For the Fermat cubic surface \(x_0^3+x_1^3+x_2^3+x_3^3=0\), find the equations of the 27 lines explicitly, and verify their incidence relations. What is the group of automorphisms of this surface?
Solution
Let \[ S=V(F)\subseteq\PP_k^3, \qquad F=x_0^3+x_1^3+x_2^3+x_3^3, \] with \(k\) algebraically closed.
There is one characteristic qualification. If \[ \operatorname{char}k=3, \] then \[ F=(x_0+x_1+x_2+x_3)^3, \] so the displayed scheme is a nonreduced triple plane, not the nonsingular cubic surface to which the 27-line theory applies. We therefore assume through step 8 that \[ \operatorname{char}k\ne3. \] The automorphism calculation is then separated according as \(\operatorname{char}k=2\) or not.
Choose a primitive cube root of unity \[ \omega\in k, \qquad \omega^3=1, \qquad \omega\ne1, \] and read all subscripts \(a,b,c,d\) below in \(\ZZ/3\).
1. The surface \(S\) is nonsingular.
Since \(\operatorname{char}k\ne3\), \[ \frac{\partial F}{\partial x_i}=3x_i^2. \] All four partial derivatives vanish only when \[ x_0=x_1=x_2=x_3=0, \] which is not a point of projective space. Hence \(S\) is nonsingular.
2. For \(a,b\in\ZZ/3\), define three families of lines\[ \begin{aligned} A_{ab}&: \quad x_0+\omega^a x_1=0, \qquad x_2+\omega^b x_3=0,\\ B_{ab}&: \quad x_0+\omega^a x_2=0, \qquad x_1+\omega^b x_3=0,\\ C_{ab}&: \quad x_0+\omega^a x_3=0, \qquad x_1+\omega^b x_2=0. \end{aligned} \]All \(27\) of these lines lie on \(S\).
On \(A_{ab}\) one has \[ x_0=-\omega^a x_1, \qquad x_2=-\omega^b x_3. \] Therefore \[ x_0^3+x_1^3 = -\omega^{3a}x_1^3+x_1^3 =0 \] and similarly \[ x_2^3+x_3^3=0. \] Thus \(F\) vanishes identically on \(A_{ab}\). The same calculation, using the other two pairings of the four coordinates, proves that every \(B_{ab}\) and \(C_{ab}\) lies on \(S\).
3. The \(27\) lines in step 2 are pairwise distinct. Hence they areexactly all the lines on \(S\).
Within one family, the two exponents are recovered from the two defining linear equations, so different pairs \((a,b)\) give different lines.
No \(A\)-line equals a \(B\)-line: the point \[ [-\omega^a:1:0:0]\in A_{ab} \] lies on no \(B_{cd}\), because the equation \[ x_0+\omega^c x_2=0 \] would force its nonzero coordinate \(x_0\) to vanish. By symmetry no two lines from different families coincide.
Thus step 2 gives \(3\cdot3^2=27\) distinct lines. A nonsingular cubic surface has exactly \(27\) lines FE-SRFCUBIC, so these are all of them.
4. For two distinct lines in the same family,\[ \boxed{ A_{ab}\cap A_{cd}\ne\varnothing \iff a=c\text{ or }b=d, } \]and the identical criterion holds in the \(B\)- and \(C\)-families.
Suppose first that \[ a=c, \qquad b\ne d. \] The two equations involving \(x_2,x_3\) then force \[ x_2=x_3=0, \] while the common first equation gives the point \[ [-\omega^a:1:0:0]. \] Thus the lines meet. If instead \(b=d\) and \(a\ne c\), they meet at \[ [0:0:-\omega^b:1]. \]
If \(a\ne c\) and \(b\ne d\), the first pair of equations forces \(x_0=x_1=0\) and the second pair forces \(x_2=x_3=0\), so there is no projective point of intersection. The other two families are obtained by permuting the coordinates.
5. Lines from the \(A\)- and \(B\)-families satisfy\[ \boxed{ A_{ab}\cap B_{cd}\ne\varnothing \iff a+d\equiv b+c\pmod3.} \]
On a common point we may write, from the equations of \(A_{ab}\) and \(B_{cd}\), \[ x_2=-\omega^b x_3, \qquad x_1=-\omega^d x_3. \] Then the two expressions for \(x_0\) are \[ x_0 = -\omega^a x_1 = \omega^{a+d}x_3 \] and \[ x_0 = -\omega^c x_2 = \omega^{b+c}x_3. \] A nonzero common point therefore exists exactly when \[ \omega^{a+d}=\omega^{b+c}, \] which is the displayed congruence. When it holds, taking \(x_3=1\) gives the unique point of intersection.
6. Lines from the \(A\)- and \(C\)-families satisfy\[ \boxed{ A_{ab}\cap C_{cd}\ne\varnothing \iff c\equiv a+b+d\pmod3.} \]
The equations give \[ x_2=-\omega^b x_3 \] and then \[ x_1=-\omega^d x_2=\omega^{b+d}x_3. \] Consequently the \(A\)-equation gives \[ x_0=-\omega^{a+b+d}x_3, \] whereas the \(C\)-equation gives \[ x_0=-\omega^c x_3. \] These are compatible exactly under the displayed congruence.
7. Lines from the \(B\)- and \(C\)-families satisfy\[ \boxed{ B_{ab}\cap C_{cd}\ne\varnothing \iff a+b\equiv c+d\pmod3.} \]
Put \(x_2=t\). The \(C\)-equation gives \[ x_1=-\omega^d t, \] and comparison with \[ x_1=-\omega^b x_3 \] gives \[ x_3=\omega^{d-b}t. \] Now \[ x_0=-\omega^a t \] from \(B_{ab}\), while \(C_{cd}\) gives \[ x_0 = -\omega^c x_3 = -\omega^{c+d-b}t. \] Compatibility is therefore equivalent to \[ a+b\equiv c+d\pmod3. \]
8. The formulas in steps <1>4–<1>7 give the complete incidenceconfiguration of the \(27\) lines. In particular every line meets exactly\(10\) of the other lines and is skew to the remaining \(16\).
Fix \(A_{ab}\). By step 4 it meets exactly four other \(A\)-lines: two with the same first exponent and two with the same second exponent.
For each \(c\in\ZZ/3\), step 5 determines a unique \(d\) for which \(A_{ab}\) meets \(B_{cd}\). Thus it meets exactly three \(B\)-lines. Likewise, for each \(d\), step 6 determines a unique \(c\), so it meets exactly three \(C\)-lines.
Hence \(A_{ab}\) meets \[ 4+3+3=10 \] other lines. Coordinate symmetry gives the same count for every \(B\)- and \(C\)-line. Since there are \(26\) other lines in total, each line is skew to \(16\). The preceding congruences decide every possible pair, so they give the full incidence relations.
9. Every automorphism of the smooth cubic surface \(S\) is induced by aprojective linear transformation of \(\PP^3\).
Adjunction for the cubic hypersurface gives \[ K_S = (K_{\PP^3}+S)|_S = \OO_S(-1). \] Thus \[ -K_S=\OO_S(1). \] Every automorphism of \(S\) preserves the canonical class, hence the complete anticanonical linear system.
The given embedding \[ S\hookrightarrow\PP^3 \] is precisely the complete anticanonical embedding: restriction of linear forms gives \[ H^0(\PP^3,\OO_{\PP^3}(1)) \xrightarrow{\sim} H^0(S,\OO_S(1)), \] as follows from the cubic hypersurface sequence twisted by \(\OO(1)\). Therefore the action on \[ H^0(S,-K_S) \] extends the automorphism uniquely to an element of \(\PGL_4(k)\).
Assume first that \[ \operatorname{char}k\ne2,3. \]
10. Any projective linear automorphism preserving \(S\) permutes the fourcoordinate hyperplanes.
Let \(A\in\GL_4(k)\) represent a projective automorphism of \(S\). Since the homogeneous ideal of \(S\) is generated by the irreducible cubic \(F\), there is \(c\in k^\times\) such that \[ F(Ax)=cF(x). \]
The Hessian matrix of \(F\) is \[ \operatorname{Hess}(F) = \operatorname{diag}(6x_0,6x_1,6x_2,6x_3), \] so its determinant is \[ \operatorname{hess}(F) = 6^4x_0x_1x_2x_3. \] For a linear change of variables, the Hessian matrix transforms by \[ \operatorname{Hess}(F\circ A)(x) = A^T\operatorname{Hess}(F)(Ax)A. \] Taking determinants and using \(F\circ A=cF\) shows that \(A\) preserves the zero locus of \(\operatorname{hess}(F)\).
Because \(6\ne0\), that zero locus is exactly \[ V(x_0x_1x_2x_3) = \bigcup_{i=0}^3V(x_i). \] Its four irreducible components are the four coordinate hyperplanes. Therefore \(A\) permutes them.
11. If \(\operatorname{char}k\ne2,3\), then\[ \boxed{ \Aut(S) \cong (\mu_3^4/\mu_3)\rtimes\Sigma_4 \cong (\ZZ/3)^3\rtimes\Sigma_4, } \]and hence\[ \boxed{|\Aut(S)|=27\cdot24=648.} \]
By step 10, a matrix representing an automorphism is monomial: after a permutation of the coordinates it is diagonal, \[ \operatorname{diag}(\lambda_0,\lambda_1,\lambda_2,\lambda_3). \] The identity \[ F(Ax)=cF(x) \] then says \[ \lambda_0^3 = \lambda_1^3 = \lambda_2^3 = \lambda_3^3 = c. \] Modulo a common scalar, the four diagonal factors therefore lie in \(\mu_3\), with simultaneous multiplication by the same cube root acting trivially in projective space. Thus the diagonal subgroup is \[ \mu_3^4/\mu_3\cong(\ZZ/3)^3. \]
Conversely, every such diagonal transformation preserves \(F=0\), and every permutation of the four coordinates preserves \(F\). Coordinate permutations normalize the diagonal subgroup, so the full group is the displayed semidirect product. Its order is \[ 3^3\cdot4!=648. \]
Now assume \[ \operatorname{char}k=2. \]
12. A projective linear transformation represented by\(A\in\GL_4(k)\) preserves \(S\) if and only if, after multiplying \(A\) by ascalar, one has\[ \boxed{A^{(2)T}A=I,} \]and every entry of this normalized matrix lies in \(\FF_4\).
In characteristic \(2\), \[ F(x) = x^{(2)T}x, \] where \(x^{(2)}\) denotes coordinatewise squaring. Hence \[ F(Ax) = x^{(2)T}A^{(2)T}Ax. \] As in step 10, preservation of \(S\) is equivalent to \[ F(Ax)=cF(x) \] for some \(c\in k^\times\). The monomials \(x_i^2x_j\) are linearly independent, so this equality is equivalent to \[ A^{(2)T}A=cI. \]
Choose \(t\in k^\times\) with \[ t^3c=1 \] and replace \(A\) by the projectively equivalent matrix \(tA\). We then have \[ A^{(2)T}A=I. \] Transposing gives \[ A^TA^{(2)}=I, \] while squaring the first matrix identity gives \[ A^{(4)T}A^{(2)}=I. \] Comparison yields \[ A^{(4)T}=A^T, \] hence \[ A^{(4)}=A. \] Thus every entry is fixed by the fourth-power Frobenius, so it belongs to \(\FF_4\).
Conversely, every matrix over \(\FF_4\) satisfying \(A^{(2)T}A=I\) preserves \(F\) exactly.
13. In characteristic \(2\),\[ \boxed{ \Aut(S) \cong U_4(2)/\mu_3, } \]where\[ U_4(2) = \{A\in\GL_4(\FF_4):A^{(2)T}A=I\}. \]Moreover\[ \boxed{|\Aut(S)|=25920.} \]
Step 12 identifies the normalized linear stabilizer of \(F\) with the finite unitary group \(U_4(2)\). Two normalized matrices define the same projective transformation exactly when they differ by a scalar \[ \lambda I \] with \[ \lambda^3=1. \] These form the central subgroup \(\mu_3\), proving the group identification.
It remains to count \(U_4(2)\). For the standard Hermitian form \[ \langle v,w\rangle=v^{(2)T}w \] on \(\FF_4^n\), every nonzero coordinate \(u\) satisfies \[ u^3=1. \] Thus a vector has norm \(1\) exactly when it has an odd number of nonzero coordinates. The numbers of unit vectors in dimensions \(1,2,3,4\) are therefore \[ \begin{aligned} N_1&=3,\\ N_2&=\binom21 3=6,\\ N_3&=\binom31 3+\binom33 3^3=36,\\ N_4&=\binom41 3+\binom43 3^3=120. \end{aligned} \]
Hermitian Gram–Schmidt shows that \(U_n(2)\) acts transitively on unit vectors, with stabilizer \(U_{n-1}(2)\) on the orthogonal complement. Hence \[ |U_4(2)| = N_4N_3N_2N_1 = 120\cdot36\cdot6\cdot3 = 77760. \] Dividing by the central subgroup of order \(3\) gives \[ |\Aut(S)| = \frac{77760}{3} = 25920. \]
14. Q.E.D.
Steps <1>2–<1>3 give the explicit \(27\) lines, and steps <1>4–<1>8 verify all their incidence relations. Step 9 reduces the automorphism problem to projective linear algebra. Steps <1>10–<1>11 compute the classical automorphism group of order \(648\) in characteristic different from \(2,3\), while steps <1>12–<1>13 give the exceptional characteristic-\(2\) group of order \(25920\). In characteristic \(3\) the displayed equation is a triple plane, as noted at the start, so the smooth-cubic question does not apply.