Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to find the manifold associated with a Lie Group?

  1. Nov 13, 2013 #1
    I have troubles formulating this question properly. So I will explain it through one example.

    If we consider the Lie group R=SO(2) of rotations on the plane, we know that we can find a manifold on which the group SO(2) acts regularly: this manifold is the unit circle in ℝ2. In fact, there is a bijection between points of the circle and elements of SO(2), and furthermore by letting SO(2) act on one arbitrary point x of the circle, the point x "orbits" smoothly along the circle.

    Similarly if we consider the Lie group T of translations on the plane, we know that T acts regularly on ℝ2. In fact there is a 1-to-1 mapping between elements of T and the points of ℝ2, and if we choose one point x in ℝ2, the action of T on x will make x orbit smoothly across the manifold ℝ2.

    Now the question is: if I consider a group G=RT (i.e. the group of rotations followed by translations), is it possible to find some manifold M on which G acts regularly?

    In the above case, G is clearly a Lie group of dimension 3, but it does not act regularly on the points of the manifold [itex]M=[0,2\pi)\times \mathbb{R}^2[/itex] because R and T are not commutative (T is only a normal subgroup of G), and there is no 1-to-1 correspondence between the points on the orbit of a point x[itex]\in[/itex]M and the elements of G.
  2. jcsd
  3. Nov 13, 2013 #2


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    Maybe this will help.

    SO(n) acts on the (n-1)-sphere by rotations. This action is transitive but has fixed points. However, since rotations are isometries, their differentials determine an action of SO(n) on the unit tangent sphere bundle of the the (n-1)-sphere. This action is transitive and has no fixed points. So SO(n) is diffeomorphic to the tangent sphere bundle of the (n-1)-sphere.

    For instance, SO(3) is diffeomorphic to the unit circle bundle of the 2-sphere which in turn is diffeomorphic to the projective space , RP[itex]^{3}[/itex]. The 3-sphere, is diffeomorphic to the Lie group of unit quaternions and is a two fold cover of projective 3-space. It is also the Lie group, Spin(3). It is also a circle bundle over the 2-sphere known as the Hopf bundle.

    SO(4) is diffeomorphic to the unit 2-sphere bundle of the 3-sphere which in turn is diffeomorphic to the Cartesian product,


    Not sure about the tangent 3-sphere bundle over the 4-sphere but as I recall, 3-sphere bundles over the 4-sphere were studied by Milnor to find exotic 7-spheres.
    Last edited: Nov 13, 2013
  4. Nov 13, 2013 #3
    um... from your discussion on the diffeomorphisms of SO(n), I can't still understand if it is possible or not to find a manifold on which the group G=RT of "roto-translations" would act regularly.

    To be honest, I don't even know if the question I am asking is a difficult one, or one that can be potentially answered with some calculations.
  5. Nov 13, 2013 #4


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    I thought once the rotation part was taken care of the rest would follow. Maybe I was hasty.

    Let's see. The pure translations are a normal subgroup of the roto-translations and the quotient group is the pure rotations. So the action of the rotations on the tangent circle bundle to the sphere can be extended to the roto-translations.

    So what happens if you let the roto-translations act on the tangent sphere bundle of the (n-1) sphere cross R^n by the usual action on the R^n c and by the differential of the rotation on the unit sphere bundle?

    So (T,M).(v,w)= (dM(v), T + M.w) where dM is the differential of the rotation,M.

    This is a well defined action since it is an action on each factor.

    It is transitive since

    (v,w) is mapped to (x,y) by dM(v) + y-M.w where M is the rotation that maps v to x.

    Have I misunderstood the problem?
  6. Nov 13, 2013 #5
    Hi Lavinia, and thanks for your help.
    I don't think you have misunderstood the problem.

    I have two questions regarding your last post:

    1) when you talk about "tangent sphere bundle", you mean the same concept as the "unit tangent bundle"?

    2) I am not sure I understand the notation (T,M).(v,w)
    Can you explain it, please?
  7. Nov 13, 2013 #6


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    Right - the bundle of unit tangent vectors. Each fiber is an (n-1)-sphere. Such a bundle is called a sphere bundle.

    (T.M) denotes the group element that first rotates by M the translates by T. This notation using pairs is common when the group is a semi-direct product as I think is true in the case of roto-translations.

    The group law in this notation is (T.M).(S,N) = (T + M.S, MN)
  8. Nov 14, 2013 #7
    Hi Lavinia. Thanks. I think that little by little I am starting to follow you :)
    I still have few doubts when you say:
    Let's stick with the roto-translations in ℝ2 for now.
    Why should we use the tangent sphere bundle on S1? I am new to this concept, and perhaps this is what is causing me headaches, and I explain why: the tangent spaces to the circle S1 are essentially tangent lines to the circle, but the set of unit-vectors in each tangent space is just a point-pair (similar to the vectors (±1,0) on the real line). What would be the tangent sphere bundle of S1 like? Two circles? :confused:

    I tried to follow your discussion and for the case of roto-translations in ℝ2 but I wonder why we couldn't simply forget about differentials and tangent sphere bundles and just define an action on S1×ℝ2 as:

    [tex](R,T)\cdot (\mathbf{p}, \mathbf{x}) = (R\cdot \mathbf{p},\; T\cdot \mathbf{x})[/tex]

    where T is a translation, R is a rotation, [itex]\mathbf{p}\in S^1[/itex], [itex]\mathbf{x}\in\mathbb{R}^2[/itex]
    which is actually trivial...:redface:

    For any two pairs (p,x) and (q,y) in S1×ℝ2, there is only one (R,T) in G such that (R,T).(p,x)=(q,y), so the action would be regular.
    Is there anything wrong with this?
    Last edited: Nov 14, 2013
  9. Nov 18, 2013 #8


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    Yes. SO(2) acts transitively and without fixed points on the unit circle. But for higher dimensions you need the unit tangent sphere bundle.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook