Contraction as Adjoint of Wedging Let $V$ be an $n$-dimensional vector space. Given $\phi^1\wedge \cdots \wedge \phi^k\in \bigwedge ^k(V^*)$ and $v_1\wedge\cdots\wedge v_k\in \bigwedge^k(V)$, we write
$$
\langle \phi^1\wedge \cdots\wedge \phi^k, v_1\wedge \cdots \wedge v_k\rangle
=
\det[\phi^i(v_j)]
$$
where $[\phi^i(v_j)]$ denotes the $k\times k$ matrix whose $i,j$-th entry is $\phi^i(v_j)$.
I was able to show that this is well defined.
I need to show that following:

Given $v\in V$, $\omega\in \bigwedge^k(V^*)$, and $Y\in \bigwedge^{k-1}(V)$, we have
  $$
\langle i_v\omega, Y \rangle = \langle \omega, v\wedge Y\rangle
$$
  where $i_v\omega$ is a covariant $k-1$ tensor on $V$ defined as $i_v\omega(u_1, \ldots, u_{k-1})=\omega(v, u_1, \ldots, v_k)$.

One possible approach would be by expressing both $\omega$ and $Y$ using a basis. But that would be very heavy (and inelegant?)
Can anybody help me with this?
Thanks.
 A: Clifford algebra can make this manipulation simpler.  To use it here, first define the geometric product of vectors.  This is traditionally denoted by juxtaposition: i.e. the geometric product of a vector $a$ and $b$ is denoted $ab$.  Given vectors $a, b, c$, the geometric product has the following properties:
$$aa \equiv \langle a,a\rangle \implies ab = \langle a, b \rangle + a \wedge b, \quad a(bc) = (ab)c$$
This does use a metric (denoted by angled brackets), but in the case that one multiplies vectors and forms only, the result is nonmetrical.  Hence, we need not impose any metric here in order to find the correct result.

Now, introduce the following notation for the grade of a geometric product.  Again, given two vectors $a, b$, we have
$$ab = \langle a, b \rangle + a \wedge b = [ab]_0 + [ab]_2$$
The square brackets here denote the grade projection operator.  (Nb. the standard notation is angled brackets, but I avoid these here in deference to the inner product.)
This notation can be used generally.  Indeed, see that the inner product of a $k$-vector $V_k$ and and a $k$-form $\Phi^k$ is equivalently
$$\langle \Phi^k , V_k \rangle = [(\Phi^k)^\dagger V_k]_0 = [(\phi^k \wedge \phi^{k-1} \wedge \ldots \wedge \phi^1) V_k ]_0$$
The dagger here ($\dagger$) is called the reversion operator.  This is also a standard practice for ensuring that the ordering of products is correct in clifford algebra, such that, say, $\langle \phi^1 \wedge \phi^2, v_1 \wedge v_2 \rangle$ has the correct sign.
Equivaently, the interior product can be written as
$$i_v \omega = [v \omega]_{k-1} $$

With both the inner product and interior product operators written in terms of the geometric product, we can now analyze the original problem:
$$\langle i_v \omega, Y \rangle = [ [\omega^\dagger v]_{k-1} Y]_0$$
We can remove the inner set of square brackets--only the grade $k-1$ term can contribute a scalar to the overall product.
$$\langle i_v \omega, Y \rangle = [ \omega^\dagger v Y]_0$$
Here we use the associativity of the geometric product to group $v Y$.  To form a scalar when multiplied by $\omega^\dagger$, which is a $k$-form, we can restrict ourselves to considering only $k$-vectors formed by the product $v Y$.  This is naturally just, and only, $v \wedge Y$.  In other words, we conclude that
$$\langle i_v \omega, Y \rangle = [\omega^\dagger v Y]_0 = [\omega^\dagger [vY]_k ]_0 = \langle \omega, [vY]_k \rangle_0 = \langle \omega, v \wedge Y \rangle$$
The result is an entirely basis-free manipulation, powered by the geometric product (from clifford algebra) and its properties, particularly its associativity.  The use of grade projection (square brackets here) allows us to consider only multivectors of particular grades, and to write those expressions otherwise in terms of the geometric product.
A: Assume that $\{e_i\}$ is a basis for $V$. Note that by linearity we
suffice to show that $$ v=e_{i_1},\ (i_v\phi^1\wedge\cdots \wedge
\phi^k, e_{i_2} \wedge\cdots \wedge e_{i_{k}} )=( \phi^1\wedge\cdots
\wedge \phi^k,v\wedge e_{i_2} \wedge\cdots \wedge e_{i_{k}} )$$
Proof : $$ i_v\phi^1\wedge\cdots \wedge \phi^k ( e_{i_2}, \cdots ,
e_{i_{k}} )= \phi^1\wedge\cdots \wedge \phi^k ( v,e_{i_2}, \cdots ,
e_{i_{k}} ) $$ $$= {\rm det}\ (a_{ml}) \ (a_{ml} = \phi^m(e_{i_l}) )
$$
$$ = (\phi^1\wedge\cdots \wedge \phi^k , v\wedge  e_{i_2}\wedge  \cdots
\wedge e_{i_{k}} ) $$
