From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: stable trivial fibration
Date: 2 Oct 1997 15:32:07 GMT

In article <01bccf28$b77dd5d0$6a03e9c1@l44-1>,
Sergey N. Maximoff <maximoff@cacr.ioc.ac.ru> wrote:
>What is stably trivial bundle?
>Wich properties does it have? 

A fibre bundle F is stably trivial if the direct sum bundle  F + E  is
trivial for some trivial bundle  E.  

The classic example is the tangent bundle  T  of a sphere  S in  R^n: if  E  
is the (1-dimensional) normal bundle on the sphere, then on the one hand  E
is trivial, since there is an obvious nonvanishing section known as "out"!
On the other hand,  T + E  is also trivial; it's just the restriction to  S
of the tangent bundle of  R^n.

Of course, trivial bundles are stably trivial, but the converse is not
true as this example shows (for most  n). One cannot even find a single
nonvanishing vector field on the sphere in  R^3, let alone finding two
linearly independent ones; thus  T  is nontrivial when  n=3  in the
previous paragraph. (It is a celebrated topic to determine the values of
n  for which this  T  is trivial, namely  n=1, 2, 4, and 8.)

I'm not sure which properties you want to know about. The stably trivial
bundles are equivalent to zero in the appropriate Grothendieck group, and
so are difficult to detect using functors defined on that group. I
guess that gives them the property of stealth!

By the way, a purist might object to your conflating of the terms
"fibration" and "fibre bundle", since the former can be defined
categorically ("for all maps  f  there is a map  g...") while the
latter is defined geometrically ("for all points  p  there is a neighborhood
U..."); fibre bundles are fibrations (in the category of smooth manifolds)
but not conversely. (Peter May has corrupted my thinking forever.)

dave