On this page I'll explain why there are no lines on the Cayley surface other than the ones that are mentioned on the introductory page and on the main equation page (and also shown in the figure that includes the lines).

•   Six of the lines join pairs of nodes.  These are the only lines on the Cayley surface that are contained in one of the coordinate planes.  Hence, we can restrict consideration to lines that are not contained in any coordinate plane.
   
•   Now let's consider a line on the Cayley surface that goes through one of the nodes.  The node  (1:0:0:0) is typical.  We may work in the affine open set where  w ≠ 0,  with affine coordinates  x,y,z.  The implicit equation of this affine piece of the surface is:
   
              xy + xz + yz + xyz = 0,
   
  and the line can be given parametrically as  t ---> (at, bt, ct), where  (a,b,c) ≠ (0,0,0).  Since we're considering a line on the Cayley surface, the following equation must be satisfied identically:
   
              (ab + ac + bc)t ² + abct ³ = 0.
   
  It follows that both coefficients on the left side of the equation must be zero.  In particular,  abc = 0,  so that  a, b, or c  must be zero.  This implies that a line on the surface passing through one of the nodes must be contained in one of the coordinate planes.
   
•   Finally, we consider a line on the Cayley surface which does not go through one of the nodes (and is not contained in one of the coordinate planes.)  Such a line must intersect each of the coordinate planes.  Since our line lies on the Cayley surface and intersects the plane  z = 0,  it intersects one of the lines  L_wz, L_xz, L_yz.  We consider the (typical) case where the line under consideration intersects the line  L_yz (y = z = 0). 
   
    Our line also must intersect the plane  x = 0,  and thus one of the lines  L_wx, L_xy, L_xz.  However, a line which intersects both  L_xy  and  L_yz  and does not go through the node  (1:0:0:0)  would have to be contained in the plane  y = 0.  Since we are considering a line that is not contained in one of the coordinate planes, we conclude that our line cannot intersect the line  L_xy  (x = y = 0).  Similarly, we can show that our line cannot intersect the line  L_xz.  It follows, therefore, that our line meets both  L_wx  and  L_yz.  (Note that these two lines form a pair of skew lines, and are opposite edges of the coordinate tetrahedron.)
   
  In particular, our line has the following kind of projective parametrization:
   
              (s:t) ---> (as, bs, ct, dt),
   
  where  a, b, c,  and  d  all are nonzero.  We substitute this into (the projective version of) the implicit equation.  After a little bit of simplifying we find that:
   
              ab(c + d)s + (a + b)cdt = 0     ∀ (s:t).
   
  This leads to the conclusion that  a + b = 0  and  c + d = 0.  Therefore, our line intersects  L_wx  and  L_yz  at  (0:0:1:-1)  and  (1:-1:0:0)  respectively.  This means that the line under consideration is one of the three lines contained in the intersection of the Cayley surface and the tritangent plane.  Since those lines are already on our list, the proof is complete.

Updated on October 27, 2010.