Let $X$ be a surface as in the title. Rick Miranda said that $X$ is a Steiner cubic in $\mathbb{P}^4$, and the cover map is projection. Invariants of $X$ can be computed directly, $p_g(X)=0,K^2_X=8,e(X)=4$.

My question is,

Question: What is a Steiner cubic? Why $X$ is a Steiner cubic?

The followings are what I know: Given $a,b,c,d\in H^0(\mathbb{P}^2,\mathcal{O}(1))$, $X$ is locally defined by \begin{cases} F(z,w)=z^2-aw-bw-2(a^2-bd) \\ G(z,w)=zw+dz+aw+ad-bc\\ H(z,w)=w^2-cz-dw-2(d^2-ac) \end{cases} Does, in fact, $X$ globally defined as an intersection of three quadrics in $\mathbb{P}^4$?

  • 1
    $\begingroup$ A locally complete intersection of three quadrics in $\mathbb{P}^4$ is a curve, not a surface. $\endgroup$
    – abx
    May 9, 2023 at 12:31
  • $\begingroup$ Maybe I should not say it is a locally complete intersection. I mean that $X$ is just like a two-dimensional version of a twisted cubic curve in $\mathbb{P}^3$ which is an intersection of three quadrics. $\endgroup$
    – Mobius
    May 9, 2023 at 12:47
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    $\begingroup$ So there are no triple covers in the question? $\endgroup$
    – Will Sawin
    May 9, 2023 at 13:35
  • $\begingroup$ Usually, Steiner surface is defined as the image of a regular liner projection of the Veronese surface into $\mathbb{P}^3$, see mathworld.wolfram.com/SteinerSurface.html $\endgroup$
    – Sasha
    May 9, 2023 at 18:03

1 Answer 1


I find the answer, see "Triple planes with p_g=q=0" Proposition 3.1 $X$ is a cubic scroll $S(1,2)\subset\mathbb{P}^4$, the image of Hirzebruch surface $\mathbb{F}_1$ under the linear system $|c_0+2f|$, where $c_0$ is the section with $c_0^2=-1$ and $f$ is the class of fiber.


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