I was trying to disprove the fact that there exist a global tangent frame on the Klein bottle, i.e. two global vector fields everywhere indipendent. Since my background involves no charachteristic classes but only basic differential geometry, I tried to tackle it by reducing to the following claim, which I am starting to suspect is not true:

Given a manifold $M$ of dimension $m$, if there exists a global tangent frame $(X_{1},...,X_{m})$ then $M$ is orientable

I am using the following definiton of oriented manifold: we need to choose an orientation on every tangent space $T_{q}M$ such that for each point there exists a neighbourhood and a chart such that the chart has orientation preserving differential for each point in the neighbourhood. I tried to prove this fact in a rather obvious way, that is to define a smooth function for each $p \in M$ from a small connected neighborhood $U_{p}$ with a chart $\psi$ $f : U_{p} \rightarrow \mathbb{R}$ defined in a point $q \in U_{p}$ as the determinant of the matrix having as columnes the vectors $d_{q}\psi(X_{i}(q))$. Then you naturally define the orientation of the tangent space for the points $q \in U_{p}$ as that of the frame, and the chart as either $\psi$ if $f > 0$ on $U_{p}$ or as $f$ composed with a reflection otherwise. However, to prove that this is a good definition (that is, is coherently defined for points belonging to two neighbourhoods $U_{p_1}$ and $U_{p_2}$) I seem to need in every way some propoerty equivalent to the orientability of $M$, and so I am stuck.

To sum it up, my question is: Is the approach described above correct? If not, what is an elementary proof of the fact that there is no frame of the Klein bottle?

Thanks in advance

  • $\begingroup$ Your claim is true. The converse holds for $m=3$ but is nor true in general. For the proof, you can consider a dual algebraic basis. $\endgroup$ – Dog_69 May 7 '18 at 0:00
  • $\begingroup$ Are you suggesting me to use the fact that a manifold is orientable if and only is there exists a volume form, and that $dX_{1} \wedge \ldots \wedge dX_{n}$ is such a form? Because I would have liked to stick to the definition of orientation not involving volume forms: however, did you mean that? $\endgroup$ – Dario RANCATI May 7 '18 at 0:11
  • $\begingroup$ No no. In fact I have said you the converse is not true in general, so haven't if and only if but only if. And a proof for your claim was taking a dual basis. For your question, I don't know how ti proceed. As you, I don't know anything about homology and cohomology, only some differential geometry $\endgroup$ – Dog_69 May 7 '18 at 8:19

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