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I have seen the following result in some notes in algebraic geometry.

Let $S: \{ f = 0 \} \subset \mathbb{P}^3$ be a nonsingular cubic surface, $A = \left( \frac{\partial^2f}{\partial x_i \partial x_j} \right)_{i,j = 0,\dots,3}$, $H = \det(A)$ the Hessian of $f$ and $B = \text{adj}(A)$.

We set $$ R = \sum_{i, j = 0}^{3} B_{i,j} \ \frac{\partial H}{\partial x_i} \frac{\partial H}{\partial x_j} $$

and $$ T = \sum_{i, j = 0}^{3} A_{i,j} \ B_{i,j}. $$

Then 27 lines are the intersection of $S$ with $F : \{ R - 4 HT = 0 \}$.

What is the name of this result? Where can one find the proof?

Is it true for singular cubic surfaces (if we count lines with multiplicity)?

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    If it is too much singular like the triple of hyperplane section in $\mathbb{P}^3$, then I'm afraid there is no way, even if you count multiplicities. – Mihail Mar 11 '18 at 19:27
  • @Mihail How do you know $27$ cannot be achieved by a count of multiplicities ? – Rene Schipperus Mar 21 '18 at 00:00
  • @ReneSchipperus Taken $\mathbb{P}^3$ with coordinates $[x,y,z,t]$, consider the cubic surface $S={x^3=0}$. There are infinite lines on $S$. – Mihail Mar 21 '18 at 08:17
  • @Mihail That can definitely happen, but there are many theorems that say that IF the dimension is zero, then the number of points, counted with multiplicity is constant. I think that is how the question in this case needs to be phrased. Now there are surfaces with a finite number of lines, less than 27, but how to see no multiplicity definition can give 27 in all finite cases ? – Rene Schipperus Mar 21 '18 at 12:16

1 Answers1

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This is a result of Salmon and Clebsch. You can find it in the fourth edition of Salmon's A treatise on the analytic geometry of three dimensions (1882). (See https://archive.org/details/treatiseanalytic00salmrich). It is on p. 510 and more details starting from p. 558. Clebsch' proof uses the symbolic method. I am not aware of a modern proof.

It is true for singular surfaces containing a finite number of lines. The statement in general is that the ideal generated by the cubic form defining the surface and this form of degree nine describes all points lying on all lines of the surface. If the surface is a cone (eg if it consists of three planes) the Hessian vanishes identically and so does $R-4HT$. Therefore the zero locus of the said ideal is the whole surface, as it should be.

Jan Stevens
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