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Lie algebra question

  1. Aug 19, 2007 #1


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    I'm reading about gauge theory and the text goes through some stuff about Lie groups and algebras rather quickly. I tried to prove one of the things they state without proof and got stuck.

    Suppose that M and N are manifolds and [itex]\phi:M\rightarrow N[/itex] is a diffeomorphism. Then we can define a function

    [tex]\phi(p)_*:T_pM\rightarrow T_{\phi(p)}N[/tex]

    for each [itex]p\in M[/itex].

    A Lie group is both a group and a manifold. We can use any member g of a group G to construct two diffeomorphisms [itex]\rho_g[/itex] and [itex]\lambda_g[/itex] that map G onto itself:


    The Lie algebra associated with the Lie Group is defined as the tangent space at the identity element, with a Lie bracket that will be defined below. Let's use the notation [itex]\mathfrak{g}=T_eG[/itex]

    We can use either right or left multiplication to map the Lie algebra onto the tangent space at any other point g:

    [tex]\rho_g(e)_*:\mathfrak{g}\rightarrow T_gG[/tex]
    [tex]\lambda_g(e)_*:\mathfrak{g}\rightarrow T_gG[/tex]

    Let's simplify the notation a bit:


    We can use these maps to construct two vector fields [itex]X_L^\rho[/itex] and [itex]X_L^\lambda[/itex] for each vector L in the Lie algebra:


    Either of these two vector fields can be used to define a Lie bracket on the Lie Algebra:


    (I assume that anyone who can help me with this already knows the definition of the commutator of two vector fields, which is used on the right).

    The claim I haven't been able to prove is that these two definitions of the Lie bracket are equivalent, i.e. that it doesn't matter if we define it using right or left multiplication. So my question is, can someone help me prove that?

    A few observations:

    [tex][K,L](f)=K(X_L^\rho f )-L(X_K^\rho f)[/tex]
    [tex][K,L](f)=K(X_L^\lambda f)-L(X_K^\lambda f)[/tex]

    [tex](X_L^\rho f)(h)=L(f\circ\rho_h)[/tex]
    [tex](X_K^\lambda f)(h)=L(f\circ\rho_h)[/tex]


    What am I missing? I have a feeling it's something simple.
    Last edited: Aug 19, 2007
  2. jcsd
  3. Aug 30, 2007 #2
    Hi Fredrik,
    I also tried to prove the equivalence of the left and right Lie bracket. But unfortunately it’s just not true – the right Lie bracket is the reverse of the left one. You can find a note on
    http://planetmath.org/encyclopedia/LieGroup.html [Broken]​

    The main idea is to consider the inversion map

    [tex] \phi:G\to G,\,g\mapsto g^{-1}\;.[/tex]

    Using your notation and [itex]X_g[/itex] for a tangentvector at [itex]g\in G[/itex] we have

    [tex] \phi(p)_*(X_g) = -g^{-1}\,X_g\,g^{-1}\;, [/itex]

    and for your right-invariant vectorfield [itex]X_L^\rho|_g=Lg[/itex] we get

    [tex](\phi_*X_L^\rho)|_g := \phi(g^{-1})_*(X_L^\rho|_{g{-1}}) = -g\,Lg^{-1}g = -gL = -X_L^\lambda|_g = X_{-L}^\lambda|_g[/tex]

    (Here we used the [itex]\phi[/itex]-transformation [itex]\phi_*X[/itex] of a vectorfield [itex]X[/itex]. In the next step we also need, that the [itex]\phi[/itex]-transformation of vectorfields is a Liealgebra-homomorphism of the set of vectorfields, i.e. [itex][\phi_*X,\,\phi_*Y] = \phi_*([X,Y])[/itex])


    [tex] [K,L] _\lambda :=[X_K^\lambda,X_L^\lambda]_e = [-\phi_*X_K^\rho,-\phi_*X_L\rho]_e = (\phi_*([X_K^\rho,X_L^\rho]))_e = -[X_K^\rho,X_L^\rho]_e=:-[K,L]_\rho = [L,K]_\rho \;,[/tex]

    i.e. the left Lie bracket is the negative of the right Lie bracket, or you can say the left Lie bracket is the reverse of the right Lie bracket.
    The result is somewhat surprising (at least to me), because this means, that if you have a Lie group, which is a matrix Lie group, the Lie algebra bracket [itex][X,Y] = XY-YX[/itex] is related to the left Lie bracket, not to the right, even though it first looked somewhat symmetric…

    A question:
    I don’t understand your ‘observations’ except the last two. There you have an unusual mixture of Lie algebra elemente like [itex]K[/itex] and [itex]L[/itex] with vectorfields and functions. What does [itex] L(f\circ\rho_h)[/itex] or [itex] K(X_L^\rho f )[/itex] mean?
    Last edited by a moderator: May 3, 2017
  4. Aug 30, 2007 #3


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    Thank you for the reply. I haven't been here for a week because it seemed that no one was going to answer.

    [itex] L(f\circ\rho_h)[/itex] is just the real number you get when [itex]L\in\mathfrak{g}=T_eG[/itex] acts on the function [itex]f\circ\rho_h:G\rightarrow\mathbb{R}[/itex].

    [itex]X_L^\rho f[/itex] is the map from G into [itex]\mathbb{R}[/itex] defined by [itex]g\mapsto X_L^\rho|_g f[/itex].

    This is the notation I would use to define the commutator [X,Y] of two vector fields X and Y:


    This holds for all g, so the same fact can be expressed this way instead:


    This equation says exactly the same thing, except that here I'm expressing the identity using vector fields (such as [tex][X,Y][/itex]) instead of tangent vectors (such as [itex][X,Y]_g[/itex]).

    I hope that helps. If there's anything else you want me to explain, just let me know. You probably don't need to pay any attention to the results I called "observations" though. Those are just results I obtained while trying to show that the two definitions of the Lie Bracket are equivalent, and they may not be useful at all.

    I have changed my mind about where it's appropriate to put the asterisk in an expression involving the "push-forward" function. E.g. I think it makes more sense to write [itex]\phi_*(g)X_g[/itex] than [itex]\phi(g)_*X_g[/itex], but from now on I'll just write [itex]\phi_*X_g[/itex]. I believe that's what most people do, and it isn't very helpful to keep the "g" anyway.

    I have tried to show this, but it seems to me that it can't be true. Maybe I misunderstood something. The left-hand side acting on a function f is


    But the right-hand side of your identity acting on the same function f is


    The same vector, except for the sign, acting on a different function...I don't see how the results can be the same.

    I tried to show this without using the other identity, and this is the closest I've been able to get:

    [tex]\phi_*X_L^\rho|_g(f) = X_L^\rho|_g(f\circ\phi) = \rho_{g*}L(f\circ\phi) = L(f\circ\phi\circ\rho_g) [/tex]

    [tex]=L(f\circ\lambda_{g^{-1}}\circ\phi) = \phi_*L(f\circ\lambda_{g^{-1}}) = (\lambda_{g^{-1}*}(\phi_*L))(f) = X_{\phi_*L}^\lambda|_{g^{-1}}(f)[/tex]
    Last edited: Aug 30, 2007
  5. Aug 31, 2007 #4
    Thanks for reminding me to think of [itex]L[/itex] and [itex]K[/itex] as operators instead just as tangentvectors. For a moment I thought only vectorfields can act on functions, so I got confused.

    Here you have to keep track of the position you’re taking the derivatives:

    [tex] (\phi_*X)_g(f) := X_{\phi^{-1}(g)}(f\circ\phi)[/tex]

    On the LHS we take the derivative of [itex]f[/itex] in [itex]g[/itex]. Since on the RHS [itex]\phi[/itex] maps [itex]g[/itex] first to [itex]\phi(g)[/itex] we have to use [itex] X_{\phi^{-1}(g)} [/itex] instead of [itex] X_g [/itex].
    You might better see this by considering [itex]\phi_*X[/itex] as a vectorfield not as an operator: [itex]\phi[/itex] is a diffeomorphism of [itex]G[/itex] and the derivative [itex]d_g\phi = \phi_*|_g[/itex] maps [itex]T_gG[/itex] into [itex]T_{\phi(g)}G[/itex], so define

    [tex] (\phi_*X)_g := d_{\phi^{-1}(g)}\phi(X_{\phi^{-1}(g)}) [/tex]

    to receive a vector tangent at [itex]g\in G[/itex]. The two definitions coincide as you can check e.g. in local coordinates.
    (All of this clearly can be done for an arbitrary manifold [itex]M[/itex] instead of the Liegroup [itex]G[/itex].)

    Now to prove [itex] \phi(p)_*(X_g) = -g^{-1}\,X_g\,g^{-1}[/itex] for the inversion map [itex]\phi(g) = g^{-1}[/itex] we have to compute [itex]d_{\phi^{-1}(g)}\phi[/itex]. For this consider

    [tex] m\circ(id,\phi): G\to G\times G\to G: g\mapsto (g,g^{-1})\mapsto g\,g^{-1} \, [/tex]

    where [itex]m(g,h) = g\,h[/itex] is just the product map of [itex]G[/itex]. Since [itex] g\,g^{-1} = e[/itex] we have [itex]m\circ(id,\phi) = \text{const}[/itex], so

    [tex] 0=d_g\big(m\circ(id,\phi)\big)(X_g) = \big(d_{(g,g^{-1})}m\big)\circ\big(d_g(id,\phi)\big)(X_g) = d_{(g,g^{-1})}m\big(X_g,\,d_g\phi(X_g)\big) = d_g\rho_{g^{-1}}(X_g) + d_{g^{-1}}\lambda_g\big(d_g\phi(X_g) \big) = X_g\,g^{-1} + g\, d_g\phi(X_g)\;.[/tex]

    Here I used the chainrule and that the differential of the product map can be written as

    [tex] d_{(g,h)}m: T_gG\times T_hG\to T_{gh}G: (X_g,\,Y_h)\mapsto d_g\rho_h(X_g) + d_h\lambda_g(Y_h) = X_g\,g + h\,Y_h\;.[/tex]

    With this we finally get

    [tex] d_g\phi(X_g) = -g^{-1}\,X_g\,g^{-1} \;.[/tex]

    So far I haven't tried to prove this by using the 'operator-viewpoint'. It might be easier.
  6. Sep 3, 2007 #5


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    I still don't get it, but it's not your fault. I just don't really understand product manifolds very well. I'll have to work through some of the basics thoroughly before I'll be able to understand this.

    When I said that [itex]\phi_*X_g=-g^{-1}X_g g^{-1}[/itex] can't be true, I didn't consider the possibility that the left-hand side was supposed to be [itex](\phi_*X)_g[/itex]. I didn't even think about how to define [itex]\phi_*[/itex] acting on a vector field until after I had finished writing post #3.

    It wasn't too hard to figure out how to define it and use the definition and to prove the identity [itex][\phi_*X,\,\phi_*Y] = \phi_*([X,Y])[/itex] that you used earlier, so at least I understand something.

    You have probably given me enough information to understand this. Now I just need to make the effort to think hard about some of the things I still don't understand. I appreciate all the help you've given me.

    Edit: I have now understood the things I didn't understand before about product manifolds, so I hope I'm going to figure out the rest tomorrow. I also read #2 again, and I see that you made it clear that [itex]\phi_*[/itex] was acting on a vector field. I don't know how I could have missed that before.
    Last edited: Sep 3, 2007
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