Lie group actions and submanifolds

  • Thread starter mnb96
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  • #1
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
morphism
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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.
 
  • #3
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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.
 
  • #4
morphism
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Fix a p in R^2 and consider the map G -> R^2 sending g to gp.
 
  • #5
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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?
 
  • #6
morphism
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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).
 
  • #7
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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?
 
  • #8
morphism
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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:).
 

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