nLab
vectorial bundle

Idea

A vectorial bundle is a 2\mathbb{Z}_2-graded vector bundle EE of finite rank, equipped with an odd endomorphism h:EEh : E \to E. Homomorphisms of vectorial bundles are such that the endomorphism hh acts like canceling parts of the even and odd degree of EE against each other.

This way vectorial bundles lend themselves to the description of K-theory. In particular, they allow a geometric model for twisted K-theory.

Definition

For XX a topological space, the category VectrBund(X)VectrBund(X) of vectorial bundles on XX has

  • as objects (EhE)(E \stackrel{h}{\to} E) finite rank Hermitean 2\mathbb{Z}_2-graded vector bundles EXE\to X equipped with a self-adjoint endomorphism hh of odd degree. In matrix calculus

    E=(E 0 E 1) E = \left( \array{ E_{0} \\ E_{1} } \right)
    h=(0 h 10 h 01 0) h = \left( \array{ 0 & h_{10} \\ h_{01} & 0 } \right)
  • as morphisms ϕ:(E,h)(E,h)\phi : (E,h) \to (E',h) equivalence classes of morphisms ϕ:EE\phi : E \to E' of vector bundles such that

    E ϕ E h h E ϕ E, \array{ E &\stackrel{\phi}{\to}& E \\ \downarrow^{h} && \downarrow^{h'} \\ E &\stackrel{\phi}{\to}& E } \,,

    where two such maps are regarded as equivalent, ϕϕ\phi \sim \phi', already if they coincide on the kernel of h x 2h^2_x for each point xx.

In particular, we have the following two important special cases:

  • the case that h=0h = 0 – in this case all eigenvalues of all h x 2h_x^2 are zero. and hence maps ϕ,ϕ:(E,0)(E,0)\phi, \phi' : (E,0) \to (E',0) represent the same morphism precisely if they are actually equal as morphisms ϕ,ϕ:EE\phi, \phi' : E \to E' of vector bundles.

    (Notice that there is only the 0-morphism (E,0)(E,h)(E,0) \to (E',h') for h0h' \neq 0.)

    This yields a canonical inclusion

    VectBund(X)VectrBund(X) VectBund(X) \hookrightarrow VectrBund(X)

    by sending E(E0E)E \mapsto (E \stackrel{0}{\to} E).

  • the case that E=(V V)E = \left( \array{V \\ V}\right) and h=(0 Id Id 0)h = \left( \array{ 0 & Id \\ Id & 0 } \right)

    Here E x <μ<1=0E_x|_{\lt \mu \lt 1} = 0 and hence two morphisms ϕ,ϕ:(E,h)(E,h)\phi, \phi' : (E,h) \to (E',h') are identified already if they agree on the 0-vector. In other words, all morphisms out of such (E,h)(E,h) are identified. In particular they are all equal to the 0-morphism to (0,0)(0,0). Therefore the bundles of this form represent the 0-element.

Definition

Say two vectorial bundles (E,h)(E,h), (E,h)(E',h') on XX are concordant if there is a vectorial bundle on X×[0,1]X \times [0,1] which restricts to them at either end, respectively.

Let (E,h) = (E,h)^{\vee} = be the degree-reversed bundle to (E,h)(E,h).

Lemma

There is a concordance

EE 0. E \oplus E^\vee \to 0 .

References

The definition of vectorial bundles is due to Furuta. It is recalled and applied to the study of K-theory and twisted K-theory in

  • K. Gomi, Twisted K-theory and finite-dimensional approximation (arXiv)

Revised on July 21, 2009 10:30:55 by Urs Schreiber (134.100.222.156)