nLab
vectorial bundle

Idea

A vectorial bundle is a ${\mathbb{Z}}_2$-graded vector bundle $E$ of finite rank, equipped with an odd endomorphism $h:E\to E$. 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 K-theory. In particular, they allow a geometric model for twisted K-theory.

Definition

For $X$ a topological space, the category $\mathrm{VectrBund}(X)$ 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

  $$E=\left(\begin{array}{c}{E}_0\\ E_1\end{array}\right)$$E = \left( \array{ E_{0} \\ E_{1} } \right)
  $$h=\left(\begin{array}{cc}0& h_{10}\\ h_{01}& 0\end{array}\right)$$h = \left( \array{ 0 & h_{10} \\ h_{01} & 0 } \right)

• as morphisms $\varphi :(E,h)\to (E\prime ,h)$ equivalence classes of morphisms $\varphi :E\to E\prime$ of vector bundles such that

  $$\begin{array}{ccc}E& \stackrel{\varphi }{\to }& E\\ {\downarrow }^h& & {\downarrow }^{h\prime }\\ E& \stackrel{\varphi }{\to }& E\end{array},$$\array{ E &\stackrel{\phi}{\to}& E \\ \downarrow^{h} && \downarrow^{h'} \\ E &\stackrel{\phi}{\to}& E } \,,

  where two such maps are regarded as equivalent, $\varphi \sim \varphi \prime$, already if they coincide on the kernel of $h_x^2$ 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 $\varphi ,\varphi \prime :(E,0)\to (E\prime ,0)$ represent the same morphism precisely if they are actually equal as morphisms $\varphi ,\varphi \prime :E\to E\prime$ of vector bundles.

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

  This yields a canonical inclusion

  $$\mathrm{VectBund}(X)\hookrightarrow \mathrm{VectrBund}(X)$$VectBund(X) \hookrightarrow VectrBund(X)

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

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

  Here $E_x{\mid }_{<\mu <1}=0$ and hence two morphisms $\varphi ,\varphi \prime :(E,h)\to (E\prime ,h\prime )$ 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 $(E,h)$, $(E\prime ,h\prime )$ 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

$$E\oplus E^{\vee }\to 0.$$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)