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.

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.