I'm having trouble understanding why the following logic is wrong. Let $E$ be an orientable rank $2$ vector bundle over a smooth manifold $M$. Endow it with an orientation which we call the positive orientation. Let $-E$ be the same vector bundle but with the reverse orientation. The map $f:E\rightarrow-E$, $x\mapsto Ax$ where $A=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ is a vector bundle isomorphism, which reverses the orientation of the vector bundle.
A similar construction holds in higher even dimensions. But one should not be able to construct an orientation-reversing vector bundle isomorphism on a general even-dimensional vector bundle. When the vector bundle is odd dimensional, the antipodal map is an orientation reversing vector bundle isomorphism, and I'm just trying to understand why this other one does not work more generally. Please help me with this confusion.
Best Answer
Unless the bundle is trivial, the map that you defined is not well defined. A $2$-dimensional vector bundle $E$ over the manifold $M$ is defined by a trivislization $(U_i)_{i\in I}$ such that the restriction $E_i$ of $E$ to $U_i$ is isomorphic to $U_i\times \mathbb{R}^2$ and coordinates change $g_{ij}:U_i\cap U_j\times\mathbb{R}^2\rightarrow U_i\cap U_j\times\mathbb{R}^2$ defined by $(x,y)\rightarrow (x,u_{ij}(x)(y))$. You can define the map you want on $U_i\cap U_j\times \mathbb{R}^2$, but for this map to be defined globally, you need $A$ to commutes with $u_{ij}$ which is not always true.