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- Newsgroups: sci.math
- Path: sparky!uunet!enterpoop.mit.edu!galois!riesz!jbaez
- From: jbaez@riesz.mit.edu (John C. Baez)
- Subject: Re: more math puzzles
- Message-ID: <1992Dec21.185416.4483@galois.mit.edu>
- Sender: news@galois.mit.edu
- Nntp-Posting-Host: riesz
- Organization: MIT Department of Mathematics, Cambridge, MA
- References: <24341@galaxy.ucr.edu> <1992Dec18.101409.4666@black.ox.ac.uk> <1992Dec21.103233.723@black.ox.ac.uk>
- Date: Mon, 21 Dec 92 18:54:16 GMT
- Lines: 42
-
- In article <1992Dec21.103233.723@black.ox.ac.uk> mbeattie@black.ox.ac.uk (Malcolm Beattie) writes:
- >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.
-
- >>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 :-) ...
- >Surely there must be a nicer proof than this?
-
- I'll relent and give Hirsch's one-sentence proof. If the tangent bundle
- of the long line were trivial, the long line would admit a Riemannian
- metric, hence be metrizable, but it's not.
-
- Perhaps some clarification might help those who aren't differential
- geometry junkies. The long line is not metrizable, i.e. it does not
- admit a metric inducing its standard topology, essentially because it's
- way too long. It's also not paracompact. Usually one only considers
- paracompact smooth manifolds. Paracompact smooth manifolds always admit
- a Riemannian metric and hence, if they are connected, are metrizable.
- But this fails without paracompactness. If the tangent bundle of the
- long line were trivial, all the tensor bundles would be, so one could
- find a constant section of the bundle of (0,2) tensors (or is it (2,0)
- tensors - I refuse to learn this convention!) giving the long line a
- Riemannian metric - a contradiction.
-
- If one has never shown the following, one should: every sum of
- uncountably many positive real numbers diverges. In a heuristic sort of
- way this is why one might expect the long line not to be metrizable:
- it's built out of uncountably many intervals, and if all of these had
- positive length one could find two points that were infinitely far away
- because there were uncountably many intervals between them. That's no
- proof, mind you.
-
- I don't feel I understand this all as well as I might. Someone with
- good German might try Koch and Puppe, Differenzbar Strukturen auf
- Mannifaltigkeiten ohne abzahlbare Basis, Arch. D. Math. 9 (1960) p. 104.
-