Concrete Example Illustrating the Interior Product Let $V$ be a finite-dimensional vector space, let $v \in V$ and let $\omega$ be an alternating $k$-tensor on $V$, i.e., $\omega \in \Lambda^{k}(V)$. Then, the interior product of $v$ with $w$, denoted by $i_{v}$, is a mapping
$$
i_{v}:\Lambda^{k}(V)\rightarrow \Lambda^{k-1}(V)
$$
determined by
$$
(i_v \omega)(v_1, \dots, v_{k-1}) = \omega(v, v_1, \dots, v_{k-1}).
$$
My understanding of this, which is probably far from complete, is that the interior product basically provides a mechanism to produce a $k-1$-tensor from a $k$ tensor relative to some fixed vector $v$. I'm trying to understand however what the interior product actually means and how it is used in practice. Therefore, my question is, Can anyone provide example(s) illustrating computations and/or physical examples that will shed light on its purpose? 
Also, the interior product seems to be somewhat (inversely?) related to the exterior product in that an exterior product takes a $p$-tensor and a $q$ tensor and makes a $p+q$ tensor and therefore is an "expansion". The interior product, on the other hand, is a contraction but always produces a tensor of degree one less than you started out with. So, secondly, What is the precise relation between the interior and exterior products?
Unfortunately, the Wikipedia page is of little help here and I can't find a reference that clearly explains these things.
 A: Here is a partial answer to my own question, specifically the part asking for a concrete example illustrating computational aspects of the interior product. 
Let the vector space in question be $\mathbb{R}^3$ endowed with the ordered basis $(e_1, e_2, e_3)$ and let $e^1, e^2, e^3$ be the relative cobasis. Here, we can think of the cobasis as just ordinary vectors that satisfy $e^i(e_j) = \delta^i_j$ or consider them as linear functionals on the dual space $(\mathbb{R}^n)^*$ that satisfy the same relations.
Now, suppose  $\omega \in \Lambda^{2}(\mathbb{R}^3)$ is given by $\omega = e^1 \wedge e^2$ and let $v = e_1$. Then, for any vector $x \in \mathbb{R}^3$ we can then compute the interior product as follows:
$$
(i_v \omega)(x) = (e^1 \wedge e^2)(e_1, x) = e^1(e_1)e^2(x) - e^1(x)e^2(e_1) = e^2(x)
$$
Therefore $i_v \omega = e^2$
Next, keep the same $\omega$ but let $v = e_2$. Then,
$$
(i_v \omega)(x) = (e^1 \wedge e^2)(e_2, x) = e^1(e_2)e^2(x) - e^1(x)e^2(e_2) = -e^1(x)
$$
So $i_v \omega = -e^1$
Finally, it is also easy to see by inspection that if $v =e_3$ then $i_v \omega = 0$
Computations for other values of $\omega$ proceed similarly.
A: I don't think this will completely satisfy your questions, but I think the interior product is a neat way to induce orientations. To give an orientation on an $n$-manifold with boundary $M$ is the same as giving a nowhere-vanishing $n$-form $\Omega$. If $H \subset M$ is a hypersurface and $N$ is a transverse vector field along $H$ (so $N\colon H \to TM$, such that $N_x \in T_xM$ and $T_xM = N_x + T_xH$ for $x \in H$), then $i_N\Omega$ restricts to an orientation form on $H$. If $H = \partial M$, then taking $N$ to be an outward-pointing vector field along $\partial M$ gives the usual orientation used in Stokes's theorem.
I don't have my copy with me, but a lot of this should be in Lee's Introduction to Smooth Manifolds.
A: Let me give another illustration of how the interior and exterior products are related. This particular case, however, works not on differential geometry, but requires Riemannian geometry. 
Given a metric $g$, denote by $\langle,\rangle$ the extension of its inner (not interior) product to forms. The metric $g$ induces an identification between the vector space $V$ and its dual $V^*$, via the operators $v\mapsto v^\flat$, where 
$$ v^\flat(w) = \langle v,w\rangle $$
($v^\flat \in V^*$ is a linear functional on $V$, and here we define it by its action on $w\in V$)
Then we have the nice property for $\eta\in\Lambda^{k-1}(V),\tau \in \Lambda^k(V)$, and $v\in V$ that
$$ \langle v^\flat \wedge \eta, \tau\rangle = \langle \eta,(i_v)\tau \rangle $$
showing how the interior and exterior products are actually adjoint with respect to the metric inner product. 
A similar statement can be made by appealing to the Hodge-star operator associated to a Riemannian metric. Up to a constant multiplier $C$ (whose form depends a bit on your conventions, and which depends on the dimension and the degree of the forms), you have that
$$ (i_v)\tau = C *(x^\flat\wedge *\tau) $$
where $*$ is the Hodge star operator. 
A: Lie derivative, of course! It's even mentioned in the interior product wiki page. As for the relation to the exterior product, IMO you should look more on the exterior derivative for comparison.
But if you insist, for any $1$-form $\alpha$ and $k$-form $\omega$ and any vector $x$ we have $i_x (\alpha \wedge \omega) = \alpha(x) \omega$.
UPD: from purely algebraic standpoint, it is useful to consider Lie coalgebra structure on $V^*$: a linear mapping $\mathrm{d}: V^* \to \bigwedge^2 V^*$ extended to a graded anti-derivation. It is easy to see that by defining $\omega([x, y]) = \mathrm{d}\omega(x, y)$ we get Lie algebra structure on $V$.
So, interesting things must arise when we extend some $f: \bigwedge^2 V^* \to V^*$ to a graded antiderivation, right? Well, I can't answer that question, but I'm pretty sure that was the motivation :)
A: $
\newcommand\form[1]{\langle#1\rangle}
\newcommand\R{\mathbb R}
\newcommand\doub\mathfrak
\newcommand\Ext{{\textstyle\bigwedge}}
\newcommand\MVects[1]{\mathop{{\textstyle\bigwedge}^{\!#1}}}
\newcommand\Blades[1]{\mathop{{\textstyle\bigwedge}^{\!(#1)}}}
\newcommand\lintr{\mathbin{{\lrcorner}}}
\newcommand\rintr{\mathbin{{\llcorner}}}
\newcommand\lcontr{\mathbin{{\rfloor}}}
\newcommand\rcontr{\mathbin{{\lfloor}}}
\newcommand\rev\widetilde
\newcommand\Cl{\mathrm{Cl}}
\newcommand\End{\mathrm{End}}
%\newcommand\pperp{\mathrel{{}_\circ\kern-0.75pt{\perp}_\circ}}
\newcommand\pperp{\mathrel\top}
\newcommand\G{\mathcal G}
\newcommand\Hyp{\mathcal H}
\newcommand\pstar{\mathop{*}}
\newcommand\dd{\mathrm d}
$It does not seem to me that any of the answers, including the accepted answer, actually answer the question.
Rather than alternating tensors, we will focus on the exterior algebra; these are not the same thing in non-zero characteristic, and in any case the exterior algebra viewpoint is both more appropriate and very geometric. Since you ask for applications, I will focus on $\R$ and at times assume the presence of a metric (i.e. symmetric bilinear form) denoted by $\form{\cdot,\cdot}_V : V\times V \to \R$. I will not necessarily assume that this form is non-degenerate.
Setup
Contractions
Before getting into applications, we make clear out conventions and notations by considering the natural form
$$
  \form{v^* + v, w^* + w} = v^*(w) + w^*(v)
$$
on $\doub W = V^*\oplus V$. This form is non-degenerate, and in fact comes directly from the sequence of natural isomorphisms
$$
  (V^*\oplus V)^* \cong (V^*)^*\oplus V^* \cong V\oplus V^* \cong V^*\oplus V.
$$
We extend $\form{\cdot,\cdot}$ to all of $\Ext\doub W$ such that
$$
  \form{v^*_k\wedge v^*_{k-1}\wedge\dotsb\wedge v^*_1,\;
        v_1\wedge v_2\wedge\dotsb\wedge v_l}
    = \delta_{kl}\det\Bigl(v^*_i(v_j)\Bigr)_{i,j=1}^k.
$$
This extension is very natural, but I won't get into that here. Often you will see the convention of using $1,2,\dotsc, k$ instead of $k, k-1,\dotsc,1$; while it can be argued that the former is more geometric, the latter tends to be easier to work with algebraically especially together with what will be our conventions for the interior product. Note that $\Ext\doub W$ contains $\Ext V^*$ and $\Ext V$ as subalgebras, and in fact $\Ext\doub W \cong \Ext V^*\mathbin{\widehat\otimes}\Ext V$ as algebras where $\widehat\otimes$ is a graded tensor product.
Since $\form{\cdot,\cdot}$ is non-degenerate, we can now define the left and right interior products ${\lintr}, {\rintr} : \Ext\doub W\times\Ext\doub W \to \Ext\doub W$ as the adjoints of the exterior product:
$$
  \form{\doub X\wedge\doub Y, \doub Z} = \form{\doub X, \doub Y\lintr\doub Z},\quad
  \form{\doub X, \doub Y\wedge\doub Z} = \form{\doub X\rintr\doub Y, \doub Z}.
$$
$i_v$ corresponds to $v\lintr$ where $\lintr$ is restricted to $V\times\Ext V^*$. It is easy to see from this definition that
$$
  \form{\doub X, \doub Y} = \form{\doub X\lintr\doub Y}_0 = \form{\doub X\rintr\doub Y}_0
$$
where $\form\cdot_0$ is the scalar (i.e. grade $0$) part of a multivector. These products are closely related: if $\rev{\doub X}$ is the reversal anti-automorphism (or main anti-involution) of $\doub X$ which reverses all exterior products and is the identity on vectors, then
$$
  \rev{\doub X\lintr\doub Y} = \rev{\doub Y}\rintr\rev{\doub X}.
$$
For this reason we will generally only state properties of the left interior product. Some key properties to be aware of are

*

*$\doub v\lintr\doub w = \form{\doub v,\doub w}$ for $\doub v,\doub w \in \doub W$.


*$\doub X\lintr\doub Y \in \MVects{k-j}\doub W$ if $k\geq j$ and $\doub X \in \MVects k\doub W$ and $\doub Y \in \MVects j\doub W$, and is zero if $k < j$. In particular, if $k = j$ we get a scalar.


*The interior products make $\Ext\doub W$ into an $\Ext\doub W$-module:
$$
  \doub X\wedge\doub Y\lintr \doub Z = \doub X\lintr\doub Y\lintr\doub Z.
$$


*Vectors give anti-derivations:
$$
  \doub v\lintr\doub X\wedge\doub Y = (\doub v\lintr\doub X)\wedge\doub Y + \widehat{\doub X}\wedge(\doub v\lintr\doub Y)
$$
where $\doub v \in \doub W$ and $\widehat{\doub X}$ is the grade involution (or main involution) which is the homomorphism that negates all vectors.
We use the convention that $\wedge$ binds tighter than $\lintr$ or $\rintr$, and that $\lintr$ is right-associative and $\rintr$ is left-associative.
A metric $\form{\cdot,\cdot}_V : V\times V \to \R$ induces a linear map $\flat : V \to V^*$;
the universal property of $\Ext V$ extends this to a homomorphism $\Ext V \to \Ext V^*$.
This allows us to extend $\form{\cdot,\cdot}_V$ to $\Ext V$ via
$$
  \form{X, Y}_V = \form{X^\flat, Y}
$$
and to define the left and right contractions on $X, Y \in \Ext V$ as
$$
  X\lcontr Y = X^\flat\lintr Y,\quad X\rcontr Y = X\rintr Y^\flat.
$$
These contractions will be the form in which we apply the interior products.
While they are adjoints of the exterior product under $\form{\cdot,\cdot}_V$,
if this form is degenerate then they are not the unique adjoints.
Clifford Algebras
Once we have a quadratic form $Q(v) = \form{v, v}_V$ we also immediately get a Clifford algebra $\Cl(Q)$. In short, this is the freest associative algebra generated by $V$ subject to the relations $v^2 = \form{v, v}_V$ for all $v \in V$. The contractions on $\Ext V$ are merely the manifestation of $\Cl(V)$ in $\Ext V$.
To elaborate a bit, we define an additional Clifford product $V\times\Ext V \to \Ext V$ via
$$
  vX = v\lcontr X + v\wedge X.
$$
This gives us a map $v \mapsto v(\cdot) : V \to \End(\Ext V)$ into the linear endomorphisms of $\Ext V$, and the universal property of Clifford algebras extends this to an algebra homomorphism $\psi : \Cl(Q) \to \End(\Ext V)$. The map $X \mapsto \psi(X)(1)$ is then a linear isomorphism $\Cl(Q) \cong \Ext V$. To illustrate, the Clifford product of $u + vw = u + \form{v,w}_V + v\wedge w$ for $u, v, w \in V$ and $X \in \Ext V$ would be
$$\begin{aligned}
  (u + vw)X
&= uX + vwX
\\
&= u\lcontr X + u\wedge X + v(w\lcontr X + w\wedge X)
\\
&= u\lcontr X + u\wedge X + v\lcontr w\lcontr X + v\wedge(w\lcontr X) + v\lcontr w\wedge X + v\wedge w\wedge X
\\
&= u\lcontr X + u\wedge X + v\wedge x\lcontr X + v\wedge(w\lcontr X) + \form{v,w}_VX - w\wedge(v\lcontr X) + v\wedge w\wedge X.
\end{aligned}$$
This is all to say that the interpretation of the contractions and the interpretation of the associated Clifford algebra are strongly linked. We will not speak too much about Clifford algebras because we are interested in only the contractions, but they will be mentioned briefly again. It is safe to say, however, that if you find yourself working with contractions and exterior products all at once then there is certainly a Clifford algebra waiting for your attention as well.
Applications
The following are just some examples of the applications of $\lcontr$ and $\rcontr$.
The Join Geometry of $\Ext V$
For $X \in \Ext V$, we will be interested in the kernel of $v \mapsto X\wedge v$ over $V$; denote this by
$$
  [X]^\wedge = \{v \in V \;:\; v\wedge X = 0\}.
$$
The exterior algebra can be characterized as the most general associative algebra where
$$\begin{aligned}
  &v_1\wedge\dotsb\wedge v_k = 0 &&\text{if }\{v_1,\dotsc,v_k\}\text{ is linearly independent},
\\
  &[v_1\wedge\dotsb\wedge v_k]^\wedge = \mathrm{span}\{v_1,\dotsc,v_k\} &&\text{otherwise}.
\end{aligned}$$
Note that $[0]^\wedge = V$. We will call such an element $v_1\wedge\dotsb\wedge v_k$ a blade and denote the set of $k$-blades by $\Blades kV$. In this way we identify subspaces of $V$ with blades. The blades $\Blades nV = \MVects nV$ where $n = \dim V$ play a particularly special role and are often called pseudoscalars since they are one-dimensional.
If $X, Y$ are blades, then in general we have
$$\begin{aligned}
  &X\wedge Y = 0 &&\text{if }\exists v \in V.\: v \in [X]^\wedge\cap[Y]^\wedge,
\\
  &[X\wedge Y]^\wedge = [X]^\wedge\oplus[Y]^\wedge &&\text{otherwise},
\end{aligned}$$
so $X\wedge Y = 0$ detects if $X$ and $Y$ intersect, and if not then $X\wedge Y$ models the "join" (i.e. span) of subspaces.
It is in this framework that we may interpret the contraction. Suppose we have a metric on $V$ and that $X, Y$ are blades representing subspaces which are non-degenerate with respect to the metric; then
$$\begin{aligned}
  &X\lcontr Y = 0 &&\text{if }\exists v \in [X]^\wedge.\: v\perp[Y]^\wedge,
\\
  &[X\lcontr Y]^\wedge = P_Y([X]^\wedge)^\perp\cap[Y]^\wedge &&\text{otherwise},
\end{aligned}$$
where $P_Y$ is the orthogonal projection onto $[Y]^\wedge$ and $\perp$ denotes orthogonality or the orthogonal complement. In words, $X\lcontr Y = 0$ detects whether or not $X$ is partially orthogonal to $Y$, and if not then $X\lcontr Y$ computes the unique subspace of $Y$ which is orthogonal to $P_Y(X)$. We can get exactly the projection by orthogonalizing again:
$$
  P_Y(v) = (v\lcontr Y)\lcontr Y^{-1} = (v\lcontr Y)Y^{-1}
$$
where $Y^{-1} = Y/Y^2 = Y/\form{Y, Y}_V$ is the Clifford inverse, and in the last equality we may replace the contraction with a Clifford product.
If we contract a pseudoscalar then the projection is trivial since pseudoscalars represent the entirety of $V$. So choose a unit pseudoscalar $I$, i.e. $\form{I, I}_V = \pm1$; there are exactly two of these, the other being $-I$. This amounts to a choice of orientation. We see that
$$
  [X\lcontr I]^\wedge = [X]^{{\wedge}\perp}.
$$
This is quite like the Hodge star. Actually
$$
  \rev X\lcontr I = \rev XI
$$
is exactly the Hodge star. This should make clear the following duality, valid for any pseudoscalar and written with the Clifford product to reduce clutter:
$$
  X\wedge(YI) = X\lcontr Y\,I,\quad X\lcontr(YI) = X\wedge Y\,I.
$$
Here we use the convention that the Clifford product binds weaker than any other product.
The Meet Geometry of $\Ext V^*$
If we were to replace $V$ with $V^*$ in the above, we would get nothing interesting. So we look at what $V^*$ says about $V$: just as a vector represents an oriented line with magnitude, we will interpret covectors as oriented hyperplanes with magnitude. In particular, $v^* \in V^*$ represents the hyperplane $\ker(v^*)$. But recall that $v^*(v) = \form{v^*, v} = v^*\rintr v$. This suggests that we define
$$
  [X^*]^{\lintr} = \{v \in V \;:\; v\lintr X^* = 0\}.
$$
Rather than the joins of lines like in $\Ext V$, the blades of $\Ext V^*$ represent intersections of hyperplanes:
$$\begin{aligned}
  &v^*_1\wedge\dotsb\wedge v^*_k = 0 &&\text{if }\{v^*_1,\dotsc,v^*_k\}\text{ is linearly dependent},
\\
  &[v^*_1\wedge\dotsb\wedge v^*_k]^{\lintr} = \bigcap_{i=1}^k[v^*_i]^{\lintr} &&\text{otherwise}.
\end{aligned}$$
While the sum of two oriented, weighted lines
may be familiar via the sum of two vectors,
the sum of two hyperplanes is probably less familiar.
Suppose that $v, w \in V$ have equal weight
and that $v^*, w^* \in V^*$ have equal weight
(defining this precisely requires something norm-like, like a quadratic form).
Just as $v+w$ is a line
contained in the span of $v, w$ "half way between" them,
$v^* + w^*$ is a hyperplane
which contains the intersection of $v^*, w^*$
and is "half way between" them.
Just as $v+\alpha w$ is "closer" to $w$ for $\alpha > 1$,
$v^*+\alpha w^*$ is "closer" to $w^*$.
These two facts should give sufficient intuition for the sum of two hyperplanes.
Let
$$
  \Hyp = \{[v^*]^{\lintr} \;:\; v^* \in V^*\},\quad
  \Hyp_S = \{H \in \Hyp \;:\; S \subseteq H\}
$$
denote the set of hyperplanes of $V$
and the set of hyperplanes containing a subspace $S \subseteq V$,
respectively.
In general, for blades $X^*$ and $Y^*$ we have
$$\begin{aligned}
  &X^*\wedge Y^* = 0 &&\text{if }\exists H\in\Hyp.\: [X^*]^{\lintr}\cup[Y^*]^{\lintr} \subseteq H,
\\
  &[X^*\wedge Y^*]^{\lintr} = [X^*]^{\lintr}\cap[Y^*]^{\lintr} &&\text{otherwise}.
\end{aligned}$$
Thus $X^*\wedge Y^* = 0$ tests whether $X^*$ and $Y^*$
are contained in a common hyperplane, and if not then $X^*\wedge Y^*$
models the "meet" (i.e. intersection) of subspaces.
Now suppose we have a metric on $V^*$.
We will say that hyperplanes corresponding to orthogonal covectors
are $*$-orthogonal, written $\pperp$.
A subspace $S \subseteq V$ is $*$-orthogonal to $H \in \Hyp$,
written $S\pperp H$, if $K \pperp H$ for every $K \in \Hyp_S$.
The $*$-orthogonal complement $S^{\pperp}$ of $S$
is the intersection of all such hyperplanes.
We then say that $S\pperp T$ for subspaces $S, T$
if $T^{\pperp} \subseteq S$ or $S^{\pperp} \subseteq T$.
Now let $X^*, Y^* \in \Ext V^*$ represent non-degenerate subspaces of $V$.
We have
$$\begin{aligned}
  &X^*\lcontr Y^* = 0 &&\text{if }\exists H\in\Hyp_{X^*}.\: H \pperp [Y^*]^{\lintr},
\\
  &[X^*\lcontr Y^*]^{\lintr} = P^*_{Y^*}([X^*]^{\lintr})^{\pperp}\oplus[Y^*]^{\lintr} &&\text{otherwise}.
\end{aligned}$$
$P^*_{Y^*}$ is the $*$-orthogonal projection onto $[Y^*]^{\lintr}$.
Of course we have
$$
  P^*_{Y^*}(v^*) = (v^*\lcontr Y^*)\lcontr (Y^*)^{-1}.
$$
This is defined by analogy to orthogonal projection:
given a non-degenerate subspace $S$,
for any oriented, weighted line (vector) $v \in V$
there are unique $v_S \in S$ and $v_\perp \in S^\perp$
such that $v = v_S + v_\perp$;
$v_S$ is then the orthogonal projection of $v$ onto $S$.
Thus, for any oriented, weighted hyperplane $v^* \in V^*$
there are unique $v^*_S, v^*_{\pperp}$ with $S \subseteq [v^*_S]^{\lintr}$
and $S^{\pperp} \subseteq [v^*_{\pperp}]^{\lintr}$
such that $v^* = v^*_S + v^*_{\pperp}$;
$v^*_S$ is then the $*$-orthogonal projection of $v^*$ onto $S$.
To project an arbitrary subspace $T$ onto $S$,
project each hyperplane that contains $T$ and then take their intersection:
$$
  P^*_S(T) = \bigcap_{H\in\Hyp_T}P^*_S(H).
$$
As a simple example of $*$-orthogonal projection,
consider $\R^3$ with the standard metric.
This induces a metric on $(\R^3)^*$ via non-degeneracy.
Consider a plane $\pi$ and an non-orthogonal line $l$.
The orthogonal projection $P_\pi(l)$
can be visualized as rotating $l$ through the small angle it makes with $\pi$
in such a way that $l\oplus\pi^\perp$ is constant throughout the motion, until $l$ lies inside $\pi$.
The $*$-orthogonal projection $P^*_l(\pi)$
can be visualized as rotating $\pi$ through the small angle it makes with $l$
in such a way that $l^{\pperp}\cap \pi = l^\perp\cap \pi$ is constant throughout the motion, until $\pi$ contains $l$.
The Poincaré Isomorphisms
Recall with a non-degenerate metric on $V$
and a chosen orientation represented with unit pseudoscalar $I \in \MVects nV$
that the Hodge star $\Ext V \to \Ext V$ is
$$
  \star X = \rev X\lcontr I.
$$
Without such a non-degenerate metric, we of course do not have a Hodge star.
However, the Hodge star actually descends from a more fundamental duality.
Choose any non-zero $I \in \MVects nV$
and choose the unique $I^* \in \MVects nV^*$
such that $\form{\rev I^*, I} = 1$.
We can then define the Poincaré isomorphisms
$I_\#, I^*_\# : \Ext V \to \Ext V^*$
and $I^\#, I^\#_* : \Ext V^* \to \Ext V$ by
$$\begin{aligned}
  I_\#(X) &= \rev X\lintr I^*, & I_\#^*(X) &= I^*\rintr\rev X,
\\
  I^\#(X^*) &= \rev X^*\lintr I, & I^\#_*(X^*) &= I\rintr\rev X^*.
\end{aligned}$$
There is essentially only one isomorphism: as the notation suggests,
the pairs $I_\#, I_\#^*$ and $I^\#, I^\#_*$ are dual,
and additionally the pairs $I^\#, I_\#^*$ and $I_\#, I^\#_*$ are inverses.
These isomorphisms are grade reversing, taking $k$-vectors to $(n-k)$-vectors.
It is easy to see that if we replace $I$ by $\alpha I$
for nonzero scalar $\alpha$
then each of these scales by $\alpha$ or $1/\alpha$.
In the presence of a non-degenerate metric
we get an isomorphism $\flat : V \to V^*$
which then extends to an algebra isomorphism $\Ext V \to \Ext V^*$;
the Hodge star $\star : \Ext V \to \Ext V$ is exactly
$$
  \star X = I^\#(X^\flat),\quad
  \star^{-1}X = [I_\#^*(X)]^\sharp
$$
where $\sharp = \flat^{-1}$.
Note that even if $\flat$ is not an isomorphism
the composition $I^\#\circ\flat$ still makes sense.
The Poincaré isomorphisms are geometry preserving in the sense that
$$
  [I_\#(X)]^{\lintr} = [X]^{\wedge}
$$
and similary for the others.
Thus we can transfer the meet geometry of $\Ext V^*$ to $\Ext V$
(or similarly the join geometry of $\Ext V$ to $\Ext V^*$) by defining
$$
  X\vee Y = I^\#(I_\#^*(X)\wedge I_\#^*(Y)).
$$
(Both the pairs $I^\#, I_\#^*$ and $I^\#_*, I^\#$
give the same definition of $\vee$).
Generically, we call such a $\vee$ the regressive product;
in the context of $\Ext V$ we call it the meet product,
and in the context of $\Ext V^*$ we call it the join product.
In terms of the Hodge star (when available)
$$
  X\vee Y = \star(\star^{-1}X\wedge \star^{-1}Y).
$$
Finally, the Poincaré isomorphisms show us that regardless of degeneracy
there is a sense in which a metric on $V$ induces a metric on $V^*$.
A metric $\form{\cdot,\cdot}_V$ on $V$
extends to a metric $(X, Y) \mapsto \form{\rev X, Y}_V$
on $X, Y \in \MVects {n-1}V$; then
$$
  (v^*, w^*) \mapsto \form{\rev{I^\#(v^*)}, I^\#(w^*)}_V
$$
is a metric on $V^*$.
While this metric is dependent on $I$,
all such metrics are equivalent as quadratic forms
and the induced notion of $*$-orthogonality is independent of $I$.
When $\form{\cdot,\cdot}_V$ is non-degenerate
this induced metric has the same signature;
however, in the presence of degeneracy it becomes quite different.
The $*$-metric induced by a metric of signature $(p,q,r)$
(number of positive-, negative-, zero-squaring orthogonal basis elements)
has degeneracy $p+q$ if $r = 1$ and is completely degenerate if $r > 1$.
Rotational Intertia
A metric of signature $(n,0,0)$ is, of course,
natural for linear Euclidean geometry.
In physics, the (rotational, origin-relative) inertia tensor
is a linear function on bivectors $\mathcal I : \MVects 2V \to \MVects 2V$
which takes an angular velocity bivector $B$
to the corresponding angular momentum bivector $\mathcal I(B)$
for a given rigid body.
If the rigid body is composed of particles with position vectors $x_i$
and masses $m_i$, then the inertia takes the form
$$
  \mathcal I(B) = \sum_i m_ix_i\wedge(B\rcontr x_i).
$$
This can easily be extended to a continuous body
by replacing the sum with an integral.
See Section 3.4.4 of Geometric Algebra for Physicists
by Chris Doran and Anthony Lasenby for more detail.
Projective Euclidean Geometry
A metric of signature $(n,0,1)$ on $V^*$
turns out to be the natural setting for $n$D projective Euclidean geometry.
This construction goes by the name of Plane-based Geometric Algebra
or less commonly Projective Geometric Algebra,
but both shorten to PGA and this acronym is the most common.
In PGA, as one would expect hyperplanes are represented by covectors in $V^*$;
however, we now get all hyperplanes of Euclidean space
and not just those that pass through a specific origin point.
There is also a unique hyperlane, the hyperplane at infinity,
corresponding to the span of covectors $v^*$ such that $\form{v^*,v^*} = 0$.
We then use the exterior product
to form all the other flat elements of Euclidean
as intersections of hyperplanes.
Thus a point is an $n$-vector, a line is a $(n-1)$-vector, etc.
Restrict now to the case $n=3$ and let $\pi, l, P$
be a plane, a line, and a point as represented in the exterior algebra,
and assume none of them are at infinity.
Here are some geometric interpretations of $\lcontr$:

*

*$\pi\lcontr P$ is the unique line orthogonal to $\pi$ passing through $P$.

*$(\pi\lcontr P)\lcontr P^{-1}$ (using the Clifford inverse)
is the unique plane parallel to $\pi$ passing through $P$.

*$l\lcontr P$ is the unique plane orthogonal to $l$ passing through $P$.

*$(l\lcontr P)\lcontr P^{-1}$
is the unique line parallel to $l$ passing through $P$.

*$\pi\lcontr l = 0$ if $\pi$ is orthogonal to $l$,
and otherwise is the unique plane $*$-orthogonal to $\pi$ containing $l$.

*$(\pi\lcontr l)\lcontr l^{-1}$
is the $*$-orthogonal projection of $\pi$ onto $l$
with their intersection as origin;
if they do not intersect, then it is the unique plane parallel to $\pi$
containing $l$.

The regressive product $\vee$ is also significant and is maybe more immediately intuitable.
Let $P, Q$ be points, $l, m$ be lines, and $\pi$ be a plane.

*

*$P\vee Q = 0$ if $P$ and $Q$ are the same point, and otherwise is the line joining them.

*$P\vee l = 0$ if $P$ lies on $l$, and otherwise is the plane joining them.

*$P\vee\pi$ is the orthogonal distance from $P$ to $\pi$.

*$l\vee m = 0$ if $l$ and $m$ lie in a common plane, and more generally it is proportional to $\sin\theta$ where $\theta$ is the angle $l$ and $m$ would make if they were translated so as to intersect.

For a treatment of rigid-body mechanics using PGA, see May the Forque be with You! by Leo Dorst and Steven De Keninck. For more on PGA itself and its relationship with projective geometry, see Geometry, Kinematics, and Rigid Body Mechanics in Cayley-Klein Geometries by Charles Gunn.
Codifferentials
Briefly, a differential form $\omega$ on a differentiable manifold $M$
is defined by $\omega_x \in (\Ext T_xM)^* \cong \Ext T^*_xM$ for each point $x \in M$,
where $T_xM$ and $T^*_xM$ are respectively the tangent and cotangent spaces at $x$.
Given coordinates $x^i$, the exterior derivative of $\omega$ is simply
$$
  \dd\omega = \nabla\wedge\omega = \sum_i\dd x^i\wedge\frac{\partial\omega}{\partial x^i}.
$$
If we have a metric $\form{\cdot,\cdot}_M$ and hence a Hodge star, then the codifferential is
$$
  {\star^{-1}}\dd{\star}\omega
= (\nabla\wedge(\rev\omega I)\,I^{-1})^\sim
= (\nabla\lcontr\rev\omega\,II^{-1})^\sim
= \dot\omega\rcontr\dot\nabla.
$$
Here, $I \in \MVects nT^*M$ is the chosen volume form, $I^{-1}$ is its point-wise Clifford inverse,
Clifford products are taken point-wise, and the overdots $\dot\omega$ and $\dot\nabla$ clarify what $\nabla$ is being applied to.
We can take this a step further; since
$$
  \dot\omega\rcontr\dot\nabla = \nabla\lcontr\widehat\omega
$$
using the Clifford product we may write
$$
  \dd\omega + {\star^{-1}}\dd{\star}\widehat\omega
= \nabla\wedge\omega + \nabla\lcontr\omega = \nabla\omega.
$$
The expression $\nabla\omega$ is to be understood with the Clifford product.
Since the two terms on the LHS have different grades we can recover both from $\nabla\omega$.
A: This isn't an answer, but it certainly should be relevant to this discussion.  Andrew McInerney's excellent new book "First Steps in Differential Geometry" discusses the interior product of a (o, k) tensor field (always alternating according to the book's conventions) on pp. 169 - 171.  
He gives 2 examples: 4.6.17 and 4.6.18
If any one understands how the result follows in 4.6.18 which "the reader may verify"  please inform me.
