I study at a university where it is fashionable to provide a rather poor quality of teaching and claim that it is a world-leading institution. For example, lecture notes are often sloppy in their notation and sometimes expressions that haven't been used before are introduced without any explanation.
So here is a proposition from my notes:
"We just need to prove that it [the interior product] is well-defined and independent of the way we do that, which motivates the following abstract proof:
Proof We defined $\wedge^p V$ as the dual space of the space of alternating p-multilinear forms on V. If M is a (p-1)-multilinear form on $V$ and $\epsilon$ a linear form on V then
$$(\epsilon M)(v_1, ..., v_p) = \epsilon(v_1) M(v_2, ..., v_p) - \epsilon(v_2) M(v_1, v_3, ..., v_p) + ...$$ is an alternating p-multilinear form. So if $\alpha \in \wedge^p V$ we can define $i_{\epsilon} \alpha \in \wedge^{(p-1)}V$ by
$$ (i_\epsilon \alpha)(M) = \alpha(\epsilon M)$$
Take $V= T^{*}, \epsilon = v \in V^{*} = (T^{*})^{*} = TX$, gives the interior product. The first equation above gives the formula for working out interior products. $\blacksquare$
I have three questions, mostly about interpretation:
(1) Given a smooth manifold N, we defined TN = $\{(x,v): x \in N, v \in T_x N\}$ and analogously for $T^*X$ with $T_x^*N$ in place of $T_xN$. We have shown that these are in fact vector bundles of dimension $2 dim\ N$. So I suspect when my lecturer wrote $V=T^*$ what he meant was $V=T^* N$ for some given manifold $N$. Is this correct?
(2) Why is $(T^*)^* = T$? If $V = T^*N$ (viewed either as a manifold or vector bundle) then how can you take the dual of this cotangent bundle?
(3) This proof implicitly relies on the assumption that vector fields $\epsilon \in C^\infty(TN)$ are linear forms on $T^{*}N$. We defined vector fields as sections (ie smooth right inverses) of the projection $\Pi: TX \rightarrow X$. Why can these be viewed as linear forms on $V = T^*N$.