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Lie group actions and submanifolds 
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#1
Feb2812, 04:15 PM

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Hello,
Let's suppose that I have a Lie group G parametrized by one real scalar t and acting on ℝ^{2}. Is it generally correct to say that the orbits of the points of ℝ^{2} under the group action are onedimensional submanifolds of ℝ^{2}, because G is parametrized by one single scalar? If so, how can I prove this statement? Thanks. 


#2
Feb2812, 04:35 PM

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What if the action has a fixed point at p? Will the orbit through p be onedimensional?
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 0dimensional. 


#3
Feb2812, 04:38 PM

P: 626

thanks a lot!
you are right. I am just thinking of the action of the rotation group SO(2) on ℝ^{2}; clearly the point at (0,0) will remain unchanged, hence its orbit is 0dimensional. 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
Feb2812, 04:45 PM

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Lie group actions and submanifolds
Fix a p in R^2 and consider the map G > R^2 sending g to gp.



#5
Feb2812, 04:49 PM

P: 626

I see...
I guess all I have to do is to prove that by letting G act on ℝ^{2}, the space ℝ^{2} will be partitioned into equivalence classes (=the orbits), and that follows from the very fact that G is a group...maybe? 


#6
Feb2812, 04:59 PM

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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
Feb2812, 05:14 PM

P: 626

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
Feb2812, 05:34 PM

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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! ). 


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