Re: Prove that Right Trivialization is smooth?

On 26 Aug., 18:41, Jannick Asmus <jannick.n...@xxxxxx> wrote:
On 26.08.2007 18:16, orbif...@xxxxxxxxxxxxxxxxxx wrote:

Hi!

Is there an elegant way to prove that Right Trivialization is smooth?

Yes, but it needs some basic knowledge in how objects in linear algebra
paste together to smooth maps.

So, let G be a Lie Group and denote the Right Multiplication with g
\in G by R_g: R_g(h) := hg (for all h \in G).

Now consider the map:

r: TG ---> G x *g*, v_g |------> (g,T_gR_{g^{-1}} v_g)

where TG is the tangent bundle and *g* = T_eG denotes the Lie-Algebra
of G; the Tangent map at point p is denoted by T_g.

So how can I see that r (and its inverse) are smooth maps (without
working in charts...).

Let v1,...,vn a basis of *g*. Then the map Xm : G -> TG, g ->
T_e(R_g)(vm) defines a vector field for each m.

Ok, I see this if I work in coordinates.

There are 1-forms
w1,...,wn on G which are pointwise a dual basis to X1,..,Xn. Then the
second component of r(v) is sum_m wm(v)vm. This is easily seen to be
smooth map.


This (and the your second proof) essentially is the proof of the
general statement, that a vector bundle is trivial iff there is a
global section, applied to my concrete problem. I think it's a nice
idea to do so.

I thougth it might be possible to write the Right Trivialization as a
composition of "well known" smooth maps. With "well known" I mean maps
like natural projection, tangent map and so on, which are usefull in
many other situations, too. But I don't manage this in a nice way.

Next guess the inverse map. Again choose a basis v1,...,vn of *g*. Now
take a dual basis f1,...,fn in the space of linear forms on *g*. Use
again the vector fields X1,...,Xn to separate the dependence of the
inverse map of r on g in G and v in T_eG.


There is an alternative to the second proof if you apply the terms of
vector bundles. But basically it is the same.

Could you show me?

best regards
Sabine

.

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