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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?

Gold
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3 Answers3

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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.

rschwieb
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I don't know if it will be "geometric" enough for you, but the exterior algebra $\Lambda (V)$ is exactly the associated graded algebra of the Clifford algebra $\operatorname{Cl}(V,Q)$ for any quadratic form $Q$.

In more detail: Clifford algebra is a filtered algebra, meaning it has a natural decomposition into a set of increasing subspaces:

$$F_0 \subset F_1 \subset F_2 \subset \dots \subset F_d = \operatorname{Cl}(V,Q)$$

such that $$F_n \cdot F_m \subset F_{n+m}$$

Here, $F_n$ is the subspace spanned by products of at most $n$ vectors from $V$.

For any filtered algebra $A$, we can form the associated graded algebra $G(A)$ by taking a direct sum of quotient spaces

$$G_n = F_n / F_{n-1}$$ $$G(A) = \bigoplus G_k$$

($F_{-1}$ is understood to be the zero space). The algebra $G(A)$ is graded in the sense that

$$G_n \cdot G_m \subset G_{n+m}$$

One can check that the multiplication on the original filtered algebra $A$ uniquely defines a multiplication on $G(A)$, and that for the Clifford algebra $A=\operatorname{Cl}(V,Q)$ the associated graded algebra is indeed isomorphic to $G(A)=\Lambda(V)$.

lisyarus
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From the geometric algebra point of view the situation is the following. The defining feature of the geometric product (the product in a Clifford algebra) is that for vectors $v, w$ we have

$$vw = v \cdot w + v \wedge w$$

where $v \cdot w$ is the dot product and $v \wedge w$ is a version of the exterior product. These are respectively the symmetric and the antisymmetric parts of the geometric product, in the sense that they can be recovered from the geometric product by symmetrizing resp. antisymmetrizing:

$$v \cdot w = \frac{vw + wv}{2}, v \wedge w = \frac{vw - wv}{2}.$$

Now observe that the quadratic form only enters into this definition via the symmetric part, the part involving the dot product (which for a general quadratic form is the associated bilinear form). So, one thing you can do, and one thing physicists do when working with geometric algebra, is to equip the geometric / Clifford algebra with this antisymmetrized product $v \wedge w$, in addition to the full geometric product.

With the antisymmetrized product, the geometric / Clifford algebra is now just isomorphic to the exterior algebra, in the obvious way. Antisymmetrization has the effect of removing the dot product part of the geometric product, which is the part that knows about lengths, so the result is the "pure linear algebra" of the exterior product.

Qiaochu Yuan
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