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Bases in dual space

  1. May 15, 2010 #1
    I'm a beginner at differential geometry.
    I have a problem about dual space. I understand why we use [tex]\left\{\frac{\partial}{\partial x^{\mu}}\right\} [/tex] as the bases in vector space, but I have no idea why
    we use [tex] \left\{ dx^{\mu} \right\} [/tex] as the bases of dual space. What is the reason
    of using it?
     
  2. jcsd
  3. May 15, 2010 #2

    lavinia

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    I may not understand you question but

    [tex]
    \left\{ dx^{\mu} \right\}
    [/tex]

    is the dual basis. This is easy to check.
     
  4. May 15, 2010 #3
    Yea, if you work out the "d" operator applied to a coordinate function x^i, you see that the dx^i are identical to the covector basis a^i, where a^i are just the functions such that a^i(e_j) = delta_ij.
     
  5. May 17, 2010 #4
    The vectors {v_1*,..,v_n*} are the vectors that satisfy the condition:

    v_i*(v_i)=1

    v_i*(v_j)=0

    Given a basis {v_1,..,v_n} .

    You can look at the del/delx_j as positions , by using the isomorphism between

    vector fields/derivations/directional derivatives, and the direction of the

    directional derivatives. Then, if you use the standard (directional) bases

    (1,0,0,.)=e_j (e_j is a vector with 1 in the j-th coordinate and is 0 everywhere

    else) , the dx_i's are linear maps that project onto the i-th coordinate,

    so that dx_i(e_i)=1 , and dx_i(e_j)=0
     
  6. May 17, 2010 #5
    Perhaps it would be helpful if you gave us your definition of dual space. With every definition I've ever seen, the answer to your question is "by definition".
     
  7. May 18, 2010 #6

    HallsofIvy

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    Here's the definition I would use: If V is a vector space then the dual space is the set of all linear functionals from V to its underlying scalar field with addtion defined by (f+ g)(v)= f(v)+ g(v) and scalar multiplication by af(v)= f(av).

    When we use [tex]\{\frac{\partial}{\partial x^\mu}\}[/tex] as a basis for the vector space, we represent the dual space basis as [itex]dx^\nu[/itex] because the linear functional is really [tex]\int \frac{\partial }{\partial x^\mu} dx^\nu[/tex].
     
  8. May 18, 2010 #7
    The action of a dual f on a vector v is: [tex]f_i v^i [/tex] where the index i is summed over the dimension of the vector space.

    So how would it go when you write it in functional form like you did. Would [tex]\int \frac{\partial }{\partial x^\mu} dx^\nu[/tex] be equal to
    [tex]\int v^\mu\frac{\partial }{\partial x^\mu} [dx^\nu f_\nu][/tex]
    or
    [tex]\int [dx^\nu f_\nu] v^\mu\frac{\partial }{\partial x^\mu} [/tex]
     
  9. May 19, 2010 #8
    Thanks to you all, I got it^^
     
  10. May 22, 2010 #9

    haushofer

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    I must say I've never seen this integral sign before. Can you elaborate on that a little more?
     
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