Defining covectors when the basis is oblique Given a $2$-dimensional vector space with an oblique unit length basis, say, $(f_1, f_2)$, what is the dual vector or covector corresponding to $f_1$, call it $\hat f_1$? 
There appear to me to be two possible definitions, with $v=(a, b)_f$, written in the given basis.


*

*Definition 1: $\hat f_1 :(a, b)_f \mapsto a$

*Definition 2: $\hat f_1 :(a, b)_f \mapsto \textrm{(the projection of the vector $(a, b)_f$ on $(1, 0)_f$)}.$


I note that definiton 2 seems to require a metric, but that can be inferred from the usual dot product in an orthonormal basis.
I have seen both definitions.  For orthogonal bases they are the same, but not for oblique bases.  
Which is it?

It is the first definition: Given a basis $\mathcal F = (f_i)$ of a finite-dimensional vector space $\Bbb V$, the dual basis of the dual space, $\Bbb V^*$, is the unique basis $(\hat f{}^i)$  of $\Bbb V^*$ such that $$\hat f{}^i(f_j) = \left\{\begin{array}{rl}1,& i = j \\ 0,& i \neq j\end{array}\right.$$
In particular, if $v = \pmatrix{a\\b}_{\mathcal F}$, so that $v = a f_1 + b f_2$, then $$\hat f{}^1(v) = \hat f{}^1(a f_1 + b f_2) = a \hat f{}^1(f_1) + b \hat f{}^1(f_2) = a (1) + b(0) = a .$$
Like you observe, this definition does not require an inner product.
Note that the two definitions coincide in the case that the given basis $\mathcal F$ is orthogonal with respect to the inner product used.
