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One is represented by a dot product, the other by a cross product. The "inner product collapses two co-ordinate vectors into a scalar, the exterior product seems to expand them in a multilinear (manifold)? The inner product seldom has "cancellation," the exterior product has a lot of cancellation (between the same differential forms).

Are they just two opposite sides of the same coin? If not, why do they seem to be so "parallel"?

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The exterior product can be defined with no extra assumptions on a vector space $V$, whereas the inner product is extra structure on $V$. It seems to me that the "opposite" of the exterior product ought to be the interior product. – Qiaochu Yuan Jul 11 '11 at 15:58
One should not that the cross product also depends on having an inner product; in a three-dimensional space the cross product of two vectors $v$ and $w$ is $*(v \wedge w)$, where $\wedge$ is the exterior product (independent of the inner product) and $*$ is the Hodge-star operatior (which depends on the inner product). – Gunnar Þór Magnússon Jul 11 '11 at 16:45
@Gunnar: the cross product depends in addition on an orientation. – Qiaochu Yuan Jul 11 '11 at 19:00
@Qiaochu: don't we get one for free with the inner product? – Gunnar Þór Magnússon Jul 12 '11 at 7:34
@Gunnar: no. Without an orientation there are two choices for the Hodge star which you implicitly choose if you define the Hodge star on a basis (you're implicitly choosing the one compatible with the order you've implicitly chosen on the basis). The problem is that the exterior product gives you a pairing $V \times \Lambda^2(V) \to \Lambda^3(V)$ but we do not get a canonical isomorphism $\Lambda^3(V) \cong \mathbb{R}$ unless we choose an orientation. – Qiaochu Yuan Jul 12 '11 at 13:57
up vote 6 down vote accepted

Don't make to much from the words "inner", "interior" and "exterior"; none of them is the opposite to another of them. It is a fact that in connection with a vector space (maybe provided with a scalar product) there are various natural bilinear (or multilinear) functions.

The first of them (called "opération extérieure" by Bourbaki) is the map $F\times V\to V\ $ ($F$ being the ground field) $(\alpha, x)\mapsto \alpha x$. If the ground field $f$ is $\ ={\mathbb R}$ or $\ ={\mathbb C}$ then $V$ can be provided with a ${\it scalar\ product}\ $ $\bullet:\ V\times V\to F$. This scalar product is a symmetric bilinear function with the extra property that $x\bullet x >0$ for all $x\ne 0$ and is often called the ${\it inner\ product}\ $ on $V$.

Independently of $\bullet$ one can set up the following: If $F$ is any ground field of characteristic $\ne2$ then there is a certain algebraic construction called ${\it exterior\ product}\ $ $\wedge$, which is skew-multilinear on the cartesian powers $V^r$, $\ 1\leq r\leq{\rm dim}(V)$.

Now combine the two constructions: If $V$ is a real vector space provided with a scalar product, and if ${\rm dim}(V)=3$, then any parallelepiped spanned by three vectors $p$, $q$, $x\in V$ has a natural volume $\mu(p,q,x)$. The volume form $\mu(\cdot,\cdot,\cdot)$ is trilinear and gives negative values for "negatively oriented" triples $p$, $q$, $x$; it is uniquely determined by the condition that the volume of an a priori chosen orthonormal basis $(e_1,e_2,e_3)$ should be $=1$. Given this volume form $\mu(\cdot,\cdot,\cdot)$ the exterior product $p\wedge q$ of two vectors $p$, $q\in V$ can be identfied with a unique vector $r\in V$ where $r$ is attached to $p$ and $q$ by the identity $$\mu(p,q,x)=r\bullet x\qquad(x\in V)\ .$$ This vector $r$ is then called the vector product or the exterior product of the given vectors $p$ and $q$.

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But isn't the interior product on the exterior algebra of an inner product space roughly the adjoint of the exterior product (after identifying the vector space with its dual)? – Willie Wong Jul 12 '11 at 9:58

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