Let $(M^2,g)$ be a noncompact orientable Riemannian surface without boundary. Let $A \in \Gamma(\operatorname{Sym}(TM))$ be a section of the bundle of symmetric endomorphisms of $TM$, that is, for each $p \in M$, the linear map $A\_p : T\_p M \to T\_p M$ is symmetric with respect to the inner product $g\_p$. Assume that the eigenvalues $\lambda\_1(p), \lambda\_2(p) \in \mathbb{R}$ of $A\_p$ are distinct for all $p \in M$.



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> Do there exist two smooth globally defined unit vector fields $E\_1, E\_2$ on $M$ such that $E\_i(p)$ is an eigenvector of $A\_p$ associated with $\lambda\_i(p)$ for $i=1,2$ and all $p \in M$?
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> Does the topology of $M$ impose any restrictions on the existence of such vector fields?
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Global choice of eigenvectors on an open surface
Not necessarily. To construct a counter-example, start from the other direction. Suppose that the tangent bundle of $M$ can be split as the direct sum $TM = L\_1\oplus L\_2$ where $L\_1$ and $L\_2$ are non-trivial smooth line bundles. Then you can easily construct a metric $g$ on $M$ such that $L\_1$ and $L\_2$ are $g$-orthogonal. Let $A:TM\to TM$ be the linear transformation that is the identity on $L\_1$ and minus the identity on $L\_2$. Then $A$ is symmetric with respect to $g$, but the two eigenbundles of $A$ have no non-trivial sections.


A simple example of such a splitting is to let the (orientable) cylinder $M$ be the quotient of the plane $\mathbb{R}^2$ by the translation $(x,y)\mapsto (x{+}\pi,y)$, and let $L\_1$ be the bundle spanned by the vector field $Z = \cos x\,\partial\_x + \sin x\,\partial\_y$ (which is only defined up to a sign on $M$), while $L\_2$ is spanned by the vector field $W = -\sin x\,\partial\_x + \cos x\,\partial\_y$ (again defined only up to a sign on $M$).


