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If $P$ and $Q$ are two sufficiently general points on a cubic surface, the line between them intersects the cubic surface at a unique third point, $f(P,Q)$. This gives a binary operation on (generic) points on a cubic surface.

What identities does this binary operation satisfy?

For instance, it clearly satisfies the identites $f(P,Q) =f(Q,P)$ and $f(P,f(P,Q) ) = Q$.

For an elliptic curve, we can easily find the answer to the analogous question using the fact that $f(P,Q)=P^{-1}Q^{-1}$ in some abelian group. This also tells us the answer for identities with at most three variables - since any three points lie on a plane, the question reduces to the elliptic curve case. But what about four variables? Are there any identities with four or more variables that hold on a cubic surface but are not formally implied by the three variable identities?

Lastly, these identities clearly define a variety in the sense of universal algebra? What properties does this variety have? What does a free algebra on $n$ elements look like?

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    $\begingroup$ There is a book by Manin on, more or less, this subject: Cubic forms. $\endgroup$ May 9, 2013 at 4:50
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    $\begingroup$ (You should get the second edition) $\endgroup$ May 9, 2013 at 4:56
  • $\begingroup$ What is f(P,P)? I presume it will be P if you are serious about looking at the identities. Gerhard "Ask Me About System Design" Paseman, 2013.05.08 $\endgroup$ May 9, 2013 at 5:31
  • $\begingroup$ $f(P,P)$ is undefined. On an elliptic curve, it has a definite value, which is usually not $P$, but each point on a cubic surface is on many elliptic curves. $\endgroup$
    – Will Sawin
    May 9, 2013 at 15:37
  • $\begingroup$ Manin's book is really about the composition law modulo certain equivalence relations on points, known as "admissible" equivalence relations (to ensure that the composition is always defined). These relations are very coarse, so I don't think it is too relevant here. I posted a practically identical question before I realized that this one exists, sorry for the duplicate. $\endgroup$
    – Gro-Tsen
    Nov 23, 2017 at 15:49

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