Free summands of stably free modules

TL;DR

This work translates the algebraic question of when a stably free module $P$ with $Poxplus R eq R^n$ contains a free summand into an $ ext{A}^1$-homotopy problem about sections of a Stiefel-variety projection $ ho:V_r(m{ ext{A}}^n) o V_1(m{ ext{A}}^n)$. By mixing $ ext{A}^1$-homotopy theory with realization techniques (complex and real), the authors reduce obstructions to classical homotopy groups via a comparison of motivic and topological data, using recent computations of stable motivic homotopy sheaves and the Freudenthal suspension theorem in this setting. They obtain explicit divisibility-type criteria: over characteristic zero fields, a stably free module of type $(24n,24n-1)$ must have a free summand of rank $2$, and under certain real-closed or quadratically closed base fields, a free summand of rank $3$ follows; these results extend to schemes and vector bundles. The methods combine geometric Raynaud–James–Adams-type obstructions with modern $ ext{A}^1$-homotopy, yielding concrete, field-dependent decompositions for stably free modules with large trivial summands and giving a robust template for analogous questions over more general bases.

Abstract

Let $R$ be a commutative ring. One may ask when a general $R$-module $P$ that satisfies $P \oplus R \cong R^n$ has a free summand of a given rank. M. Raynaud translated this question into one about sections of certain maps between Stiefel varieties: if $V_r(\mathbb{A}^n)$ denotes the Stiefel variety $\textrm{GL}(n) / \textrm{GL}(n-r)$ over a field $k$, then the projection $V_r(\mathbb{A}^n) \to V_1(\mathbb{A}^n)$ has a section if and only if the following holds: any module $P$ over any $k$-algebra $R$ with the property that $P \oplus R \cong R^n$ has a free summand of rank $r-1$. Using techniques from $\mathbb{A}^1$-homotopy theory, we characterize those $n$ for which the map $V_r(\mathbb{A}^n) \to V_1(\mathbb{A}^n)$ has a section in the cases $r=3,4$ under some assumptions on the base field. We conclude that if $P \oplus R \cong R^{24m}$ and $R$ contains a field of characteristic $0$, then $P$ contains a free summand of rank $2$. If $R$ contains a quadratically closed field of characteristic $0$, or the field of real numbers, then $P$ contains a free summand of rank $3$. The analogous results hold for schemes and vector bundles over them.

Theorems & Definitions (39)

• Theorem : Raynaud
• Theorem
• Remark 1.1
• Proposition 2.1
• proof
• Remark 3.1
• Proposition 3.2
• proof
• Corollary 3.3
• proof