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

Lie group actions and submanifolds

  1. Feb 28, 2012 #1
    Hello,

    Let's suppose that I have a Lie group G parametrized by one real scalar t and acting on ℝ2.
    Is it generally correct to say that the orbits of the points of ℝ2 under the group action are one-dimensional submanifolds of ℝ2, because G is parametrized by one single scalar?

    If so, how can I prove this statement?

    Thanks.
     
  2. jcsd
  3. Feb 28, 2012 #2

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    What if the action has a fixed point at p? Will the orbit through p be one-dimensional?

    If your G is acting smoothly on R^2, you can at least say that the orbit through each point is an immersed submanifold of R^2; it will generally be either 1- or 0-dimensional.
     
  4. Feb 28, 2012 #3
    thanks a lot!

    you are right. I am just thinking of the action of the rotation group SO(2) on ℝ2; clearly the point at (0,0) will remain unchanged, hence its orbit is 0-dimensional.

    Do you have any hint to suggest in order to prove these facts? I mean, proving that the orbits are immersed submanifolds of R^2.
     
  5. Feb 28, 2012 #4

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    Fix a p in R^2 and consider the map G -> R^2 sending g to gp.
     
  6. Feb 28, 2012 #5
    I see...
    I guess all I have to do is to prove that by letting G act on ℝ2, the space ℝ2 will be partitioned into equivalence classes (=the orbits), and that follows from the very fact that G is a group...maybe?
     
  7. Feb 28, 2012 #6

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    That's kind of besides the point. The image of the map I wrote down is precisely the orbit through p. So all that remains is to show that the map is an immersion - this will prove that the orbit is an immersed submanifold (by definition).

    Be careful to note that while each orbit itself is an immersed submanifold, the orbit space R^2/G with the quotient topology need not even be Hausdorff (let alone a manifold).
     
  8. Feb 28, 2012 #7
    ok...
    everything is almost clear. The only piece I am missing is how to prove that a mapping is an immersion (I am not familiar with this definition). Am I supposed to consider the mapping from the smooth manifold G (the Lie group parametrized by t) to the orbit of a point in R^2, take their derivatives with respect to t, and show that the map is injective?
     
  9. Feb 28, 2012 #8

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    I lied earlier - the map G -> R^2 isn't necessarily an immersion (e.g. if p is a fixed point). What we should be looking at is the induced map G/G_p -> R^2, where G_p = {g in G | gp=p} is the isotropy subgroup at p.

    To rigorously prove that this map is an immersion you need to know a thing or two about differential geometry (in particular you need to know what "immersion" means! :smile:).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Lie group actions and submanifolds
  1. Lie groups (Replies: 1)

  2. Group actions (Replies: 1)

  3. On Group Actions ! (Replies: 6)

  4. Lie group (Replies: 11)

Loading...