I am new to this differential geometry business. I am trying to understand the concept of the dual space and dual vectors. I found this great answer on another stack exchange post here, but I have a few questions.
I know that if two sets of vectors $V^{\mu}$ and $W_{\nu}$ are duals of each other, then $V^{\mu}W_{\nu} = \delta^{\mu}_{\;\;\nu}$. So let us say that $V = (1,x,x^2)$ and $W = (a,b,c)$, then the inner product defines $a + bx + cx^2 = 0$. Which is now an element of $\mathbb{R}$.
So here lies my confusion. I would say that $W$ is a dual vector, however in the answer I linked, he defines the process of bringing $V$ to $\mathbb{R}$ as a map. So is the map the actual dual vector $W$? Or is the map defined through the inner product of $V$ and $W$? I know that dual vectors are also called 1-forms. Is that because of this mapping process? Can a regular vector be defined as a 1-form in relation to its dual vector?
I hope all this make sense!
The dual space of a vector space, is the space of all linear functionals (that is, maps from the vector space to its scalar field that satisfy Linearity).
It is a non-trivial fact that for finite dimensional vector spaces, the dual space of the dual space of that vector space is isomorphic to that vector space (structurally the same space; this result does not hold in general for infinite dimensional vector spaces).
Note: Linearity is the property that the function 'commutes' with scalars, and 'distributes' over vector addition.
The one of the main theorems about dual spaces is this:
Given a pair of vector spaces, $V,W$ and a linear map, $T:V\rightarrow W$, then there exists a corresponding linear map, $T^{*}:W^{*}\rightarrow V^{*}$ called the transpose or dual or adjoint of that linear map. (e.g. https://en.wikipedia.org/wiki/Transpose_of_a_linear_map)