Motivic obstruction to rationality of a very general cubic hypersurface in $\mathbb P^5$

TL;DR

This paper investigates the indecomposability of transcendental motives of surfaces and uses this to provide evidence that very general cubic hypersurfaces in P^5 are not rational, linking motivic properties to rationality questions.

Contribution

It introduces the notion of integral indecomposability of transcendental motives and proves indecomposability for certain surfaces, proposing a conjecture that implies non-rationality of very general cubic hypersurfaces in P^5.

Findings

Transcendental motives of certain surfaces are integrally indecomposable.

Indecomposability of motives for specific surfaces like Fermat sextic.

Conjecture linking motivic indecomposability and non-rationality of cubic hypersurfaces.

Abstract

Let $S$ be a smooth projective surface over a field. We introduce the notion of integral decomposability and, respectively, the opposite notion of integral indecomposability, of the transcendental motive $M^2_{\rm tr}(S)$. If the transcendental motive is indecomposable rationally, then it is indecomposable integrally. For example, $M^2_{\rm tr}(S)$ is rationally, and hence integrally indecomposable if $S$ is an algebraic $K3$-surface whose motive is known to be finite-dimensional. In the paper we prove that $M^2_{\rm tr}(S)$ is integrally indecomposable when $S$ is the self-product of a smooth projective curve having enough morphisms onto an elliptic curve with complex multiplication. This applies, for example, when $S$ is the self-product of the Fermat sextic in $\mathbb P^2$. Some refinement of the same technique yields that $M^2_{\rm tr}(S_6)$ is integrally indecomposable, where $S_6$ is the Fermat sextic in $\mathbb P^3$. This suggests a conjecture saying that the transcendental motive of any smooth projective surface is integrally indecomposable. We prove in the paper that if this motivic integral indecomposability conjecture is true, and if the motive of any smooth projective surface is finite-dimensional, then a very general cubic hypersurface in $\mathbb P^5$ is not rational.