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Thread: Group Action

  1. #1 Group Action 
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    I have a slight problem. The subject is rather specialized; any attempt on my part to sugar the pill would take half a book, so I simply throw myself on the mercy of those who have half an idea of what I am talking about.

    So we let be a Lie group and a manifold. Then the group (left) action is given by

    One sets so that . This all fine with me.

    Now I am told told that the vector field on "generates" this action, which I am not quite getting. Help!

    Worse, my text now says something like "let us realize the algebra by means of the fields induced by the mapping ". Note the muddle (in my mind) of the active "generates" and passive "induced by"

    And they go to some pains to point out that they use the notation to denote a field, i.e. an element in the set to distinguish it from the element .

    Now I am well aware that the elements close into a real Lie algebra, so it seems we might have TWO distinct algebras - one as the vector space of fields on and the other as the vector space at the identity of .

    Is this right? Is it possible? I doubt it (since has the product topology). And in any case, even if there ARE 2 algebras, surely it's wrong to conflate them in this way?

    Please help (I have more questions, but later for that)


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  3. #2 Re: Group Action 
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    Quote Originally Posted by Guitarist
    I have a slight problem. The subject is rather specialized; any attempt on my part to sugar the pill would take half a book, so I simply throw myself on the mercy of those who have half an idea of what I am talking about.

    So we let be a Lie group and a manifold. Then the group (left) action is given by

    One sets so that . This all fine with me.

    Now I am told told that the vector field on "generates" this action, which I am not quite getting. Help!

    Worse, my text now says something like "let us realize the algebra by means of the fields induced by the mapping ". Note the muddle (in my mind) of the active "generates" and passive "induced by"

    And they go to some pains to point out that they use the notation to denote a field, i.e. an element in the set to distinguish it from the element .

    Now I am well aware that the elements close into a real Lie algebra, so it seems we might have TWO distinct algebras - one as the vector space of fields on and the other as the vector space at the identity of .

    Is this right? Is it possible? I doubt it (since has the product topology). And in any case, even if there ARE 2 algebras, surely it's wrong to conflate them in this way?

    Please help (I have more questions, but later for that)
    I think a bit more explanation of the setting is needed.

    What is the relationship between G and M other than the given fact that there is a (presumably smooth) G-action on M ? Is it possible that M is in fact G ?

    Is Xm an arbitrary vector field on M or this there something more to it ? In short, where did Xm come from ?

    What book are you reading ?


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  4. #3 Re: Group Action 
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    Quote Originally Posted by DrRocket
    I think a bit more explanation of the setting is needed.
    Well, the general setting is that the manifold is symplectic. I had not thought it necessary for for first question, but perhaps I was wrong?

    What is the relationship between G and M other than the given fact that there is a (presumably smooth) G-action on M ?
    Yes, smoothness is assumed. In fact, as per my construction,
    Is it possible that M is in fact G ?
    Not in the present case, no. I DO understand the left (right) action of a group on itself

    Is Xm an arbitrary vector field on M or this there something more to it ? In short, where did Xm come from ?
    I guess that was part of my question. All I am told is that there is a vector field that generates the action

    What book are you reading ?
    As always this one. The relevant p.173 seems to be viewable
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  5. #4  
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    Without knowing the full context of what you're reading, here are my two cents:

    Suppose that the exponential map is onto, so every g in G is exp(X) for some X in Lie(G). Then, given any g in G, there is a natural vectorfield on M associated to the diffeomorphism f_g : M--->M (where f_g means translation by g).

    Namely, since g = exp(X), consider the 1-parameter subgroup in G generated by X. This subgroup acts on M, and gives rise to the following vectorfield:



    Note that if we move along the flow of X_M from time t=0 to t=1, this is the same as multiplying by g.
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  6. #5  
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    Quote Originally Posted by salsaonline
    Without knowing the full context of what you're reading, here are my two cents:

    Suppose that the exponential map is onto, so every g in G is exp(X) for some X in Lie(G). Then, given any g in G, there is a natural vectorfield on M associated to the diffeomorphism f_g : M--->M (where f_g means translation by g).

    Namely, since g = exp(X), consider the 1-parameter subgroup in G generated by X. This subgroup acts on M, and gives rise to the following vectorfield:



    Note that if we move along the flow of X_M from time t=0 to t=1, this is the same as multiplying by g.
    That makes sense, and there are Lie groups for which the exponential map is onto. Simply connected nilpotent Lie groups for instance (the Heinsenberg group being one).

    But the construction is the production of a vector field from a group action (of a special class of Lie groups) rather than production of a group action starting from an essentially arbitrary vector field, which is what I thought guatarist had implied. I have no idea how to do the latter.

    I suspect that you have put your finger on the gist of the problem, that this sort of thing is what the author of the book had in mind and that the idea is not so general as originally represented.
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  7. #6  
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    I thank you both for your help. I have to go to Brest for a few days, so don't take my silence as ingratitude
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