Hasse principle and its failure for a special class of plane cubics

Question

Let $X$ be a variety defined over a number field $K$. The Hasse principle, or the local-to-global principle, asserts that $X(K) \ne \emptyset$ if and only if for each completion $K_v$ of $K$, we have $X(K_v) \ne \emptyset$. This is known to hold for every quadric hypersurface, by the work of Hasse and Minkowski. In general, this does not hold.

For curves, there is a famous example of Selmer: the plane cubic curve $C: 3x^3 + 4y^3 + 5z^3 = 0$ has a point over $\mathbb{F}_p$ for every prime $p$ which can be lifted to a point in $\mathbb{Q}_p$, and as a cubic curve it has a real point. Nevertheless, it has no rational points.

By now it is known, due to the following paper of Bhargava which builds upon the previous work of Bhargava and Shankar on computing the average sizes of $\text{Sel}_2(E), \text{Sel}_3(E), \text{Sel}_5(E)$, that a positive proportion of plane cubics satisfy the Hasse principle (that is, they either fail to have a local point at some completion or they have a rational point), and a positive proportion fails the Hasse principle (that is, they have a local point everywhere but no global point).

However, there is a different question that perhaps should have been inspired by Selmer's original example. The Selmer curve is an example of a generalized Fermat cubic curve $ax^3 + by^3 + cz^3$, and even more generally, an example of a plane cubic where the Arnholdt invariant $S$ vanishes, or in the language of this paper of Bhargava and Shankar, the $I$-invariant. Such ternary cubic forms are also characterized by the property that their Hessian determinant (also a ternary cubic form) splits into three linear factors over $\mathbb{C}$.

My question is: what is the behaviour of the Hasse principle among generalized Fermat cubics $ax^3 + by^3 + cz^3$ with $a,b,c \in \mathbb{Z}$ varying over a box? What about plane cubics with vanishing $I$-invariant?

Answer 1

The case of plane cubics is actually easier than the family $ax^3 + by^3 + cz^3$. The reason being that a positive proportion of plane cubics are everywhere locally soluble, when ordered by the height of their coefficients.

However the problem, which one would naively expect to be simpler, of determining the number of Selmer curves which are everywhere locally soluble, turns out to be incredibly difficult. The number of such curves with coefficients bounded by $T$ was shown to be $$\ll \frac{T^3}{( \log T)}, \quad T \to \infty,$$ in T. Browning, R. Dietmann - Solubility of Fermat equations. This upper bound is expected to be sharp, but is far from being proved.

So we don't even understand the distribution of everywhere locally soluble Selmer curves, let alone those which fail/satisfy the Hasse principle.