Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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"?

share|improve this question
6  
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
2  
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 Magnusson 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 Magnusson Jul 12 '11 at 7:34
1  
@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

1 Answer 1

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 $q$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.