Tangent surface of a rational curve of degree 4

The figure shows the tangent surface of a particular rational curve of degree 4. Thus, the surface is the union of the tangent lines of that curve. The curve is given parametrically by:
t ---> (x,y,z) = (t, t2, t4),
so that the surface is parametrically by:
(t,u) --> (t+u, t2 + 2tu, t4 + 4t3u).

The 4th degree rational curve is lightly sketched in dark blue on the surface.

Viewing suggestions:
    Starting from the home position, try one or more of the following
(i) horizontal mouse moves [rotation about the vertical axis],
(ii) vertical mouse moves [rotation about the horizontal axis in the viewing plane], or
(iii) long diagonal mouse moves.
In the portion shown here, we have   -1 < t < 1,  while the range of u-values varies with  t  in such a way that a tangent line segment of length = 2  is shown for each value of  t.  The midpoint of each tangent line segment is at a point of our degree 4 rational curve.
  • ¿ How is this surface different from the tangent surface of the twisted cubic?  The most visible difference is that the surface crosses itself -- and this is something that does not happen for the tangent surface of the twisted cubic. Thus, for t ≠ 0, the tangent line to our 4th degree rational curve at  (x,y,z) = (t,t2,t4)  intersects the tangent line at  (-x,y,z) = (-t,(-t)2,(-t)4).  {The intersection of the two tangent lines occurs at  (0,-t2,-3t4).}  As  t  varies, we obtain a line of intersection points. Two sheets of the surface cross each other along this line.
  • For comparison:  click here to view  the tangent surface of the twisted cubic.
  • ¿ Is this curve a generic projection?  Of course, the answer depends on what we mean by "generic". Our curve certainly is a projection of the rational normal curve in 4-space, given parametrically {in the affine version} by  t --> (t,t2,t3,t4).  And it certainly is non-singular; moreover, these properties extend to the projective closure. So far, so good.
        On the other hand, the projective closure does not have a well defined osculating plane at its (unique) point at infinity.  {Proof omitted.} To see why this could disqualify our curve from being a generic projection, we just have to observe that if a non-singular curve in n-space, with n ≥ 4, has a well defined osculating plane at every point, then its projection from a sufficiently general center is a non-singular curve in 3-space which has a well defined osculating plane at every point.  {Indeed, the union of the osculating planes of the curve in n-space is a 3-dimensional variety, and thus not all of n-space when  n ≥ 4.}

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I made the figure on this page by substituting my own data in a Geometry Center webpage.

Prof. Joel Roberts
School of Mathematics
University of Minnesota
Minneapolis, MN 55455

Office: 351 Vincent Hall
Phone: (612) 625-1076
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e-mail: roberts@math.umn.edu