Tangent bundle of the long line - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T07:17:33Z http://mathoverflow.net/feeds/question/35087 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35087/tangent-bundle-of-the-long-line Tangent bundle of the long line Zack 2010-08-10T07:26:20Z 2010-08-10T15:14:09Z <p><strong>Question:</strong> 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. </p> <p><strong>Motivation:</strong> 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?</p> <p>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.</p> <p>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 $L$.</p> http://mathoverflow.net/questions/35087/tangent-bundle-of-the-long-line/35113#35113 Answer by BS for Tangent bundle of the long line BS 2010-08-10T13:42:25Z 2010-08-10T14:17:48Z <p>The long line has the property that there is no continuous self map $f:L\to L$ such that $f(x)>x$ (or <code>$f(x)&lt;x$</code>) 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, <em>if possible</em>, 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 <a href="http://en.wikipedia.org/wiki/Peano_theorem" rel="nofollow">Peano's existence theorem</a>, and the fact that it furnishes a unique maximal (in the order sense) solution on any compact time interval.</p> <p>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.</p> <p>EDIT: I realize that this only proves that the tangent bundle of $L$ is not <em>fibrewise</em> homeomorphic to $L\times\mathbb{R}$. I hope this is what you meant.</p> http://mathoverflow.net/questions/35087/tangent-bundle-of-the-long-line/35123#35123 Answer by Andreas Blass for Tangent bundle of the long line Andreas Blass 2010-08-10T15:14:09Z 2010-08-10T15:14:09Z <p>Lots of information about this is available in a paper by Peter Nyikos:</p> <p>Various smoothings of the long line and their tangent bundles. Adv. Math. 93 (1992), no. 2, 129--213. </p>