2. A Flat Möbius Band in R³

The usual instructions for making a Möbius band say: Take a long rectangular piece of paper, give it a half twist, and glue the ends together. Mathematically, such a model (shown below) can be gotten by moving a line segment around a circle, giving it only a half turn as it traverses the full circle.

However, coordinate computations (say, using superK) show that this Möbius band is not flat. (In fact, its Gaussian curvature is everywhere negative: K<0.) Though it may not be evident, the half-twist has distorted the flat paper.

How can a Möbius band in R³ manage to be flat?

To build one, start with an abstract Möbius band, that is, a closed rectangle in the Euclidean plane R² with opposite ends identified in reverse (as shown here). This band is certainly flat.

Our goal is to find in R³ an isometric copy B of an abstract Möbius band; in other words, an isometric imbedding of the abstract model into R³. Then B also will be flat.

In the construction we use the following commands for a rotation of R², a translation of R³ by (a,b,c), and a rotation of R³ around the z-axis.

r2[a_][u_,v_]:={Cos[a]*u-Sin[a]*v, Sin[a]*u+Cos[a]*v}

tr[a_,b_,c_][{u_,v_,w_}]:={u+a,v+b,w+c}

r3z[a_][{u_,v_,w_}]:={Cos[a]*u-Sin[a]*v,Sin[a]*u+Cos[a]*v,w}

Note: Recall that for Mathematica, {r₁,..., r_n} always denotes a list, not a set. So if the r_i are real numbers, then {r₁,..., r_n} is an n-vector, that is, an element of Rⁿ.

Suppose that, like those above, functions f and g are vector-valued. In a composite function Composition[g,f], the function g is applied to the values of f, so g must be defined on vectors--hence the braces ("curly brackets") in these definitions. No vector-valued function will feed into r2, and in such cases, no braces are needed.

We build the Möbius band B by assembling three parts. All three have essentially the same shape, namely, a rectangle that has been bent--on the bias--through 180^o.

To do this, we first bend an entire plane halfway around a cylinder, and then insert a strip in the plane at an angle off vertical--with result as shown here.

The bent plane is defined using the Mathematica command Which that consists of a condition, the input if that condition holds, then another condition, another input, and so on.

Building Part 1

    bend[{u_,v_}]:= Which[v<-Pi/2,{u,v+Pi/2,-2},

      v^2<Pi^2/4,{u,Cos[v],-1+Sin[v]},

      v>Pi/2,{u,-v+Pi/2,0}]

To check that this works we could plot it by:

    ParametricPlot3D[bend[{u,v}],{u,-3,3}],{v,-5,5},

      PlotPoints->{15,25}, Boxed->False, Axes->None]

The strip described above is given by a composite function that first rotates the plane by Pi/6=30^o, then applies bend:

    pone[u_,v_]:= Composition[bend,r2[Pi/6]][u,v]

Then Part 1 of the Möbius band is plotted by

    part1=ParametricPlot3D[pone[u,v],{u,-1.7,1.7},{v,-5,5},

      PlotPoints->{15,25}]

The map bend is of the ordinary Euclidean plane into R³ (it has E=1,F=0,G=1), so the restriction to the strip is also an isometric imbedding. In particular, this proves that Part 1 really is flat.

Building Part 2

The second part of B is another climbing turn to the left. It is easily produced by rotating Part 1 through 120^o, then attaching it to Part 1 by a suitable translation.

The rotation of Part 1 is given by

    preptwo[u_,v_]:= Composition[r3z[2Pi/3],pone][u,v]

The translation must carry preptwo[0,-5] to pone[0,5]. So we substitute

    vtr1=N[pone[0,5]-preptwo[0,-5]

into the translation command tr to get

    ptwo[u_,v_]:= Composition[tr[vtr1], preptwo][u,v]

Then Part 2 is plotted by

    part2=ParametricPlot3D[ptwo[u,v],{u,-1.7,1.7},{v,-5,5},

      PlotPoints->{15,25}, Boxed->False, Axes->None]

If all has gone well,    Show[part1,part2]     will indeed show the union of Parts 1 and 2. Note that its planar regions are on three different horizontal levels.

Building Part 3

The third part of B has to be somewhat different: As the figure above shows, Part 1 rose from z=-2 to z=0, and Part 2 rose from z=0 to z=+2. Thus to complete the band, Part 3 must be a descending turn (to the left) that drops from z=+2 to z=-2.

We can get a command that will plot the shape of Part 3 (though not its location) by simply changing the z coordinate of pone. Multiplying z by -2 will produce a descending turn (to the left) from z=4 to z=0. Subtracting 2 will make the drop go from +2 to -2.

    prep3[u_,v_]:= {pone[u,v][[1]], pone[u,v][[2]],

     -2(1 + pone[u,v][[3]])}

Now as before, we rotate and translate to fit Part 3 onto the earlier parts. This time the rotation of prep3 is through 240^o (or -120^o).

    rprep3[u_,v_]:= Composition[r3z[-2Pi/3],prep3][u,v]

The required translation carries prep3[0,5] to pone[0.-5], hence is given by substituting

    vtr2=N[pone[0,-5]-prepthree[0,5]]

into the translation command. Finally, we define

    pthree[u_,v_]:= Composition[tr[vtr2], prep3][u,v]

Then Part 3 is given by

    part3=ParametricPlot3D[pthree[u_,v_],

      {u,-1.7,1.7},{v,-5,5}, PlotPoints->{15,25}]

Assembly

If all has continued to go well, the three parts will fit together to form a flat Möbius band.

    flatmb=Show[part1,part2,part3]

       Show[xflatmb, Boxed->False, Axes->None,

       ViewPoint->{-1.3,2.4,2}, ImageSize->360]

The new viewpoint {-1.3,2.4,2} replaces Mathematica's default viewpoint {1.3,-2.4,2}, and the figure has also been slightly enlarged.

Thanks to UCLA's Bob Edwards for the scheme of this page and to Peter van Summeren (peter_van_summeren@hotmail.com) for corrections and suggestions.

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