# Trying to understand dual space / dual vectors / 1-forms

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!

• You don't need an inner product to define the dual space $V^*$ which is the space of all linear maps (= 1-forms) $V\to \mathbb R$. Just by definition if $\alpha\in V^*$ and $v\in V$, then $\alpha(v)\in \mathbb R$, but it's not an inner product (moreover, $\alpha$ and $v$ lies in different spaces). But if $V$ is finite-dimensional, you can use an inner product $g:V\times V\to\mathbb R$ to define an isomorphism $V\to V^*$, namely $v\mapsto g(v,-)$. – Canis Lupus Nov 5 '16 at 16:32

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)