Differential Geom.: Dual Space Bases for Beginners

[nklohit]

I'm a beginner at differential geometry.
I have a problem about dual space. I understand why we use $$\left\{\frac{\partial}{\partial x^{\mu}}\right\}$$ as the bases in vector space, but I have no idea why
we use $$\left\{ dx^{\mu} \right\}$$ as the bases of dual space. What is the reason
of using it?

 

[lavinia]

I may not understand you question but

$$\left\{ dx^{\mu} \right\}$$

is the dual basis. This is easy to check.

 

[mordechai9]

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.

 

[Bacle]

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

 

[zhentil]

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".

 

[HallsofIvy]

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 $$\{\frac{\partial}{\partial x^\mu}\}$$ as a basis for the vector space, we represent the dual space basis as $dx^\nu$ because the linear functional is really $$\int \frac{\partial }{\partial x^\mu} dx^\nu$$.

 

[RedX]

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 $$\{\frac{\partial}{\partial x^\mu}\}$$ as a basis for the vector space, we represent the dual space basis as $dx^\nu$ because the linear functional is really $$\int \frac{\partial }{\partial x^\mu} dx^\nu$$.

The action of a dual f on a vector v is: $$f_i v^i$$ 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 $$\int \frac{\partial }{\partial x^\mu} dx^\nu$$ be equal to
$$\int v^\mu\frac{\partial }{\partial x^\mu} [dx^\nu f_\nu]$$
or
$$\int [dx^\nu f_\nu] v^\mu\frac{\partial }{\partial x^\mu}$$

 

[nklohit]

Thanks to you all, I got it^^

 

[haushofer]

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 $$\{\frac{\partial}{\partial x^\mu}\}$$ as a basis for the vector space, we represent the dual space basis as $dx^\nu$ because the linear functional is really $$\int \frac{\partial }{\partial x^\mu} dx^\nu$$.

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