Assume that $M$ is the long line. Is $TM$, the tangent bundle, isomorphic to $T^{*}(M)$, the cotangent bundle?
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No (for any differentiable structure on the long line, there are many). Since $TM$ is a line bundle, an isomorphism $TM\cong T^*M$ is necessarily symmetric, i.e. given by a global section of $\mathrm{Sym}^2T^*M$, locally of the form $f(t)dt^2$. Since $M$ is connected the form is either positive or negative, thus provides a riemannian metric, hence a distance, which does not exist on $M$.
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$\begingroup$ The metric gives a local distance, but since the long line is not path-connected, the local distance doesn't compare two arbitrary points. $\endgroup$ Commented Dec 18, 2013 at 5:42
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$\begingroup$ Ah, I did not realize that even a very large "manifold" with a Riemannian metric could be topologically metrized. I was thinking of the obvious definition where you measure the infimum length of piecewise-smooth paths between two points, as computed by arc length, and which obviously doesn't work for a non-path-connected space. $\endgroup$ Commented Dec 18, 2013 at 5:58
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1$\begingroup$ Triviality of the cotangent bundle implies triviality of the tangent bundle (its dual), hence $T M\cong T^*M$ and you can apply my answer. $\endgroup$– abxCommented Dec 18, 2013 at 8:21
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2$\begingroup$ Although this question is 3 years old, just a comment to one of the comments above: The Long Line IS path connected. Any two points are contained in a chart which is diffeomorphic to a unit interval. The 1-point compactification of the Long Line is NOT path connected but since this it not a manifold, it is not interesting. $\endgroup$– TomCommented Nov 7, 2016 at 10:35