# Tangent bundle of a Lie Group is trivial

• Jan 16th 2011, 02:00 PM
Sogan
Tangent bundle of a Lie Group is trivial
Hello,

Let G be a Lie Group and g his Lie Algebra, that is the tangent space at $e \in G$ I have to show that the map $\phi$: G x g -> TG, $\phi(h,X)=dL_h[X]$, (whereas
$dL_h[X](f)=X(f \circ L_h)$ with L_h left multiplication )
is a diffeomorphism and a linear isomorphism on each fibre.

I'm really hopeless! I don't know how i can show this property's of $\phi$.

Do you know some books or links, where i can read the proof? Or can you please explain it to me?

Regards
• Jan 16th 2011, 08:23 PM
xxp9
Since from construction the tangent bundle is locally diffeomorphic to the product of the base manifold G and the identity fiber g, you need only to show that the map is globally one to one. This is easily verified by the fact that $dL_h \cdot dL_{h^{-1}} = id$.
• Jan 17th 2011, 11:41 AM
Sogan
Hello,

thank you for your help! I could show that the map is bijective. But i don't see why it is diffeomorphic, even locally diffeomorphic?

Now i have to show that $\phi$ is a linear isomorphism on each fiber.
I suppose that fiber doesn't mean $\phi^{-1}(V_h)$, because $\phi$ is bijective. So what does fiber means here?

Regards
• Jan 17th 2011, 06:06 PM
xxp9
A bundle is DEFINED to be locally trivial, that is, locally diffeomorphic to the product of the base space and any of the fiber spaces( since all the fibers are isomorphic). In your case, the base space is G and the fiber at identity is g, the tangent plane. So \phi is DEFINED to be locally diffeomorphic.
To see it is a isomorphism between two fibers T_a and T_b, you only need to show that dL_a * dL_b^{-1} is an isomorphism. Intuitively, you can always translate a vector from T_b to g( the tangent plane at identity), then translate to T_a.