Finding $\operatorname{Aut}(X)$ where $X$ is the Fermat cubic surface in $\mathbb{P}^3$.

Question

I am trying to understand the automorphism group of the Fermat Cubic surface $$x_0^3 + x_1^3 + x_2^3 + x_3^3 = 0$$ in $\mathbb{P}^3$ to solve Hartshorne's exercise V.4.16.

The way I thought to go about this was to find a subgroup, and argue that the order of this subgroup is as large as the whole group to conclude that this subgroup is actually the whole thing.

To this end, $\operatorname{Aut}(S)$ has a subgroup isomorphic to $S_4$ obtained by permuting the variables, and in each varable we can act my multiplication by a third root of unity. These will yield $G \subset \operatorname{Aut}(X)$ where this is the subgroup generated by the aforementioned automorphisms. It's not hard to show that $G$ is an extension of $S_4$ and $\mathbb{F}_3^3$, so $|G| = 648$.

Now to conclude, I would like to show that if $X$ is a cubic surface, then $|\operatorname{Aut}(X)| \leq 648$. This is where I'm stuck. There should be some way to look at the action of $\operatorname{Aut}(X)$ on the configuration of the $27$ lines but it's not clear to me how to do so, since this configuration is quite complicated, and the whole automorphism group of the configuration is much too large to be useful. (It's $W(E_6)$ which has order $51,840$.)

Thanks!

Answer 1

Let me first point out that this is only true when $\operatorname{char} k \neq 2,3$ - in characteristic 3, the surface isn't even reduced, while in characteristic 2, we have that the automorphism group is of order $25920$ (!).

We'll start by showing that every automorphism of a smooth cubic surface comes from an automorphism of the projective space. Note that the canonical bundle of a smooth cubic surface $X$ in $\Bbb P^3$ is $\mathcal{O}(-1)$, so the embedding $X\to\Bbb P^3$ is given by $-K_X$. As $K_X$ and it's powers are preserved under automorphisms, this shows any automorphism of the Fermat cubic surface comes from an automorphism of the ambient projective space. Now you're reduced to a linear algebra problem if you want, but we can do better.

I originally learned the following from Dolgachev's Automorphisms of cubic surfaces in positive characteristic. For a projective hypersurface $X=V(f)\subset\Bbb P^n$, define the Hessian hypersurface $Hess(X)=\det \{ \frac{\partial^2 f}{\partial x_i \partial x_j}\}$. This definition is coordinate-independent: under a projective coordinate transformation given by an invertible matrix $A$, the Hessian matrix transforms like $H\mapsto A^tHA$. So the Hessian hypersurface of the Fermat cubic surface must be preserved under automorphisms. For the Fermat cubic surface, this hypersurface is cut out by $6^4x_0x_1x_2x_3$, and so the automorphisms are (generated by) permutations of coordinates and multiplication of coordinates by constant factors, assuming $\operatorname{char} k >3$. In order for a multiplication of coordinates by constant factors to give an automorphism, we must have that the multiplication matrix is projectively equivalent to one with third roots of unity on the diagonal.