# Bases in dual space

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
Gold Member
I may not understand you question but

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

is the dual basis. This is easy to check.

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.

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

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
Homework Helper
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 $$\int \frac{\partial }{\partial x^\mu} dx^\nu$$.

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
$$\int [dx^\nu f_\nu] v^\mu\frac{\partial }{\partial x^\mu}$$

Thanks to you all, I got it^^

haushofer
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 $$\int \frac{\partial }{\partial x^\mu} dx^\nu$$.