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Importance of Lie algebra in differential geometry 
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#1
Mar1412, 04:55 AM

P: 145

Hi all,
I've been wondering about this for some time. While I am only familiar with the basics of differential geometry, I have come across the Lie bracket commutator in a few places. Firstly, what is the intuitive explanation of the Lie bracket [X,Y] of two vectors, if there is one? In terms of vectors being differential linear operators along specific curves? Another specific example I came across is the invariant formula for the exterior derivative: http://en.wikipedia.org/wiki/Exterio...ariant_formula Is the Lie bracket here just a useful notation, or something deeper? I'm assuming deeper, so could someone explain why? What does it mean to have a different algebra (structure constants)? Do they relate to the curvature of the space? Any other general comments on how the Lie bracket is related to differential geometry would be much appreciated! Ianhoolihan 


#2
Mar1412, 07:35 AM

P: 134

Hi,
this is discussed in Lee's book on smooth manifolds. Get hold of it if you can. He says that lie bracket is geometricaly interpreted as lie derivative, which is derivation of one vector field by another, in similar way you would have directional derivation of vector field by a single vector in R^n. 


#3
Mar1412, 10:00 AM

P: 134




#4
Mar1412, 02:00 PM

P: 145

Importance of Lie algebra in differential geometry
Thanks Alesak,
The link was useful, and I will look further into Lee's book. Ianhoolihan 


#5
Mar1412, 03:02 PM

Sci Advisor
P: 2,671

As you said yourself, a vector is a differential operator, and so the Lie bracket, when thought of in these terms, is simply a measure of the amount by which these operators don't commute.
If you consider vectors to be tangent vectors to curves, then the Lie bracket can be thought of infinitesimally as the "closer of parallelograms". Basically, two vector fields will have nonzero Lie bracket if their integral curves don't mesh along lines of constant affine parameter. I.e. if you move along the integral curve of one vector field, you're not moving along the constant affine parameter of the other vector field like you would if you were moving along coordinate lines (which is why coordinate bases have 0 Lie bracket). If I start at point A, and move a "distance" epsilon along a vector field X (where distance is measured simply by affine parameter), and then I move along the vector field Y a distance epsilon, and then I move in the opposite direction a distance epsilon along X, and move in the opposite direction a distance epsilon along Y, then I find that I don't get back to where I started in general. In the infinitesimal case, one can think of the Lie bracket as the vector (arrow) pointing from where you ended up to where you started (or the other way around, I can't remember the sign convention). 


#6
Mar1412, 03:51 PM

P: 145

The Lie derivative then (I'm assuming that is what you've just described by "Lie bracket") can be operated on any vectors. But, if they are the tangent vectors to coordinate curves, does one then get some sense of curvature of the manifold? I.e. [itex][e_a,e_b]={f_{ab}}^c e_c[/itex] gives the structure constants. Can these be related somehow to the curvature? (On second thought, if the manifold itself has curvature, should it matter whether or not I use a coordinate curve or not?). Thanks Matterwave for your help. I've actually been reading up on the questions you've asked and the responses from Ben Niehoff  good questions, good answers. Ianhoolihan 


#7
Mar1512, 12:41 AM

Sci Advisor
P: 2,671

By integral curves, I mean the curves which are tangent to your vector field at every point. Since vector fields are differential operators, one can integrate them to obtain integral curves.
By mesh, I meant the integral curves of two different vector fields intersecting each other. The curvature of a manifold is additional structure that isn't present in the mere differential structure of a manifold. You can't get a sense of curvature until you define a connection. 


#8
Mar1512, 01:17 AM

Sci Advisor
P: 2,671

I should note that if you are using the LeviCivita connection, then the structure constants do appear in the Christoffel symbols. But that has more to do with what type of basis you're using than anything else. The structure constants for a holonomic basis are 0.



#9
Mar1512, 04:39 AM

P: 145

Another thing: it is always talked about how Lie algebras are a way of expressing the continuous symmetries of objects. How does this work? For rotations for example, does [itex][J_a,J_b]=i{\epsilon_{ab}}^c J_c[/itex] express all there is to know about rotational symmetry? Ianhoolihan 


#10
Mar1512, 04:57 AM

Sci Advisor
P: 2,671

No, the Lie algebra gives you some of the structure, but not all of it. It won't give you, for example, the topological structure of whatever Lie group you are looking at.
You can think of Lie Algebras as the generator of infinitesimal group actions, which is how Lie himself thought about them. The Lie algebra, if you will, are the "elements of the Lie group infinitesimally close to the identity". If you want to express the full symmetry of the problem, you need to use the Lie group, but under many circumstances, using the Lie algebra is "good enough". Since Lie algebras are linear vector spaces, they are a lot easier to work with than a general Lie group which may have nontrivial topology. 


#11
Mar1512, 08:11 AM

Sci Advisor
P: 1,716

But when there is a LeviCivita connection on the manifold, then the Lie bracket expresses geometric relationships. For instance, the Lie bracket expresses the failure of the covariant derivative to commute. This is the equation, [itex]\nabla_{X}[/itex]Y  [itex]\nabla_{Y}[/itex]X = [X,Y] For a surface, the Riemannian geometry can be expressed in terms of a Lie algebra on the tangent space to the bundle of unit vectors  the so called tangent circle bundle. On the tangent circle bundle, there are three global vector fields, V,E[itex]_{1}[/itex] and E[itex]_{2}[/itex] The E's are horizontal and V is vertical. V is the derivative of the action of the rotation group on the fiber circles and the E's are defined in terms of the Riemannian metric. These three vector fields form a Lie algebra. The relations in this algebra are [V,E[itex]_{1}[/itex]] = E[itex]_{2}[/itex] [V,E[itex]_{2}[/itex]] = E[itex]_{1}[/itex] [E[itex]_{1}[/itex],E[itex]_{2}[/itex]] = K[itex]\circ[/itex][itex]\pi[/itex] Here, K is the Gauss curvature and [itex]\pi[/itex] is the projection map onto the surface. These relations are dual to the usual structure equations expressed in terms of differential forms. Interestingly, the last equation says that the Gauss curvature measures the failure of the horizontal distribution to be involutive. Here is a problem for you that might help you get an intuition of the geometric meaning of the Lie bracket. let (s,t) be a positively oriented orthonormal frame and define the 1 forms w[itex]_{1}[/itex](v) = <s,v> and w[itex]_{2}[/itex](v) = <t,v> where <,> is the Riemannian metric. Find a formula that expresses the exterior derivatives of the w's in terms of the Lie bracket of s and t. Use this to compute the length of the covariant derivative of s with respect to itself. What does this mean about the Lie bracket, [s,t], when s is tangent to a geodesic? 


#12
Mar1512, 08:28 PM

P: 145

But then, if you look at the next section http://en.wikipedia.org/wiki/Exterio...ariant_formula, it's nonzero(?) I realise I must be wrong somewhere (probably that I need a coordinate basis above), but I'm not sure why, so I'd appreciate someone reconciling the two, and showing what's right. I do feel a little stupid, but such is life. 


#13
Mar1512, 08:55 PM

Sci Advisor
P: 1,716

For a levi_civita connection the entire geometry on a surface can be expressed in term of a Lie algebra. And the answer to your question as to whether the curvature was related to the Lie derivative is yes. I strongly recommend reading Singer and Thorpe's book on elementary topology and geometry. It covers this Lie algebra approach to the Riemannian geometry of surfaces. It is just a great book and is written for beginners. For the problem I suggested there is a formula for the exterior derivative of a 1 form that will help. dw(x,y) = x.w(y)  y.w(x)  w([x,y]) ( i may be off by a factor of 1/2 but no matter) If you do not know this formula, try to prove it. 


#14
Mar1512, 09:51 PM

P: 145

As for your problem: I didn't know how to use the above formula in the covariant derivative  I just get some nasty coordinate thing. What (non coordinate?) formula would you prod me toward in this case? Cheers 


#15
Mar1512, 10:02 PM

Sci Advisor
P: 1,716

i leave that up to you. Note that dw1(s,t) =  w1([s,t]). 


#16
Mar1512, 10:50 PM

P: 145

Cheers 


#17
Mar1612, 05:30 AM

P: 134

OT, yesterday I got this book from library and it looks very good. It avoids differential geometry, but perhaps it is a plus.



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