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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Following my previous question Relation between cross-product and outer product where I learnt that the Exterior Product generalises the Cross Product whereas the Inner Product generalises the Dot Product, I was wondering if the simple map that I drew below is at all an accurate representation of the links between these different products? Vertical lines denote generalisation-specification, horizontal lines denote "in opposition to". I'm just trying to get a quick overview before I dive in. Thanks enter image description here

share|cite|improve this question
What does "interior product" mean? What does "outer product" mean? What do the lines on your diagram signify? – Chris Eagle Aug 13 '12 at 11:26
Hi Chris, the context and motivation is in the link in my question. Sorry for being unclear. – Ryan Aug 13 '12 at 12:07
The terms "exterior" and "outer" product are overloaded, and you're going to get clashing answers unless you are specific about what you mean. Actually I have never heard of an interior product... – rschwieb Aug 13 '12 at 12:25
Inner product generalizes dot product. Outer product is a particular case of tensor product and not related to scalar product. Cross product is a particular case of exterior product. Interior product is a particular case of exterior product. It is not well known and quite unnatural, because it depends on the choice of some vector. – Norbert Aug 13 '12 at 12:30
@Nobert Thanks for clarifying. The wikipedia page on Interior Product which implies that IP is opposed to EP but it must've meant just the naming. Anyway. – Ryan Aug 13 '12 at 13:51
up vote 2 down vote accepted

I'm not sure what you're reading, but it certainly contains a lot of terms talked about in Clifford algebra (or "geometric algebra").

Clifford algebras were born of a synthesis of inner product spaces and Grassmann's exterior algebras, both of which have geometric applications.

A Clifford algebra is constructed from an inner product space $(V,Q)$ by generating an associative algebra (whose product is a descendant of the tensor product in the tensor algebra for $V$). These are compatible in a sense made clear in the Wiki.

Suppose we are in a real inner product space, and let's denote the algebra multiplication with $\otimes$ and the inner product with $\cdot$. One identity that holds for parallel vectors $v,w$ is $v\otimes w=v\cdot w$. Part of the reason this is true is that $v\otimes w=w\otimes v$ when $v,w$ are parallel.

When they are not parallel, then there is a skew component to $v\otimes w$. This can be retrieved explicitly by computing $\frac{1}{2}(v\otimes w-w\otimes v):=v\wedge w$. With this notation, $v\otimes w=v\cdot w+v\wedge w$ for all vectors $v,w$. This $\wedge$ is called the exterior or outer product in this algebra.

They key thing to know is that this inner and this outer product don't uniquely extend to the whole algebra (well the outer product has a slightly more natural extension than the inner product). They are not "natural" to the algebra really: they are mostly just a notational convenience with some algebraic properties that make them easy to view as products.

The cross product arises too in the discussion of Clifford algebras, but it seems to be regarded as second class to the outer product. I've seen this attributed to the cross-product somehow not carrying the correct information for physical applications.

I went and read a little and found out that "interior product" looks a lot like an important extension of the inner product in a Clifford algebra which is sometimes called the contraction.

In summary, I think these various products might become less mysterious if you read some of these linked articles and see where they are used. It's going to be especially enlightening when you find out where they are all defined (in the vector space $V$ or in an algebra containing $V$.)

share|cite|improve this answer
Thanks for your input! – Ryan Aug 13 '12 at 12:30

The relation between the inner product of vectors and the interior product is that if you have a metric tensor (and thus a canonical relation between vectors and covectors = $1$-forms), the inner product of two vectors is the interior product of one of the vectors and the $1$-form associated with the other one. That is, if $g$ is the metric tensor, then the inner product of the vectors $v$ and $w$ is $g(v,w)$, and the $1$-form $\omega$ associated with $v$ is defined by $\omega(w)=g(v,w)$. Then it is obvious that $\iota_w(\omega) = \omega(w) = g(v,w)$.

The outer product is, as noted in answer to the other question you referred to, related to the tensor product. Indeed, if we associate row vectors with $1$-forms and column vectors with vectors, then we can write (using Einstein summation convention) the outer product of the vector $w = w^ie_i$ and the $1$-form $\omega = \omega_i\,e^i$ as the $(1,1)$ tensor $M = w^i\omega_j e_i\otimes e^j$ which describes an object that maps vectors to vectors. Its relation to the inner product is that you get the inner product of $w$ and $\omega$ by contracting the two indices of $M$ (which in the language of matrices corresponds to the trace of $M$).

The exterior product is related to the tensor product in that the exterior product of two forms (a form is a skew-symmetric tensor of type $(0,p)$) is just the antisymmetrization of the tensor product.

The cross product is a speciality of the three-dimensional space; here the space of $2$-forms has the same dimension as the space of $1$-forms; indeed, given a metric, the hodge star maps between them. Since the metric also allows to associate vectors and $1$-forms, you can define the cross product of $v$ and $w$ by the following procedure: Determine the $1$-forms corresponding to $v$ and $w$, calculate their exterior product (which is a $2$-form), apply the Hodge star to the result (which, given that we are in three dimensions, again results in a $1$-form), and finally determine the vector corresponding to that $1$-form.

share|cite|improve this answer

Your Answer


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.