A Flat Klein Bottle Isometrically Imbedded in R⁵

Reference: C. Tompkins, "A flat Klein bottle isometrically imbedded in Euclidean 4-space," Bull. Amer. Math. Soc. 47 (1941), 508.

As Tompkins knew, his example in R⁴ is not one-to-one, so in current terminology it's only an isometric immersion. We describe it first, then modify it to get an imbedding in R⁵.
The scheme is this:   For each number u, consider the plane in R⁴ spanned by the orthogonal unit vectors

(cos(u), sin(u), 0, 0)   and   (0, 0, cos(u/2), sin(u/2)).

As u increases from 0 to 2 this plane makes a smooth circuit in R⁴ -- and though the second vector above doesn't return to its original value, the plane does.

Now draw an ellipse in each of these planes by defining

x(u,v) = cos(v) (cos(u), sin(u), 0, 0) + 2 sin(v) (0, 0, cos(u/2), sin(u/2)

    = (cos(u) cos(v), sin(u) cos(v), 2 cos(u/2) sin(v), 2 sin(u/2) sin(v)).

Here the 2's compensate for the half angles, for with them, superK shows at once that the image of x is flat.

(Aside: The map x is not isometric, for we compute E=1, F=0, G=(5+3 cos 2v)/2, while x isometric requires G=1. Since EG-F²=G is strictly positive, the map x is regular, and since G depends solely on v, only elementary formulas for Gaussian curvature are needed to show K=0.)

The map x will produce a mapping of the Klein bottle B into R⁴ provided that its restriction to the square D = [0,2] × [0,2] satisfies the boundary conditions suggested in the figure, namely,

x(u, 2) = x(u, 0)     and     x(2,v) = x(0, 2-v).

It's easy to see that these equations do hold.

If x is one-to one on the half-open square D* =[0,2) × [0,2), then it is a imbedding of B in R⁴. Unfortunately, one-to-one fails, but only along the u-parameter curves v=0 and v=, namely

x₀(u) = x(u,0) = (cos(u), sin(u), 0, 0),

    x (u) = x(u,) = (-cos(u), -sin(u), 0, 0).

where 0 u < 2. Explicitly, the first half of x₀ matches the last half of x, and vice versa. These crossings occur when sin(v)=0, which suggests that they can be eliminated by adding cos(v) as a fifth component:

xx(u,v) = (cos(u) cos(v), sin(u) cos(v), 2 cos(u/2) sin(v), 2 sin(u/2) sin(v), cos(v)).

In fact, superK shows that the image of xx is still flat, and the following argument shows that it is one-to-one.

Assume that xx(u*,v*)=xx(u,v) for some numbers u*,v*,u,v in [0, 2). We must show that u*=u, v*=v. The preceding vector equation in xx splits into five scalar equations:

    (1)   cos(u*) cos(v*) = cos(u) cos(v)

    (2)   sin(u*) cos(v*) = sin(u) cos(v)

    (3)   cos(u*/2) sin(v*) = cos(u/2) sin(v)

    (4)   sin(u*/2) sin(v*) = sin(u/2) sin(v)

    (5)   cos(v*) = cos(v)

Squaring (3) and(4) and adding gives sin²(v*) = sin²(v).

  Case 1: The number cos(v*)=cos(v) is zero. Then v and v* can only have values /2 or 3/2.

• If v*=v, then sin(v*)=sin(v), which is necessarily nonzero. Hence by (3) and (4), u*=/2=u.
• If v* and v are different, say v=/2 and v*=3/2, then sin(v)=1 and sin(v*)=-1. Now (3) and (4) show that u*/2 and u/2 differ by a nonzero multiple of , a representative case being u*/2= u/2+. Hence u*=u+2, which is impossible since by assumption, 0 u < 2.

  Case 2: The number cos(v*)=cos(v) is nonzero. Then (1) and (2) show that u*=u. Next, (3) and (4) give sin(v*)=sin(v). Again since 0 u,u* < 2, we conclude that v*=v.

Thus the image of xx is the required flat Klein bottle in R⁵.

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