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.

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Just nitpicking: you can identify elements of $\bigwedge^k$ with alternating k-tensors only in characteristic 0. – Alexei Averchenko Jul 8 '11 at 15:32
@Alexei the question is tagged differential geometry :-) – Willie Wong Jul 9 '11 at 15:37
For a physical example of the use of the interior product: that is how one can geometrically define the "electric" and "magnetic" parts of an electromagnetic field given by the Faraday tensor. It also shows conveniently how electric and magnetic fields change under Lorentz transformations. – Willie Wong Jul 9 '11 at 16:04

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.

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Very nice example – ItsNotObvious Jul 9 '11 at 21:34

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.

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Thanks for the reference; I've looked it up and it appears to be Proposition 13.12 on p 337. I have some work to do to completely understand the example but it's a start. – ItsNotObvious Jul 8 '11 at 17:38
No problem. To me it's kind of an uninspiring example, since the usefulness of orientation forms at that point is that they spit out a sign. I wonder if there's something involving volume forms. – Dylan Moreland Jul 8 '11 at 18:10

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 :)

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Of course? I'm glad that others find it to be so obvious. Notice though that Wikipedia page discusses the interior product within the context of manifolds and differential forms. Although I tagged the question with "differential geometry", the definition I provided, which comes from Jeff Lee's Manifolds and Differential Geometry, makes no reference to smooth infrastructure. Therefore, I'm hoping for an explanation in the more elementary context of multilinear algebra. – ItsNotObvious Jul 8 '11 at 15:46
I didn't mean it's obvious, I meant it's useful for computing the Lie derivative. Is the interior product even used that much outside differential geometry? – Alexei Averchenko Jul 8 '11 at 15:48
As I'm just learning the concept I can't really say how it's used, but since it can be defined in a completely algebraic manner I would think that it would have some sort of meaning in that context. If its only purpose is to relate the exterior/Lie derivatives then I can accept that, but would be surprised if this was the case – ItsNotObvious Jul 8 '11 at 15:55
I wouldn't call it its whole purpose, no. – Alexei Averchenko Jul 8 '11 at 16:01
I'm a bit puzzled by your answer. Did you read the question at all? You don't seem to address it (except for something in the second paragraph). – t.b. Jul 8 '11 at 16:12

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.

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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.

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