Geometric Intuition about the relation between Clifford Algebra and Exterior Algebra

It is common to see a relation being established between the Clifford Algebra and the Exterior Algebra of a vector space. Recently reading some texts written by Physicists I've seem applications of Clifford Algebras on which it seems that the author is dealing with the usual exterior algebra somehow. It seems, in that case, that this relation is quite often used when appliying Clifford Algebras on Physics and Geometry.

Searching for this a little I've found the Clifford Algebra page on Wikipedia, where we can find a section "Relation to the exterior algebra". It is then said:

Given a vector space $V$ one can construct the exterior algebra $\bigwedge (V)$, whose definition is independent of any quadratic form on $V$. It turns out that if $K$ does not have characteristic $2$ then there is a natural isomorphism between $\bigwedge(V)$ and $\mathcal{Cl}(V,Q)$ considered as vector spaces. This is an algebra isomorphism if and only if $Q = 0$. One can thus consider the Clifford algebra $\mathcal{Cl}(V,Q)$ as an enrichment of the exterior algebra on $V$ with a multiplication that depends on $Q$ (one can still define the exterior product independent of $Q$).

Now I want to get some geometrical intuition on this.

I know that the exterior algebra is the algebra inside of which all $k$-vectors can be manipulated together for all possible values of $k$. I also know that a $k$-vector has a direct geometrical meaning: it is a piece of oriented $k$-dimensional subspace.

Also I know that one quadratic form $Q$ also has one geometric meaning: it might represent some sort of length defined on $V$.

Now, what is the idea here? That when we construct $\bigwedge (V)$ we somehow are not accounting for the idea of "measure" of those pieces of $k$-dimensional subspace, because no idea of length was defined in $V$ and then $\mathcal{Cl}(V,Q)$ is the result of carrying the idea of length introduced in $V$ by $Q$ to all the $k$ vectors and hence to the algebra where we manipulate them together?

I feel this is not quite the right intuition yet.

What is the correct geometrical intuition behind the relation between the clifford algebra and the exterior algebra?

Solution 1:

There are two relationships along these lines.

The one you are referring to, I think, is the linear isomorphism between a Clifford algebra of $(V,Q)$ and the exterior algebra for $V$ mentioned in the wiki article. This lets you apply some intuition from exterior algebra inside the Clifford algebra.

This linear isomorphism lets you identify a linear copy of the exterior algebra inside the Clifford algebra.

The other interesting relationship, which I don't think is the same phenomenon you are describing, is that the exterior algebra for $V$ is exactly the Clifford algebra for $V$ with the constantly zero bilinear form.