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Question: Is the tangent bundle of the long line $L$ homeomorphic to $L\times\mathbb R$? I'd guess that the answer doesn't depend on choice of differentiable structure, but maybe it does.

Motivation: One night at dinner, someone brought up a puzzle involving infinitely many prisoners standing in a line, and someone asked if there was a physical reason that the collection of prisoners had to be countable. In other words, might (one of) the directions in the physical universe be modeled after the long line?

The answer to that question is no: the universe has a metric, but the long line has no Riemannian structure. The standard explanation for this is that a Riemannian manifold is metrizable, and a non-paracompact space isn't. Without using fancy theorems, one could instead suppose that $L$ was Riemannian and look at the exponential map starting at a point $x$ going in the increasing direction. This is an increasing function from $\mathbb R$ to $L$, so it converges to some point $y$. Then the exponential map from $y$ downwards reaches $x$ in finite time, contradiction.

In any case, the basic result is that $L$ is not Riemannian, so its tangent bundle must be nontrivial, but only in the differential sense. One could try to instead consider a continuous metric (if the tangent bundle were indeed topologically trivial), but this wouldn't give rise to an exponential map, nor, as far as I know, a metric space structure on $L$.

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  • $\begingroup$ Maybe there's something I don't get, but can't we assert that the line being contractible, any vector bundle on it must be trivial? $\endgroup$
    – Henri
    Commented Aug 10, 2010 at 9:21
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    $\begingroup$ @ Henri : the long line isn't contractible. See en.wikipedia.org/wiki/Long_line_(topology) for a list of properties and some references. $\endgroup$
    – BS.
    Commented Aug 10, 2010 at 11:58
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    $\begingroup$ The long line is weakly contractible (all its homotopy groups are zero) but not contractible (there is no [0,1]-parametrized homotopy between the identity map and the zero map). $\endgroup$
    – André Henriques
    Commented Aug 10, 2010 at 16:09

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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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  • $\begingroup$ I think this second edit brings up an important point. When the OP says $L \times \mathbb{R}$ do they mean as a vector bundle (with the obvious action on the $\mathbb{R}$ factor)? or do they mean merely as a topological space? These are not obviously equivalent for the long line. So really it looks like there are two questions here. $\endgroup$
    – Chris Schommer-Pries
    Commented Aug 10, 2010 at 15:09
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    $\begingroup$ Indeed, this proves that they are not the same as topological fiber bundles (not only as vector bundles). As to general homeomorphism, I can only prove that this would imply the existence in $TL^+$ ($L^+$ the half ray) of an homeomorphic copy of $L^+$, disjoint from the zero section and with proper projection to $L^+$. I don't know how to exclude this. $\endgroup$
    – BS.
    Commented Aug 10, 2010 at 15:52
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Lots of information about this is available in a paper by Peter Nyikos:

Various smoothings of the long line and their tangent bundles. Adv. Math. 93 (1992), no. 2, 129--213.

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