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Finding when covector disappears

  1. Oct 4, 2011 #1
    1. The problem statement, all variables and given/known data
    Find all the points [itex] p \in \mathbb R^2 [/itex] such that [itex] df_p =0 [/itex] where
    [tex] df = \frac{ (y^2-x^2) dx - 2xy dy }{(x^2+y^2)^2} [/tex]

    3. The attempt at a solution
    I figure the way this should be done is by solving the differential equation derived from
    [tex] (y^2-x^2) dx - 2xy dy =0 .[/tex]
    It's either that or just find when the coefficients are identically zero. I can do either quite easily once I know for sure which I should be doing. Thoughts?

    Edit: My only issue is that if I solve the differential equation, I will technically have a constant that I can't get rid of. Is there a canonical way of setting such a constant? Or is the solution the whole one-parameter family dictated by possible choices of the constant?
     
  2. jcsd
  3. Oct 4, 2011 #2

    Dick

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    This a part of some larger problem, right? A covector is a linear functional on vectors, right? dx and dy are linearly independent covectors. The only way a*dx+b*dy can vanish is if a and b are both zero.
     
  4. Oct 4, 2011 #3
    Yeah sorry. For some reason I actually knew that but my brain died.

    Indeed, for any covector field to be identically zero the components must be zero in any coordinate chart. Since [itex] df_p(X) =0, \forall X \in T_pM [/itex] we can, in particular, simply choose elements of the tangent basis, which forces the components to be zero.

    I don't know why that slipped my mind. I appreciate the input though!
     
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