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The implicit equation and the singular locus:
 
    We continue to work 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,  and the singular locus of the Cayley surface is the set of points where  F and all four of its partial derivatives are equal to zero.  The partial derivatives are:
 
      ^∂F⁄_∂w  = xy + xz + yz;     ^∂F⁄_∂x  = wy + wz + yz;
 
      ^∂F⁄_∂y  = wx + wz + xz;     ^∂F⁄_∂z  = wx + wy + xy.
 
First of all, if a point  (w:x:y:z)  is one of the points  (1:0:0:0:0),  (0:1:0:0),  (0:0:1:0),  or  (0:0:0:1)  then we check easily that  F  and all of its partial derivatives are equal to zero at  (w:x:y:z).

    Conversely, suppose that  (w:x:y:z)  is a singular point of the Cayley surface. We calculate:
 
      F - w ^∂F⁄_∂w  = xyz.
This shows that  xyz = 0  at a singular point.  Assuming (for instance) that  z = 0,  we evaluate the other partial derivatives at  (w:x:y:0):
 
      ^∂F⁄_∂w  = xy;     ^∂F⁄_∂x  = wy;     ^∂F⁄_∂y  = wx.
 
Since  (w:x:y:0)  is a singular point, it follows that  wx = wy = xy = 0,  and therefore that two of the three coordinates  w, x, y  are equal to zero (along with  z being equal to zero).  Explicitly, this shows that any singular point with  z = 0  is one of the points  (1:0:0:0),  (0:1:0:0),  (0:0:1:0).  By doing similar calculations in the cases  x = 0  and  y = 0,  we find that the only singular points of the Cayley surface are these three points and  (0:0:0:1).  In other words, the singular points are exactly the vertices of the coordinate tetrahedron.