 Direct sum of vector bundles

For two vector bundles $E,F$ over the same base space $X$ and the same scalar field, their direct sum of vector bundles is a bundle with fibersIts total space consists of pairs $(e,f)$ lying above the same $x$, with the topology inherited from the fiber product. If $E$ and $F$ have ranks $r$ and $s$, simultaneous local trivializations give fiber $k^{r+s}$, and their transition functions of a vector bundle combine in block-diagonal matrices. Thus the ranks add. This differs from the tensor product of vector bundles, whose ranks multiply.^[ref]

A geometric example comes from the unit sphere $S^2\subset {\mathbb{R}}^3$. At each $x$, tangent vectors form the plane perpendicular to $x$ and normal vectors form the line spanned by $x$. Their direct sum is all of ${\mathbb{R}}^3$, continuously in $x$. Consequently $T{S}^2\oplus {\epsilon }^1\cong {\epsilon }^3$, where ${\epsilon }^j$ denotes the trivial vector bundle of real rank $j$. Explicitly, $(v,t)\mapsto v+tx$ is the fiberwise isomorphism. Adding one trivial direction can therefore trivialize a nontrivial tangent bundle; the Hairy ball theorem prevents $T{S}^2$ itself from being trivial.