Following my previous question Relationship 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 have drawn 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.
Best Answer
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.