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Let $V$ be a general smooth projective cubic hypersurface. Doing literally as in case of cubic curves we define a relation on $V\times V\times V$: $(x,y,z)$ satisfy it iff $x+y+z$ is an intersection of $V$ with a line. Contrary to the one-dimensional case this relation is not a graph of a binary operation ($x^2$ is not defined or two points may lie on the line contained in $V$). From now on let us consider only pairs $(x,y)$ for which $z$ is uniquely defined (that is $(x,y,z)$ are on a line and there is one such $z$, let's denote it by $x\circ z$). Let's chose some $u$ (that will be the 'unit') and define the product as usual $xy=u\circ (x\circ y)$. This partially defined product however may be non-associative in the following sense: there are $x,y,z$ such that $$x(yz)\neq(xy)z$$ where $yz$,$xy$, $x(yz)$ and $(xy)z$ are uniquely defined.

Are there any examples of such non-associative triples?

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Here is an explicit example over the rationals.

Consider the diagonal Clebsch cubic surface given by $\sum_{i=0}^4 X_i = 0$ and $\sum_{i=0}^4 X_i^3 = 0$. Let me take the point $u := (0:0:0:1:-1)$ so that $x \mapsto u\circ x$ takes $(X_0:X_1:X_2:X_3:X_4)$ to $(X_0:X_1:X_2:X_4:X_3)$ provided $X_3 + X_4 \neq 0$.

Let $x := (1, 0, -1, -1, 1)$ and $y := (1, -2, 0, -1, 2)$ and $z := (1, 0, 3, -1, -3)$. Then we have $xy = (1, 2, -2, 0, -1)$ and $(xy)z = (18, 26, -11, -28, -5)$ whereas $yz = (5, -6, 6, 0, -5)$ and $x(yz) = (8, -18, 25, -22, 7)$. These are, indeed, different.

To check my calculations, all you need to do is check that the points in question do satisfy the equations of the Clebsch cubic, that the announed triplets are indeed aligned (e.g., $x$, $y$ and $x\circ y = u\circ(xy)$ are indeed linearly dependent), and that no two lie on one of the well-known lines of the Clebsch cubic. (Note that $x$ is on a line with $u$, but you didn't forbid this and we don't need to compute $u\circ x$.)

I tried to find an example with smaller heights, but points of small height have a strong tendency to lie on the lines. I didn't try very hard, though.

PS: I can provide a bit of Sage code to compute compositions on the Clebsch cubic if you want (it's probably not very robust, but it allowed me to find this example).

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  • $\begingroup$ Great, thank you! Although I was looking for a hypersurface example, your computation demonstrates non-associativity on 3d Fermat cubic $\sum_{i=0}^4 X_i^3 = 0$ as well. But I'm still not sure about 2-dimensional cubic hypersurface. $\endgroup$ Nov 15, 2017 at 16:32
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    $\begingroup$ @DmitryK My example is $2$-dimensional: the Clebsch diagonal cubic surface is defined in the ($3$-dimensional) hyperplane $\sum_{i=0}^4 X_i=0$ inside $\mathbb{P}^4$ for symmetry, but by seeing this hyperplane as $\mathbb{P}^3$, it is a smooth surface in $\mathbb{P}^3$. $\endgroup$
    – Gro-Tsen
    Nov 15, 2017 at 16:45

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