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The long line has the property that there is no continuous self map $f:L\to L$ such that $f(x)>x$ (or $f(x)<x$) for all $x\in L$. Indeed, $f^n(0)$ is an increasing sequence, hence converges to some $x$, which is a fixed point. So if there were an everywhere nonzero tangent vector field (for some differentiable structure on $L$), integrating it, if possible, would give a contradiction. Integration is possible for any locally lipschitz vector field, by the usual argument plus the fact that any map $\mathbb{R}\to L$ has relatively compact image. But it would remain to "regularize" a continuous vector field, which seems not possible in the usual way. Instead, one may try to use Peano's existence theorem, and the fact that it furnishes a unique maximal (in the order sense) solution on any compact time interval.

EDIT: if you are willing to accept using differential forms, any non vanishing continuous vector field $v$ on $L$ would give a dual continuous $1$-form $\alpha$ such that $\alpha(v)=1$. But then, integrating $\alpha$ would give an injective (monotone) function $f:L\to\mathbb{R}$, which is absurd. This is sort of "square root" of the riemannian metric argument, and is much simpler.

EDIT: I realize that this only proves that the tangent bundle of $L$ is not fibrewise homeomorphic to $L\times\mathbb{R}$. I hope this is what you meant.

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The long line has the property that there is no continuous self map $f:L\to L$ such that $f(x)>x$ (or $f(x)<x$) for all $x\in L$. Indeed, $f^n(0)$ is an increasing sequence, hence converges to some $x$, which is a fixed point. So if there were an everywhere nonzero tangent vector field (for some differentiable structure on $L$), integrating it, if possible, would give a contradiction. Integration is possible for any locally lipschitz vector field, by the usual argument plus the fact that any map $\mathbb{R}\to L$ has relatively compact image. But it would remain to "regularize" a continuous vector field, which seems not possible in the usual way. Instead, one may try to use Peano's existence theorem, and the fact that it furnishes a unique maximal (in the order sense) solution on any compact time interval.

EDIT: if you are willing to accept using differential forms, any non vanishing continuous vector field $v$ on $L$ would give a dual continuous $1$-form $\alpha$ such that $\alpha(v)=1$. But then, integrating $\alpha$ would give an injective (monotone) function $f:L\to\mathbb{R}$, which is absurd. This is sort of "square root" of the riemannian metric argument, and is much simpler.

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The long line has the property that there is no continuous self map $f:L\to L$ such that $f(x)>x$ (or $f(x)<x$) for all $x\in L$. Indeed, $f^n(0)$ is an increasing sequence, hence converges to some $x$, which is a fixed point. So if there were an everywhere nonzero tangent vector field (for some differentiable structure on $L$), integrating it, if possible, would give a contradiction. Integration is possible for any locally lipschitz vector field, by the usual argument plus the fact that any map $\mathbb{R}\to L$ has relatively compact image. But it would remain to "regularize" a continuous vector field, which seems not possible in the usual way. Instead, one may try to use Peano's existence theorem, and the fact that it furnishes a unique maximal (in the order sense) solution on any compact time interval.