Bases in dual space

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  • #1
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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?
 

Answers and Replies

  • #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.
 
  • #3
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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.
 
  • #4
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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
 
  • #5
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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".
 
  • #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].
 
  • #7
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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].
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]
 
  • #8
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Thanks to you all, I got it^^
 
  • #9
haushofer
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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].
I must say I've never seen this integral sign before. Can you elaborate on that a little more?
 

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