I want to describe the $27$ lines on the Fermat surface and have found the information below. I don't understand the last part. How is it possible to go from $9$ different lines to $27$ different lines by changing the projective coordinates. Can somebody explain how you get the three last equations?

Lemma 11.1. The Fermat cubic $X = V_p(x_0^3+x_1^3+x_2^3+x_3^3)\subset\mathbb P^3$ contains exactly 27 lines.

Proof. Up to a permutation of coordinates, every line in $\mathbb P_3$ is given by two linear equations of the form $x_0 = a_2x_2 + a_3x_3$ and $x_1 = b_2x_2 + b_3x_3$ for suitable $a_2,a_3,b_2,b_3 \in \mathbb C$.

Such a line lies in X if and only if $(a_2x_2 +a_3x_3)^3 + (b_2x_2 +b_3x_3)^3 +x_2^3+x_3^3 = 0$ as a polynomial in $\mathbb C[x_2, x_3]$

So by comparing coefficients, if and only if

$$a_2^3 +b_2^3=-1 \dots (1)$$

$$a_3^3 +b_3^3=-1 \dots (2)$$ $$a_2^2a_3=-b_2^2b_3 \dots (3)$$ $$a_2a_3^2=-b_2b_3^2 \dots (4)$$

If $a_2,a_3,b_2,b_3$ are all non-zero, then $(3)^2/(4)$ gives $a_2^3=-b_2^3$ in contradiction to $(1)$

Hence for a line in the cubic at least one of these numbers must be zero.

Again after possibly renumbering the coordinates we may assume that $a_2 = 0$.

Then $b_2^3=-1$ by $(1)$; $b_3 = 0$ by $(3)$; and $a_3^3=-1$ by $(2)$

Conversely, for such values of $a_2,a_3,b_2,b_3$ the above equations all hold, so that we really obtain a line in the cubic.

We thus obtain $9$ lines in $X$ by setting $b_2 = −\omega^j; a_3=-\omega^k$ for $0 \le j, k \le 2$ with $\omega = exp(\frac {2\pi i}3)$, a primitive third root of unity.

So finally, by allowing permutations of the coordinates we find that there are exactly the following $27$ lines on $X$: $$x_0 +x_3\omega^k = x_1 +x_2\omega^j=0: 0 \le j, k \le 2$$ $$x_0 +x_2\omega^k = x_3 +x_1\omega^j = 0: 0 \le j, k \le 2$$ $$x_0 +x_1\omega^k = x_3 +x_2\omega^j = 0: 0 \le j, k \le 2$$


  • $\begingroup$ This question needs some copy-editing to make it legible. ${}\qquad{}$ $\endgroup$ – Michael Hardy May 14 '15 at 14:30
  • $\begingroup$ I'm guessing they're nine different lines each with multiplicity $3$, and a small perturbation makes three lines that are identical into three lines that are distinct. ${}\qquad{}$ $\endgroup$ – Michael Hardy May 14 '15 at 14:31
  • 2
    $\begingroup$ @MichaelHardy: The Fermat cubic is smooth, so the 27 lines should be distinct. $\endgroup$ – Daniel Hast May 14 '15 at 15:23
  • $\begingroup$ @MichaelHardy The co-ordinates are renumbered to give $a_2=0$ in the analysis, so permutations of the solution are allowed. $\endgroup$ – Mark Bennet May 14 '15 at 15:27
  • $\begingroup$ @DanielHast : Fortunately I said "guessing". $\endgroup$ – Michael Hardy May 14 '15 at 16:39

You get the first equation as the nine lines in the analysis. Working back with the values discovered the lines become $$x_0=-\omega^kx_3$$together with (two equations for each line) $$ x_1=-\omega^jx_2$$

Gathering terms together on the same side, this makes the first set of nine lines - the first line in the solution.

Then the equivalent lines with co-ordinates permuted (the original curve is symmetric) also lie in the cubic. And that gives the other two sets of lines. Note that multiplying through the equations by powers of $\omega$ you can bring any equation of a line in the cubic into one of the three given forms.


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