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Lie algebra of the diffeomorphism group of a manifold.

  1. May 12, 2010 #1
    I have seen it mentioned in various places that the Lie algebra of the diffeomorphism group of a manifold M is identifiable with the Lie algebra of all vector fields on M, but I have not found a demonstration of this. I can show that the map

    [tex] \rho: Lie(Diff(M)) \to Vect(M), ~~~ \rho(X)_p = \frac{d}{dt}\vert_{t=0} \exp(tX) \cdot p[/tex]

    is surjective but am having difficulty showing injectivity. Here Vect(M) is vector fields on M, p is in M and . denotes the action of Diff(M) on M. Is this even the right way to see the correspondence between Lie(Diff(M)) and Vect(M)?

  2. jcsd
  3. May 12, 2010 #2


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    The diffeomorphism group is generated by the infinitesimal diffeomorphisms, and the infinitesimal diffeomorphisms are vector fields.
  4. May 12, 2010 #3
    Ah, I think I was being dense: the Lie algebra of a Lie group is identifiable with one-parameter subgroups. But vector fields correspond to one-parameter subgroups of diffeomorphisms, so vector fields correspond to the Lie algebra of the diffeomorphism group.
  5. May 12, 2010 #4
    Are we specifying that M is compact here?
  6. May 13, 2010 #5


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    A naive question: clearly there are diffeomorphisms that are not generated by vector fields Are you excluding these?
  7. May 13, 2010 #6
    I guess so, as we need that for the vector fields to be complete.
  8. May 13, 2010 #7
    Maybe there are diffeomorphisms not generated by vector fields but the diffeomorphism group doesn't correspond to vector fields, its Lie algebra does (since its Lie algebra is identifiable with one parameter subgroups, all of which are generated by vector fields).

    Actually, is it the case that there are always diffeomorphisms that aren't the result of following a flow?
  9. May 13, 2010 #8
    What's the flow associated to reflection of R^2 in the x-axis?

    But I think the general flavor of this is that we want to say that if two diffeomorphisms are close enough, they're isotopic, and hence they're isotopic through the flow of a time-dependent vector field. At the infinitesimal level, this corresponds to a vector field.
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