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Simple definition of a Lie algebra?

  1. Nov 23, 2011 #1
    Is this statement correct as it stands?

    The Lie algebra of a Lie group G consists of extending the group operation of G to the elements of the tangent space of the identity element of G.

    or does it need to be qualified in some way?
     
  2. jcsd
  3. Nov 23, 2011 #2

    Fredrik

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    No, that's not accurate. The Lie bracket on the tangent space at the identity has a lot of properties that the group multiplication doesn't.

    What you actually need to do is to identify vectors in the tangent space at the identity with vector fields, and then define the Lie bracket of two vectors as the commutator of the corresponding vector fields. I'll quote myself from another thread (with a few LaTeX edits):
    A Lie group is a group that's also a manifold. A Lie algebra is a vector space V, with a function [itex](x,y)\mapsto [x,y][/itex] from V×V into V, that's bilinear and satisfies the Jacobi identity: [x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0. (So it's like a distributive multiplication operation, but instead of being associative, it satisfies the Jacobi identity). Such a function is called a Lie bracket.


    I'm going to describe how to define a Lie bracket on the vector space [itex]\mathfrak g=T_eG[/itex] of a Lie group G. (That's the tangent space at the identity element). We need a few definitions first. Suppose that M and N are manifolds and that [itex]\phi:M\rightarrow N[/itex]. If f is a real-valued function on N, then [itex]f\circ\phi[/itex] is a real-valued function on M, called the pullback of f. We can use the function [itex]\phi[/itex] to define a function [itex]\phi_*:T_pM\rightarrow T_{\phi(p)}N[/itex], called a pushforward. Recall that tangent vectors at a point are defined as "derivations" at that point, so to specify a tangent vector at [itex]\phi(p)[/itex], we need to specify its action on an arbitrary real-valued function f: [tex]\phi_* v(f)=v(f\circ\phi).[/tex] A vector field is a function that takes each point p (in some open subset of a manifold) to a tangent vector at p. If X is a vector field, the corresponding tangent vector at p is written as Xp. For every real-valued function f, the function that takes p to Xpf is written as Xf. The commutator of two vector fields X and Y is another vector field, defined by [tex][X,Y]_pf=X_p(Yf)-Y_p(Xf).[/tex] A pushforward of a vector field is defined in terms of the pushforwards of tangent vectors:
    [tex](\phi_*X)_{\phi(p)}=\phi_*X_p.[/tex] Recall that a Lie group G is also a group. For each g in G, we define left multiplication by g as the function [itex]\lambda_g:G\rightarrow G[/itex] defined by [tex]\lambda_g(h)=gh.[/tex] Now we have all the tools we need. Suppose that [itex]K,L\in\mathfrak g[/itex]. Then [itex](\lambda_g)_*K\in T_g G[/itex]. We define the left-invariant vector field corresponding to K by
    [tex]X^K_g=(\lambda_g)_*K[/tex] and the Lie bracket on [itex]\mathfrak g[/itex] by
    [tex][K,L]=[X^K,X^L]_e.[/tex] I hope it's obvious that I have omitted some minor technicalities (e.g. I never mentioned that "left multiplication by g" is required to be a smooth function by the definition of "Lie group"). If you're wondering if I could have used right multiplication instead of left, the answer is yes. The result would have been a Lie algebra that's isomorphic to this one.
     
    Last edited: Nov 23, 2011
  4. Nov 23, 2011 #3
    To a first order approximation, the group multiplication is just the vector space addition on the tangent space at the identity. That is commutative, so passing to just the tangent space loses most of the interesting information about the group. So, to study what you lost, you can see what happens to the commutators. The infinitesimal version of those is the Lie bracket.
     
  5. Nov 23, 2011 #4
    Fredrik, thank you for the detailed answer.


    Is it not the case that, given two left-invariant vector fields on a Lie group, the value of

    [tex][X^K,X^L]_e[/tex]

    (where e is the identity) is [tex]KL-LK[/tex] where we use the group "multiplication" (no derivatives needed)? By "extending the group operation" to the tangent vectors, I mean applying the coordinate representation of the group operation to the components of the tangent vectors.

    E.g., any of the groups consisting of nXn matrices have tangent vectors which are also nXn matrices and regular matrix multiplication applies to both. For left-invariant vector fields we have [tex][X^K,X^L]_e=KL-LK[/tex], right? So in what sense can the Lie bracket on the tangent space of the identity have properties that group multiplication doesn't when the Lie bracket is identical to the group multiplication commutator on that space?
     
    Last edited: Nov 23, 2011
  6. Nov 23, 2011 #5

    Fredrik

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    I don't have time to think about it right now, but I suppose that something like that could work.

    In general, there isn't even an addition operation defined on the group, so the group multiplication can't be bilinear or satisfy the Jacobi identity.
     
  7. Nov 23, 2011 #6
    Thank you again, guys. This has been helpful.
     
  8. Nov 24, 2011 #7

    HallsofIvy

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    If you mean L and K to be members of the Lie group, no that is not true. In fact, as Fredrik points out, such an operation is not defined- we have only one operation on a group not two as is required here. In general, the commutator of two vectors in the Lie algebra is, in your notation, [itex]X^KX^L- X^LX^K[/itex]. That is we multiply the vectors together, which we can do because this is an "algebra" so multiplication is defined, in both orders, then subtract, which we can do because this is a vector space.

     
  9. Nov 24, 2011 #8
    No, L and K are members of the vector space tangent to the identity of the Lie group e, so they necessarily have an addition operation. So given left-invariant vector fields X^L and X^K for which [tex]X^L|_e=L[/tex] and [tex]X^K|_e=K[/tex], I believe the following holds true

    [tex][X^K,X^L]_e=KL-LK[/tex]

    where the LHS is the Lie bracket of the vector fields but where multiplication on the RHS means "take the coordinate representation of the group operation and applied it to the components of K and L". I know this relation is true if the group operation is matrix multiplication and pretty sure it holds in general.
     
  10. Nov 27, 2011 #9
    No, it doesn't hold in general. I finally showed that relation above only works if the coordinates of gh for g,h contained in G depend linearly on the coordinates of both and g and h, as they do in matrix multiplication. In the general case then, the Lie bracket at the identity cannot be expressed in terms of "extending the group operation" to the components of the tangent space. So the answer to my original question is "No, it isn't that simple" and we need all of the structure described in Fredrik's posts.

    "pretty sure" always gets you in trouble in math.
     
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