How to find 27 lines on Cayley's nodal cubic surface

Question

Cayley's nodal cubic surface is given by the equation: $$wxy+xyz+yzw+zwx=0$$ And I want to find all 27 lines on it, but I can only find 9 lines, they are 6 lines of the form:$x=y=0$, and 3 lines of the form:$x+y=z+w=0$. So what are the other 18 lines? How can I find them?

Answer 1

By symmetry we can assume any such line is given by $x = az + bw$, $y = cz + dw$. Thus $$acz^3 + (ad + bc + ac + a + c)z^2w + (ad + bc + bd + b + d)zw^2 + bdw^3$$ must vanish identically in $(z, w)$, so each coefficient must be $0$. From $ac = bd = 0$, accounting for symmetry, there are two cases:

• $a = b = 0 \implies c = 0$ and $d = 0$, which gives $x = y = 0$.
• $a = d = 0 \implies bc + c = bc + b = 0$, so $b = c$ and $b^2 + b = 0$. So we are either in the previous case or $b = c = -1$, which gives the line $x + w = y + z = 0$.

Accounting for permutations the $9$ lines you have found are exactly the ones on this surface. The Cayley-Salmon theorem does not apply since this cubic surface is not smooth (at $[1:0:0:0]$ and its permutations).