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Prove a polynomial is a submersion

  1. Sep 23, 2009 #1

    Office_Shredder

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    1. The problem statement, all variables and given/known data

    Given a polynomial p(z) with complex coefficients, consider the map [itex] z \rightarrow p(z)[/itex] from the complex plane to the complex plane, where the complex plane is a manifold. Prove this is a submersion at all but finitely many points

    2. Relevant equations
    A submersion is one where the derivative is surjective


    3. The attempt at a solution
    I'm not really sure here.... I'm supposed to treat the complex plane as a 2 dimensional manifold right? In which case, how do I calculate the derivative of this map? It seems like taking partial derivatives of a complex polynomial is a terribly tedious and ineffective thing to do. Intuitively, it's obviously a submersion everywhere except where p'(z)=0, but I'm not sure how to treat the objects here in order to prove that
     
    Last edited: Sep 23, 2009
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  3. Sep 23, 2009 #2

    Dick

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    The differential map of p(x) at x=a is dp(y)=p(a)+p'(a)*y, isn't it? Isn't that surjective unless p'(a)=0? I'm not sure what's confusing you. Why worry about explicit partial derivatives?
     
  4. Sep 23, 2009 #3

    Office_Shredder

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    Well, that's where my confusion is. In the course we're treating everything as a real manifold; that means that the complex plane is really R2. And p(z) is a map p(x,y) which happens to be easy to express in terms of complex numbers. But to "really" calculate the derivative of that map, I think I should have to calculate the partial derivative of the real part of p(z) with respect to x and y, and the partial derivative of the imaginary part of p(z) with respect to x and y, to show that it's surjective.

    But now it occurs to me... p(z) is holomorphic! So if p(x,y) = U(x,y)+iV(x,y) then Ux = Vy and Uy = -Vx.

    So the determinant of the matrix of partial derivatives is

    Ux[/sup]2 + Uy[/sup]2 = Vx[/sup]2 + Vy2

    So for that to be 0, all the partial derivatives must be 0, which means p'(z)=0. Hooray!
     
  5. Sep 23, 2009 #4

    Dick

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    Sure, that's it in terms of partial derivatives. But you know the derivative map is surjective without even thinking hard if you know complex numbers, a linear map f(z)=a+bz maps the complex plane surjectively onto the complex plane if b is not zero. Because it's invertible. Just another way to think about it.
     
  6. Sep 23, 2009 #5

    Office_Shredder

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    I know that, but I don't see a way to prove its derivative map is actually
    p(a) + p'(a)y

    without looking at its partial derivatives (since we only have the partial derivative definition of derivative to work with at the moment)
     
  7. Sep 23, 2009 #6

    Dick

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    I sort of get that. But wouldn't it be strange if the complex derivative didn't coincide with the manifold derivative for an analytic function? There's really only one fundamental definition of a 'derivative'. I'm not suggesting you rewrite the proof or anything. The only thing is that a nonanalytic function might have a manifold derivative without having a complex derivative. But if it has both, they coincide. Just trying to say that your original notion that p'(z) not equal to 0 is a correct condition.
     
  8. Sep 23, 2009 #7

    Office_Shredder

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    Ok, that makes sense. Thanks for the help
     
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