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