From: mbeattie@black.ox.ac.uk (Malcolm Beattie)
Newsgroups: sci.math
Subject: Re: more math puzzles
Date: 21 Dec 92 10:32:33 GMT
Organization: Oxford University Computing Service, 13 Banbury Rd, Oxford, U
Originator: mbeattie@black

In article <1992Dec18.101409.4666@black.ox.ac.uk> mbeattie@black.ox.ac.uk (Malcolm Beattie) writes:
>In article <24341@galaxy.ucr.edu> baez@ucrmath.ucr.edu (john baez) writes:
>
>>3)  Prove that the tangent bundle of the long line is nontrivial.
>>
>>The answers to the above 3 are in Hirsch's "Differential Topology."  (Topologists 
>>on the net will be delighted to learn that I finally obtained this book and
>>will stop pestering them with questions whose answers are in it.)  
>
>First, the answer. Let L be the long line (more of which, later
>in this post.) Assume for a contradiction that TL is trivial.
>Then TL has a nowhere-vanishing section. Integrate this to get
>a monotonic map from the reals to L. Any such map from the
>reals to L must eventually become constant at some point of L
>and the section therefore vanishes at that point. Contradiction.

Rubbish :-) John Rickard <jrickard@eoe.co.uk> has mailed me
and pointed out that this is the wrong way around: the result
is that any map at all (not necessarily monotonic) from L to the
reals must eventually become constant. I think I can patch up
the proof as follows: integrate the vector field defined by the
section as before. We get a (strictly) monotonic map \lambda
from the reals to L which can't be onto. Look at the long half
line L' where the action is. Let \alpha be the smallest ordinal
such that the image of \lambda avoids the \alpha copy of the
reals in L'. Since \alpha is countable, the part of L' from 0
up to and including the \alpha copy of the reals is homeomorpic to
the reals and so \lambda, being strictly monotonic, must be
surjective. But \lambda avoids the \alpha copy of the reals:
contradiction.

Surely there must be a nicer proof than this?

>
>The clue to answering question (3) is to consider maps
>from L to the reals (R) and maps from R to L.
>Because L is very `long', any map from to R must wriggle a
                                       L
>lot---there's no room to do anything else---and you can show
>that any smooth map from L to R must have uncountably many points
>where its derivative vanishes. I think you can prove more than
>this but I'm not sure. I'm an algebraic topologist not a
>set-theoretic one.

As John points out, the `more than this' is that the map has to
be eventually constant. Somebody sketched the proof for me once,
but I can't remember how it goes. Anyone?

>Conversely, any map from R to L can't
>cover too much of L because L is too long. One thing one
>can show is that any monotonic map from R to L must
>eventually become constant because R isn't long enough
^^^^^^^^^^^^^^^^^^^^^^^^^^^
This is rubbish.
>to have a nice surjection onto L.

--Malcolm




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Malcolm Beattie <mbeattie@black.ox.ac.uk> | I'm not a kernel hacker
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