• Introduction
• Singularities
• Duality map
• Dual parametrization
• Dual equation
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The duality map:
 
    We work once again with the standard equation of the Cayley surface.  If we set  F(w,x,y,z) = wxy + wxz + wyz + xyz,  then the standard equation is  F(w,x,y,z) = 0.  The duality map, from the Cayley surface to the dual projective space is given by:
 
      (w:x:y:z) ---> (^∂F⁄_∂w: ^∂F⁄_∂x: ^∂F⁄_∂y: ^∂F⁄_∂z)
 
The basic idea is that each smooth point of the surface is mapped to the point of the dual projective space that is represented by the coordinate vector of the equation of the tangent plane at the point.  The duality map is defined on the complement of the singular locus, i.e., everywhere except at the four nodes.

    We will begin by showing that each line that joins two of the nodes is blown down to a single point by the duality map.  Recall that the partial derivatives are:
 
      ^∂F⁄_∂w  = xy + xz + yz;     ^∂F⁄_∂x  = wy + wz + yz;
 
      ^∂F⁄_∂y  = wx + wz + xz;     ^∂F⁄_∂z  = wx + wy + xy.
 
The case of the line  L_yz (y = z = 0)  is typical.  Indeed, at any point of this line we have:
 
      ^∂F⁄_∂w  = 0;     ^∂F⁄_∂x  = 0;     ^∂F⁄_∂y  = wx;     ^∂F⁄_∂z  = wx.
 
Therefore, every point of the line  L_yz  is mapped to the point  (0:0:1:1),  except for the two nodes  (1:0:0:0) and (0:1:0:0)  where the duality map is undefined.  Similar considerations apply to the lines that join other pairs of nodes:  we find that the duality map blows down the six edges of the coordinate tetrahedron to six distinct points of the dual projective space.

    To make further progress, we use a version of the duality map which was shown to me by Sevín Recillas.  This version is valid at points of the Cayley surface where all four coordinates are nonzero.  The idea is to divide the implict equation by  wxyz,  so that we work with the implicit equation  G(w,x,y,z) = 0,  where  G(w,x,y,z) = ¹⁄_w + ¹⁄_x + ¹⁄_y + ¹⁄_z.  This leads to the following version of the duality map:
 
      (w:x:y:z) ---> (^∂G⁄_∂w: ^∂G⁄_∂x: ^∂G⁄_∂y: ^∂G⁄_∂z) = (¹⁄_w² : ¹⁄_x² : ¹⁄_y² : ¹⁄_z² ).
 
It is immediately clear that the duality map is finite-to-one on the open set where  wxyz ≠ 0,  although this version does not clearly exhibit the fact that the duality map is birational (onto its image).  To see where the map can fail to be one-to-one, we work in the affine open set  w ≠ 0,  with affine coordinates  x,y,z.  The affine version of the implicit equation is  1 + ¹⁄_x + ¹⁄_y + ¹⁄_z = 0.  Given two points with the same image,  say  (x₁,y₁,z₁)  and  (x₂,y₂,z₂),  we must have  x₂ = ±x₁,  y₂ = ±y₁,  and  z₂ = ±z₁.  If all three signs are negative, then we have:
 
      1 + ¹⁄_x1 + ¹⁄_y1 + ¹⁄_z1 = 0     and     1 − ¹⁄_x1 − ¹⁄_y1 − ¹⁄_z1 = 0.  
 
Adding these two equations leads to the obvious contradiction  2 = 0,  so that this case is impossible.  If just one sign is negative, say  x₂ = −x₁,  y₂ = y₁,  and  z₂ = z₁,  then:
 
      1 + ¹⁄_x1 + ¹⁄_y1 + ¹⁄_z1 = 0     and     1 - ¹⁄_x1 + ¹⁄_y1 + ¹⁄_z1 = 0.  
 
Subtracting these two equations leads to the conclusion that  ²⁄_x1 = 0,  which also is impossible.  Finally, suppose that two signs are negative and one sign is positive,  say  x₂ = -x₁,  y₂ = -y₁,  and  z₂ = z₁.  Then:
 
      1 + ¹⁄_x1 + ¹⁄_y1 + ¹⁄_z1 = 0     and     1 - ¹⁄_x1 - ¹⁄_y1 + ¹⁄_z1 = 0.  
 
Adding these two equations, respectively subtracting them, leads to the conclusion that:
 
      1 + ¹⁄_z1 = 0     and     ¹⁄_x1 + ¹⁄_y1 = 0.  
 
Thus, the points  (x,-x,-1)  and  (-x,x,-1)  are sent to the same image by the duality map.  In other words the duality map is generically two-to-one along the line  z = -1,  x + y = 0.  The projective closure of this line is given by the equations  w + x = 0,  y + z = 0;  this is one of the lines in the tritangent plane. 

    Our conclusion, therefore, is that the duality map is generically two-to-one along each of the three lines in the tritangent plane that lie on the Cayley surface, and one-to-one at all other points of the Cayley surface where  wxyz ≠ 0.  It is fairly easy to check that each of these three lines is mapped to a line on the dual surface, and that all three of the lines on the dual surface pass through the point  (1:1:1:1)  in the dual projective space.

    Thus, there have to be two ramification points on each of the three lines in the tritangent plane.  Explicitly, the ramification points are:
 
      (1:-1:0:0)     and    (0:0:1:-1);
      (1:0:-1:0)     and     (0:1:0:-1);
      (1:0:0:-1)     and     (0:1:-1:0).
 
To check this claim and find the images of these points, we have to use the original version of the duality map.