The definition of vector bundles seems to be split in the mathematical community: some sources insist that the rank of each fibre is the same, whereas some don't ask for this requirement.
I was wondering, how does the semi-group of isomorphism classes of vector bundles over a (compact Hausdorff) space depend on the definition one considers. In particular, what effect does it have on the group K(X)?
Under the assumption of a globally constant rank vector bundles are special cases of fibre bundles. Recall that in a fibre bundle all fibres are homeomorphic to a common "fibre model" $F$. Anyway, in my opinion most authors do not require that all fibres have the same rank. Prominent examples are
The definition allowing variable rank is of course more general. It resembles the definition of a covering map which we can regard as a "generalized fiber bundle" with discrete fibers having not necessarily constant cardinality.
Wikipedia writes
The local trivializations show that the function $x ↦ k_x$ is locally constant, and is therefore constant on each connected component of $X$. If $k_x$ is equal to a constant $k$ on all of $X$, then $k$ is called the rank of the vector bundle, and $E$ is said to be a vector bundle of rank $k$. Often the definition of a vector bundle includes that the rank is well defined, so that $k_x$ is constant.
Given a vector bundle over $X$, we always get a partition of $X$ into finitely many or countably many pairwise disjoint open sets $X_k$ on which all fibres have the same rank $k$ (because the map $\rho : X \to \mathbb N, \rho(x) = k_x$, is locally constant and thus continuous). Each $X_k$ is the union of components of $X$ and $X$ is the topological sum of the $X_k$. Unfortunately, this partition explicitly depends on the vector bundle. In other words, if we consider different vector bundles, we may get different such partitions.
The situation is much nicer if all components of $X$ are open in $X$. In this case $X$ is the topological sum of its components and we may study vector bundles componentwise which shows that conceptionally we do not get something really new. Note that the partition into components does not depend on any vector bundle over $X$ and may be finer than the above bundle-specific partition. However, we shall see later that even in that nice case the $K$-groups based on the constant rank and the variable rank approaches are formally different.
When are all components of $X$ open? There are two main types of examples:
Spaces with finitely many components.
Locally connected spaces (e.g. manifolds and CW-complexes). See About locally path-connected spaces.
If $X$ has infinitely many components and is not locally connected, the situation becomes complicated because there is no universal "topological sum decomposition" of $X$.
If one works with the variable rank definition, one usually restricts to compact base spaces $X$. The benefit of compactness is that the fibre rank $k_x$ is bounded. This allows to exhibit each fibre bundle as a direct summand of a trivial bundle of a sufficiently large rank $N$.
If $X$ has only finitely many components (which is true if $X$ compact and locally connected), then the variable rank approach produces $$K(X) \approx K(C_1) \oplus \ldots \oplus K(C_m) \tag{1}. $$ For $m > 1$ the constant rank approach produces a smaller group simply because the rank of the vector bundles cannot vary on different components. In fact, the elements of $K(X)$ are formal differences $[E] - [F]$ with vector bundles $E, F$, and we have $[E] - [F] = [E'] - [F']$ iff $E \oplus F' \oplus G \approx E' \oplus F \oplus G$ for some bundle $G$. In the variable rank approach we may choose bundles $E$ and $F$ such that the rank difference is $i$ on component $C_i$. Then it is impossible that $[E] - [F] = [E'] - [F']$ with constant rank bundles $E', F'$.
Formula $(1)$ shows that the variable rank approach is superior to the constant rank approach.
Anyway, if $X$ has infinitely many components (as e.g. the Cantor set), then the variable rank approach does not produce such a nice decomposition of $K(X)$ as above. Things get really complicated in this case.
