [r0r000r(4)] + [r00r00r(4)] + 12 [r0r(4)r000] + [r00r000r000] + 12 [r000r00r000]= 12 [r0r000r(4)]= 12 ������ 1 �1 00 3 0z0(0) z000(0) z(4)(0) ������ = 0, z(4)(0) = a1 + 16a2 + 81a3 = 0: (4)Combining the statements (3) and (4) we have the family of center curves givenin the theorem.

Sides of the M�obius Strip 9The solutions for (N (0); N (�)) = (0; 1) are the same as those we alreadyhave found. This can easily be seen as follows. First, change the parameter tto u+ � leading torxy(u+ �) = �� sinu;� sinu� 12 sin 2u� :Let the xy-plane undergo the linear transformation described by the matrix� �1 0�2 1 �. Then we obtain~rxy(u) = �sinu; sinu� 12 sin 2u� :Using the parameter uwe now have to �nd the solutions ful�lling (N (0); N (�)) =(1; 0), which we already have found. �The �rst curve which is considered in the article [6] is the curve of degree 2given byr1(t) = (sin t; sin t(1� cos t); (1� cos t)2)= � 003=2�+ � 00�2� cos t+ � 001=2� cos 2t+ � 110� sin t + � 0�1=20 � sin 2t:G. Schwarz concludes that this curve cannot produce a M�obius strip since fort = 0 the rectifying developable is singular. But as the degree of the curve r1is 2 it follows immediately from Theorem 7 that r1 cannot produce a M�obiusstrip.The way that G. Schwarz succeeds in �nding a at, analytic, and algebraicM�obius strip is by raising the degree of the third coordinate. He proves in [6]that the curve r2(t) = (sin t; sin t(1� cos t); (1� cos t)3)is the center curve of a at, analytic, and algebraic M�obius strip. We observethat the curve found by G. Schwarz is the one from Theorem 8 with a2 = 32 andb1 = b2 = b3 = 0, sincez(t) = (1� cos t)3 = 32 �53 � 52 cos t+ cos 2t� 16 cos 3t� :4 Disconnectedness of the FamilyWe will now give a topological argument that explains why any family of centercurves of M�obius strips obtained as in Section 3 cannot be connected. Givenan embedding of the M�obius strip into euclidean 3-space we will now focus onits center curve and its boundary curve. The linking number (see e.g. [2], pp.79-80) between the center curve and the boundary curve of an embedding ofthe M�obius strip gives a natural invariant for embeddings of the M�obius strip.We shall now describe this invariant.

10 Thomas Randrup and Peter R�genThe linking number between two oriented curves is invariant under shiftof the orientation of both curves. Hence, orienting both the center curve andthe boundary curve, from a choice of a direction in which the embedding istraversed, makes the linking number between the center curve and the boundarycurve well-de�ned as an invariant attached to embeddings of the M�obius strip.This invariant will always equal an odd integer (and it equals twice the intuitivetwisting number of each embedding). See [4] for related work where we use thetwisting number and the knot described by the center curve to classify the classof closed strips in euclidean 3-space.A re ection of euclidean 3-space in a plane changes the orientation of eu-clidean 3-space. Hence, this invariant for embeddings of the M�obius stripchanges sign if the embeddings are re ected in a plane. So, since this invariantdoes not equal zero an embedding can never lie in the same connected compo-nent of a family of embeddings as its mirror image.Finally, a change of sign on the parameter space in our family of centercurves of M�obius strips corresponds exactly to a re ection in the plane givenby z = 0 (cf. Theorem 8). This is a topological explanation why the parameterspace in our family of center curves of M�obius strips has to be disconnected.5 Another M�obius Strip ConstructionWe will point out how the work of C. Chicone and N.J. Kalton in [1] on atand analytic M�obius strips is connected to the construction of at and analyticM�obius strips presented above. The main concern in [1] is with axes of ruleddevelopable surfaces with the property that they are orthogonal to the rulings.The following theorem and remark describe these orthogonal axes.Theorem 9 On a ruled surface containing a M�obius strip whose center curve,r : [0; T [! R3, only has one point with zero curvature there exists one uniquecurve, y, orthogonal to the rulings, q, which closes after one traversion. Thecurve y : [0; T [!R3 is given byy(s) = r(s) + f(s) qjqjwhere s is the arc length of the curve r, �(0) = 0, andf(s) = 12 Z Ts �p�2 + �2du� Z s0 �p�2 + �2 du! :All other curves which are orthogonal to the rulings closes after two traversions.Proof: Since q never is parallel to the tangent of the center curve, a curveorthogonal to the rulings of a ruled surface can be written on the formy(s) = r(s) + f(s) qjqj :

Sides of the M�obius Strip 11Choosing a normalized ruling q = �t+�bp�2+�2 , we obtain y(s) = r(s) + f(s)q. Theorthogonality of y and the ruling gives0 = y0 � q = t � q+ f 0(s):Hence, f(s) = � Z s0 �p�2 + �2 du+ f(0):Since the curve might not close after one traversion of the ruled surface we putindices on the maps according to the number of traversions. The curve y is con-tinuous (which automatically, by orthogonality, ensures that y is continuouslydi�erentiable) if fi(0) = �fi�1(T ); i 2Z:Hence, the curve y closes after, at most, two traversions of the ruled surface. Aswith the closed geodesic curves on a M�obius strip (cf. page 1) there is exactlyone of these orthogonal curves which closes after one traversion of the ruledsurface. For y to close after one traversion it is necessary and su�cient thatf(T ) = �f(0), i.e.,f(T ) = � Z T0 �p�2 + �2 du+ f(0) = �f(0)The following calculation completes the proof.f(s) = � Z s0 �p�2 + �2du+ 12 Z T0 �p�2 + �2 du= 12 Z Ts �p�2 + �2 du� Z s0 �p�2 + �2du! : �Remark 10 It is obvious, that the description of the orthogonal axes in The-orem 9 also apply if the center curve has more than one isolated zero for itscurvature ful�lling Theorem 5. But then the function f in Theorem 9 has tobe discontinuous, whenever, the integer N from Theorem 3 is odd, leading to amore complicated formula for the function f .Again, since an analytic center curve of a M�obius strip has isolated zerosfor its curvature, we conclude that the total ruled developable of this analyticcenter curve contains one curve orthogonal to the rulings which closes after onetraversion. Henceforward, we shall refer to this curve as the orthogonal curve.All other curves which are orthogonal to the rulings closes after two traversions.This orthogonal curve is closely connected with at and analytic M�obiusstrips through the following theorem by C. Chicone and N.J. Kalton [1].

12 Thomas Randrup and Peter R�genTheorem 11 (C. Chicone and N.J. Kalton) Let ra : R! R3 be a closedsmooth (respectively analytic) curve in R3 with non-vanishing curvature. Thenra is an axis of a ruled developable M�obius strip with smooth (respectively ana-lytic) rulings everywhere orthogonal to ra if and only if the total torsion of rais an odd multiple of �.As mentioned in Section 4, the linking number between the center curve andthe boundary curve of an embedding of the M�obius strip is an invariant attachedto the embedding. Since this linking number equals an odd integer and sincethe total torsion of the orthogonal curve divided by � is an odd integer; we arelead to ask if there is a connection between the total torsion of the orthogonalcurve and the linking number between the center curve and the boundary curve.Actually, a curve satisfying Theorem 11 is an axis not only of one ruleddevelopable M�obius strip, but of a family of ruled developable M�obius strips.Let ra be an axis as described in Theorem 11 parametrized by arc length, s,then for all k 2 R the rulingsrr(s) = cos�� Z s0 �du+ k�n+ sin�� Z s0 �du+ k�balong ra gives at ruled non-orientable surfaces.There are two problems with the construction used by C. Chicone and N.J.Kalton. One problem is that the orthogonal curve of a at analytic M�obiusstrip might have isolated zeros for its curvature. This problem can, however, betreated as we have treated the analogous problem for center curves of at andanalytic M�obius strips. The other problem is that the unique center curve ofthe ruled surface might have self-intersections, whereby, the ruled surface willnot contain an embedding of the M�obius strip isomorphic to a rectangle. Inother words, their construction does not give control of the unfolding. In thisconnection one has to be aware of the fact that it is hard to �nd (i.e. solve thegeodesic equation) the center curve from the orthogonal curve.AcknowledgmentsIt is our pleasure to thank J. Gravesen and S. Markvorsen for suggesting theproblems and for fruitful discussions during our work and writing.

Sides of the M�obius Strip 13References[1] Chicone C. & Kalton N.J., Flat Embeddings of the M�obius Strip in R3,Preprint, Department of Mathematics, University of Missouri, Columbia,(1984).[2] Flanders H., Di�erential Forms, Mathematics in Science and Engineering,Academic Press, New York, (1963).[3] Graustein W.C., Di�erential Geometry, Dover Publications, Inc., NewYork, (1966).[4] Randrup T. & R�gen P., How to Twist a Knot, Master of Science Thesis,Mathematical Institute, Technical University of Denmark, (January 1995).[5] Schwarz G., The Dark Side of the Moebius Strip, Amer. Math. Monthly,Vol. 97, No. II, (1990), pp. 890-897.[6] Schwarz G., A pretender to the title \Canonical Moebius Strip", Paci�c J.Math., Vol. 143, No. 1, (1990), pp. 195-200.Thomas Randrup and Peter R�genMathematical InstituteTechnical University of DenmarkBuilding 303DK-2800 LyngbyDenmarkE-mail addresses:[email protected]@mat.dtu.dk