Vector bundle from an idempotent matrix - Imported Knowlpedia Content

Definition

Let $M$ be a smooth manifold, let $\mathbb F=\mathbb R$ or $\mathbb C$ , and let

p\in M_n\!\left(C^\infty(M,\mathbb F)\right) \qquad\text{satisfy}\qquad p^2=p.

Here $C^\infty(M,\mathbb F)$ is the algebra of smooth $\mathbb F$ -valued functions. The vector bundle defined by $p$ is

E_p := \{(x,v)\in M\times\mathbb F^n:p(x)v=v\} = \coprod_{x\in M}\operatorname{im}p(x).

Because a smooth idempotent has locally constant rank, these images form a smooth vector subbundle of the trivial bundle. Its complementary subbundle is $E_{1-p}$ , and $E_p\oplus E_{1-p}=M\times\mathbb F^n$ .

Sections and projective modules

A smooth section of $E_p$ is precisely a smooth function $s:M\to\mathbb F^n$ satisfying $ps=s$ . Hence

\Gamma(E_p)\cong p\,C^\infty(M,\mathbb F)^n.

This module is a direct summand of the free module $C^\infty(M,\mathbb F)^n$ , so it is a finitely generated projective module. Conversely, every finitely generated projective module is the image of an idempotent endomorphism of a finite free module. This is the concrete reconstruction step in the Serre–Swan correspondence; compare Swan, §§1–3.

Geometry of the construction

The kernel of $p(x)$ equals the image of $1-p(x)$ , giving a smooth complement even when $p$ is not self-adjoint. If $p$ is self-adjoint for the standard inner product, then $p(x)$ is the orthogonal projection onto $E_{p,x}$ .

For $M=S^m\subset\mathbb R^{m+1}$ , the matrix

p(x)=I-xx^{\mathsf T}

is a smooth self-adjoint idempotent whose image is $T_xS^m$ . Thus the tangent bundle of a sphere appears directly as an image bundle.

Scope and near-misses

The construction itself works on any smooth manifold. The familiar equivalence between smooth vector bundles and finitely generated projective $C^\infty(M)$ -modules is most often stated for compact $M$ ; noncompact variants require care about finite generation and the chosen function algebra.

A smooth matrix with varying rank does not define a vector bundle by taking images. For example, $q(x)=[x]$ on $\mathbb R$ has jumping image dimension at $0$ , and it fails the decisive axiom $q^2=q$ . The term “projector” here means idempotent, not necessarily orthogonal.

References

Richard G. Swan, “Vector Bundles and Projective Modules,” Transactions of the American Mathematical Society 105 (1962), 264–277. DOI record. Relevant: §§1–3, section modules and the vector-bundle/projective-module correspondence.
Max Karoubi, K-Theory: An Introduction, Springer, 1978. DOI record. Relevant: Chapter I, vector bundles, projective modules, and idempotents.