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Associating a Lie Algebra with a Lie Group

  1. Nov 1, 2011 #1

    Matterwave

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    I've been trying for many hours to wrap my head around this problem. Schutz, in his Geometrical Methods of Mathematical Physics book goes through great lengths defining a left-translation map on a Lie Group G, and then defining left-invariant vector fields on G, and then he goes on to say that "The left-invariant vector fields form a Lie Algebra...this is called the Lie Algebra of G".

    I don't understand several things.

    First of all, why define "the Lie Algebra of G" this way? Obviously since G is a manifold, then the tangent spaces at every point in G ARE Lie Algebras; every tangent space is a vector space, and the vector space obviously is endowed with the Lie Bracket operation which takes vectors to vectors and obviously is anti-commutative and obeys the Jacobi Identities. Why go through this convoluted way of finding a Lie Algebra? Why not just pick a tangent space (say the tangent space at the identity element e) and call that "the Lie Algebra of G"?

    Secondly, his statement "The left-invariant vector fields form a Lie Algebra...this is called the Lie Algebra of G", has me puzzled on what a Lie Algebra really is. I thought a Lie Algebra is any vector space which is endowed with a multiplication rule (the Lie bracket) which satisfies the 2 requirements of anti-commutivity and Jacobi-identity. A vector field does not live in any 1 vector space, it is a rule which defines at each point on the manifold a vector which lives in the tangent space to that point on the manifold. Therefore, wouldn't Shutz's statement that the vector FIELDS form a Lie Algebra, be more accurate if he said that they formed an infinity of Lie Algebras, with each Lie Algebra residing at each point on the manifold? I'm picturing here that every point P on the manifold G has it's own Lie Algebra.

    Thirdly, Schutz mentions that every vector at e induces a unique left-invariant vector field. This statement is fine; however, what does this statement have to do with anything? Why is e special in the sense of defining Lie Algebras?

    I've spent a long time trying to understand this to no avail...someone help please!
     
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  3. Nov 1, 2011 #2

    quasar987

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    Hello,

    1&3) Your definition wouldn't work because the Lie bracket is not defined pointwise. In other words, it is not defined for tangent vectors, it is defined for vector fields: if you take Xe, Ye two vectors at e, [Xe,Ye] is not defined. You could say, well pick extensions X and Y of Xe and Ye to vector fields around e, and set [Xe,Ye]:=[X,Y]e, but this would be ill-defined also because the result would not be independent of the choice of the extensions X and Y. However, there is a natural choice for the extensions: pick X and Y the unique left-invariant extensions of Xe and Ye! Now this is well-defined and what you get is isomorphic to the Lie algebra defined in terms of of the left invariant vector fields on G, and it is indeed sometimes how the Lie algebra of G is defined. This is the content of Schutz's remark making up your third interrogation.

    2) The set of vector fields on a manifold forms a vector space: take X,Y two such vector field. Define a third one X+Y by the rule (X+Y)(p)=X(p)+Y(p), and similarly for scalar multiplication. The resulting vector space is infinite dimensional. If on a Lie group G you restrict to left-invariant vector field, you get a linear subspace of finite dimension dim(G) [indeed, it is isomorphic to TeG!] which is closed under the Lie bracket operation. That is, the Lie bracket of two left-invariant v.f is still a left invariant v.f. This of course is just the naturality property of the Lie bracket: [F*X,F*Y]=F*[X,Y] for F a diffeomorphism, here to be taken as "left multiplication by g"
     
    Last edited: Nov 1, 2011
  4. Nov 1, 2011 #3

    lavinia

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    The vector fields on a manifold are a Lie algebra under the bracket operation but at a given point there is no natural structure. On a Lie group though there is a natural bracket at a point. One just takes the Lie bracket of the unique left invariant vector fields that fit the two vectors at the point.
     
  5. Nov 1, 2011 #4

    Matterwave

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    Ok I will think harder about this problem and come back with questions, thanks.
     
  6. Nov 1, 2011 #5
    Actually, there is a way to just define the Lie algebra on structure on the tangent space to the identity. Classically, if you have a matrix Lie group, the tangent space is just a collection of matrices and you take commutators [A, B] = AB - BA as the Lie bracket (from power series considerations, you can see that this is an infinitesimal version of the commutators aba^(-1)b^(-1) in group theory--the general idea is that the Lie Algebra concepts are infinitesimal forms of group concepts). If you want a definition that works for an abstract Lie group, you can proceed as follows.

    There's an obvious representation of the Lie group on the Lie Algebra, given by the differential at the identity of the conjugation map x → x g x^-1 (there's one such map from the Lie group to itself, fixing the identity). This representation is a map Ad : G → Aut (L), where L is the tangent space at the identity. This is called the adjoint representation of the group. You can take the differential of that map at the identity to get a map from L to the tangent space at Aut (L), which can be identified with End (L) (since Aut (L) is just an open subset of End (L)). Call this map ad : L → End(L).

    So, ad is a gadget that eats two guys in L and spits out an element of L. Turns out it's a Lie bracket. And it's equivalent to the vector field definition.

    This is a bit convoluted, but it if you think about it carefully, it's actually fairly intuitive. It captures the idea that the Lie algebra deals with infinitesimal symmetry. You have the adjoint representation of the group, which is the group acting on the small elements, and then you just want to look at what really small group elements are doing. The "small" group elements are Lie algebra elements. If A is a small group element, it acts on B by [A,B].

    By the way, the left-invariant vector-fields actually form a finite-dimensional vector space because they can be identified with the tangent space at the identity. And not just at each point. Vector fields are things you can add and multiply by scalars. What's a vector space? In a nutshell, it's a place where you can add stuff together and multiply by scalars.
     
    Last edited: Nov 1, 2011
  7. Nov 2, 2011 #6

    Matterwave

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    Ok, so I think I've been able to (sort of) wrap my head around this definition. I now wonder how do we practically obtain the Lie algebra of a Lie group? It's all nice to say that the left-invariant vector fields V(P) form the Lie Algebra, but in practice how do we find this?

    I know that what I should do is find the basis "vectors" for the Tangent space at the identity, because each "vector" gives me a unique left-invariant vector field. But I don't even know how to obtain this.

    Schutz gives only 1 example of him doing this. He shows one "curve" on the manifold GL(n,R) passing through the identity matrix as: diag(1+exp(lambda),1,1,1...), and then says that one vector of the tangent space is then diag(1,0,0,0...).

    I get this since we can say that the tangent vector to that curve is simply diag(exp(lambda),0,0,0...) and that for lambda=0, we get diag(1,0,0,0...). I even get why he put the exponential there to keep the determinant from vanishing.

    His next statement is then something like "Obviously, the Lie Algebra of GL(n,R) consists of all nxn arbitrary (real) matrices". How do you make this deduction? He only found 1 tangent vector!

    How would you do this practically for some random group like O(n), for example.

    So far, I only get how to find 1 specific element of the Lie Algebra corresponding to one particular curve through the identity. Trying to figure out all the other curves that pass through the identity...doesn't seem like a good practical method to find a basis.

    Also, vice versa. Is there a nice way of constructing the associated Lie Group from a Lie Algebra? Schutz gives a theorem that says you can always do this, but does not show any method of doing this, at least as far as I can tell.
     
    Last edited: Nov 2, 2011
  8. Nov 2, 2011 #7

    quasar987

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    It is permitted to look at other books, you know! :)
     
  9. Nov 2, 2011 #8
    That's essentially the fact that I mentioned earlier. That the tangent space Aut(L) is End(L).

    GL(n, R) is an open subset of all the n by n matrices containing the identity. So, the tangent space can be identified with all the n by n matrices. Basically, it's just R^(n^2). So, it's just like if you had an open set U in R^n. The tangent space at some point in U can be identified with R^n.

    Any real matrix Lie group will be a subgroup of GL(n, R), so the Lie algebra will be a subalgebra of the n by n matrices with commutators giving the Lie bracket.

    Yes, but you will only get the connected component of the identity. You exponentiate. This is enough to fill up a neighborhood of the identity and you go from there.
     
  10. Nov 2, 2011 #9

    Matterwave

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    Yea, I have tried, but it seems to me that other books are even more abstract and rigorous and confuses me even more. Do you have a good recommendation?

    @Homeomorphic, does this mean that we can't construct disconnected groups (like GL(n,R)) from its Lie Algebra alone?
     
  11. Nov 2, 2011 #10

    George Jones

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    For theoretical physicists, a good comprise between physics-style and math-style presentations of math might be Fecko's book. Unfortunately, I think that it is at its pedagogically worst in its first chapter. I like Fecko.

     
  12. Nov 2, 2011 #11

    Matterwave

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    I guess when I have time, I may take a look at that, thanks for the recommendation. =]
     
  13. Nov 2, 2011 #12

    quasar987

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    I was going to recommend Lee or Tu's book which also appear in the above George Jones quote.
     
  14. Nov 2, 2011 #13
    Yes, and that's not very surprising because they have the same Lie Algebra as the connected component of the identity, call it U.

    The argument is that you can exponentiate and the inverse function theorem tells you it's a diffeomorphism near the identity. Then, you do one of those connectedness arguments to show that you get the whole connected component. It's clear that it has to be an open set because you can hit that little neighborhood by g to get a neighborhood of any g in the set that you can exponentiate to. Then, I guess you could argue by continuity that it's also a closed set. It's open and closed, so it's a connected component.

    As far as books go, I like Fulton and Harris: Representation theory, but I have heard John Stillwell's Naive Lie Theory is really good. It's supposed to be pretty elementary. I haven't read it, but I would if I had time.
     
  15. Nov 2, 2011 #14

    Matterwave

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    Ok, thanks for all the recommendations, I'll look into it. =D
     
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