A ruled surface of degree 4

Degree 4 ruled: Tangent surface: Other:  

Equation:
 
    As discussed on another page, the tangent surface of the twisted cubic is given parametrically by:

(t,u) --> (t+u, t² + 2tu, t³ + 3t²u).
 
And as discussed elsewhere on that same page, this leads to the relations:
 
y - x² = - u²    and    z - 3xy + 2x³ = 2u³,
 
and therefore to the following version of the implicit equation:
 
4(y - x²)³ + (z - 3xy + 2x³)² = 0.
 
We can interpret this as saying that we obtain this implicit equation (of the tangent surface) by substituting  Y = y - x²  and  Z = z - 3xy + 2x³   into the equation of the planar cusp curve  Z² = - 4.  {Note that the terms of degrees 5 and 6 cancel in the implicit equation of the tangent surface, so that it actually is an equation of degree 4.}

    Now, we have a planar node curve with a similar equation:

Z² = Y² - 4,
 
and we also can substitute  Y = y - x²  and  Z = z - 3xy + 2x³  into the equation of this curve.  If we do this,  we obtain the following equation of a surface:
 
(z - 3xy + 2x³)² = (y - x²)² - 4(y - x²)³.
 
Exactly as with the implicit equation of the tangent surface, the terms of degrees 5 and 6 cancel, so that we have a surface of degree 4.  And since the equations  y ‑ x² = z - 3xy + 2x³ = 0  define a nonsingular space curve (namely the twisted cubic), it follows that the singular locus of our surface is the twisted cubic

    Our next objective is to find a parametrization of this surface.  We begin by observing that the planar node curve has the following parametrization:

(Y,Z) = (¼ - u², 2u(¼ - u²)).
 
{To see how this was obtained, divide the equation of the node curve by  4Y².  This yields  (Z2Y)² = 14 - Y,  which we use to express  Y  in terms of  u := Z2Y.} Using this parametrization of the curve, we obtain a surface parametrization by solving the following equations:
 
y - x² = ¼ - u²         z - 3xy + 2x³ = 12u - 2u³.
 
We find the following parametrization:
 
(x,u) --> (x, x² - u² + 14, x³ + 34x - 3xu² + 12u - 2u³).
 
Note, in particular, that for  u = ±½  the parametrization yields a point of the twisted cubic.  One also can use this parametrization to show that our surface (this affine model, anyway) has a nonsingular normalization.

    This parametrization ( with some re-scaling) was used in constructing the figure. The rectangular parameter region  - 58x,u58 is shown, with some trimming at two of the corners.

    Our final objective is to show how the parametrization can be modified in order to verify that our surface is indeed ruled.  We set  t = x + u,  so that  x = t - u.  Substituting this into the parametrization and doing some calculation (mainly taking advantage of various cancellations), we obtain the following parametrization:

 
(t,u) --> (t - u, t² - 2tu + 14, t³ + 34t - 3t²u - 14u).
 
Since all three coordinates are given by expressions that are linear in  uit follows that the surface is ruled.
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The Java files used in this page were downloaded from the Geometry Center webpage.
I generated the geometric data for this figure in May 2010.
Update completed on July
7, 2010.

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

Office: 531 Vincent Hall
Phone: (612) 626-9531
Dept. FAX: (612) 626-2017
e-mail: roberts@math.umn.edu
http://www.math.umn.edu/~roberts