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Question on Lie Algebras

  1. Dec 22, 2009 #1
    Let:

    [tex]g_{j}(t)[/tex] be a curve in a group G, which goes through the identity element; g_j(t=0) = identity.

    and:

    [tex]\xi_{j}=\frac{d}{dt}g_{j}(t)\right|_{t=0}[/tex]

    We know that:

    [tex]\xi_{j}{\in}Lie(G)[/tex]

    Why can we say:

    1) [tex]hg(t)h^{-1}[/tex] (h is an element of the Group)

    is also a curve in the group, which goes through the identity element, ie. g(t=0)=identity? [As an aside - how would you even go about doing this transformation - I mean if g(t) is a curve (for example g(t)=2t+4t^3), how can you combine this function with h and h^-1, which are, say, SU(2) matrices?]

    2) [tex]g_{2}(t)\xi_{1}g_{2}(t)^{-1}{\in}Lie(G)[/tex]?

    I mean, these look a bit like similarity transformations - can someone clarify why these statements are true?

    Thanks.
     
    Last edited: Dec 23, 2009
  2. jcsd
  3. Dec 22, 2009 #2

    George Jones

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    Do mean g_j(t = 0) = identity?
    As a function, the curve g_j in the group G maps what set to what set?

    As a function, g(t) = 2t+4t^3 maps what set to what set?
     
  4. Dec 23, 2009 #3
    yes (i have corrected OP)

    I am not sure if g_j is a map as such. Infact the definition of g_j(t) is very loose - it is any curve that simply goes through the identity of the group.

    Well, from parameter t to the function g(t), but the parameter is arbitrary, it doesn't matter what it is (with the exception of the constraint mentioned in the OP that d/dt og it evaluated at t=0 must be the tangent space to G at the identity)
     
  5. Dec 23, 2009 #4

    Fredrik

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    A curve in a set S is a function from an interval of the real numbers into S, so if g is a curve in a Lie group G, then so is [itex]t\mapsto hg(t)h^{-1}[/itex] (since g(t) and h are both members of G).

    An expression like [itex]gAg^{-1}[/itex] where g is a member of the Lie group and A is a member of the Lie algebra clearly makes sense if we're dealing with a group of matrices (because then g and A are both matrices). If the Lie group isn't a matrix Lie group, then we can still make sense of it, by defining that notation to mean

    [tex]\lambda_g_*\rho_{g^{-1}}_*A[/tex]

    where [itex]\lambda_g[/itex] is left multiplication by g and [itex]\rho_{g^{-1}}[/itex] is right multiplication by [itex]g^{-1}[/itex]. Look up "pushforward" or "push-forward" if you don't know what the * means.
     
  6. Dec 23, 2009 #5
    Thanks for the reply.

    Okay, I am convinced that:

    [tex]
    g_{2}(t)\xi_{1}g_{2}(t)^{-1}{\in}Lie(G)
    [/tex]

    May I ask a few more questions: why is it also the case that:

    [tex]
    [g_{2}(t)\xi_{1}g_{2}(t)^{-1}-\xi]{\in}Lie(G)
    [/tex]

    and can someone please explain these statements:

    "by smoothness

    [tex]
    \underbrace{lim}_{t\rightarrow0}\frac{1}{t}[g_{2}(t)\xi_{1}g_{2}(t)^{-1}-\xi]{\in}Lie(G)
    [/tex]

    this implies that [tex][\xi_{1},\xi_{2}]{\in}Lie(G)[/tex]"

    Thanks.
     
  7. Dec 23, 2009 #6

    Fredrik

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    Because a Lie algebra is a vector space, and vector spaces are always closed under addition. If x and y are members of a vector space V, then so is x-y=x+(-y).

    What does the underbrace thing mean? Is it just a limit or something else? And what does "this" refer to? Ah, is it a new sentence, so it refers to the sentence before it? An uppercase T would have helped. I think they mean that the smoothness of the function

    [tex]\phi=\lambda_{g_2(t)}\circ\rho_{g_2(t)^{-1}}[/tex]

    implies that its pushforward is a map from [itex]T_eG[/itex] into [itex]T_{\phi(e)}G=T_eG[/itex], i.e. from [itex]\mathfrak g[/itex] into [itex]\mathfrak g[/itex].
     
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