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Lie groups and non-vanishing vector fields

  1. Aug 10, 2011 #1
    I'm trying to understand why a Lie group always has a non-vanishing 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.
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  3. Aug 10, 2011 #2

    Ben Niehoff

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    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 1-to-1 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 left-invariant, since the vector field is mapped into itself under the left action of the group.
  4. Aug 11, 2011 #3
    Thank you very much for your elaborate answer. However, some things are not clear to me yet.

    I don't really understand what free means, but I do believe that a this mapping is bijective (which is true for any group, isn't it, not just Lie groups?) and I can imagine that it is differentiable since the mapping takes an analytic function (meaning the representation matrices are analytic functions of the group parameters) and creates another analytic function, though that is perhaps not a very rigorous way of looking at it. However, I do not understand why it must therefore induce a map on the tangent space.
  5. Aug 11, 2011 #4

    Ben Niehoff

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    Transitive means that any given two points on M are connected by some group multiplication. That is, for any [itex]g_1, g_2 \in M[/itex], there exists some [itex]h \in M[/itex] such that [itex]g_2 = h \cdot g_1[/itex].

    Free 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 1-to-1. 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.
  6. Aug 11, 2011 #5
    I managed to get a hold on this pushforward map in terms of the group representation. Thanks for your help!
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