Re: Prove that Right Trivialization is smooth?

On 26.08.2007 18:16, orbifold@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. 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.

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.

HTH.

Best wishes,
J.
.

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