What is the difference between a dual vector and a reciprocal vector? I am familiar with the concept of a dual space $V^*$ as the set of all linear functionals $\tilde{\omega}: V \rightarrow \mathbb{R}$. The inner product on $V$ is usually used to define the dual of a vector $\vec{v}\in V$, that is the dual vector $\tilde{v}$ is the unique linear functional which satisfies
\begin{equation}
\tilde{v}(\vec{w}) := \langle \vec{v}, \vec{w}\rangle,
\end{equation}
and the dual basis satisfies $\tilde{e}^\alpha(\vec{e}_\beta) = \delta^\alpha_\beta $.
I am used to thinking about the dual space as a distinct object to the original vector space - that vectors in $V$ cannot be expressed as linear combinations of dual vectors. However, in their book General Relativity: An Introduction for Physicists, the authors Hobson, Efstathiou and Lasenby write that any vector may be written as either a sum of basis vectors or as basis covectors:

$$\vec{v} = v^{\alpha}\vec{e}_\alpha = v_\alpha\vec{e}^\alpha $$

Is this reasonable? 
If it is not reasonable, then suppose for some basis $\{\vec{e}_\alpha\}$ of a $3D$ vector space, a set of reciprocal vectors $\{\vec{e}^\alpha\}$ are defined as 
$$
\vec{e}^1= \frac{\vec{e}_2\times \vec{e}_3}{\vec{e}_1\cdot( \vec{e}_2\times\vec{e}_3)}
$$
etc. (this can easily be extended to higher dimensions). These vectors satisfy $\langle{\vec{e}^\alpha}, \vec{e}_\beta\rangle = \delta^\alpha_\beta$, so act as dual vectors, but are still members of $V$. Is there any problem with the quoted equation if these reciprocal vectors are used? Are reciprocal vectors covectors?
 A: With respect to your question "Are reciprocal vectors covectors" I asked the same question about 6 months ago on a number of forums. No one could give me an answer. I took some time but finaly conviced my self that the answer is yes but had no second proof backing me up. I have just found a paper at the link;
https://www.iucr.org/__data/assets/pdf_file/0017/13193/4.pdf
Which says YES Reciprocal vectors are the covectors of the Real space. See Section 3. Of course I cannot vouch for the credentials of the Author so will leave that with you.
Hope this helps.
Paul
A: For the first part: no, it's not reasonable: any vector $\vec{v}\in V$ can be written $v^\alpha\vec{e}_\alpha$ for some basis $\left\{\vec{e}_\alpha\right\}$, and any vector $\tilde{u} \in V^*$ can be written $\tilde{u}= u_\alpha \tilde{e}^\alpha$, but these are not elements of the same vector space: you can't add them or equate them.
Note that the dual space exists whether or not there is an inner product: it's just the space of linear functions $V \to \mathbb{R}$. And in particular given some basis $\left\{\vec{e}_\alpha\right\}$ for $V$ there is a uniquely-defined dual basis for $V^*$, $\left\{\tilde{e}^\alpha\right\}$, defined by $\tilde{e}^\alpha(\vec{e}_\beta) = \delta^\alpha{}_\beta$: none of this depends on an inner product.
What an inner product gives you is a 1-1 map between vectors and covectors.  Given an inner product $\langle\_,\_\rangle$, then you can define $\tilde{v} = \langle\vec{v},\_\rangle$.  This in particular gives you a 1-1 relationship between the elements of a basis on $V$ and one on $V^*$: as opposed to their being merely uniquely-defined dual basis without an inner product you can now identify elements of the two bases with each other.
And of course this 1-1 mapping causes people to get lazy and say that $\vec{v}$ is the same as $\tilde{v} = \langle\vec{v},\_\rangle$ but it's not.
In the second part, I think that what you're doing is just a special case of a slightly clever change of basis, and it's easier to understand the general case (for me anyway) because it avoids all the cross-product stuff.
Given two bases $\left\{\vec{e}_\alpha\right\}$ and $\left\{\vec{e}'_\alpha\right\}$, then the relation between them is
$$\vec{e}'_\alpha = \Lambda_\alpha{}^\beta\vec{e}_\beta$$
The corresponding rule for the indices of a vector $\vec{v}$ is
$$v'^\alpha = v^\beta\left(\Lambda^{-1}\right)_\beta{}^\alpha$$
But, given some inner product we can express as a tensor in the usual way, and use it in the usual way to find the components of $\tilde{v}$ in the dual basis:
$$v_\alpha = v^\beta g_{\beta\alpha}$$
Oh, but now we can do a disgusting trick: choose a new basis such that 
$$\left(\Lambda^{-1}\right)_\beta{}^\alpha = g_{\beta\alpha}$$
This is fine although it looks horrible: the metric must be nonsingular, so I can simply pick its components as the components of the change-of-basis matrix $\Lambda$.  In particular remember that $\Lambda$ isn't a tensor.
And now I get:
$$
\begin{align}
v'^\alpha &= v^\beta\left(\Lambda^{-1}\right)_\beta{}^\alpha\\
          &= v^\beta g_{\beta\alpha}&&\quad\text{yes, this is OK}\\
          &= v_\alpha&&\quad\text{as is this}
\end{align}
$$
Well, this is just an artifact of picking a suitable change of basis, and it's sort-of the same thing as noticing that $g^{\alpha\beta} = \left(g^{-1}\right)_{\alpha\beta}$ as matrices.
But what it does not mean is that the basis I constructed above is a basis for covectors.  I think it's just an interesting property of the metric: perhaps someone who has thought about this harder can say more.
