NetNews Usenet Archive 1992 #30/NN_1992_30.iso/spool/sci/math/17141

Internet Message Format | 1992-12-20 | 5.9 KB

Path: sparky!uunet!mcsun!uknet!comlab.ox.ac.uk!mbeattie
From: [e-mail] (Malcolm Beattie)
Newsgroups: sci.math
Subject: Re: more math puzzles
Message-ID: <[e-mail]>
Date: 18 Dec 92 10:14:09 GMT
References: <[e-mail]>
Organization: Oxford University Computing Service, 13 Banbury Rd, Oxford, U
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Originator: mbeattie@black

In article <[e-mail]> [e-mail] (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.

Now for an explanation of `the long line' for those that aren't
familiar with it.

Gluing two real intervals together is easy: place the two intervals
side by side and glue the right hand end of the left copy to the
left hand end of the right copy.

Gluing a copy of [0,\infty) (call it B) onto the right hand side
of another copy of [0,\infty) (call this copy A) is almost as easy:
small neighbourhoods are just as usual around all points other than
0 ]in B.

                 A                                B
[0--1--2--3--4--...----------------)  [0--1--2--3--4--...--------)

We want the 0 \in B to be `near' the right hand (`infinite')
end of A so we declare an open neighbourhood base of 0 \in B
to be of the form: (t,\infty) union [0,s)
                     in A            in B

What we get is homeomorphic to the reals, which is not very
interesting. Before we start gluing infinitely many copies
together, we need a digression on ordinals.

Ordinals are special well-ordered sets.
More precisely, you can assign an ordinal ord(\alpha) to
any well-ordered set \alpha (the notation omits mentioning
the order) such that ord(\alpha) = ord(\beta) if and only
if the (well-ordered) sets \alpha and \beta are *order*-isomorphic.
Recall that a well-ordered set is a set with an ordering
relation in which any two elements can be compared and for
which any non-empty subset has a least element.
A good way of thinking about ordinals is that they `locally'
look like the naturals, i.e. given any member \beta of an
ordinal, the ordering proceeds with successors
\beta, \beta + 1, \beta + 2, ...
but don't you dare try to go `backwards.'
Yes, Virginia, you can do arithmetic with infinite ordinals
but don't make unwarranted assumptions: ordinal addition is
not even commutative.
An example of an ordinal is the set of natural numbers
(called \omega when thought of as an ordinal.)
Members of ordinals are also ordinals (this works because
you the natural number n `really is' the set {0,1,2,...,n-1}.

The class of all ordinals satisfies the well-ordering property
(you don't need posh set theory to do this, ZFC suffices, but
it's easier to state with the word `class' not `property'.)
So consider the (non-empty) class of all uncountable ordinals
and take the least one. This smallest uncountable ordinal is
called \aleph_1. One more thing before we get back to the
topology: each ordinal \alpha is either
(1) zero
(2) a successor ordinal (i.e. \alpha = \beta + 1 for some
    ordinal \beta)
(3) a limit ordinal (e.g. \omega)
Now back to the topology.

Notice that you can give each ordinal a topology by declaring
intervals (\alpha,\beta) to be open for each \alpha and \beta
in the ordinal. Although the ordinal looks `discrete' at first
sight, the interesting things happen at limit ordinals.
Recall that any point of an ordinal is an ordinal itself.
At a limit ordinal point \lambda, a neighbourood looks
discrete to its right but certainly not to the left:
it `looks back'. In other words, for each \alpha < \lambda,
(\alpha,\lambda) is open.

Create the long half-line as follows. Take a copy of [0,\infty)
for each member of \aleph_1 and put them all side by side.
This corresponds to the product [0,\infty) \times \aleph_1
viewed with the Hebrew lexical ordering common to the world
or ordinals. This means that the \aleph_1 coordinate takes
precedence. For each `left hand end' of a copy of [0,\infty),
let its open neighbourhood be given by `looking back' to
preceding ordinals. This formalises the idea that moving to the
left from the point 0 in a `successor ordinal copy' of [0,\infty)
jumps you into the copy of [0,\infty) corresponding to the
previous ordinal. This formalises the notion of gluing these
copies of [0,\infty) side by side.
You get the `long line' by gluing two copies of the
`long half line' back to back.

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
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. 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
to have a nice surjection onto L.

This is getting far too long so I'll stop here. Sorry if
this is badly explained, boring or wasted bandwidth.

--Malcolm
--
Malcolm Beattie <[e-mail]> | I'm not a kernel hacker
Oxford University Computing Services      | I'm a kernel hacker's mate
13 Banbury Road, Oxford, OX2 6NN (U.K.)   | And I'm only hacking kernels
Tel: [phone] Fax: [phone]   | 'Cos the kernel hacker's late