A book I am reading declares the following equality on page 9. For a vector subspace $W \subset V$, the author declares $W = (W^*)^\perp$
I assume (maybe incorrectly) that orthogonal complement $\perp$ here is wrt to the "pairing"
$\langle \cdot , \cdot \rangle : V \times V^* \rightarrow \mathbb{R}, ~(v^i e_i, w_j e^j) \mapsto v^i w_j$
That is, I assume $(W^*)^\perp= \{v \in W: \langle v, w \rangle = 0 ;\forall w \in W^*\}$
Now if $f_1, ..., f_p$ is a basis for $W$ then let $f^1, ..., f^p$ be its dual base $f^j(f_i)=\delta_i^j$. Then $W^*$ should be the span of the dual base and we should have by definition $\langle f_i, f^j \rangle = f^j(f_i) = \delta_i^j$, which would imply the opposite; that $(W^*)^\perp= \{v^i f_i : \langle v^i f_i, w_j f^j \rangle = 0 ;\forall w_j f^j \in W^*\} = W^c$
What am I missing or getting wrong here?
I believe that this question is not answered here
Update
The author states this (presumably - its very terse) to justify that that when $\sigma \in \wedge (W^*)$, then since $x \neg$ is an anti-derivation for $x \in W$, we have $x \neg \phi = 0$, which is saying (by definition of $x \neg$ as adjoint to $e(x)$)
$\langle x \wedge \eta, \sigma\rangle =\langle e(x) \eta, \sigma \rangle = \langle \eta, x \neg 0\rangle = 0$ for every $\eta \in \wedge(V)$, where this is $\langle \cdot , \cdot \rangle: \wedge(V) \times \wedge(V^*) \rightarrow \mathbb{R}$
So in particular he appears to be saying that $\langle x \wedge \eta, \sigma\rangle = 0$ for every $\eta \in \wedge(V)$ and every $\sigma \in \wedge(V^*)$ is justified by whatever he is meaning by $W = (W^*)^\perp$