
#1
Aug1011, 08:15 PM

P: 392

I'm trying to understand why a Lie group always has a nonvanishing vector field. I know that one can somehow generate one by taking a vector from the Lie algebra and "moving it around" using the group operations as a mapping, but the nature of this map eludes me.




#2
Aug1011, 10:11 PM

Sci Advisor
P: 1,562

Since the points of the manifold are elements of the Lie group, they inherit a notion of multiplication from the group multiplication. Hence if [itex]g_1[/itex] and [itex]g_2[/itex] are two points on the manifold, we can use them to find a third point
[tex]h = g_1 \cdot g_2[/tex] where [itex]\cdot[/itex] is the group multiplication. Now if M is the Lie group manifold, let's define a map [itex]\varphi_h : M \rightarrow M[/itex] that takes [itex]g \mapsto h \cdot g[/itex] for some given [itex]h[/itex]. This map is called the left action by h, since it multiplies with h on the left. In particular, we see that [itex]\varphi_h[/itex] carries the identity to h: [tex]\varphi_h(e) = h \cdot e = h[/tex] Now, since the Lie group is free and transitive, the map [itex]\varphi_h[/itex] is 1to1 and onto. It shouldn't be hard to see that [itex]\varphi_h[/itex] is also differentiable; hence it is a diffeomorphism. Therefore it also induces a map on the tangent space, the pushforward [itex]\varphi_{h*} : T_gM \rightarrow T_{h \cdot g}M[/itex], given by the Jacobian matrix in some suitable system of coordinates. Since the Lie algebra is simply the tangent space at the identity, we can use [itex]\varphi_{h*}[/itex] to map it onto the tangent space at h. This is how we "move vectors around" on the manifold. In particlar, a vector field [itex]X : M \rightarrow TM[/itex] for which [tex]X(h) = \varphi_{h*}( X(e) )[/tex] is called leftinvariant, since the vector field is mapped into itself under the left action of the group. 



#3
Aug1111, 05:47 AM

P: 392

Thank you very much for your elaborate answer. However, some things are not clear to me yet.




#4
Aug1111, 03:29 PM

Sci Advisor
P: 1,562

Lie groups and nonvanishing vector fieldsFree means that the action of the group on M has no fixed points aside from the identity. So, transitivity implies that the map [itex]\phi_h[/itex] is onto, and free implies that the map [itex]\phi_h[/itex] is 1to1. Together these facts tell us that [itex]\phi_h[/itex] is a bijection. If the group action by left multiplication were not both free and transitive, then [itex]\phi_h[/itex] would not be a bijection. However, it turns out that the left action of a group on itself is always free and transitive. If you have any differentiable map between manifolds [itex]f : M \rightarrow N[/itex], then it induces a linear map between their tangent spaces [itex]df : T_xM \rightarrow T_{f(x)}N[/itex] which is given by the matrix of partial derivatives of f. 



#5
Aug1111, 04:40 PM

P: 392

I managed to get a hold on this pushforward map in terms of the group representation. Thanks for your help!



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