nLab vectorial bundle

Idea

A vectorial bundle (Gomi 08) is a $\mathbb{Z}_2$-graded vector bundle $E$ of finite rank, equipped with an odd endomorphism $h \;\colon\; E \to E$. The homomorphisms of vectorial bundles are such that the endomorphism $h$ acts like canceling parts of the even and odd degree of $E$ against each other.

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

Definition

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

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

  $$E = \left( \array{ E_{0} \\ E_{1} } \right)$$
  $$h = \left( \array{ 0 & h_{10} \\ h_{01} & 0 } \right)$$

• as morphisms $\phi : (E,h) \to (E',h)$ equivalence classes of morphisms $\phi \colon E \to E'$ of vector bundles such that

  $$\array{ E &\stackrel{\phi}{\longrightarrow}& E \\ {}^{\mathllap{h}}\big\downarrow && \big\downarrow^{\mathrlap{h'}} \\ E &\stackrel{\phi}{\longrightarrow}& E } \,,$$

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

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

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

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

  This yields a canonical inclusion

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

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

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

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

Definition

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

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

Lemma

There is a concordance

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

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

• Kiyonori Gomi, Twisted K-theory and finite-dimensional approximation (arXiv:0803.2327)