k
is zero for all u in [0,
] and v in [0,2].
Reference: T. Otsuki, "Surfaces in the 4-dimensional Euclidean space isometric to a sphere," Kodai Math. Sem. Rep. 18 (1966), 101-115.
The scheme is this: In the following usual parametrization of the standard "round" unit 2-sphere S in R3,
x(u,v) = (sin u cos v, sin u sin v, cos u),
replace the height function cos u by two functions, thus:
xx(u,v) = (sin u cos v, sin u sin v, f(u), g(u))
(For a sphere of arbitrary radius r>0, just scalar multiply x and xx by r.) The first two coordinates in both x and xx describe circles of radius sin u lying in parallel planes. In R3, parallel planes are rather dull, but in R4 they can move around more freely. (Dropping one dimension, compare parallel lines in R2 with parallel lines in R3.)
The parametrization x of the round sphere S has
E=1, F=0, G=r2 sin2u
If f and g are chosen so that xx has these same values, then intrinsically the image of xx will be an isometric copy of the standard sphere. A computation by machine or by hand will show that the following functions work
f(u) = (4/3) cos3(u/2), g(u) =(4/3) sin3(u/2)
(As a check, superK shows that xx now has K=1.)
We still have to show that the Otsuki sphere doesn't lie in any 3-dimensional plane (as a round sphere, by definition, must). Note that the point p=(0,0,1,0) is in the sphere. Hence the sphere lies in a 3-plane if and only if there is a nonzero constant vector k=(a,b,c,d) -- the normal to the plane -- such that
the dot product (xx(u,v) - p)k
is zero for all u in [0,] and v in [0,2].
But it is not hard to show that this condition implies k=0. In fact, writing it out in full,
we see that setting u=0 kills c. Then setting u= kills d. This leaves only
sin u (a cos v + b sin v) = 0 for all u,v.
Hence a=b=0.